#### 5.1.1 Finite length string

##### 5.1.1.1 [334] General solution for both ends ﬁxed. Domain is $$0\dots L$$

problem number 334

Solve for $$u(x,t)$$ for $$t>0$$ and $$0<x<L$$ $u_{tt} = c^2 u_{xx}$ With boundary condition both ends ﬁxed \begin{align*} u(0,t) &= 0 \\ u(L,t) &=0 \end{align*}

And initial conditions \begin{align*} u(x,0) &= f(x) \\ u_t(x,0) &= g(x) \\ \end{align*}

Mathematica

Failed Surprising it could not solve this standard wave PDE on string

Maple

$u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac{1}{\pi \,ncL}\sin \left ({\frac{n\pi \,x}{L}} \right ) \left ( \pi \,\cos \left ({\frac{\pi \,nct}{L}} \right ) \int _{0}^{L}\!\sin \left ({\frac{n\pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}xnc+L\sin \left ({\frac{\pi \,nct}{L}} \right ) \int _{0}^{L}\!\sin \left ({\frac{n\pi \,x}{L}} \right ) g \left ( x \right ) \,{\rm d}x \right ) } \right )$

Hand solution

Solving for $$t>0,0<x<L$$$u_{tt}=c^{2}u_{xx}$ With BC \begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end{align*}

And initial conditions\begin{align*} u\left ( x,0\right ) & =f\left ( x\right ) \\ u_{t}\left ( x,0\right ) & =g\left ( x\right ) \end{align*}

Let $$u=X\left ( x\right ) T\left ( t\right )$$. The PDE becomes\begin{align*} \frac{T^{\prime \prime }X}{c^{2}} & =X^{\prime \prime }T\\ \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}

Where $$\lambda$$ is separation constant. Hence the eigenvalue ODE is \begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

From the boundary conditions, we see that $$\lambda >0$$ is the only possible value. Therefore the solution to the above ODE is$X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right )$ Since $$X\left ( 0\right ) =0$$ then $$A=0$$ and the solution becomes $$X\left ( x\right ) =B\sin \left ( \sqrt{\lambda }x\right )$$. Since $$X\left ( L\right ) =0$$ then for non trivial solution we want $$\sqrt{\lambda }L=n\pi$$ or $\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots$ Hence the eigenfunctions are $\Phi _{n}\left ( x\right ) =\sin \left ( \frac{n\pi }{L}x\right ) \qquad n=1,2,3,\cdots$ The time ODE now becomes$T^{\prime \prime }+c^{2}\left ( \frac{n\pi }{L}\right ) ^{2}T=0$ Which has the solution$T\left ( t\right ) =D_{n}\cos \left ( c\frac{n\pi }{L}t\right ) +E_{n}\sin \left ( c\frac{n\pi }{L}t\right )$ Therefore the complete solution becomes$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( D_{n}\cos \left ( c\frac{n\pi }{L}t\right ) +E_{n}\sin \left ( c\frac{n\pi }{L}t\right ) \right ) \Phi _{n}\left ( x\right ) \tag{1}$$ At $$t=0$$ the above becomes$f\left ( x\right ) =\sum _{n=1}^{\infty }D_{n}\Phi _{n}\left ( x\right )$ Applying orthogonality gives\begin{align*} \int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =D_{n}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac{L}{2}D_{n} \end{align*}

Hence$$D_{n}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag{2}$$ Taking time derivative of (1) gives$u_{t}\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( -c\frac{n\pi }{L}D_{n}\sin \left ( c\frac{n\pi }{L}t\right ) +E_{n}c\frac{n\pi }{L}\cos \left ( c\frac{n\pi }{L}t\right ) \right ) \Phi _{n}\left ( x\right )$ At $$t=0$$ the above becomes$g\left ( x\right ) =\sum _{n=1}^{\infty }E_{n}c\frac{n\pi }{L}\Phi _{n}\left ( x\right )$ Applying orthogonality gives\begin{align*} \int _{0}^{L}g\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =E_{n}c\frac{n\pi }{L}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac{L}{2}E_{n}c\frac{n\pi }{L}\\ & =\frac{1}{2}E_{n}cn\pi \end{align*}

Hence$$E_{n}=\frac{2}{cn\pi }\int _{0}^{L}g\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag{3}$$ Using (2,3) in (1) gives the ﬁnal solution as\begin{align*} u\left ( x,t\right ) & =\frac{2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{n\pi }{L}s\right ) ds\right ) \cos \left ( c\frac{n\pi }{L}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \\ & +\frac{2}{c\pi }\sum _{n=1}^{\infty }\frac{1}{n}\left ( \int _{0}^{L}g\left ( s\right ) \sin \left ( \frac{n\pi }{L}s\right ) ds\right ) \sin \left ( c\frac{n\pi }{L}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

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##### 5.1.1.2 [335] both ends ﬁxed, inital position zero (special case)

problem number 335

Solve for $$u(x,t)$$ for $$t>0$$ and $$0<x<L$$ $u_{tt} = c^2 u_{xx}$ With boundary condition both ends ﬁxed \begin{align*} u(0,t) &= 0 \\ u(L,t) &=0 \end{align*}

And initial conditions \begin{align*} u(x,0) &= f(x) \\ u_t(x,0) &= g(x) \\ \end{align*}

Using the following values \begin{align*} L&=10\\ c&=2\\ f(x)&=0\\ g(x)&= \frac{8 x (L-x)^2}{L^3} \end{align*}

Mathematica

$\left \{\left \{u(x,t)\to \underset{n=1}{\overset{\infty }{\sum }}\frac{160 \left (2+(-1)^n\right ) \sin \left (\frac{n \pi t}{5}\right ) \sin \left (\frac{n \pi x}{10}\right )}{n^4 \pi ^4}\right \}\right \}$

Maple

$u \left ( x,t \right ) =\sum _{n=1}^{\infty }160\,{\frac{ \left ( \left ( -1 \right ) ^{n}+2 \right ) \sin \left ( 1/10\,n\pi \,x \right ) \sin \left ( 1/5\,n\pi \,t \right ) }{{\pi }^{4}{n}^{4}}}$

Hand solution

Solving the wave PDE on string with both ends ﬁxed $u_{tt}=c^{2}u_{xx}\qquad t>0,x>0$ With BC \begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end{align*}

And initial conditions\begin{align*} u\left ( x,0\right ) & =f\left ( x\right ) =0\\ u_{t}\left ( x,0\right ) & =g\left ( x\right ) =\frac{8x\left ( L-x\right ) ^{2}}{L^{3}} \end{align*}

Using $$c=2,L=10$$.

The general problem PDE was solved in 5.1.1.1 on page 1563 and the solution is\begin{align*} u\left ( x,t\right ) & =\frac{2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\right ) \cos \left ( c\frac{n\pi }{L}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \\ & +\frac{2}{c\pi }\sum _{n=1}^{\infty }\frac{1}{n}\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\right ) \sin \left ( c\frac{n\pi }{L}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Substituting the speciﬁc values given above into this solution gives$u\left ( x,t\right ) =\frac{1}{\pi }\sum _{n=1}^{\infty }\frac{1}{n}\left ( \int _{0}^{10}\frac{8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac{n\pi }{10}x\right ) dx\right ) \sin \left ( 2\frac{n\pi }{10}t\right ) \sin \left ( \frac{n\pi }{10}x\right )$ But $$\int _{0}^{10}\frac{8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac{n\pi }{10}x\right ) dx=\frac{160\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}}$$, hence the solution becomes$u\left ( x,t\right ) =\frac{1}{\pi ^{4}}\sum _{n=1}^{\infty }\frac{160\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{4}}\sin \left ( \frac{n\pi }{5}t\right ) \sin \left ( \frac{n\pi }{10}x\right )$ Animation is below

Source code used for the above

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##### 5.1.1.3 [336] both ends ﬁxed, inital velocity zero (special case)

problem number 336

Solve for $$u(x,t)$$ for $$t>0$$ and $$0<x<L$$ $u_{tt} = c^2 u_{xx}$ With boundary condition both ends ﬁxed \begin{align*} u(0,t) &= 0 \\ u(L,t) &=0 \end{align*}

And initial conditions \begin{align*} u(x,0) &= f(x) \\ u_t(x,0) &= g(x) \\ \end{align*}

Using the following values \begin{align*} L&=10\\ c&=2\\ f(x)&=\frac{8 x (L-x)^2}{L^3}\\ g(x)&= 0 \end{align*}

Mathematica

$\left \{\left \{u(x,t)\to \underset{n=1}{\overset{\infty }{\sum }}\frac{32 \left (2+(-1)^n\right ) \cos \left (\frac{n \pi t}{5}\right ) \sin \left (\frac{n \pi x}{10}\right )}{n^3 \pi ^3}\right \}\right \}$

Maple

$u \left ( x,t \right ) =\sum _{n=1}^{\infty }32\,{\frac{ \left ( \left ( -1 \right ) ^{n}+2 \right ) \sin \left ( 1/10\,n\pi \,x \right ) \cos \left ( 1/5\,n\pi \,t \right ) }{{n}^{3}{\pi }^{3}}}$

Hand solution

Solving the wave PDE on string with both ends ﬁxed $u_{tt}=c^{2}u_{xx}\qquad t>0,x>0$ With BC \begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end{align*}

And initial conditions\begin{align*} u\left ( x,0\right ) & =f\left ( x\right ) =\frac{8x\left ( L-x\right ) ^{2}}{L^{3}}\\ u_{t}\left ( x,0\right ) & =g\left ( x\right ) =0 \end{align*}

Using $$c=2,L=10$$.

The general problem PDE was solved in 5.1.1.1 on page 1563 and the solution is\begin{align*} u\left ( x,t\right ) & =\frac{2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\right ) \cos \left ( c\frac{n\pi }{L}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \\ & +\frac{2}{c\pi }\sum _{n=1}^{\infty }\frac{1}{n}\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\right ) \sin \left ( c\frac{n\pi }{L}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Substituting the speciﬁc values given above into this solution gives$u\left ( x,t\right ) =\frac{1}{5}\sum _{n=1}^{\infty }\left ( \int _{0}^{10}\frac{8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac{n\pi }{10}x\right ) dx\right ) \cos \left ( \frac{n\pi }{5}t\right ) \sin \left ( \frac{n\pi }{10}x\right )$ But $$\int _{0}^{10}\frac{8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac{n\pi }{10}x\right ) dx=\frac{160\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}}$$, hence the solution becomes$u\left ( x,t\right ) =32\sum _{n=1}^{\infty }\frac{\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}}\cos \left ( \frac{n\pi }{5}t\right ) \sin \left ( \frac{n\pi }{10}x\right )$ Animation is below

Source code used for the above

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##### 5.1.1.4 [337] both ends ﬁxed but domain is $$-\pi \dots \pi$$. zero intial position, non zero initial velocity

problem number 337

Solve for $$u(x,t)$$ for $$t>0$$ and $$-\pi <x<\pi$$ $u_{tt} = c^2 u_{xx}$ With boundary condition \begin{align*} u(-\pi ,t) &= 0 \\ u(\pi ,t) &=0 \end{align*}

And initial conditions \begin{align*} u(x,0) &= 0 \\ u_t(x,0) &= \sin (x)^2 \\ \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) ={\frac{1}{315\,\pi \,c} \left ( 315\,\sum _{n=5}^{\infty }32\,{\frac{\sin \left ( 1/2\,n \left ( x+\pi \right ) \right ) \sin \left ( 1/2\,cnt \right ) \left ( \left ( -1 \right ) ^{n}-1 \right ) }{\pi \,{n}^{2}c \left ({n}^{2}-16 \right ) }}\pi \,c-320\,\cos \left ( 3/2\,x \right ) \sin \left ( 3/2\,ct \right ) +1344\,\cos \left ( x/2 \right ) \sin \left ( 1/2\,ct \right ) \right ) }$

Hand solution

Solve

$u_{tt}=c^{2}u_{xx}$ With BC \begin{align*} u\left ( -\pi ,t\right ) & =0\\ u\left ( \pi ,t\right ) & =0 \end{align*}

And initial conditions\begin{align*} u\left ( x,0\right ) & =0\\ u_{t}\left ( x,0\right ) & =\sin ^{2}\left ( x\right ) \end{align*}

Let $$\xi =x+\pi$$. When $$x=-\pi ,\xi =0$$ and when $$x=\pi ,\xi =2\pi$$. In terms of $$\xi$$, the new pde in $$U\left ( \xi ,t\right )$$ becomes

$U_{tt}=c^{2}U_{\xi \xi }$ With BC \begin{align*} U\left ( 0,t\right ) & =0\\ U\left ( 2\pi ,t\right ) & =0 \end{align*}

And initial conditions\begin{align*} U\left ( \xi ,0\right ) & =0\\ U_{t}\left ( \xi ,0\right ) & =\sin ^{2}\left ( \xi \right ) \end{align*}

Let $$U=X\left ( \xi \right ) T\left ( t\right )$$. The PDE becomes\begin{align*} \frac{T^{\prime \prime }X}{c^{2}} & =X^{\prime \prime }T\\ \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}

Where $$\lambda$$ is separation constant. Hence the eigenvalue ODE is \begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( 2\pi \right ) & =0 \end{align*}

From the boundary conditions, we see that $$\lambda >0$$ is the only possible value. Therefore the solution to the above ODE is$X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }\xi \right ) +B\sin \left ( \sqrt{\lambda }\xi \right )$ Since $$X\left ( 0\right ) =0$$ then $$A=0$$ and the solution becomes $$X\left ( \xi \right ) =B\sin \left ( \sqrt{\lambda }\xi \right )$$. Since $$X\left ( 2\pi \right ) =0$$ then for non trivial solution we want $$\sqrt{\lambda }2\pi =n\pi$$ or $\lambda _{n}=\left ( \frac{n}{2}\right ) ^{2}\qquad n=1,2,3,\cdots$ Hence the eigenfunctions are $X_{n}\left ( \xi \right ) =\sin \left ( \frac{n}{2}\xi \right ) \qquad n=1,2,3,\cdots$ The time ODE now becomes\begin{align*} T_{n}^{\prime \prime }+c^{2}\lambda _{n}T_{n} & =0\\ T_{n}^{\prime \prime }+c^{2}\left ( \frac{n}{2}\right ) ^{2}T_{n} & =0\\ T_{n}^{\prime \prime }+\frac{c^{2}n^{2}}{4}T_{n} & =0 \end{align*}

Which has the solution$T_{n}\left ( t\right ) =D_{n}\cos \left ( \frac{cn}{2}t\right ) +E_{n}\sin \left ( \frac{cn}{2}t\right )$ Therefore the complete solution becomes$$U\left ( \xi ,t\right ) =\sum _{n=1}^{\infty }\left ( D_{n}\cos \left ( \frac{cn}{2}t\right ) +E_{n}\sin \left ( \frac{cn}{2}t\right ) \right ) \sin \left ( \frac{n}{2}\xi \right ) \tag{1}$$

Switching back to $$x$$ the above becomes

$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( D_{n}\cos \left ( \frac{cn}{2}t\right ) +E_{n}\sin \left ( \frac{cn}{2}t\right ) \right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right ) \tag{1A}$$

At $$t=0$$ the above becomes\begin{align*} 0 & =\sum _{n=1}^{\infty }D_{n}\sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right ) \\ D_{n} & =0 \end{align*}

The solution (1A) simpliﬁes to

$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }E_{n}\sin \left ( \frac{cn}{2}t\right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right ) \tag{2}$$

Taking time derivative of (2) gives$u_{t}\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( E_{n}\frac{cn}{2}\cos \left ( \frac{cn}{2}t\right ) \right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right )$ At $$t=0$$ the above becomes$\sin ^{2}\left ( x\right ) =\sum _{n=1}^{\infty }E_{n}\frac{cn}{2}\sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right )$ Applying orthogonality gives\begin{align*} \int _{-\pi }^{\pi }\sin ^{2}\left ( x\right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right ) dx & =E_{n}\frac{cn}{2}\int _{-\pi }^{\pi }\sin ^{2}\left ( \frac{n}{2}\left ( x+\pi \right ) \right ) dx\\ & =E_{n}\frac{\pi cn}{2} \end{align*}

For the LHS, for $$n$$ even $$\int _{-\pi }^{\pi }\sin ^{2}\left ( x\right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right ) dx=0$$. Hence $$E_{n}=0$$ for all $$n$$ even. For $$n$$ odd \begin{align*} \int _{-\pi }^{\pi }\sin ^{2}\left ( x\right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right ) dx & =\frac{16\left ( \cos \left ( n\pi \right ) -1\right ) }{\left ( n^{2}-16\right ) n}\\ & =\frac{16\left ( \left ( -1\right ) ^{n}-1\right ) }{\left ( n^{2}-16\right ) n} \end{align*}

But $$n$$ is odd, hence the above simpliﬁes more to

$\int _{-\pi }^{\pi }\sin ^{2}\left ( x\right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right ) dx=\frac{-32}{\left ( n^{2}-16\right ) n}$

Therefore\begin{align} E_{n} & =\frac{2}{n\pi c}\frac{-32}{\left ( n^{2}-16\right ) n}\nonumber \\ & =\frac{-64}{n^{2}\pi c\left ( n^{2}-16\right ) }\nonumber \end{align}

Therefore the ﬁnal solution (2) now becomes$u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\frac{-64}{n^{2}\pi c\left ( n^{2}-16\right ) }\sin \left ( \frac{cn}{2}t\right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right )$

The following is an animation of the solution for $$c=2$$. Using $$c=2$$ then the solution above becomes

$u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\frac{-32}{n^{2}\pi \left ( n^{2}-16\right ) }\sin \left ( nt\right ) \sin \left ( \frac{n}{2}\left ( x+\pi \right ) \right )$

Source code used for the above

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##### 5.1.1.5 [338] both ends ﬁxed but domain is $$-1 \dots 1$$. intial position is an impulse, zero initial velocity

problem number 338

Problem 6.3.31 Introduction to Partial Dﬀerential Equations by Peter Olver, ISBN 9783319020983.

