#### 5.1.2 Semi-inﬁnite domain

##### 5.1.2.1 [365] Left end ﬁxed, (general case)

problem number 365

Added July 12, 2019 Solve for $$u(x,t)$$ with $$t>0$$ and $$x>0$$ $u_{tt} = c^2 u_{xx}$ With boundary conditions \begin{align*} u(0,t) &= 0 \end{align*}

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

Mathematica

$\left \{\left \{u(x,t)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{1}{2} (f(x-c t)+f(c t+x)) & x>c t \\ \frac{1}{2} (f(c t+x)-f(c t-x)) & x\leq c t \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \}$

Maple

sol=()

Hand solution

Solving on semi-inﬁnite domain for $$u(x,t)$$$$\frac{\partial ^{2}u}{\partial t^{2}}=c^{2}\frac{\partial ^{2}u}{\partial x^{2}}\qquad 0<x<\infty ,t>0\tag{1}$$ With BC \begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( \infty ,t\right ) & <\infty \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*}

Seperation of variables method

Let $$u\left ( x,t\right ) =X\left ( x\right ) T\left ( t\right )$$. The PDE in (1) becomes$\frac{T^{\prime \prime }}{c^{2}T}=\frac{X^{\prime \prime }}{X}=-\lambda$ Hence \begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( \infty \right ) & <\infty \end{align*}

It is clear that $$\lambda$$ can not be negative because it gives a solution that blows up. For $$\lambda =0$$, the solution is $$X\left ( x\right ) =Ax+B$$ and because $$X\left ( 0\right ) =0$$ this implies $$B=0$$. Hence solution is $$X\left ( x\right ) =Ax$$. And this blows up as $$x$$ increases unless $$A=0$$. Hence $$\lambda =0$$ is not valid eigenvalue. Therefore $$\lambda >0$$. Let $$\lambda =\alpha ^{2},\alpha >0$$ and the solution becomes$X\left ( x\right ) =A_{\alpha }\cos \left ( \alpha x\right ) +B_{\alpha }\sin \left ( \alpha x\right )$ At $$x=0$$ the above gives$0=A_{\alpha }$ Therefore the solution becomes$$X_{\alpha }\left ( x\right ) =B_{\alpha }\sin \left ( \alpha x\right ) \qquad \alpha >0\tag{1}$$ The time domain ODE becomes\begin{align*} T^{\prime \prime }+c^{2}\alpha ^{2}T & =0\\ T & =C_{\alpha }\cos \left ( \alpha ct\right ) +D_{\alpha }\sin \left ( \alpha ct\right ) \\ T^{\prime }\left ( t\right ) & =-c\alpha C_{\alpha }\sin \left ( \alpha ct\right ) +c\alpha D_{\alpha }\cos \left ( \alpha ct\right ) \end{align*}

And at $$t=0,T^{\prime }\left ( 0\right ) =0$$, hence the above becomes$0=c\alpha D_{\alpha }$ Which means $$D_{\alpha }=0$$. Therefore$$T=C_{\alpha }\cos \left ( \alpha ct\right ) \qquad \alpha >0\tag{2}$$ From (1,2) the complete solution is therefore$$u\left ( x,t\right ) =\int _{0}^{\infty }A_{\alpha }\cos \left ( \alpha ct\right ) \sin \left ( \alpha x\right ) d\alpha \tag{3}$$ Where $$A_{\alpha },C_{\alpha }$$ are merged into one constant. Now the last initial condition is applied, which is $$u\left ( x,0\right ) =f\left ( x\right )$$ to the above which gives$f\left ( x\right ) =\int _{0}^{\infty }A_{\alpha }\sin \left ( \alpha x\right ) d\alpha$ Hence$A_{\alpha }=\frac{2}{\pi }\int _{0}^{\infty }f\left ( x\right ) \sin \left ( \alpha x\right ) dx$ Using the above in (3) gives the ﬁnal solution as$u\left ( x,t\right ) =\frac{2}{\pi }\int _{0}^{\infty }\left ( \int _{0}^{\infty }f\left ( s\right ) \sin \left ( \alpha s\right ) ds\right ) \cos \left ( \alpha ct\right ) \sin \left ( \alpha x\right ) d\alpha$ D’Alambert’s formula method

For the half line, the D’Alambert’s is given by, using $$v_{0}\left ( x\right ) =u_{t}\left ( x,0\right )$$ as the initial velocity$u\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2}\left ( f\left ( x+ct\right ) +f\left ( x-ct\right ) \right ) +\frac{1}{2c}\int _{x-ct}^{x+ct}v_{0}\left ( s\right ) ds & & x>ct\geq 0\\ & & \\ \frac{1}{2}\left ( f\left ( x+ct\right ) -f\left ( ct-x\right ) \right ) +\frac{1}{2c}\int _{ct-x}^{x+ct}v_{0}\left ( s\right ) ds & & x\leq ct \end{array} \right .$ Hence, since $$u_{t}\left ( x,0\right ) =0$$ in this problem$u\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2}\left ( f\left ( x+ct\right ) +f\left ( x-ct\right ) \right ) & & x>ct\geq 0\\ & & \\ \frac{1}{2}\left ( f\left ( x+ct\right ) -f\left ( ct-x\right ) \right ) & & x\leq ct \end{array} \right .$

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##### 5.1.2.2 [366] Left end ﬁxed with speciﬁc initial position

problem number 366

Taken from Mathematica DSolve help pages.

Solve for $$u(x,t)$$ initial value wave PDE on inﬁnite domain with $$t>0$$ and $$x>0$$. $u_{tt} = c^2 u_{xx}$ With initial conditions \begin{align*} u(x,0) &= \sin ^2(x) \hspace{20pt} \pi <x< 2\pi \\ \frac{\partial u}{\partial t}(x,0) &= 0 \end{align*}

And boundary conditions $$u(0,t)=0$$

Mathematica

$\left \{\left \{u(x,t)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{1}{2} \left (\left (\begin{array}{cc} \{ & \begin{array}{cc} \sin ^2(c t-x) & \pi <x-c t<2 \pi \\ 0 & \text{True} \\\end{array} \\\end{array}\right )+\left (\begin{array}{cc} \{ & \begin{array}{cc} \sin ^2(c t+x) & \pi <c t+x<2 \pi \\ 0 & \text{True} \\\end{array} \\\end{array}\right )\right ) & x>c t\geq 0 \\ \frac{1}{2} \left (\left (\begin{array}{cc} \{ & \begin{array}{cc} \sin ^2(c t+x) & \pi <c t+x<2 \pi \\ 0 & \text{True} \\\end{array} \\\end{array}\right )-\left (\begin{array}{cc} \{ & \begin{array}{cc} \sin ^2(c t-x) & \pi <c t-x<2 \pi \\ 0 & \text{True} \\\end{array} \\\end{array}\right )\right ) & 0\leq x\leq c t \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \}$

Maple

$u \left ( x,t \right ) =1/2\,\cases{\cases{0&ct+x\leq \pi \cr \left ( \sin \left ( ct+x \right ) \right ) ^{2}&ct+x<2\,\pi \cr 0&2\,\pi \leq ct+x\cr }-\cases{0&ct-x\leq \pi \cr \left ( \sin \left ( ct-x \right ) \right ) ^{2}&ct-x<2\,\pi \cr 0&2\,\pi \leq ct-x\cr }&x<ct\cr \cases{0&ct+x\leq \pi \cr \left ( \sin \left ( ct+x \right ) \right ) ^{2}&ct+x<2\,\pi \cr 0&2\,\pi \leq ct+x\cr }+\cases{0&-ct+x\leq \pi \cr \left ( \sin \left ( ct-x \right ) \right ) ^{2}&-ct+x<2\,\pi \cr 0&2\,\pi \leq -ct+x\cr }&ct<x\cr }$

Hand solution

Solving on semi-inﬁnite domain \begin{align} u_{tt} & =c^{2}u_{xx}\qquad t>0,x>0\tag{1}\\ u\left ( 0,t\right ) & =0\nonumber \\ u\left ( x,0\right ) & =f\left ( x\right ) =\sin ^{2}\left ( x\right ) \qquad \pi <x<2\pi \nonumber \end{align}

With $$k>0$$ and $$u\left ( x,t\right ) <\infty$$ as $$x\rightarrow \infty$$. This means $$u\left ( x,t\right )$$ is bounded. The general solution to the above PDE was given in problem 5.1.2.1 on page 1717 as (using the D’Alambert’s solution and not the Fourier integral solution)$u\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2}\left ( f\left ( x+ct\right ) +f\left ( x-ct\right ) \right ) & & x>ct\geq 0\\ & & \\ \frac{1}{2}\left ( f\left ( x+ct\right ) -f\left ( ct-x\right ) \right ) & & x\leq ct \end{array} \right .$ But $$f\left ( x\right ) =\sin ^{2}\left ( x\right )$$ and the above becomes$u\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2}\left ( \sin ^{2}\left ( x+ct\right ) +\sin ^{2}\left ( x-ct\right ) \right ) & & x>ct\geq 0\\ & & \\ \frac{1}{2}\left ( \sin ^{2}\left ( x+ct\right ) -\sin ^{2}\left ( ct-x\right ) \right ) & & x\leq ct \end{array} \right .$ But here $$f\left ( x\right )$$ is restricted to $$\pi <x<2\pi$$. Hence the above solution is modiﬁed as follows$u\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}\left \{ \begin{array} [c]{ccc}\frac{1}{2}\sin ^{2}\left ( x+ct\right ) & & \pi <x+ct<2\pi \\ 0 & & \text{otherwise}\end{array} \right . +\left \{ \begin{array} [c]{ccc}\frac{1}{2}\sin ^{2}\left ( x-ct\right ) & & \pi <x-ct<2\pi \\ 0 & & \text{otherwise}\end{array} \right . & & x>ct\geq 0\\ & & \\ \left \{ \begin{array} [c]{ccc}\frac{1}{2}\sin ^{2}\left ( x+ct\right ) & & \pi <x+ct<2\pi \\ 0 & & \text{otherwise}\end{array} \right . -\left \{ \begin{array} [c]{ccc}\frac{1}{2}\sin ^{2}\left ( ct-x\right ) & & \pi <ct-x<2\pi \\ 0 & & \text{otherwise}\end{array} \right . & & 0<x\leq ct \end{array} \right .$