Solve

\begin{align*} u_{tt} & =u_{xx}\\ u\left ( -1,t\right ) & =0\\ u\left ( 1,t\right ) & =0\\ u\left ( x,0\right ) & =\delta \left ( x\right ) \\ \frac{\partial u\left ( x,0\right ) }{\partial t} & =0 \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=1}^{\infty }\sin \left ( 1/2\,n\pi \right ) \sin \left ( 1/2\,n\pi \, \left ( x+1 \right ) \right ) \cos \left ( 1/2\,n\pi \,t \right )$

Hand solution

Since the boundary conditions are at $$x=-1$$ and at $$x=1$$, it is a little easier to solve this by ﬁrst shifting the boundaries to $$x=0$$ and $$x=2$$. This is done by transformation. Let$z=x+1$ When $$x=-1$$ then $$z=0$$ and when $$x=1$$ then $$z=2$$. The PDE in terms of $$z$$ remains the same but the B.C. are shifted. Hence we want to solve for $$v\left ( z,t\right )$$ in\begin{align*} v_{tt} & =v_{zz}\\ v\left ( 0,t\right ) & =0\\ v\left ( 2,t\right ) & =0 \end{align*}

No need to worry about initial conditions now, since we will transform back to $$x$$ before applying initial conditions and therefore will use the original initial conditions. This PDE is now solved by separation. Let $$v=Z\left ( z\right ) T\left ( t\right )$$. Substituting into the PDE gives\begin{align*} T^{\prime \prime }Z & =Z^{\prime \prime }T\\ \frac{T^{\prime \prime }}{T} & =\frac{Z^{\prime \prime }}{Z}=-\lambda \end{align*}

This gives the boundary value ODE \begin{align} Z^{\prime \prime }+\lambda Z & =0\tag{1}\\ Z\left ( 0\right ) & =0\nonumber \\ Z\left ( 2\right ) & =0\nonumber \end{align}

And the time ODE$$T^{\prime \prime }+\lambda T=0\tag{2}$$ Solving (1). From the boundary conditions we know only $$\lambda >0$$ is an eigenvalue. Hence for $$\lambda >0$$ the solution is $Z\left ( z\right ) =A\cos \left ( \sqrt{\lambda }z\right ) +B\sin \left ( \sqrt{\lambda }z\right )$ At $$z=0$$ this gives $$A=0$$. Hence the solution now becomes $$Z\left ( z\right ) =B\sin \left ( \sqrt{\lambda }z\right )$$. At $$z=2$$ the above gives $$0=B\sin \left ( 2\sqrt{\lambda }\right )$$. For non-trivial solution we want $$\sin \left ( 2\sqrt{\lambda }\right ) =0$$ which implies $$2\sqrt{\lambda }=n\pi$$ or$\lambda _{n}=\left ( \frac{n\pi }{2}\right ) ^{2}\qquad n=1,2,3,\cdots$ And the corresponding eigenfunctions$Z_{n}\left ( z\right ) =\sin \left ( \frac{n\pi }{2}z\right ) \qquad n=1,2,3,\cdots$ The time ODE (2) now becomes$T^{\prime \prime }+\left ( \frac{n\pi }{2}\right ) ^{2}T=0$ Which has solution$T_{n}\left ( t\right ) =A_{n}\cos \left ( \frac{n\pi }{2}t\right ) +B_{n}\sin \left ( \frac{n\pi }{2}t\right )$ Hence the complete solution is$v\left ( z,t\right ) =\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( \frac{n\pi }{2}t\right ) +B_{n}\sin \left ( \frac{n\pi }{2}t\right ) \right ) \sin \left ( \frac{n\pi }{2}z\right )$ We are now ready to switch back from $$z$$ to $$x$$. Since $$z=x+1$$ then the above becomes$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( \frac{n\pi }{2}t\right ) +B_{n}\sin \left ( \frac{n\pi }{2}t\right ) \right ) \sin \left ( \frac{n\pi }{2}\left ( x+1\right ) \right ) \tag{3}$$ Now we apply initial conditions to ﬁnd $$A_{n},B_{n}$$. At $$t=0,u\left ( x,0\right ) =\delta \left ( x\right )$$. Hence the above gives$\delta \left ( x\right ) =\sum _{n=1}^{\infty }A_{n}\sin \left ( \frac{n\pi }{2}\left ( x+1\right ) \right )$ Multiplying both sides by $$\sin \left ( \frac{m\pi }{2}\left ( x+1\right ) \right )$$ and Integrating gives$\int _{-1}^{1}\delta \left ( x\right ) \sin \left ( \frac{m\pi }{2}\left ( x+1\right ) \right ) dx=\sum _{n=1}^{\infty }A_{n}\int _{-1}^{1}\sin \left ( \frac{n\pi }{2}\left ( x+1\right ) \right ) \sin \left ( \frac{m\pi }{2}\left ( x+1\right ) \right ) dx$ By orthogonality of $$\sin$$ functions only term survives and the above simpliﬁes to\begin{align*} \int _{-1}^{1}\delta \left ( x\right ) \sin \left ( \frac{m\pi }{2}\left ( x+1\right ) \right ) dx & =A_{m}\overset{1}{\overbrace{\int _{-1}^{1}\sin ^{2}\left ( \frac{m\pi }{2}\left ( x+1\right ) \right ) dx}}\\ & =A_{m} \end{align*}

But $$\int _{-1}^{1}\delta \left ( x\right ) \sin \left ( \frac{m\pi }{2}\left ( x+1\right ) \right ) dx=\sin \left ( \frac{m\pi }{2}\right )$$ since that is where $$x=0$$. The above reduces to$A_{n}=\sin \left ( \frac{n\pi }{2}\right ) \qquad n=1,2,3,\cdots$ The solution (1) becomes$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \sin \left ( \frac{n\pi }{2}\right ) \cos \left ( \frac{n\pi }{2}t\right ) +B_{n}\sin \left ( \frac{n\pi }{2}t\right ) \right ) \sin \left ( \frac{n\pi }{2}\left ( x+1\right ) \right ) \tag{4}$$ Taking time derivatives$\frac{\partial }{\partial t}u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( -\frac{n\pi }{2}\sin \left ( \frac{n\pi }{2}\right ) \sin \left ( \frac{n\pi }{2}t\right ) +\frac{n\pi }{2}B_{n}\cos \left ( \frac{n\pi }{2}t\right ) \right ) \sin \left ( \frac{n\pi }{2}\left ( x+1\right ) \right )$ At $$t=0$$ the above becomes$0=\sum _{n=1}^{\infty }\frac{n\pi }{2}B_{n}\sin \left ( \frac{n\pi }{2}\left ( x+1\right ) \right )$ Therefore $$B_{n}=0$$. Hence the solution (4) becomes$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\sin \left ( \frac{n\pi }{2}\right ) \cos \left ( \frac{n\pi }{2}t\right ) \sin \left ( \frac{n\pi }{2}\left ( x+1\right ) \right ) \tag{5}$$ Notice that $$\sin \left ( \frac{n\pi }{2}\right )$$ is zero when $$n$$ is even.

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##### 5.1.1.6 [339] Logan book, page 28. Both ends ﬁxed

problem number 339

This is problem at page 28, David J Logan textbook, applied PDE textbook. No initial conditions given $u_{tt} = c^2 u_{xx}$ With boundary condition \begin{align*} u(0,t) &= 0 \\ u(L,t) &= 0 \end{align*}

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$u \left ( x,t \right ) =\sum _{n=1}^{\infty }\sin \left ({\frac{n\pi \,x}{L}} \right ) \left ( \sin \left ({\frac{cn\pi \,t}{L}} \right ){\it \_C1} \left ( n \right ) +\cos \left ({\frac{cn\pi \,t}{L}} \right ){\it \_C5} \left ( n \right ) \right )$

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##### 5.1.1.7 [340] non-zero initial velocity. Both ends ﬁxed

problem number 340

Added Feb 25, 2019. Exam 1 problem, MATH 4567 Applied Fourier Analysis, University of Minnesota, Twin Cities.

Solve for $$u(x,t)$$ $u_{tt} = u_{xx} -u$ With boundary condition \begin{align*} u(0,t) &= 0 \\ u(\pi ,t) &=0 \end{align*}

And initial conditions \begin{align*} u(x,0) &= 0 \\ u_t(x,0) &= 1 \\ \end{align*}

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$u \left ( x,t \right ) =\sum _{n=1}^{\infty }-2\,{\frac{ \left ( \left ( -1 \right ) ^{n}-1 \right ) \sin \left ( nx \right ) \sin \left ( \sqrt{{n}^{2}+1}t \right ) }{\pi \,\sqrt{{n}^{2}+1}n}}$

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##### 5.1.1.8 [341] Logan book page 149)

problem number 341

This is problem at page 149, David J Logan textbook, applied PDE textbook.

$\frac{\partial ^2 u}{\partial t^2} = c^2 \frac{\partial ^2 u}{\partial x^2} + p(x,t)$ With boundary conditions \begin{align*} u(\pi ,0) &=0 \\ u(0,t) &= 0 \end{align*}

With initial conditions \begin{align*} \frac{\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= 0 \end{align*}

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$u \left ( x,t \right ) =\int _{0}^{t}\!\sum _{n=1}^{\infty } \left ( 2\,{\frac{\int _{0}^{\pi }\!\sin \left ( nx \right ) p \left ( x,\tau \right ) \,{\rm d}x\sin \left ( nx \right ) \sin \left ( cn \left ( t-\tau \right ) \right ) }{cn\pi }} \right ) \,{\rm d}\tau$

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##### 5.1.1.9 [342] Haberman 8.5.2 (a)

problem number 342

This is problem 8.5.2 (a), Richard Haberman applied partial diﬀerential equations book, 5th edition

Both ends ﬁxed end, initial position given, zero initial velocity, with source that depends on time and space.

Consider a vibrating string with time-dependent forcing: $\frac{\partial ^2 u}{\partial t^2} = c^2 \frac{\partial ^2 u}{\partial x^2} + Q(x,t)$ With boundary conditions \begin{align*} u(0,t) &=0 \\ u(L,t) &= 0 \end{align*}

With initial conditions \begin{align*} u_t(x,0) &=0 \\ u(x,0) &= f(x) \end{align*}

my hand solution in in the ﬁle feb_24_2019_4_24_pm.tex, but I need to go over my solution again to make sure it is correct.

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$u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac{1}{L}\int _{0}^{L}\!\sin \left ({\frac{\pi \,n\tau }{L}} \right ) f \left ( \tau \right ) \,{\rm d}\tau \sin \left ({\frac{\pi \,nx}{L}} \right ) \cos \left ({\frac{cn\pi \,t}{L}} \right ) } \right ) +\int _{0}^{t}\!\sum _{n=1}^{\infty } \left ( 2\,{\frac{1}{cn\pi }\int _{0}^{L}\!\sin \left ({\frac{\pi \,nx}{L}} \right ) Q \left ( x,\tau \right ) \,{\rm d}x\sin \left ({\frac{\pi \,nx}{L}} \right ) \sin \left ({\frac{cn\pi \, \left ( t-\tau \right ) }{L}} \right ) } \right ) \,{\rm d}\tau$

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##### 5.1.1.10 [343] Haberman 8.5.2 (b)

problem number 343

This is problem 8.5.2 (b), Richard Haberman applied partial diﬀerential equations book, 5th edition.

Both ends ﬁxed end, initial position given, zero initial velocity, with source that depends on time and space.

Consider a vibrating string with time-dependent forcing: $u_{tt} = c^2 u_{xx} + g(x) \cos (\omega t)$ With boundary conditions \begin{align*} u(0,t) &=0 \\ u(L,t) &= 0 \end{align*}

With initial conditions \begin{align*} u_t(x,0) &=0 \\ u(x,0) &= f(x) \end{align*}

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Maple

$u \left ( x,t \right ) ={\frac{1}{\pi \,c} \left ( -2\,\sum _{n=1}^{\infty } \left ({\frac{L\cos \left ( wt \right ) \pi \,c}{-{\pi }^{2}{c}^{2}{n}^{2}+{L}^{2}{w}^{2}}\sin \left ({\frac{\pi \,nx}{L}} \right ) \int _{0}^{L}\!\sin \left ({\frac{\pi \,nx}{L}} \right ) g \left ( x \right ) \,{\rm d}x} \right ) +2\,\sum _{n=1}^{\infty } \left ({\frac{L\pi \,c}{-{\pi }^{2}{c}^{2}{n}^{2}+{L}^{2}{w}^{2}}\sin \left ({\frac{\pi \,nx}{L}} \right ) \int _{0}^{L}\!\sin \left ({\frac{\pi \,nx}{L}} \right ) g \left ( x \right ) \,{\rm d}x\cos \left ({\frac{cn\pi \,t}{L}} \right ) } \right ) \right ) }$

Hand solution

Let $u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right )$ Where we used $$=$$ instead of $$\sim$$ above, since the PDE given has homogeneous B.C. We know that $$\phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right )$$ for $$n=1,2,3,\cdots$$ where $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}$$. Substituting the above in the given PDE gives$\sum _{n=1}^{\infty }A_{n}^{\prime \prime }\left ( t\right ) \phi _{n}\left ( x\right ) =c^{2}\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \frac{d^{2}\phi _{n}\left ( x\right ) }{dx^{2}}+Q\left ( x,t\right )$ But $$Q\left ( x,t\right ) =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \phi _{n}\left ( x\right )$$, hence the above becomes$\sum _{n=1}^{\infty }A_{n}^{\prime \prime }\left ( t\right ) \phi _{n}\left ( x\right ) =c^{2}\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \frac{d^{2}\phi _{n}\left ( x\right ) }{dx^{2}}+\sum _{n=1}^{\infty }g_{n}\left ( t\right ) \phi _{n}\left ( x\right )$ But $$\frac{d^{2}\phi _{n}\left ( x\right ) }{dx^{2}}=-\lambda _{n}\phi _{n}\left ( x\right )$$, hence$\sum _{n=1}^{\infty }A_{n}^{\prime \prime }\left ( t\right ) \phi _{n}\left ( x\right ) =-c^{2}\sum _{n=1}^{\infty }\lambda _{n}A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }g_{n}\left ( t\right ) \phi _{n}\left ( x\right )$ Multiplying both sides by  $$\phi _{m}\left ( x\right )$$ and integrating gives\begin{align*} \int _{0}^{L}\sum _{n=1}^{\infty }A_{n}^{\prime \prime }\left ( t\right ) \phi _{m}\left ( x\right ) \phi _{n}\left ( x\right ) dx & =-c^{2}\int _{0}^{L}\sum _{n=1}^{\infty }\lambda _{n}A_{n}\left ( t\right ) \phi _{m}\left ( x\right ) \phi _{n}\left ( x\right ) dx+\int _{0}^{L}\sum _{n=1}^{\infty }g_{n}\left ( t\right ) \phi _{m}\left ( x\right ) \phi _{n}\left ( x\right ) dx\\ A_{n}^{\prime \prime }\left ( t\right ) \int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx & =-c^{2}\lambda _{n}A_{n}\left ( t\right ) \int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx+g_{n}\left ( t\right ) \int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx \end{align*}

Hence$A_{n}^{\prime \prime }\left ( t\right ) +c^{2}\lambda _{n}A_{n}\left ( t\right ) =g_{n}\left ( t\right )$ Now we solve the above ODE. Let solution be $A_{n}\left ( t\right ) =A_{n}^{h}\left ( t\right ) +A_{n}^{p}\left ( t\right )$ Which is the sum of the homogenous and particular solutions. The homogenous solution is $A_{n}^{h}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt{\lambda _{n}}t\right )$ And the particular solution depends on $$q_{n}\left ( t\right )$$. Once we ﬁnd $$q_{n}\left ( t\right )$$, we plug-in everything back into $$u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right )$$ and then use initial conditions to ﬁnd $$c_{1_{n}},c_{2_{n}}$$, the two constant of integrations. Now we are given that $$Q\left ( x,t\right ) =g\left ( x\right ) \cos \left ( \omega t\right )$$. Hence$g_{n}\left ( t\right ) =\frac{\int _{0}^{L}Q\left ( x,t\right ) \phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx}=\frac{\cos \left ( \omega t\right ) \int _{0}^{L}g\left ( x\right ) \phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx}=\cos \left ( \omega t\right ) \gamma _{n}$ Where$\gamma _{n}=\frac{\int _{0}^{L}g\left ( x\right ) \phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx}$ is constant that depends on $$n$$. Now we use the above in result found in part (a)$$A_{n}^{\prime \prime }\left ( t\right ) +c^{2}\lambda _{n}A_{n}\left ( t\right ) =\gamma _{n}\cos \left ( \omega t\right ) \tag{1}$$ We know the homogenous solution from part (a). $A_{n}^{h}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt{\lambda _{n}}t\right )$ We now need to ﬁnd the particular solution. Will solve using method of undetermined coeﬃcients.

Case 1 $$\omega \neq c\sqrt{\lambda _{n}}$$ (no resonance)

We can now guess $A_{n}^{p}\left ( t\right ) =z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right )$ Plugging this back into (1) gives\begin{align*} \left ( z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \right ) ^{\prime \prime }+c^{2}\lambda _{n}\left ( z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \right ) & =\gamma _{n}\cos \left ( \omega t\right ) \\ \left ( -\omega z_{1}\sin \left ( \omega t\right ) +\omega z_{2}\cos \left ( \omega t\right ) \right ) ^{\prime }+c^{2}\lambda _{n}\left ( z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \right ) & =\gamma _{n}\cos \left ( \omega t\right ) \\ -\omega ^{2}z_{1}\cos \left ( \omega t\right ) -\omega ^{2}z_{2}\sin \left ( \omega t\right ) +c^{2}\lambda _{n}\left ( z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \right ) & =\gamma _{n}\cos \left ( \omega t\right ) \end{align*}

Collecting terms$\cos \left ( \omega t\right ) \left ( -\omega ^{2}z_{1}+c^{2}\lambda _{n}z_{1}\right ) +\sin \left ( \omega t\right ) \left ( -\omega ^{2}z_{2}+c^{2}\lambda _{n}z_{2}\right ) =\gamma _{n}\cos \left ( \omega t\right )$ Therefore we obtain two equations in two unknowns\begin{align*} -\omega ^{2}z_{1}+c^{2}\lambda _{n}z_{1} & =\gamma _{n}\\ -\omega ^{2}z_{2}+c^{2}\lambda _{n}z_{2} & =0 \end{align*}

From the second equation, $$z_{2}=0$$ and from the ﬁrst equation\begin{align*} z_{1}\left ( c^{2}\lambda _{n}-\omega ^{2}\right ) & =\gamma _{n}\\ z_{1} & =\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}} \end{align*}

Hence \begin{align*} A_{n}^{p}\left ( t\right ) & =z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \\ & =\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right ) \end{align*}

Therefore\begin{align*} A_{n}\left ( t\right ) & =A_{n}^{h}\left ( t\right ) +A_{n}^{p}\left ( t\right ) \\ & =c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt{\lambda _{n}}t\right ) +\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right ) \end{align*}

Now we need to ﬁnd $$c_{1_{n}},c_{2_{n}}$$. Since\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }\left ( c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt{\lambda _{n}}t\right ) +\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

At $$t=0$$ the above becomes\begin{align*} f\left ( x\right ) & =\sum _{n=1}^{\infty }\left ( c_{1_{n}}+\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\right ) \sin \left ( \frac{n\pi }{L}x\right ) \\ & =\sum _{n=1}^{\infty }c_{1_{n}}\sin \left ( \frac{n\pi }{L}x\right ) +\sum _{n=1}^{\infty }\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Applying orthogonality\begin{align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx & =\int _{0}^{L}\sum _{n=1}^{\infty }c_{1_{n}}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx+\int _{0}^{L}\sum _{n=1}^{\infty }\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx\\ \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx & =c_{1_{n}}\int _{0}^{L}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) dx+\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\int _{0}^{L}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) dx \end{align*}

Rearranging\begin{align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx-\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\int _{0}^{L}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) dx & =c_{1_{n}}\int _{0}^{L}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) dx\\ c_{1_{n}} & =\frac{\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx}{\int _{0}^{L}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) dx}-\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\\ & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx-\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}} \end{align*}

We now need to ﬁnd $$c_{2_{n}}$$. For this we need to diﬀerentiate the solution once.$\frac{\partial u\left ( x,t\right ) }{\partial t}=\sum _{n=1}^{\infty }\left ( -c\sqrt{\lambda _{n}}c_{1_{n}}\sin \left ( c\sqrt{\lambda _{n}}t\right ) +c\sqrt{\lambda _{n}}c_{2_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) -\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\omega \sin \left ( \omega t\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right )$ Applying initial conditions $$\frac{\partial u\left ( x,0\right ) }{\partial t}=0$$ gives$0=\sum _{n=1}^{\infty }c\sqrt{\lambda _{n}}c_{2_{n}}\sin \left ( \frac{n\pi }{L}x\right )$ Hence $c_{2_{n}}=0$ Therefore the ﬁnal solution is$A_{n}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right )$ And$u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \sin \left ( \frac{n\pi }{L}x\right )$ Where $c_{1_{n}}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx-\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}$ Case 2 $$\omega =c\sqrt{\lambda _{n}}$$ Resonance case. Now we can’t guess $$A_{n}^{p}\left ( t\right ) =z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right )$$ so we have to use $A_{n}^{p}\left ( t\right ) =z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right )$ Substituting this in $$A_{n}^{\prime \prime }\left ( t\right ) +c^{2}\lambda _{n}A_{n}\left ( t\right ) =\gamma _{n}\cos \left ( \omega t\right )$$ gives$$\left ( z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \right ) ^{\prime \prime }+c^{2}\lambda _{n}\left ( z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \right ) =\gamma _{n}\cos \left ( \omega t\right ) \tag{2}$$ But \begin{align*} \left ( z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \right ) ^{\prime \prime } & =\left ( z_{1}\cos \left ( \omega t\right ) -z_{1}\omega t\sin \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) +z_{2}\omega t\cos \left ( \omega t\right ) \right ) ^{\prime }\\ & =-z_{1}\omega \sin \left ( \omega t\right ) -\left ( z_{1}\omega \sin \left ( \omega t\right ) +z_{1}\omega ^{2}t\cos \left ( \omega t\right ) \right ) \\ & +z_{2}\omega \cos \left ( \omega t\right ) +\left ( z_{2}\omega \cos \left ( \omega t\right ) -z_{2}\omega ^{2}t\sin \left ( \omega t\right ) \right ) \\ & =-2z_{1}\omega \sin \left ( \omega t\right ) -z_{1}\omega ^{2}t\cos \left ( \omega t\right ) +2z_{2}\omega \cos \left ( \omega t\right ) -z_{2}\omega ^{2}t\sin \left ( \omega t\right ) \end{align*}

Hence (2) becomes$-2z_{1}\omega \sin \left ( \omega t\right ) -z_{1}\omega ^{2}t\cos \left ( \omega t\right ) +2z_{2}\omega \cos \left ( \omega t\right ) -z_{2}\omega ^{2}t\sin \left ( \omega t\right ) +c^{2}\lambda _{n}\left ( z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \right ) =\gamma _{n}\cos \left ( \omega t\right )$ Comparing coeﬃcients we see that $$2z_{2}\omega =\gamma _{n}$$ or $z_{2}=\frac{\gamma _{n}}{2\omega }$ And $$z_{1}=0$$. Therefore $A_{n}^{p}\left ( t\right ) =\frac{\gamma _{n}}{2\omega }t\sin \left ( \omega t\right )$ Therefore\begin{align*} A_{n}\left ( t\right ) & =A_{n}^{h}\left ( t\right ) +A_{n}^{p}\left ( t\right ) \\ & =c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt{\lambda _{n}}t\right ) +\frac{\gamma _{n}}{2c\sqrt{\lambda _{n}}}t\sin \left ( \omega t\right ) \end{align*}

We now can ﬁnd $$c_{1_{n}},c_{2_{n}}$$ from initial conditions.\begin{align} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\left ( c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt{\lambda _{n}}t\right ) +\frac{\gamma _{n}}{2c\sqrt{\lambda _{n}}}t\sin \left ( \omega t\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right ) \tag{4} \end{align}

At $$t=0$$\begin{align*} f\left ( x\right ) & =\sum _{n=1}^{\infty }c_{1_{n}}\sin \left ( \frac{n\pi }{L}x\right ) \\ c_{1n} & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx \end{align*}

Taking time derivative of (4) and setting it to zero will give $$c_{2n}$$. Since initial speed is zero then $$c_{2_{n}}=0$$. Hence$A_{n}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +\frac{\gamma _{n}}{2c\sqrt{\lambda _{n}}}t\sin \left ( \omega t\right )$ This completes the solution.

Summary of solution

The solution is given by$u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right )$ Case $$\omega \neq c\sqrt{\lambda _{n}}$$$A_{n}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right )$ And$c_{1_{n}}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx-\frac{\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}$ And$\gamma _{n}=\frac{\int _{0}^{L}g\left ( x\right ) \phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx}$ And $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,$$

Case $$\omega =c\sqrt{\lambda _{n}}$$ (resonance)$A_{n}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt{\lambda _{n}}t\right ) +\frac{\gamma _{n}}{2c\sqrt{\lambda _{n}}}t\sin \left ( \omega t\right )$ And$c_{1_{n}}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx$

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##### 5.1.1.11 [344] Both I.C. not zero

problem number 344

Taken from Maple 2018.1 improvements to PDE’s document. Solve $v_{tt} = v_{xx}$ For $$t>0$$ and $$0<x<1$$. With boundary conditions \begin{align*} v(0,t)&=0\\ v(1,0)&=0 \end{align*}

With initial conditions \begin{align*} v( x,0) & =f(x) \\ \frac{\partial v}{\partial t}(x,0) &=g(x) \\ \end{align*}

Where $$f(x)=-{\frac{{{\rm e}^{2}}x-{{\rm e}^{x+1}}-x+{{\rm e}^{1-x}}}{{{\rm e}^{2}}-1}}$$ and $$g(x)=1+{\frac{{{\rm e}^{2}}x-{{\rm e}^{x+1}}-x+{{\rm e}^{1-x}}}{{{\rm e}^{2}}-1}}$$

Mathematica

$\left \{\left \{v(x,t)\to \underset{n=1}{\overset{\infty }{\sum }}-\frac{2 \left ((-1)^{n+1} n \pi \cos (n \pi t)+\left (\left (-1+(-1)^n\right ) \pi ^2 n^2+2 (-1)^n-1\right ) \sin (n \pi t)\right ) \sin (n \pi x)}{\pi ^4 n^4+\pi ^2 n^2}\right \}\right \}$

Maple

$v \left ( x,t \right ) =\sum _{n=1}^{\infty }-2\,{\frac{\sin \left ( \pi \,nx \right ) \left ( \left ({\pi }^{2} \left ( -1 \right ) ^{n}{n}^{2}-{\pi }^{2}{n}^{2}+2\, \left ( -1 \right ) ^{n}-1 \right ) \sin \left ( n\pi \,t \right ) - \left ( -1 \right ) ^{n}\cos \left ( n\pi \,t \right ) \pi \,n \right ) }{{\pi }^{2}{n}^{2} \left ({\pi }^{2}{n}^{2}+1 \right ) }}$

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##### 5.1.1.12 [345] With constant source

problem number 345

Third example, from Maple 2018.1 improvements to PDE’s document. What_is_New_after_Maple_2018.pdf

Solve $\frac{\partial ^2 u}{\partial t^2} = c^2 \frac{\partial ^2 u}{\partial x^2} + 1$ For $$t>0$$ and $$0<x<L$$. With boundary conditions \begin{align*} u(0,t)&=0\\ u(L,0)&=0 \end{align*}

With initial conditions \begin{align*} u ( x,0) & =f(x) \\ \frac{\partial u}{\partial t}(x,0) &=g(x) \\ \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =1/2\,{\frac{1}{{c}^{2}} \left ( 2\,\sum _{n=1}^{\infty } \left ({\frac{1}{\pi \,n{c}^{2}L}\sin \left ({\frac{\pi \,nx}{L}} \right ) \left ( 2\,L\int _{0}^{L}\!\sin \left ({\frac{\pi \,nx}{L}} \right ) g \left ( x \right ) \,{\rm d}x\sin \left ({\frac{cn\pi \,t}{L}} \right ) c-\pi \,\int _{0}^{L}\!\sin \left ({\frac{\pi \,nx}{L}} \right ) \left ( -2\,f \left ( x \right ){c}^{2}+Lx-{x}^{2} \right ) \,{\rm d}x\cos \left ({\frac{cn\pi \,t}{L}} \right ) n \right ) } \right ){c}^{2}+Lx-{x}^{2} \right ) }$

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##### 5.1.1.13 [346] Logan page 213

problem number 346

This is problem at page 213, David J Logan textbook, applied PDE textbook. Both ends ﬁxed end, with source.

$u_{tt} = c^2 u_{xx}+ A x$ With boundary conditions \begin{align*} u(L,0) &=0 \\ u(0,t) &= 0 \end{align*}

With initial conditions \begin{align*} \frac{\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= 0 \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =1/6\,{\frac{1}{{c}^{2}} \left ( A{L}^{2}x-A{x}^{3}+6\,\sum _{n=1}^{\infty }2\,{\frac{{L}^{3} \left ( -1 \right ) ^{n}A}{{n}^{3}{\pi }^{3}{c}^{2}}\sin \left ({\frac{\pi \,nx}{L}} \right ) \cos \left ({\frac{cn\pi \,t}{L}} \right ) }{c}^{2} \right ) }$

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##### 5.1.1.14 [347] Telegraphy PDE

problem number 347

Both ends ﬁxed with damping Solve $u_{tt} + 2 u_t = c^2 u_{xx}$ With boundary conditions \begin{align*} u(0,t) &= 0\\ u(\pi ,0) &=0 \end{align*}

With initial conditions \begin{align*} \frac{\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= f(x) \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( \cases{2\,{\frac{{{\rm e}^{-t}} \left ( t+1 \right ) \sin \left ( x \right ) \int _{0}^{\pi }\!\sin \left ( x \right ) f \left ( x \right ) \,{\rm d}x}{\pi }}&n=1\cr{\frac{\sin \left ( nx \right ) \int _{0}^{\pi }\!\sin \left ( nx \right ) f \left ( x \right ) \,{\rm d}x \left ({{\rm e}^{- \left ( \sqrt{-{n}^{2}+1}+1 \right ) t}}\sqrt{-{n}^{2}+1}+{{\rm e}^{ \left ( -1+\sqrt{-{n}^{2}+1} \right ) t}}\sqrt{-{n}^{2}+1}-{{\rm e}^{- \left ( \sqrt{-{n}^{2}+1}+1 \right ) t}}+{{\rm e}^{ \left ( -1+\sqrt{-{n}^{2}+1} \right ) t}} \right ) }{\sqrt{-{n}^{2}+1}\pi }}&otherwise\cr } \right )$ But $$n = 1$$ should not be included.

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##### 5.1.1.15 [348] Dispersion term present (general case)

problem number 348

Solve $u_{tt}+ \gamma ^2 u(x,t) = c^2 u_{xx}$ Dispersion term $$\gamma ^2 u(x,t)$$ causes the shape of the original wave to distort with time. With $$0<x<L$$ and $$t>0$$ and with boundary conditions \begin{align*} u(0,t) &= 0\\ u(L,0) &=0 \end{align*}

With initial conditions \begin{align*} \frac{\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= f(x) \end{align*}

Mathematica

Failed Due to adding dispersion term

Maple

$u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac{1}{L}\sin \left ({\frac{\pi \,nx}{L}} \right ) \cos \left ({\frac{\sqrt{{\pi }^{2}{c}^{2}{n}^{2}+{L}^{2}{\gamma }^{2}}t}{L}} \right ) \int _{0}^{L}\!\sin \left ({\frac{\pi \,nx}{L}} \right ) f \left ( x \right ) \,{\rm d}x} \right )$

Hand solution

Solving for $$t>0,0<x<L$$$\frac{\partial ^{2}u}{\partial t^{2}}+\gamma ^{2}u=c^{2}\frac{\partial ^{2}u}{\partial x^{2}}\qquad 0<x<L,t>0$ With BC \begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end{align*}

And initial conditions\begin{align*} u\left ( x,0\right ) & =f\left ( x\right ) \\ u_{t}\left ( x,0\right ) & =0 \end{align*}

Dispersion term $$\gamma ^{2}u\,$$ causes the shape of the original wave to distort with time. Using separation of variables, Let $$u=X\left ( x\right ) T\left ( t\right ) .$$ Substituting this back in the PDE gives\begin{align*} T^{\prime \prime }X+\gamma ^{2}XT & =c^{2}X^{\prime \prime }T\\ \frac{1}{c^{2}}\left ( \frac{T^{\prime \prime }}{T}+\gamma ^{2}\right ) & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}

The eigenvalue ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

The eigenvalues are $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,\cdots$$ and the eigenfunctions are $$X_{n}\left ( x\right ) =c_{n}\sin \left ( \frac{n\pi }{L}x\right )$$. The time ODE becomes$T^{\prime \prime }+\left ( \gamma ^{2}+c^{2}\lambda _{n}\right ) T=0$ The solution is$T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt{\gamma ^{2}+c^{2}\lambda _{n}}t\right ) +B_{n}\sin \left ( \sqrt{\gamma ^{2}+c^{2}\lambda _{n}}t\right )$ Taking time derivatives gives$T_{n}^{\prime }\left ( t\right ) =-\sqrt{\gamma ^{2}+c^{2}\lambda _{n}}A_{n}\sin \left ( \sqrt{\gamma ^{2}+c^{2}\lambda _{n}}t\right ) +B_{n}\sqrt{\gamma ^{2}+c^{2}\lambda _{n}}\cos \left ( \sqrt{\gamma ^{2}+c^{2}\lambda _{n}}t\right )$ At time $$t=0$$, the above is zero (initial velocity is zero), which gives$0=B_{n}\sqrt{\gamma ^{2}+c^{2}\lambda _{n}}$ Hence $$B_{n}=0$$ and the time ODE solution becomes$T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt{\gamma ^{2}+c^{2}\lambda _{n}}t\right )$ Hence the fundamental solution is\begin{align*} u_{n}\left ( x,t\right ) & =T_{n}X_{n}\\ & =c_{n}\cos \left ( \sqrt{\gamma ^{2}+c^{2}\left ( \frac{n\pi }{L}\right ) ^{2}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Therefore the solution is$u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\cos \left ( \frac{1}{L}\sqrt{\left ( L^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac{n\pi }{L}x\right )$ $$c_{n}$$ is found from initial position. At $$t=0$$ the above becomes$f\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\sin \left ( \frac{n\pi }{L}x\right )$ Applying orthogonality gives\begin{align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx & =c_{n}\frac{L}{2}\\ c_{n} & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx \end{align*}

Hence solution is$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{n\pi }{L}s\right ) ds\right ) \cos \left ( \frac{1}{L}\sqrt{\left ( L^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac{n\pi }{L}x\right )$

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##### 5.1.1.16 [349] Dispersion term present

problem number 349

Solve $u_{tt}+ \gamma ^2 u(x,t) = c^2 u_{xx}$ Dispersion term $$\gamma ^2 u(x,t)$$ causes the shape of the original wave to distort with time. With $$0<x<\pi$$ and $$t>0$$ and with boundary conditions \begin{align*} u(0,t) &= 0\\ u(\pi ,0) &=0 \end{align*}

With initial conditions \begin{align*} \frac{\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= \sin ^2(x) \end{align*}

Mathematica

Failed Due to adding dispersion term

Maple

$u \left ( x,t \right ) =1/3\,{\frac{1}{\pi } \left ( 3\,\sum _{n=3}^{\infty }4\,{\frac{\sin \left ( nx \right ) \cos \left ( \sqrt{{c}^{2}{n}^{2}+{\gamma }^{2}}t \right ) \left ( \left ( -1 \right ) ^{n}-1 \right ) }{\pi \,n \left ({n}^{2}-4 \right ) }}\pi +8\,\sin \left ( x \right ) \cos \left ( \sqrt{{c}^{2}+{\gamma }^{2}}t \right ) \right ) }$

Hand solution

Solving for $$t>0,0<x<L$$$$\frac{\partial ^{2}u}{\partial t^{2}}+\gamma ^{2}u=\frac{\partial ^{2}u}{\partial x^{2}}\qquad 0<x<L,t>0\tag{1}$$ With BC \begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end{align*}

And initial conditions\begin{align*} u\left ( x,0\right ) & =f\left ( x\right ) \\ u_{t}\left ( x,0\right ) & =0 \end{align*}

Where now $$L=\pi ,f\left ( x\right ) =\sin ^{2}\left ( x\right )$$.

The general solution for (1) was found in problem 5.1.1.15 on page 1624 as$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{n\pi }{L}s\right ) ds\right ) \cos \left ( \frac{t}{L}\sqrt{L^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}}\right ) \sin \left ( \frac{n\pi }{L}x\right )$ Using the above speciﬁc values for this problem, the solution becomes\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \frac{2}{\pi }\int _{0}^{\pi }f\left ( s\right ) \sin \left ( \frac{n\pi }{\pi }s\right ) ds\right ) \cos \left ( \frac{t}{\pi }\sqrt{\pi ^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}}\right ) \sin \left ( \frac{n\pi }{\pi }x\right ) \\ & =\sum _{n=1}^{\infty }\left ( \frac{2}{\pi }\int _{0}^{\pi }f\left ( s\right ) \sin \left ( ns\right ) ds\right ) \cos \left ( \sqrt{\gamma ^{2}+c^{2}n^{2}}t\right ) \sin \left ( nx\right ) \end{align*}

But \begin{align*} \int _{0}^{\pi }\sin ^{2}\left ( s\right ) \sin \left ( ns\right ) ds & =\int _{0}^{\pi }\left ( \frac{1}{2}-\frac{1}{2}\cos \left ( 2s\right ) \right ) \sin \left ( ns\right ) ds\\ & =\int _{0}^{\pi }\frac{1}{2}\sin \left ( ns\right ) ds-\frac{1}{2}\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds\\ & =\frac{1-\left ( -1\right ) ^{n}}{2n}-\frac{1}{2}\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds \end{align*}

$$\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds=-\frac{2}{3}$$for $$n=1$$ and $$\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds=0$$ For $$n=2$$ and $$\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds=\frac{\left ( 1-\left ( -1\right ) ^{n}\right ) n}{n^{2}-4}\$$for $$n>2$$. Hence $\int _{0}^{\pi }\sin ^{2}\left ( s\right ) \sin \left ( ns\right ) ds=\left \{ \begin{array} [c]{ccc}\frac{1-\left ( -1\right ) }{2}+\frac{1}{3}=\frac{4}{3} & & n=1\\ \frac{1-\left ( -1\right ) ^{n}}{2n}=0 & & n=2\\ \frac{1-\left ( -1\right ) ^{n}}{2n}-\frac{1}{2}\frac{\left ( 1-\left ( -1\right ) ^{n}\right ) n}{n^{2}-4}=\frac{2}{n}\frac{\left ( -1\right ) ^{n}-1}{n^{2}-4} & & n>2 \end{array} \right .$

The solution becomes\begin{align*} u\left ( x,t\right ) & =\left ( \frac{2}{\pi }\right ) \frac{4}{3}\cos \left ( \sqrt{\gamma ^{2}+c^{2}}t\right ) \sin \left ( x\right ) +\sum _{n=3}^{\infty }\left ( \frac{2}{\pi }\frac{2}{n}\frac{\left ( -1\right ) ^{n}-1}{n^{2}-4}\right ) \cos \left ( \sqrt{\gamma ^{2}+c^{2}n^{2}}t\right ) \sin \left ( nx\right ) \\ & =\frac{8}{3\pi }\cos \left ( \sqrt{\gamma ^{2}+c^{2}}t\right ) \sin \left ( x\right ) +\frac{4}{\pi }\sum _{n=3}^{\infty }\frac{\left ( -1\right ) ^{n}-1}{n\left ( n^{2}-4\right ) }\cos \left ( \sqrt{\gamma ^{2}+c^{2}n^{2}}t\right ) \sin \left ( nx\right ) \end{align*}

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##### 5.1.1.17 [350] Dispersion term present (speciﬁc case)

problem number 350

Added July 12, 2019 Solve $u_{tt}+ \gamma ^2 u(x,t) = c^2 u_{xx}$ Dispersion term $$\gamma ^2 u(x,t)$$ causes the shape of the original wave to distort with time. With $$0<x<L$$ and $$t>0$$ and with boundary conditions \begin{align*} u(0,t) &= 0\\ u(\pi ,0) &=0 \end{align*}

With initial conditions \begin{align*} \frac{\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= f(x) \end{align*}

Using the following values \begin{align*} L &= 10 \\ \gamma &= \frac{1}{8}\\ f(x) &=\left \{ \begin{array} [c]{ccc}x-4 & & 4\leq x\leq 5\\ 6-x & & 5\leq x\leq 6\\ 0 & & \text{otherwise}\end{array} \right .\\ c &= 1 \end{align*}

Mathematica

Failed Due to adding dispersion term

Maple

$u \left ( x,t \right ) =\sum _{n=1}^{\infty }-40\,{\frac{ \left ( 16\, \left ( \cos \left ( 1/10\,\pi \,n \right ) \right ) ^{5}-16\, \left ( \cos \left ( 1/10\,\pi \,n \right ) \right ) ^{4}-12\, \left ( \cos \left ( 1/10\,\pi \,n \right ) \right ) ^{3}+12\, \left ( \cos \left ( 1/10\,\pi \,n \right ) \right ) ^{2}+\cos \left ( 1/10\,\pi \,n \right ) -1 \right ) \sin \left ( 1/10\,\pi \,n \right ) \sin \left ( 1/10\,\pi \,nx \right ) \cos \left ( 1/40\,\sqrt{16\,{\pi }^{2}{n}^{2}+25}t \right ) }{{\pi }^{2}{n}^{2}}}$

Hand solution

Solve for $$0<x<L,t>0$$$\frac{\partial ^{2}u}{\partial t^{2}}+\gamma ^{2}u=c^{2}\frac{\partial ^{2}u}{\partial x^{2}}$ Boundary conditions, $$t>0$$ (both ends ﬁxed)\begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} \frac{\partial u\left ( x,0\right ) }{\partial t} & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

Using $$L=10,\gamma =1/8,c=1$$ and initial position $f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}x-4 & & 4\leq x\leq 5\\ 6-x & & 5\leq x\leq 6\\ 0 & & \text{otherwise}\end{array} \right .$

The general solution to the above PDE was given in problem 5.1.1.15 on page 1624 as

$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{n\pi }{L}s\right ) ds\right ) \cos \left ( \frac{1}{L}\sqrt{\left ( L^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac{n\pi }{L}x\right )$

Replacing given values in the above solution results in

\begin{align} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \frac{2}{10}\int _{0}^{10}f\left ( s\right ) \sin \left ( \frac{n\pi }{10}s\right ) ds\right ) \cos \left ( \frac{1}{10}\sqrt{\left ( 100\left ( \frac{1}{8}\right ) ^{2}+n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac{n\pi }{10}x\right ) \nonumber \\ & =\frac{2}{10}\sum _{n=1}^{\infty }\left ( \int _{0}^{10}f\left ( s\right ) \sin \left ( \frac{n\pi }{10}s\right ) ds\right ) \cos \left ( \frac{1}{10}\sqrt{\frac{1}{16}\left ( \frac{100}{4}+16n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac{n\pi }{10}x\right ) \tag{1}\\ & =\frac{2}{10}\sum _{n=1}^{\infty }\left ( \int _{0}^{10}f\left ( s\right ) \sin \left ( \frac{n\pi }{10}s\right ) ds\right ) \cos \left ( \frac{1}{40}\sqrt{25+16n^{2}\pi ^{2}}t\right ) \sin \left ( \frac{n\pi }{10}x\right ) \end{align}

But

$\int _{0}^{10}f\left ( x\right ) \sin \left ( \frac{n\pi }{10}x\right ) ds=\frac{100\left ( 2\sin \left ( \frac{n\pi }{2}\right ) -\sin \left ( \frac{2n\pi }{5}\right ) -\sin \left ( \frac{3n\pi }{5}\right ) \right ) }{n^{2}\pi ^{2}}$

Hence the solution (1) becomes

\begin{align*} u\left ( x,t\right ) & =\frac{2}{10}\sum _{n=1}^{\infty }\frac{100\left ( 2\sin \left ( \frac{n\pi }{2}\right ) -\sin \left ( \frac{2n\pi }{5}\right ) -\sin \left ( \frac{3n\pi }{5}\right ) \right ) }{n^{2}\pi ^{2}}\cos \left ( \frac{1}{40}\sqrt{25+16n^{2}\pi ^{2}}t\right ) \sin \left ( \frac{n\pi }{10}x\right ) \\ & =\frac{20}{\pi ^{2}}\sum _{n=1}^{\infty }\frac{1}{n^{2}}\left ( 2\sin \left ( \frac{n\pi }{2}\right ) -\sin \left ( \frac{2n\pi }{5}\right ) -\sin \left ( \frac{3n\pi }{5}\right ) \right ) \cos \left ( \frac{1}{40}\sqrt{25+16n^{2}\pi ^{2}}t\right ) \sin \left ( \frac{n\pi }{10}x\right ) \end{align*}

Animation is below. The left one uses $$\gamma =\frac{1}{8}$$ and the right one uses larger value of $$\gamma =\frac{5}{8}$$ in order to show the eﬀect of larger dispersion.

Source code used for the above

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##### 5.1.1.18 [351] non-zero initial position

problem number 351

Added March 9, 2018. Solve $u_{tt} = 4 u_{xx}$ With boundary conditions \begin{align*} u(0,t) &= 0\\ u(\pi ,0) &=0 \end{align*}

With initial conditions \begin{align*} \frac{\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= \sin ^2(x) \end{align*}

Mathematica

$\left \{\left \{u(x,t)\to \underset{n=1}{\overset{\infty }{\sum }}\frac{4 (\cos (n \pi )-1) \cos (2 n t) \sin (n x)}{\left (n^3-4 n\right ) \pi }\right \}\right \}$ But sum should not include $$n=2$$

Maple

$u \left ( x,t \right ) =1/3\,{\frac{1}{\pi } \left ( 3\,\sum _{n=3}^{\infty }4\,{\frac{\sin \left ( nx \right ) \cos \left ( 2\,nt \right ) \left ( \left ( -1 \right ) ^{n}-1 \right ) }{\pi \,n \left ({n}^{2}-4 \right ) }}\pi +8\,\sin \left ( x \right ) \cos \left ( 2\,t \right ) \right ) }$ Handled $$n=2$$ case correctly

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##### 5.1.1.19 [352] With source

problem number 352

Solve for $$u(x,t)$$ with $$0<x<1$$ and $$t>0$$ $\frac{\partial ^2 u}{\partial t^2} = \frac{\partial ^2 u}{\partial x^2} + x e^{-t}$ With boundary conditions \begin{align*} u(0,t) &= 0\\ u(1,0) &=0 \end{align*}

With initial conditions \begin{align*} u(x,0) &= 0\\ \frac{\partial u}{\partial t}(x,0) &=1 \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=1}^{\infty }-2\,{\frac{\sin \left ( \pi \,nx \right ) \left ( \left ({\pi }^{2}{n}^{2} \left ( -1 \right ) ^{n}-{\pi }^{2}{n}^{2}+2\, \left ( -1 \right ) ^{n}-1 \right ) \sin \left ( n\pi \,t \right ) + \left ( -1 \right ) ^{n}\pi \,n \left ({{\rm e}^{-t}}-\cos \left ( n\pi \,t \right ) \right ) \right ) }{{\pi }^{2}{n}^{2} \left ({\pi }^{2}{n}^{2}+1 \right ) }}$

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##### 5.1.1.20 [353] Right end free (general case)

problem number 353

Added July 8, 2019 $u_{tt} = c^2 u_{xx}$ With boundary conditions \begin{align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end{align*}

With initial conditions \begin{align*} u(x,0) &= f(x)\\ u_t(x,0) &= g(x) \\ \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=0}^{\infty } \left ( 4\,{\frac{1}{\pi \,c \left ( 1+2\,n \right ) L}\sin \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) \left ( \cos \left ( 1/2\,{\frac{\pi \,c \left ( 1+2\,n \right ) t}{L}} \right ) \pi \,c \left ( n+1/2 \right ) \int _{0}^{L}\!\sin \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x+\sin \left ( 1/2\,{\frac{\pi \,c \left ( 1+2\,n \right ) t}{L}} \right ) \int _{0}^{L}\!\sin \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) g \left ( x \right ) \,{\rm d}xL \right ) } \right )$

Hand solution

Solving for $$0<x<L$$$\frac{\partial ^{2}u}{\partial t^{2}}=c^{2}\frac{\partial ^{2}u}{\partial x^{2}}\qquad 0<x<L,t>0$ Boundary conditions, $$t>0$$\begin{align*} u\left ( 0,t\right ) & =0\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} u_{t}\left ( x,0\right ) & =g\left ( x\right ) \\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

Separation of variables gives the eigenvalue ODE\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end{align*}

Only $$\lambda >0$$ gives non-trivial solution from the nature of the boundary conditions. Hence solution is$X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right )$ Since $$X\left ( 0\right ) =0$$ then the above gives $$0=A$$ and the solution becomes$X\left ( x\right ) =B\sin \left ( \sqrt{\lambda }x\right )$ Taking derivatives$X^{\prime }\left ( x\right ) =\sqrt{\lambda }B\cos \left ( \sqrt{\lambda }x\right )$ Since $$X^{\prime }\left ( L\right ) =0$$ the above becomes$0=\sqrt{\lambda }B\cos \left ( \sqrt{\lambda }L\right )$ Which implies $$\sqrt{\lambda }L=\frac{n\pi }{2}$$ for $$n=1,3,5,\cdots$$ or $\lambda _{n}=\left ( \frac{n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots$ Hence the eigenfunctions are$\Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots$

The time ODE now becomes$T^{\prime \prime }+c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}T=0$ Which has the solution$T\left ( t\right ) =D_{n}\cos \left ( c\frac{n\pi }{2L}t\right ) +E_{n}\sin \left ( c\frac{n\pi }{2L}t\right )$ Therefore the complete solution becomes$$u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( D_{n}\cos \left ( c\frac{n\pi }{2L}t\right ) +E_{n}\sin \left ( c\frac{n\pi }{2L}t\right ) \right ) \Phi _{n}\left ( x\right ) \tag{1}$$ At $$t=0$$ the above becomes$f\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }D_{n}\Phi _{n}\left ( x\right )$ Applying orthogonality gives\begin{align*} \int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =D_{n}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac{L}{2}D_{n} \end{align*}

Hence$$D_{n}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag{2}$$ Taking time derivative of (1) gives$u_{t}\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( -c\frac{n\pi }{2L}D_{n}\sin \left ( c\frac{n\pi }{2L}t\right ) +E_{n}c\frac{n\pi }{L}\cos \left ( c\frac{n\pi }{2L}t\right ) \right ) \Phi _{n}\left ( x\right )$ At $$t=0$$ the above becomes$g\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }E_{n}c\frac{n\pi }{2L}\Phi _{n}\left ( x\right )$ Applying orthogonality gives\begin{align*} \int _{0}^{L}g\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =E_{n}c\frac{n\pi }{L}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac{L}{2}E_{n}c\frac{n\pi }{2L}\\ & =\frac{1}{4}E_{n}cn\pi \end{align*}

Hence$$E_{n}=\frac{4}{cn\pi }\int _{0}^{L}g\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag{3}$$ Using (2,3) in (1) gives the ﬁnal solution as\begin{align*} u\left ( x,t\right ) & =\frac{2}{L}\sum _{n=1,3,5,\cdots }^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{2L}x\right ) dx\right ) \cos \left ( c\frac{n\pi }{2L}t\right ) \sin \left ( \frac{n\pi }{2L}x\right ) \\ & +\frac{4}{c\pi }\sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n}\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{n\pi }{2L}x\right ) dx\right ) \sin \left ( c\frac{n\pi }{2L}t\right ) \sin \left ( \frac{n\pi }{2L}x\right ) \end{align*}

Or

\begin{align*} u\left ( x,t\right ) & =\frac{2}{L}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \cos \left ( c\frac{\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\frac{4}{c\pi }\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) }\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \sin \left ( c\frac{\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \end{align*}

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##### 5.1.1.21 [354] Right end free, zero initial velocity (general case)

problem number 354

Added July 8, 2019 $u_{tt} = c^2 u_{xx}$ With boundary conditions \begin{align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end{align*}

With initial conditions \begin{align*} u(x,0) &= f(x)\\ u_t(x,0) &=0 \\ \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=0}^{\infty } \left ( 2\,{\frac{1}{L}\sin \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) \cos \left ( 1/2\,{\frac{\pi \,c \left ( 1+2\,n \right ) t}{L}} \right ) \int _{0}^{L}\!\sin \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x} \right )$

Hand solution

Solving for $$0<x<L$$$u_{tt}=c^{2}u_{xx}\qquad 0<x<L,t>0$ Boundary conditions, $$t>0$$\begin{align*} u\left ( 0,t\right ) & =0\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} u_{t}\left ( x,0\right ) & =g\left ( x\right ) =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

The general PDE was solved in 5.1.1.20 on page 1648 and the solution is

\begin{align*} u\left ( x,t\right ) & =\frac{2}{L}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \cos \left ( c\frac{\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\frac{4}{c\pi }\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) }\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \sin \left ( c\frac{\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \end{align*}

But here $$g\left ( x\right ) =0$$, hence the above reduces to

$u\left ( x,t\right ) =\frac{2}{L}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \cos \left ( c\frac{\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right )$

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##### 5.1.1.22 [355] Right end free, zero initial velocity (special case)

problem number 355

Added July 8, 2019 $u_{tt} = c^2 u_{xx}$ With boundary conditions \begin{align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end{align*}

With initial conditions \begin{align*} u(x,0) &= f(x)\\ u_t(x,0) &=0 \\ \end{align*}

Using the following values \begin{align*} c &= 4\\ L &=3 \\ h &= \frac{1}{10}\\ f(x) &= \left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right . \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=0}^{\infty }6/5\,{\frac{\sin \left ( 1/6\, \left ( 1+2\,n \right ) \pi \,x \right ) \cos \left ( 1/3\, \left ( 4\,n+2 \right ) t\pi \right ) \left ( \sqrt{3}\sin \left ( 1/3\,\pi \,n \right ) +\cos \left ( 1/3\,\pi \,n \right ) \right ) }{{\pi }^{2} \left ( 1+2\,n \right ) ^{2}}}$

Hand solution

Solving the wave PDE on string $u_{tt}=c^{2}u_{xx}\qquad t>0,x>0$ Boundary conditions, $$t>0$$\begin{align*} u\left ( 0,t\right ) & =0\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} u\left ( x,0\right ) & =f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right . \\ u_{t}\left ( x,0\right ) & =0 \end{align*}

Using $$c=4,L=3,h=\frac{1}{10}$$. Hence $$f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{10}x & & 0<x<1\\ \frac{1}{10} & & 1<x<3 \end{array} \right .$$

The general problem PDE was solved in 5.1.1.21 on page 1652 and the solution is$u\left ( x,t\right ) =\frac{2}{L}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) \cos \left ( \frac{\left ( 2n+1\right ) \pi }{2L}ct\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right )$ Substituting the speciﬁc values given above into this solution gives$u\left ( x,t\right ) =\frac{2}{3}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) \cos \left ( 4\frac{\left ( 2n+1\right ) \pi }{6}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right )$ But \begin{align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx & =\int _{0}^{\frac{L}{3}}f\left ( x\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx+\int _{\frac{L}{3}}^{L}f\left ( x\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac{1}{10}\int _{0}^{1}x\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx+\frac{1}{10}\int _{1}^{3}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac{18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) \end{align*}

Hence the solution becomes\begin{align*} u\left ( x,t\right ) & =\frac{2}{3}\sum _{n=0}^{\infty }\frac{18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) \cos \left ( \frac{2}{3}\left ( 2n+1\right ) \pi t\right ) \sin \left ( \frac{1}{6}\left ( 2n+1\right ) \pi x\right ) \\ & =\frac{36}{15\pi ^{2}}\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) \cos \left ( \frac{2}{3}\left ( 2n+1\right ) \pi t\right ) \sin \left ( \frac{1}{6}\left ( 2n+1\right ) \pi x\right ) \end{align*}

Animation is below

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##### 5.1.1.23 [356] Right end free, zero initial velocity, damping present (general case)

problem number 356

Added July 9, 2019 $u_{tt} + b u_t = c^2 u_{xx}$ For $$t>0$$ and $$0<x<L$$ and boundary conditions \begin{align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end{align*}

With initial conditions \begin{align*} u(x,0) &= f(x)\\ u_t(x,0) &= 0 \\ \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=0}^{\infty } \left ({\frac{1}{\sqrt{-4\, \left ( n+1/2 \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}}L} \left ( \left ( Lb+\sqrt{-4\, \left ( n+1/2 \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}} \right ){{\rm e}^{-1/2\,{\frac{ \left ( Lb-\sqrt{-4\, \left ( n+1/2 \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}} \right ) t}{L}}}}-{{\rm e}^{-1/2\,{\frac{ \left ( Lb+\sqrt{-4\, \left ( n+1/2 \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}} \right ) t}{L}}}} \left ( Lb-\sqrt{-4\, \left ( n+1/2 \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}} \right ) \right ) \int _{0}^{L}\!\sin \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x\sin \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) } \right )$

Hand solution

Solving for $$t>0,0<x<L$$$u_{tt}+bu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0$ Boundary conditions\begin{align*} u\left ( 0,t\right ) & =0\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} u_{t}\left ( x,0\right ) & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

Separation of variables gives\begin{align*} T^{\prime \prime }X+bT^{\prime }X & =c^{2}X^{\prime \prime }T\\ \frac{1}{c^{2}}\left ( \frac{T^{\prime \prime }}{T}+b\frac{T^{\prime }}{T}\right ) & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}

The eigenvalue ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end{align*}

Only $$\lambda >0$$ gives non-trivial solution from the nature of the boundary conditions. Hence solution is$X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right )$ Since $$X\left ( 0\right ) =0$$ then the above gives $$0=A$$ and the solution becomes$X\left ( x\right ) =B\sin \left ( \sqrt{\lambda }x\right )$ Taking derivatives$X^{\prime }\left ( x\right ) =\sqrt{\lambda }B\cos \left ( \sqrt{\lambda }x\right )$ Since $$X^{\prime }\left ( L\right ) =0$$ the above becomes$0=\sqrt{\lambda }B\cos \left ( \sqrt{\lambda }L\right )$ Which implies $$\sqrt{\lambda }L=\frac{n\pi }{2}$$ for $$n=1,3,5,\cdots$$ or $\lambda _{n}=\left ( \frac{n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots$ Hence the eigenfunctions are$\Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots$ The time ODE now becomes\begin{align*} \frac{1}{c^{2}}\left ( \frac{T^{\prime \prime }}{T}+b\frac{T^{\prime }}{T}\right ) & =-\lambda _{n}\\ \frac{T^{\prime \prime }}{T}+b\frac{T^{\prime }}{T} & =-c^{2}\lambda _{n}\\ T^{\prime \prime }+bT^{\prime }+c^{2}\lambda _{n}T & =0 \end{align*}

The characteristic equation $$r^{2}+br+c^{2}\lambda _{n}=0$$ has the roots $$r=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^{2}-4AC}\rightarrow r=\frac{-b}{2}\pm \frac{1}{2}\sqrt{b^{2}-4c^{2}\lambda _{n}}$$ or$r=\frac{-b}{2}\pm \frac{1}{2}\sqrt{b^{2}-4c^{2}\lambda _{n}}$ Case $$b^{2}<4c^{2}\lambda _{n}$$ for all $$n$$. This is called the underdamped case, which generates damped oscillations. The roots becomes$r=\frac{-b}{2}\pm \frac{1}{2}i\sqrt{4c^{2}\lambda _{n}-b^{2}}$ Let $$\beta _{n}=\frac{1}{2}\sqrt{4c^{2}\lambda _{n}-b^{2}}$$, then$r=\frac{-b}{2}\pm i\beta _{n}$ Hence the solution is$T_{n}\left ( t\right ) =e^{\frac{-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right )$ Therefore the complete solution becomes$$u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }e^{\frac{-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) \tag{1}$$ At $$t=0$$ the above becomes$f\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }D_{n}\Phi _{n}\left ( x\right )$ Applying orthogonality gives\begin{align*} \int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =D_{n}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac{L}{2}D_{n} \end{align*}

Hence$$D_{n}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag{2}$$ Taking time derivative of (1) gives$u_{t}\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\frac{-b}{2}e^{\frac{-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) +e^{\frac{-b}{2}t}\left ( -\beta _{n}D_{n}\sin \left ( \beta _{n}t\right ) +E_{n}\beta _{n}\cos \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right )$ At $$t=0$$ since $$g\left ( x\right ) =0$$ then the above becomes\begin{align} 0 & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac{-b}{2}D_{n}+E_{n}\beta _{n}\right ) \Phi _{n}\left ( x\right ) \nonumber \\ 0 & =\frac{-b}{2}D_{n}+E_{n}\beta _{n}\nonumber \\ E_{n} & =\frac{bD_{n}}{2\beta _{n}}\tag{3} \end{align}

Using (2,3), the solution (1) now becomes\begin{align*} u\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }e^{\frac{-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +\frac{bD_{n}}{2\beta _{n}}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }D_{n}e^{\frac{-b}{2}t}\left ( \cos \left ( \beta _{n}t\right ) +\frac{b}{2\beta _{n}}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) \end{align*}

Or$u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) e^{\frac{-b}{2}t}\left ( \cos \left ( \beta _{n}t\right ) +\frac{b}{2\beta _{n}}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right )$ But $$\beta _{n}=\frac{1}{2}\sqrt{4c^{2}\lambda _{n}-b^{2}}$$,$$\lambda _{n}=\left ( \frac{n\pi }{2L}\right ) ^{2}$$ and $$\Phi _{n}\left ( x\right ) =\sin \left ( \frac{n\pi }{2L}x\right )$$, hence the above becomes

\begin{align*} u\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{n\pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\cos \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac{n\pi }{2L}x\right ) \\ & +\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{n\pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{b\sin \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt{4c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac{n\pi }{2L}x\right ) \end{align*}

Or\begin{align*} u\left ( x,t\right ) & =\sum _{n=0}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\cos \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=0}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{b\sin \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \end{align*}

Case $$b^{2}=4c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}$$. We see that for $$n=1$$ it becomes critical, because then $$b=\frac{\pi c}{L}$$ and now and the discriminant is zero in this case. This is called the critical damped case. For $$n>1$$, it becomes underdamped, which is the above case. So we only need to ﬁnd solution for $$n=1$$. In this case, the solution to $$T^{\prime \prime }+bT^{\prime }+c^{2}\lambda _{1}T=0$$ is$T_{1}\left ( t\right ) =D_{1}e^{\frac{-b}{2}t}+E_{1}te^{\frac{-b}{2}t}$ Therefore the complete solution becomes$$u\left ( x,t\right ) =\left ( D_{1}e^{\frac{-b}{2}t}+E_{1}te^{\frac{-b}{2}t}\right ) \sin \left ( \frac{\pi }{2L}x\right ) +\sum _{n=3,5,\cdots }^{\infty }\left ( D_{n}e^{\frac{-b}{2}t}+E_{n}te^{\frac{-b}{2}t}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \tag{4}$$ For $$n=1$$, At $$t=0$$, from intial conditions (4) becomes$f\left ( x\right ) =D_{1}\sin \left ( \frac{\pi }{2L}x\right )$ By orthognailty the above gives$$D_{1}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx\tag{5}$$ Taking time derivative of (4) for $$n=1$$, gives$u\left ( x,t\right ) =\left ( \frac{-b}{2}D_{1}e^{\frac{-b}{2}t}+E_{1}\left ( e^{\frac{-b}{2}t}-\frac{b}{2}te^{\frac{-b}{2}t}\right ) \right ) \sin \left ( \frac{\pi }{2L}x\right )$ At $$t=0$$ and since $$g\left ( x\right ) =0$$, then the above becomes\begin{align} 0 & =\left ( \frac{-b}{2}D_{1}+E_{1}\right ) \sin \left ( \frac{\pi }{2L}x\right ) \nonumber \\ \frac{-b}{2}D_{1}+E_{1} & =0\nonumber \\ E_{1} & =\frac{b}{2}D_{1}\tag{6} \end{align}

Using (5,6) then (4) becomes\begin{align*} u\left ( x,t\right ) & =D_{1}\left ( e^{\frac{-b}{2}t}+\frac{b}{2}te^{\frac{-b}{2}t}\right ) \sin \left ( \frac{\pi }{2L}x\right ) +\sum _{n=3,5,\cdots }^{\infty }\left ( D_{n}e^{\frac{-b}{2}t}+E_{n}te^{\frac{-b}{2}t}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \\ & =\left ( \frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx\right ) \left ( e^{\frac{-b}{2}t}+\frac{b}{2}te^{\frac{-b}{2}t}\right ) \sin \left ( \frac{\pi }{2L}x\right ) \\ & +\sum _{n=3,5,\cdots }^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt{\lambda _{n}}s\right ) ds\right ) \left ( e^{\frac{-b}{2}t}+\frac{bt}{2}e^{\frac{-b}{2}t}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}

For the case of $$n>1$$, the solution was found in the above underdamped case. Putting all these together, gives the solution as\begin{align*} u\left ( x,t\right ) & =\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\pi }{2L}s\right ) ds\right ) \left ( e^{\frac{-b}{2}t}+\frac{b}{2}te^{\frac{-b}{2}t}\right ) \sin \left ( \frac{\pi }{2L}x\right ) \\ & +\sum _{n=2}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\cos \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=2}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{b\sin \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \end{align*}

Case $$b^{2}>4c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}$$. Will consider only the case when this is true for $$n=1$$ only. If this is true for larger $$n$$, then same solution needs to be summed for each mode. But for simplicity, will consider $$n=1$$ here. In this case, the roots are$r=\frac{-b}{2}\pm \frac{1}{2}\sqrt{b^{2}-4c^{2}\lambda _{1}}$ Where now $$b^{2}-4c^{2}\lambda _{1}$$ is positive. Hence we get $$r_{1}=\frac{-b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\lambda _{1}},r_{2}=\frac{-b}{2}-\frac{1}{2}\sqrt{b^{2}-4c^{2}\lambda _{1}}$$ or \begin{align*} r_{1} & =\frac{-b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}\\ r_{1} & =\frac{-b}{2}-\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}} \end{align*}

And the solution to $$T_{1}^{\prime \prime }+bT_{1}^{\prime }+c^{2}\lambda _{1}T=0$$ is$T_{1}\left ( t\right ) =e^{\frac{-b}{2}t}\left ( D_{1}e^{\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}+E_{1}e^{\frac{-1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\right )$ For the rest of the modes, the solution is from above$T_{n}\left ( t\right ) =e^{\frac{-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \qquad n=3,5,7,\cdots$ Hence the complete solution becomes\begin{align} u\left ( x,t\right ) & =e^{\frac{-b}{2}t}\left ( D_{1}e^{\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}+E_{1}e^{\frac{-1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\right ) \sin \left ( \frac{\pi }{2L}x\right ) \tag{7}\\ & +\sum _{n=3,5,\cdots }^{\infty }e^{\frac{-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \sin \left ( \frac{n\pi }{2L}x\right ) \nonumber \end{align}

At $$t=0$$ and for $$n=1$$, the above becomes$f\left ( x\right ) =\left ( D_{1}+E_{1}\right ) \sin \left ( \frac{\pi }{2L}x\right )$ Hence$$\left ( D_{1}+E_{1}\right ) =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx\tag{8}$$ Taking time derivative of (7) and for $$n=1$$ at $$t=0$$ it gives$0=\left ( \frac{-b}{2}\left ( D_{1}+E_{1}\right ) +\left ( \frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}D_{1}-E_{1}\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}\right ) \right ) \sin \left ( \frac{\pi }{2L}x\right )$ Hence\begin{align} \frac{-b}{2}\left ( D_{1}+E_{1}\right ) +\left ( \frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}D_{1}-E_{1}\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}\right ) & =0\nonumber \\ -E_{1}\left ( \frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}\right ) & =\left ( \frac{b}{2}-\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}\right ) D_{1}\nonumber \\ E_{1} & =\frac{-\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}D_{1}\tag{9} \end{align}

From (8,9)\begin{align*} D_{1} & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx-E_{1}\\ D_{1}-\frac{\frac{b}{2}-\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}D_{1} & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx\\ D_{1} & =\frac{\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx}{1-\frac{\frac{b}{2}-\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}}\\ & =\frac{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx \end{align*}

And therefore\begin{align*} E_{1} & =\frac{-\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\left ( \frac{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\right ) \frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx\\ & =\frac{-\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{\pi }{2L}x\right ) dx \end{align*}

For $$n>1$$ the solution is the same as the underdamped case above. Hence the complete solution becomes from (7)\begin{align*} u\left ( x,t\right ) & =e^{\frac{-b}{2}t}\left ( D_{1}e^{\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}+E_{1}e^{\frac{-1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\right ) \sin \left ( \frac{\pi }{2L}x\right ) \\ & +\sum _{n=3,5,\cdots }^{\infty }e^{\frac{-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \sin \left ( \frac{n\pi }{2L}x\right ) \end{align*}

Or\begin{align*} u\left ( x,t\right ) & =e^{\frac{-b}{2}t}\left ( \frac{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\right ) \frac{2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt{\frac{\pi }{2L}}s\right ) ds\right ) e^{\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac{\pi }{2L}x\right ) \\ & +e^{\frac{-b}{2}t}\left ( \frac{-\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\right ) \frac{2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt{\frac{\pi }{2L}}s\right ) ds\right ) e^{\frac{-1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac{\pi }{2L}x\right ) \\ & +\sum _{n=3,5,\cdots }^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{n\pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\cos \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac{n\pi }{2L}x\right ) \\ & +\sum _{n=3,5,\cdots }^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{n\pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{b\sin \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt{4c^{2}\left ( \frac{n\pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac{n\pi }{2L}x\right ) \end{align*}

Or\begin{align*} u\left ( x,t\right ) & =e^{\frac{-b}{2}t}\left ( \frac{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\right ) \frac{2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt{\frac{\pi }{2L}}s\right ) ds\right ) e^{\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac{\pi }{2L}x\right ) \\ & +e^{\frac{-b}{2}t}\left ( \frac{-\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\right ) \frac{2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt{\frac{\pi }{2L}}s\right ) ds\right ) e^{\frac{-1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac{\pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\cos \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{b\sin \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \end{align*}

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##### 5.1.1.24 [357] Right end free, zero initial velocity, damping present (special case, underdamped)

problem number 357

Added July 9, 2019 $u_{tt} + b u_t = c^2 u_{xx}$ For $$t>0$$ and $$0<x<L$$ and boundary conditions \begin{align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end{align*}

With initial conditions \begin{align*} u(x,0) &= f(x)\\ u_t(x,0) &= 0 \\ \end{align*}

Using the following values \begin{align*} L &=3 \\ c &= 4\\ f(x) &= \left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right .\\ h &=\frac{1}{10}\\ b &=\frac{1}{2} \frac{\pi c}{L} \end{align*}

Hence $$b = \frac{2 \pi }{3}$$

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=0}^{\infty }{\frac{-3/5\,i\sin \left ( 1/6\, \left ( 1+2\,n \right ) \pi \,x \right ) \left ( \sqrt{3}\sin \left ( 1/3\,\pi \,n \right ) +\cos \left ( 1/3\,\pi \,n \right ) \right ) \left ( \left ( i\sqrt{16\,{n}^{2}+16\,n+3}+1 \right ){{\rm e}^{1/3\,\pi \, \left ( i\sqrt{16\,{n}^{2}+16\,n+3}-1 \right ) t}}+{{\rm e}^{-1/3\,\pi \, \left ( i\sqrt{16\,{n}^{2}+16\,n+3}+1 \right ) t}} \left ( i\sqrt{16\,{n}^{2}+16\,n+3}-1 \right ) \right ) }{\sqrt{16\,{n}^{2}+16\,n+3}{\pi }^{2} \left ( 1+2\,n \right ) ^{2}}}$

Hand solution

Solving the wave PDE on string underdamped case $$t>0,0<x<L$$$u_{tt}+bu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0$ Boundary conditions\begin{align*} u\left ( 0,t\right ) & =0\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} u_{t}\left ( x,0\right ) & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

Using  \begin{align*} f\left ( x\right ) & =\left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right . \\ b & =\frac{2\pi }{3}\\ c & =4\\ L & =3 \end{align*}

Hence the PDE becomes $$u_{tt}+\frac{2\pi }{3}u_{t}=16u_{xx}$$. The general solution to the above PDE was given in problem 5.1.1.23 on page 1664. The eigenvalues are given as $\lambda _{n}=\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}\qquad n=0,1,2,\cdots$ And the discriminant is $$b^{2}-4c^{2}\lambda _{n}=b^{2}-4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}=b^{2}-\left ( 4\right ) \left ( 16\right ) \left ( \frac{\left ( 2n+1\right ) ^{2}\pi ^{2}}{36}\right )$$. For $$n=0$$ this gives$$\ b^{2}-\frac{16}{9}\pi ^{2}$$. But$$\ b=\frac{2\pi }{3}$$. Hence discriminant is $$\left ( \frac{2\pi }{3}\right ) ^{2}-\frac{16}{9}\pi ^{2}=-\frac{4}{3}\pi ^{2}$$. Since discriminant is negative, then this is underdamped wave with damped oscillations as the solution given from the above problem as\begin{align*} u\left ( x,t\right ) & =\sum _{n=0}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\cos \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=0}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{b\sin \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \end{align*}

Replacing given values in the above solution results in\begin{align} u\left ( x,t\right ) & =\sum _{n=0}^{\infty }\left ( \frac{2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-\pi }{3}t}\left ( \cos \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{2\pi }{3}\right ) ^{2}}t\right ) \right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \tag{1}\\ & +\sum _{n=0}^{\infty }\left ( \frac{2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{\frac{-\pi }{3}t}\left ( \frac{2\pi }{3}\frac{\sin \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{2\pi }{3}\right ) ^{2}}t\right ) }{\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{2\pi }{3}\right ) ^{2}}}\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end{align}

But \begin{align*} \int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) & =\frac{1}{10}\int _{0}^{1}x\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx+\frac{1}{10}\int _{1}^{3}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac{18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) \end{align*}

Hence the solution (1) becomes\begin{align} u\left ( x,t\right ) & =\sum _{n=0}^{\infty }\left ( \frac{2}{3}\frac{18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) \right ) e^{\frac{-\pi }{3}t}\left ( \cos \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{2\pi }{3}\right ) ^{2}}t\right ) \right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \tag{2}\\ & +\sum _{n=0}^{\infty }\left ( \frac{2}{3}\frac{18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) \right ) e^{\frac{-\pi }{3}t}\left ( \frac{2\pi }{3}\frac{\sin \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{2\pi }{3}\right ) ^{2}}t\right ) }{\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{2\pi }{3}\right ) ^{2}}}\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end{align}

Animation is below

Source code used for the above

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##### 5.1.1.25 [358] Right end free, zero initial velocity, damping present (special case, critical damped)

problem number 358

Added July 10, 2019 $u_{tt} + b u_t = c^2 u_{xx}$ For $$t>0$$ and $$0<x<L$$ and boundary conditions \begin{align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end{align*}

With initial conditions \begin{align*} u(x,0) &= f(x)\\ u_t(x,0) &= 0 \\ \end{align*}

Using the following values \begin{align*} L &=3 \\ c &= 4\\ f(x) &= \left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right .\\ h &=\frac{1}{10}\\ b &=\frac{\pi c}{L} \end{align*}

Hence $$b = \frac{4 \pi }{3}$$

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=0}^{\infty }\cases{4/5\,{\frac{{{\rm e}^{-2/3\,\pi \,t}} \left ( \pi \,t+3/2 \right ) \sin \left ( 1/6\,\pi \,x \right ) }{{\pi }^{2}}}&n=0\cr 3/10\,{\frac{ \left ( \left ( 2\,\sqrt{n}\sqrt{n+1}-i \right ){{\rm e}^{2/3\,i\pi \, \left ( 2\,\sqrt{n}\sqrt{n+1}+i \right ) t}}+ \left ( 2\,\sqrt{n}\sqrt{n+1}+i \right ){{\rm e}^{2/3\,i\pi \, \left ( -2\,\sqrt{n}\sqrt{n+1}+i \right ) t}} \right ) \sin \left ( 1/6\, \left ( 1+2\,n \right ) \pi \,x \right ) \left ( \sqrt{3}\sin \left ( 1/3\,\pi \,n \right ) +\cos \left ( 1/3\,\pi \,n \right ) \right ) }{\sqrt{n}\sqrt{n+1}{\pi }^{2} \left ( 1+2\,n \right ) ^{2}}}&otherwise\cr }$

Hand solution

Solving the wave PDE on string underdamped case $$t>0,0<x<L$$$u_{tt}+bu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0$ Boundary conditions\begin{align*} u\left ( 0,t\right ) & =0\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} u_{t}\left ( x,0\right ) & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

Using  \begin{align*} f\left ( x\right ) & =\left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right . \\ b & =\frac{4\pi }{3}\\ c & =4\\ L & =3\\ h & =\frac{1}{10} \end{align*}

Hence the PDE becomes $$u_{tt}+\frac{4\pi }{3}u_{t}=16u_{xx}$$. The general solution to the above PDE was given in problem 5.1.1.23 on page 1664. The eigenvalues are given as $\lambda _{n}=\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}\qquad n=0,1,2,\cdots$ And the discriminant is $$b^{2}-4c^{2}\lambda _{n}=b^{2}-4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}=b^{2}-\left ( 4\right ) \left ( 16\right ) \left ( \frac{\left ( 2n+1\right ) ^{2}\pi ^{2}}{36}\right )$$. For $$n=0$$ this gives$$\ b^{2}-\left ( \frac{4}{3}\pi \right ) ^{2}$$. But$$\ b=\frac{4\pi }{3}$$. Hence discriminant is zero for $$n=0$$. This means this is critically damped in ﬁrst mode. Using the the solution for this case from the above general solution as

\begin{align*} u\left ( x,t\right ) & =\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\pi }{2L}s\right ) ds\right ) \left ( e^{\frac{-b}{2}t}+\frac{b}{2}te^{\frac{-b}{2}t}\right ) \sin \left ( \frac{\pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\cos \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{b\sin \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \end{align*}

Replacing given values in the above solution results in\begin{align} u\left ( x,t\right ) & =\left ( \frac{2}{3}\int _{0}^{3}f\left ( x\right ) \sin \left ( \frac{\pi }{6}x\right ) dx\right ) \left ( e^{\frac{-2\pi t}{3}}+\frac{b}{2}te^{\frac{-2\pi t}{3}}\right ) \sin \left ( \frac{\pi }{6}x\right ) \tag{1}\\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{\frac{-2\pi t}{3}}\cos \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{4\pi }{3}\right ) ^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{\left ( \frac{4\pi }{3}\right ) \sin \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{4\pi }{3}\right ) ^{2}}t\right ) }{\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{4\pi }{3}\right ) ^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end{align}

But

$\int _{0}^{3}f\left ( x\right ) \sin \left ( \frac{\pi }{6}x\right ) dx=\frac{9}{5\pi ^{2}}$

And\begin{align*} \int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) & =\frac{1}{10}\int _{0}^{1}x\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx+\frac{1}{10}\int _{1}^{3}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac{18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) \end{align*}

Hence the solution (1) becomes

\begin{align} u\left ( x,t\right ) & =\frac{6}{5\pi ^{2}}\left ( e^{\frac{-2\pi t}{3}}+\frac{2}{3}\pi te^{\frac{-2\pi t}{3}}\right ) \sin \left ( \frac{\pi }{6}x\right ) \tag{1}\\ & +\sum _{n=1}^{\infty }\frac{12}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) e^{\frac{-2\pi t}{3}}\cos \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{4\pi }{3}\right ) ^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\frac{12}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) e^{\frac{-2\pi t}{3}}\frac{\left ( \frac{4\pi }{3}\right ) \sin \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{4\pi }{3}\right ) ^{2}}t\right ) }{\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac{4\pi }{3}\right ) ^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end{align}

Animation is below

Source code used for the above

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##### 5.1.1.26 [359] Right end free, zero initial velocity, damping present (special case, over damped)

problem number 359

Added July 11, 2019 $u_{tt} + b u_t = c^2 u_{xx}$ For $$t>0$$ and $$0<x<L$$ and boundary conditions \begin{align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end{align*}

With initial conditions \begin{align*} u(x,0) &= f(x)\\ u_t(x,0) &= 0 \\ \end{align*}

Using the following values \begin{align*} L &=3 \\ c &= 4\\ f(x) &= \left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right .\\ h &=\frac{1}{10}\\ b &=\frac{3}{2} \frac{\pi c}{L} \end{align*}

Hence $$b = 2 \pi$$

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=0}^{\infty }3/5\,{\frac{\sin \left ( 1/6\, \left ( 1+2\,n \right ) \pi \,x \right ) \left ( \sqrt{3}\sin \left ( 1/3\,\pi \,n \right ) +\cos \left ( 1/3\,\pi \,n \right ) \right ) \left ( \left ( \sqrt{-16\,{n}^{2}-16\,n+5}+3 \right ){{\rm e}^{1/3\,\pi \, \left ( \sqrt{-16\,{n}^{2}-16\,n+5}-3 \right ) t}}+{{\rm e}^{-1/3\,\pi \, \left ( \sqrt{-16\,{n}^{2}-16\,n+5}+3 \right ) t}} \left ( \sqrt{-16\,{n}^{2}-16\,n+5}-3 \right ) \right ) }{\sqrt{-16\,{n}^{2}-16\,n+5}{\pi }^{2} \left ( 1+2\,n \right ) ^{2}}}$

Hand solution

Solving the wave PDE on string underdamped case $$t>0,0<x<L$$$u_{tt}+bu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0$ Boundary conditions\begin{align*} u\left ( 0,t\right ) & =0\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} u_{t}\left ( x,0\right ) & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

Using  \begin{align*} f\left ( x\right ) & =\left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right . \\ b & =2\pi \\ c & =4\\ L & =3\\ h & =\frac{1}{10} \end{align*}

Hence the PDE becomes $$u_{tt}+2\pi u_{t}=16u_{xx}$$. The general solution to the above PDE was given in problem 5.1.1.23 on page 1664. The eigenvalues are given as $\lambda _{n}=\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}\qquad n=0,1,2,\cdots$ And the discriminant is $$b^{2}-4c^{2}\lambda _{n}=b^{2}-4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}=b^{2}-\left ( 4\right ) \left ( 16\right ) \left ( \frac{\left ( 2n+1\right ) ^{2}\pi ^{2}}{36}\right )$$. For $$n=0$$ this gives$$\ b^{2}-\left ( \frac{4}{3}\pi \right ) ^{2}$$. But$$\ b=2\pi$$. Hence discriminant is positive for $$n=0$$. This means this is critically damped in ﬁrst mode. Using the the solution for this case from the above general solution as

\begin{align*} u\left ( x,t\right ) & =e^{\frac{-b}{2}t}\left ( \frac{\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\right ) \frac{2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt{\frac{\pi }{2L}}s\right ) ds\right ) e^{\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac{\pi }{2L}x\right ) \\ & +e^{\frac{-b}{2}t}\left ( \frac{-\frac{b}{2}+\frac{1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}{\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}}\right ) \frac{2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt{\frac{\pi }{2L}}s\right ) ds\right ) e^{\frac{-1}{2}\sqrt{b^{2}-4c^{2}\left ( \frac{\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac{\pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\cos \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac{-b}{2}t}\frac{b\sin \left ( \frac{1}{2}\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt{4c^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \end{align*}

Replacing given values in the above solution results in\begin{align} u\left ( x,t\right ) & =e^{-\pi t}\left ( \frac{\pi +\frac{1}{2}\sqrt{4\pi ^{2}-64\left ( \frac{\pi }{6}\right ) ^{2}}}{\sqrt{4\pi ^{2}-64\left ( \frac{\pi }{6}\right ) ^{2}}}\right ) \frac{2}{3}\left ( \int _{0}^{3}f\left ( s\right ) \sin \left ( \sqrt{\frac{\pi }{6}}s\right ) ds\right ) e^{\frac{1}{2}\sqrt{4\pi ^{2}-64\left ( \frac{\pi }{6}\right ) ^{2}}t}\sin \left ( \frac{\pi }{6}x\right ) \tag{1}\\ & +e^{-\pi t}\left ( \frac{-\pi +\frac{1}{2}\sqrt{4\pi ^{2}-64\left ( \frac{\pi }{6}\right ) ^{2}}}{\sqrt{4\pi ^{2}-64\left ( \frac{\pi }{6}\right ) ^{2}}}\right ) \frac{2}{3}\left ( \int _{0}^{3}f\left ( s\right ) \sin \left ( \sqrt{\frac{\pi }{6}}s\right ) ds\right ) e^{\frac{-1}{2}\sqrt{4\pi ^{2}-64\left ( \frac{\pi }{6}\right ) ^{2}}t}\sin \left ( \frac{\pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{-\pi t}\cos \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-4\pi ^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\left ( \frac{2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{-\pi t}\frac{2\pi \sin \left ( \frac{1}{2}\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-4\pi ^{2}}t\right ) }{\sqrt{64\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-4\pi ^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end{align}

But

$\int _{0}^{3}f\left ( x\right ) \sin \left ( \frac{\pi }{6}x\right ) dx=\frac{9}{5\pi ^{2}}$

And\begin{align*} \int _{0}^{3}f\left ( s\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}s\right ) & =\frac{1}{10}\int _{0}^{1}x\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx+\frac{1}{10}\int _{1}^{3}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac{18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) \end{align*}

Hence the solution (1) becomes

\begin{align} u\left ( x,t\right ) & =e^{-\pi t}\left ( \frac{\pi +\sqrt{\pi ^{2}-16\left ( \frac{\pi }{6}\right ) ^{2}}}{2\sqrt{\pi ^{2}-16\left ( \frac{\pi }{6}\right ) ^{2}}}\right ) \left ( \frac{6}{5\pi ^{2}}\right ) e^{\sqrt{\pi ^{2}-16\left ( \frac{\pi }{6}\right ) ^{2}}t}\sin \left ( \frac{\pi }{6}x\right ) \tag{2}\\ & +e^{-\pi t}\left ( \frac{-\pi +\sqrt{\pi ^{2}-16\left ( \frac{\pi }{6}\right ) ^{2}}}{2\sqrt{\pi ^{2}-16\left ( \frac{\pi }{6}\right ) ^{2}}}\right ) \left ( \frac{6}{5\pi ^{2}}\right ) e^{-\sqrt{\pi ^{2}-16\left ( \frac{\pi }{6}\right ) ^{2}}t}\sin \left ( \frac{\pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\frac{12}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) e^{-\pi t}\cos \left ( \sqrt{16\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\pi ^{2}}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\frac{12}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) e^{-\pi t}\frac{\pi \sin \left ( \sqrt{16\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\pi ^{2}}t\right ) }{\sqrt{16\left ( \frac{\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\pi ^{2}}}\sin \left ( \frac{\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end{align}

Animation is below

Source code used for the above

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##### 5.1.1.27 [360] I.C. at diﬀerent times, right end free, with source

problem number 360

Added July 2, 2018. This is Example 2 (pde 10) taken from Maple document What_is_New_after_Maple_2018.pdf

Solve $-u_{tt} + u(x,t)= u_{xx} + 2 e^{-t} \left ( x - \frac{1}{2} x^2 + \frac{1}{2} t - 1 \right )$ With boundary condition \begin{align*} u(0,t) &= 0 \\ \frac{\partial u(1,t)}{\partial x} &= 0 \end{align*}

And initial conditions \begin{align*} u(x,0) &= x^2-2 x \\ u(x,1)&= u(x,\frac{1}{2}) + e^{-1} \left ( \frac{1}{2} x^2-x\right ) - \left ( \frac{3}{4} x^2- \frac{3}{2}x \right ) e^{\frac{-1}{2}} \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) =-1/2\,{{\rm e}^{-t}}x \left ( x-2 \right ) \left ( t-2 \right )$

Hand solution

Solve $$-u_{tt}+u=u_{xx}+2e^{-t}\left ( x-\frac{1}{2}x^{2}+\frac{1}{2}t-1\right ) \qquad t>0,0<x<1\tag{1}$$ Boundary conditions\begin{align*} u\left ( 0,t\right ) & =0\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=1} & =0 \end{align*}

Initial conditions, $$t=0$$\begin{align*} u\left ( x,0\right ) & =x^{2}-2x\\ u\left ( x,1\right ) & =u\left ( x,\frac{1}{2}\right ) +e^{-1}\left ( \frac{1}{2}x^{2}-x\right ) -\left ( \frac{3}{4}x^{2}-\frac{3}{2}x\right ) e^{\frac{-1}{2}} \end{align*}

Since boundary conditions are homogeneous, we can directly use eigenfunction expansion method. Let the solution be $$u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag{2}$$ Where $$\Phi _{n}\left ( x\right )$$ are the eigenfunctions of the corresponding homogeneous PDE $$-u_{tt}+u=u_{xx}$$. Using separation of variables, Let $$u=X\left ( x\right ) T\left ( t\right ) .$$ Substituting this back in $$-u_{tt}+u=u_{xx}$$ gives\begin{align*} -T^{\prime \prime }X+XT & =X^{\prime \prime }T\\ -\frac{T^{\prime \prime }}{T}+1 & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}

The eigenvalue ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( 1\right ) & =0 \end{align*}

This is known to have the eigenvalues are $$\lambda _{n}=\left ( \frac{n\pi }{2L}\right ) ^{2}=\left ( \frac{n\pi }{2}\right ) ^{2}\,$$, since $$L=1$$. This is for $$n=1,3,5,\cdots$$ and the eigenfunctions are $$\Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) =\sin \left ( \frac{n\pi }{2}x\right )$$. Therefore the solution (2) is\begin{align*} u\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \sin \left ( \frac{n\pi }{2}x\right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \end{align*}

Substituting this back into (1) gives$$-\sum _{n=1,3,5,\cdots }^{\infty }c_{n}^{\prime \prime }\left ( t\right ) \Phi _{n}\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag{3}$$ Where $$\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\left ( t\right ) \Phi \left ( x\right ) =2e^{-t}\left ( x-\frac{1}{2}x^{2}+\frac{1}{2}t-1\right )$$. By orthogonality this becomes\begin{align*} \int _{0}^{1}2e^{-t}\left ( x-\frac{1}{2}x^{2}+\frac{1}{2}t-1\right ) \Phi _{n}\left ( x\right ) dx & =b_{n}\left ( t\right ) \int _{0}^{1}\Phi _{n}^{2}\left ( x\right ) dx\\ 2e^{-t}\int _{0}^{1}\left ( x-\frac{1}{2}x^{2}+\frac{1}{2}t-1\right ) \sin \left ( \frac{n\pi }{2}x\right ) dx & =\frac{1}{2}b_{n}\left ( t\right ) \end{align*}

But $$2e^{-t}\int _{0}^{1}\left ( x-\frac{1}{2}x^{2}+\frac{1}{2}t-1\right ) \sin \left ( \frac{n\pi }{2}x\right ) dx=\frac{2e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}$$. Hence the above gives$b_{n}\left ( t\right ) =\frac{4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}$ Substituting the above into (3) gives$-\sum _{n=1,3,5,\cdots }^{\infty }c_{n}^{\prime \prime }\left ( t\right ) \Phi _{n}\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\frac{4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}\Phi _{n}\left ( x\right )$ But $$\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right )$$, hence the above simpliﬁes to\begin{align*} -c_{n}^{\prime \prime }\left ( t\right ) +c_{n}\left ( t\right ) & =-\lambda _{n}c_{n}\left ( t\right ) +\frac{4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}\\ -c_{n}^{\prime \prime }\left ( t\right ) +\left ( 1+\lambda _{n}\right ) c_{n}\left ( t\right ) & =\frac{4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}\\ c_{n}^{\prime \prime }\left ( t\right ) -\left ( 1+\frac{n^{2}\pi ^{2}}{4}\right ) c_{n}\left ( t\right ) & =-\frac{4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}} \end{align*}

The solution to this second order ODE can be found to be$c_{n}\left ( t\right ) =A_{n}e^{\frac{-\sqrt{n^{2}\pi ^{2}+4}t}{2}}+B_{n}e^{\frac{\sqrt{n^{2}\pi ^{2}+4}t}{2}}+\frac{16\left ( t-2\right ) e^{-t}}{n^{3}\pi ^{3}}$ Hence (2) becomes$$u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( A_{n}e^{\frac{-\sqrt{n^{2}\pi ^{2}+4}t}{2}}+B_{n}e^{\frac{\sqrt{n^{2}\pi ^{2}+4}t}{2}}+\frac{16\left ( t-2\right ) e^{-t}}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac{n\pi }{2}x\right ) \tag{4}$$ At $$t=0$$ the above becomes$x^{2}-2x=\sum _{n=1,3,5,\cdots }^{\infty }\left ( A_{n}+B_{n}-\frac{32}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac{n\pi }{2}x\right )$ Applying orthogonality gives$\int _{0}^{1}\left ( x^{2}-2x\right ) \sin \left ( \frac{n\pi }{2}x\right ) dx=\left ( A_{n}+B_{n}-\frac{32}{n^{3}\pi ^{3}}\right ) \frac{1}{2}$ But $$\int _{0}^{1}\left ( x^{2}-2x\right ) \sin \left ( \frac{n\pi }{2}x\right ) dx=-\frac{16}{n^{3}\pi ^{3}}$$, hence the above gives\begin{align} -\frac{32}{n^{3}\pi ^{3}} & =A_{n}+B_{n}-\frac{32}{n^{3}\pi ^{3}}\tag{5}\\ A_{n} & =-B_{n}\nonumber \end{align}

Therefore the solution (4) now becomes$$u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac{-\sqrt{n^{2}\pi ^{2}+4}t}{2}}-e^{\frac{\sqrt{n^{2}\pi ^{2}+4}t}{2}}+\frac{16\left ( t-2\right ) e^{-t}}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac{n\pi }{2}x\right ) \tag{6}$$ The second initial conditions is $$u\left ( x,1\right ) =u\left ( x,\frac{1}{2}\right ) +e^{-1}\left ( \frac{1}{2}x^{2}-x\right ) -\left ( \frac{3}{4}x^{2}-\frac{3}{2}x\right ) e^{\frac{-1}{2}}$$. At $$t=1$$ the above gives$u\left ( x,1\right ) =\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac{-\sqrt{n^{2}\pi ^{2}+4}}{2}}-e^{\frac{\sqrt{n^{2}\pi ^{2}+4}}{2}}-\frac{16e^{-1}}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac{n\pi }{2}x\right )$ At $$t=\frac{1}{2}$$ Eq (6) gives$u\left ( x,\frac{1}{2}\right ) =\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac{-\sqrt{n^{2}\pi ^{2}+4}}{4}}-e^{\frac{\sqrt{n^{2}\pi ^{2}+4}}{4}}-\frac{24e^{-\frac{1}{2}}}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac{n\pi }{2}x\right )$ Hence the second initial conditions implies \begin{multline*} \sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac{-\sqrt{n^{2}\pi ^{2}+4}}{2}}-e^{\frac{\sqrt{n^{2}\pi ^{2}+4}}{2}}-\frac{16}{n^{3}\pi ^{3}}e^{-1}\right ) \sin \left ( \frac{n\pi }{2}x\right ) \\ -\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac{-\sqrt{n^{2}\pi ^{2}+4}}{4}}-e^{\frac{\sqrt{n^{2}\pi ^{2}+4}}{4}}-\frac{24}{n^{3}\pi ^{3}}e^{-\frac{1}{2}}\right ) \sin \left ( \frac{n\pi }{2}x\right ) =e^{-1}\left ( \frac{1}{2}x^{2}-x\right ) -\left ( \frac{3}{4}x^{2}-\frac{3}{2}x\right ) e^{\frac{-1}{2}} \end{multline*} Or\begin{multline*} \sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( -2\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{2}\right ) -\frac{16}{n^{3}\pi ^{3}}e^{-1}\right ) \sin \left ( \frac{n\pi }{2}x\right ) \\ -\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( -2\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{4}\right ) -\frac{24}{n^{3}\pi ^{3}}e^{-\frac{1}{2}}\right ) \sin \left ( \frac{n\pi }{2}x\right ) =e^{-1}\left ( \frac{1}{2}x^{2}-x\right ) -\left ( \frac{3}{4}x^{2}-\frac{3}{2}x\right ) e^{\frac{-1}{2}} \end{multline*} Simplifying gives\begin{multline*} \sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( 2\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{4}\right ) -2\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{2}\right ) -\frac{16}{n^{3}\pi ^{3}}e^{-1}+\frac{24}{n^{3}\pi ^{3}}e^{-\frac{1}{2}}\right ) \sin \left ( \frac{n\pi }{2}x\right ) =\\ e^{-1}\left ( \frac{1}{2}x^{2}-x\right ) -\left ( \frac{3}{4}x^{2}-\frac{3}{2}x\right ) e^{\frac{-1}{2}} \end{multline*} Applying orthogonality gives$A_{n}\left ( 2\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{4}\right ) -2\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{2}\right ) -\frac{16}{n^{3}\pi ^{3}}e^{-1}+\frac{24}{n^{3}\pi ^{3}}e^{-\frac{1}{2}}\right ) \frac{1}{2}=\int _{0}^{1}\left ( e^{-1}\left ( \frac{1}{2}x^{2}-x\right ) -\left ( \frac{3}{4}x^{2}-\frac{3}{2}x\right ) e^{\frac{-1}{2}}\right ) \sin \left ( \frac{n\pi }{2}x\right ) dx$ But $$\int _{0}^{1}\left ( e^{-1}\left ( \frac{1}{2}x^{2}-x\right ) -\left ( \frac{3}{4}x^{2}-\frac{3}{2}x\right ) e^{\frac{-1}{2}}\right ) \sin \left ( \frac{n\pi }{2}x\right ) dx=-\frac{8e^{-\frac{1}{2}}-12e^{-1}}{n^{3}\pi ^{3}}$$, hence the above becomes\begin{align*} A_{n}\left ( 2\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{4}\right ) -2\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{2}\right ) -\frac{16}{n^{3}\pi ^{3}}e^{-1}+\frac{24}{n^{3}\pi ^{3}}e^{-\frac{1}{2}}\right ) & =-\frac{16e^{-1}}{n^{3}\pi ^{3}}+\frac{24e^{-\frac{1}{2}}}{n^{3}\pi ^{3}}\\ 2A_{n}\left ( \sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{4}\right ) -\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{2}\right ) \right ) +A_{n}\left ( -\frac{16e^{-1}}{n^{3}\pi ^{3}}+\frac{24e^{-\frac{1}{2}}}{n^{3}\pi ^{3}}\right ) & =-\frac{16e^{-1}}{n^{3}\pi ^{3}}+\frac{24e^{-\frac{1}{2}}}{n^{3}\pi ^{3}} \end{align*}

Since this is true for all $$n=1,3,5,\cdots$$ then\begin{align*} A_{n}\left ( \sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{4}\right ) -\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{2}\right ) \right ) & =0\\ A_{n} & =1 \end{align*}

Which implies $$\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{4}\right ) -\sinh \left ( \frac{\sqrt{n^{2}\pi ^{2}+4}}{2}\right ) =0$$ but this is not possible for $$n=1,3,5,\cdots$$. Something went wrong. I need to look at this again.

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##### 5.1.1.28 [361] Right end oscillates

problem number 361

Left end ﬁxed, right end oscillates, initially at rest. With source that depends on time and space.

Solve for $$u(x,t)$$ with $$0<x<\pi$$ and $$t>0$$ $\frac{\partial ^2 u}{\partial t^2} = 4 \frac{\partial ^2 u}{\partial x^2} + (1+t) x$ With boundary conditions \begin{align*} u(0,t) &= 0\\ u(\pi ,0) &=\sin (t) \end{align*}

With initial conditions \begin{align*} u(x,0) &= 0\\ \frac{\partial u}{\partial t}(x,0) &=0 \end{align*}

Mathematica

Failed

Maple

$u \left ( x,t \right ) ={\frac{1}{\pi } \left ( \sum _{n=1}^{\infty }4\,{\frac{\sin \left ( nx \right ) \left ( -1 \right ) ^{n} \left ( \left ({n}^{4}+1/4\,\pi \,{n}^{2}-\pi /16 \right ) \sin \left ( 2\,nt \right ) -1/2\,n \left ( \left ( -\pi \,{n}^{2}+\pi /4 \right ) \cos \left ( 2\,nt \right ) +\sin \left ( t \right ){n}^{2}+ \left ( t+1 \right ) \pi \, \left ( n+1/2 \right ) \left ( n-1/2 \right ) \right ) \right ) }{\pi \,{n}^{4} \left ( 4\,{n}^{2}-1 \right ) }}\pi +x\sin \left ( t \right ) \right ) }$

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##### 5.1.1.29 [362] Perioidic B.C.

problem number 362

Taken from midterm 2 sample exam. UMN Math 5587, Fall 2016. Problem 8

Solve for $$u(x,t)$$ with $$-\pi <x<\pi$$ and $$t>0$$ $u_{tt} = u_{xx}$ With boundary conditions \begin{align*} u(-\pi ,t) &= u(\pi ,t)\\ u_x(-\pi ,0) &=u_x(\pi ,t) \end{align*}

With initial conditions \begin{align*} u(x,0) &= x\\ u_t(x,0) &=0 \end{align*}

Mathematica

Failed

Maple

sol=()

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##### 5.1.1.30 [363] Mixed B.C.

problem number 363

Taken from midterm 2 sample exam. UMN Math 5587, Fall 2016. Problem 10

Solve for $$u(x,t)$$ with $$0<x<\pi$$ and $$t>0$$ $u_{tt} = u_{xx}$ With boundary conditions $$u(0,t) = u_t(\pi ,t)$$ and initial conditions \begin{align*} u(x,0) &= 0\\ u_t(x,0) &=1 \end{align*}

Mathematica

Failed

Maple

sol=()

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##### 5.1.1.31 [364] Left end ﬁxed, right end non-homogeneous Neumann BC. Zero initial conditions

problem number 364

Solve for $$u(x,t)$$ with $$0<x<L$$ and $$t>0$$ $u_{tt} = c^2 u_{xx}$ With boundary conditions $$u(0,t) = 0, u_x(L,t)=C$$ and zero initial conditions \begin{align*} u(x,0) &= 0\\ u_t(x,0) &=0 \end{align*}

For animations use $$L=10,c=1,C=5$$

Mathematica

Failed

Maple

$u \left ( x,t \right ) =\sum _{n=0}^{\infty }-8\,{\frac{L \left ( -1 \right ) ^{n}{\it C0}}{{\pi }^{2} \left ( 1+2\,n \right ) ^{2}}\sin \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) \cos \left ( 1/2\,{\frac{c \left ( 1+2\,n \right ) \pi \,t}{L}} \right ) }+x{\it C0}$

Hand solution

Let $$u\left ( x,t\right ) =v\left ( x,t\right ) +u_{E}\left ( x\right ) \tag{2}$$ $$u_{E}\left ( x\right )$$ is the steady state solution which only needs to satisfy the non-homogeneous boundary conditions. At equilibrium $$\frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=0$$ and the PDE becomes $$\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=0$$ or the ODE $$\frac{d^{2}u_{E}\left ( x,t\right ) }{dx^{2}}=0$$ with B.C. $$u_{E}\left ( 0\right ) =0,u_{E}^{\prime }\left ( L\right ) =C$$. The solution to this ODE is$u_{E}\left ( x\right ) =c_{1}x+c_{2}$ At ﬁrst B.C.$0=c_{2}$ Solution becomes $$u_{E}\left ( x\right ) =c_{1}x$$. At second B.C. $$u_{E}^{\prime }\left ( x\right ) =c_{1}=C$$. Therefore solution is$u_{E}\left ( x\right ) =Cx$ Hence$u\left ( x,t\right ) =v\left ( x,t\right ) +Cx$ $$v\left ( x,t\right )$$ is the solution to the PDE but with homogeneous B.C. Plugging (2) into (1) gives$\frac{\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}}+\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=c\left ( \frac{\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}+\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial x^{2}}\right )$ But $$\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial x^{2}}=0$$ and also $$\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=0$$, hence above becomes$\frac{\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}}=c\frac{\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}$ With $$v\left ( x,t\right )$$ having now homogeneous B.C.\begin{align*} v\left ( 0,t\right ) & =0\\ \frac{\partial v\left ( L,t\right ) }{\partial x} & =0 \end{align*}

And initial conditions given by\begin{align*} v\left ( x,0\right ) & =u\left ( x,0\right ) -u_{E}\left ( x\right ) \\ & =0-Cx\\ & =-Cx \end{align*}

And\begin{align*} \frac{\partial v\left ( x,0\right ) }{\partial t} & =\frac{\partial u\left ( x,0\right ) }{\partial t}-\frac{\partial u_{E}\left ( x\right ) }{\partial t}\\ & =0 \end{align*}

In summary, the PDE to solve for $$v\left ( x,t\right )$$ is\begin{align} \frac{\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}} & =c\frac{\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}\tag{3}\\ v\left ( 0,t\right ) & =0\nonumber \\ \frac{\partial v\left ( L,t\right ) }{\partial x} & =0\nonumber \\ v\left ( x,0\right ) & =-Cx\nonumber \\ \frac{\partial v\left ( x,0\right ) }{\partial t} & =0\nonumber \end{align}

Now we solve for PDE (3) for $$v\left ( x,t\right )$$ using separation of variables since the boundary conditions in space are now homogeneous. Let $$v\left ( x,t\right ) =X\left ( x\right ) T\left ( t\right )$$ and the PDE becomes$\frac{1}{c}T^{\prime \prime }X=X^{\prime \prime }T$ Dividing by $$XT\neq 0$$ gives$$\frac{1}{c}\frac{T^{\prime \prime }}{T}=\frac{X^{\prime \prime }}{X}=-\lambda \tag{4}$$ Where $$\lambda$$ is some real positive constant. The space ODE becomes\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end{align*}

Case $$\lambda <0$$: Let The solution is$X\left ( x\right ) =c_{1}\cosh \left ( \sqrt{\lambda }x\right ) +c_{2}\sinh \left ( \sqrt{\lambda }x\right )$ At $$x=0$$$0=c_{1}$ Hence solution becomes $X\left ( x\right ) =c_{2}\sinh \left ( \sqrt{\lambda }x\right )$ Taking derivative$X^{\prime }\left ( x\right ) =\sqrt{\lambda }c_{2}\cosh \left ( \sqrt{\lambda }x\right )$ Using second boundary conditions gives$0=\sqrt{\lambda }c_{2}\cosh \left ( \sqrt{\lambda }L\right )$ Since $$\cosh$$ is zero only when its argument is zero. But we assumed $$\sqrt{\lambda }$$ not zero here, then $$c_{2}=0$$ in only other choice. Hence this gives trivial solution. Therefore $$\lambda <0$$ is not possible.

Case $$\lambda =0$$\begin{align*} X^{\prime \prime } & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end{align*}

Solution is $$X\left ( x\right ) =c_{1}x+c_{2}$$. First B.C. gives $$0=c_{2}$$. Solution becomes $$X\left ( x\right ) =c_{1}x$$. Second B.C. gives $$c_{1}=0$$. This gives trivial solution again. Hence $$\lambda =0$$ is not possible eigenvalue.

Case $$\lambda >0$$: The solution becomes$X\left ( x\right ) =B_{1}\cos \left ( \sqrt{\lambda }x\right ) +B_{2}\sin \left ( \sqrt{\lambda }x\right )$ AT ﬁrst B.C.$0=B_{1}$ Hence solution becomes$X\left ( x\right ) =B_{2}\sin \left ( \sqrt{\lambda }x\right )$ Taking derivative$X^{\prime }\left ( x\right ) =\sqrt{\lambda }B_{2}\cos \left ( \sqrt{\lambda }x\right )$ At second B.C.$0=\sqrt{\lambda }B_{2}\cos \left ( \sqrt{\lambda }L\right )$ To avoid trivial solution, take $$\cos \left ( \sqrt{\lambda }L\right ) =0$$ or $$\sqrt{\lambda }L=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\cdots$$ or \begin{align} \sqrt{\lambda } & =\frac{\pi }{2L},\frac{3\pi }{2L},\frac{5\pi }{2L},\cdots \nonumber \\ \sqrt{\lambda _{n}} & =\left ( \frac{n\pi }{2L}\right ) \qquad n=1,3,5,\cdots \nonumber \\ & =\frac{\left ( 2n+1\right ) \pi }{2L}\qquad n=0,1,2,\cdots \tag{5} \end{align}

Hence the space solution is$$X_{n}\left ( x\right ) =B_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \tag{6}$$ Now we solve the time ODE $$T\left ( t\right )$$ from (4), which is$T^{\prime \prime }+\lambda cT=0$ The solution is$T_{n}\left ( t\right ) =D_{n}\cos \left ( \sqrt{\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt{\lambda _{n}c}t\right )$ Therefore\begin{align*} v\left ( x,t\right ) & =\sum _{n=0}^{\infty }T_{n}\left ( t\right ) X_{n}\left ( x\right ) \\ & =\sum _{n=0}^{\infty }\left ( D_{n}\cos \left ( \sqrt{\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt{\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}

Where constant $$B_{n}$$ merged with the other constants. Now At $$t=0$$$-Cx=\sum _{n=0}^{\infty }D_{n}\sin \left ( \sqrt{\lambda _{n}}x\right )$ Applying orthogonality\begin{align*} -\int _{0}^{L}Cx\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx & =D_{n}\int _{0}^{L}\sin ^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx\\ -C\int _{0}^{L}x\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx & =D_{n}\frac{L}{2}\\ -C\left ( \frac{4\left ( -1\right ) ^{n}L^{2}}{\left ( \pi +2n\pi \right ) ^{2}}\right ) & =D_{n}\frac{L}{2}\\ D_{n} & =-C\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}} \end{align*}

Therefore$v\left ( x,t\right ) =\sum _{n=0}^{\infty }\left ( -C\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \sqrt{\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt{\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$ Taking time derivative$\frac{\partial v\left ( x,t\right ) }{\partial t}=\sum _{n=0}^{\infty }\left ( C\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\frac{\left ( 2n+1\right ) \pi }{2L}\sin \left ( \sqrt{\lambda _{n}c}t\right ) +E_{n}\sqrt{\lambda _{n}c}\cos \left ( \sqrt{\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$ At $$t=0$$$0=\sum _{n=0}^{\infty }E_{n}\sqrt{\lambda _{n}c}\sin \left ( \sqrt{\lambda _{n}}x\right )$ Hence $$E_{n}=0$$. Therefore solution becomes$v\left ( x,t\right ) =\sum _{n=0}^{\infty }-C\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \sqrt{\lambda _{n}c}t\right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$ Therefore, since $$u\left ( x,t\right ) =v\left ( x,t\right ) +u_{E}\left ( x\right )$$ then$$u\left ( x,t\right ) =Cx-C\sum _{n=0}^{\infty }\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \frac{\left ( 2n+1\right ) \pi }{2L}\sqrt{c}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \tag{7}$$ This animation runs for 40 seconds for $$L=10,c=1,C=5$$. The solution becomes$$u\left ( x,t\right ) =5x-5\sum _{n=0}^{\infty }\frac{80\left ( -1\right ) ^{n}}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \frac{\left ( 2n+1\right ) \pi }{20}\sqrt{10}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{20}x\right ) \tag{7}$$

Code used for the above is

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