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##### 5.1.2.3 [367] Logan page 115, left end ﬁxed with source

problem number 367

This is problem at page 115, David J Logan textbook, applied PDE textbook.

Falling cable lying on a table that is suddenly removed. $u_{tt} = c^2 u_{xx} - g$ With boundary condition \begin{align*} u(0,t) &= 0 \end{align*}

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

Mathematica

$\left \{\left \{u(x,t)\to \frac{1}{2} g \left (\left (t-\frac{x}{c}\right )^2 \theta \left (t-\frac{x}{c}\right )-t^2\right )-c_1 \delta \left (t-\frac{x}{c}\right )\right \}\right \}$

Maple

$u \left ( x,t \right ) =1/2\,{\frac{g}{{c}^{2}} \left ({\it Heaviside} \left ( t-{\frac{x}{c}} \right ) \left ( ct-x \right ) ^{2}-{t}^{2}{c}^{2} \right ) }$

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##### 5.1.2.4 [368] Left moving boundary condition

problem number 368

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

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

Mathematica

$\left \{\left \{u(x,t)\to \begin{array}{cc} \{ & \begin{array}{cc} 0 & x>c t \\ g\left (t-\frac{x}{c}\right ) & x\leq c t \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \}$

Maple

$u \left ( x,t \right ) ={\it Heaviside} \left ( t-{\frac{x}{c}} \right ) g \left ({\frac{ct-x}{c}} \right )$

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##### 5.1.2.5 [369] moving Left end

problem number 369

Taken from Mathematica DSolve help pages. Initial value problem with a Neumann condition on the half-line. $u_{tt} = c^2 u_{xx}$ With initial conditions \begin{align*} u(x,0) &= \sin ^3(x) \\ \frac{\partial u}{\partial t}(x,0) &= 1-e^{- \frac{x}{10}} \end{align*}

And boundary conditions $$\frac{\partial u}{\partial x}(0,t)=1$$

Mathematica

$\left \{\left \{u(x,t)\to \begin{array}{cc} \{ & \begin{array}{cc} 0 & x>c t \\ g\left (t-\frac{x}{c}\right ) & x\leq c t \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \}$

Maple

$u \left ( x,t \right ) =1/2\,{\frac{\cases{ \left ( \sin \left ( ct+x \right ) \right ) ^{3}c- \left ( \sin \left ( ct-x \right ) \right ) ^{3}c+2\,ct+10\,{{\rm e}^{-1/10\,ct-x/10}}-10\,{{\rm e}^{1/10\,ct-x/10}}&ct<x\cr 10\,{{\rm e}^{-1/10\,ct-x/10}}+10\,{{\rm e}^{-1/10\,ct+x/10}}+ \left ( \sin \left ( ct-x \right ) \right ) ^{3}c+ \left ( \sin \left ( ct+x \right ) \right ) ^{3}c-20-2\,{c}^{2}t+ \left ( 2\,t+2\,x \right ) c&x<ct\cr }}{c}}$

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##### 5.1.2.6 [370] I.C. at $$t=1$$

problem number 370

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018 Solve $u_{tt} = u_{xx}$ With initial conditions \begin{align*} u(x,1) &= e^{-(x-6)^2}+e^{-(x+6)^2} \\ \frac{\partial u}{\partial t}(x,1) &= \frac{1}{2} \end{align*}

And boundary conditions $$\frac{\partial u}{\partial x}(0,t)=1$$

Mathematica

$\left \{\left \{u(x,t)\to \begin{array}{cc} \{ & \begin{array}{cc} 0 & x>c t \\ g\left (t-\frac{x}{c}\right ) & x\leq c t \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \}$

Maple

$u \left ( x,t \right ) =1/2\,{{\rm e}^{- \left ( 5-x+t \right ) ^{2}}}+1/2\,{{\rm e}^{- \left ( -x+t-7 \right ) ^{2}}}+1/2\,{{\rm e}^{- \left ( -7+x+t \right ) ^{2}}}+1/2\,{{\rm e}^{- \left ( x+t+5 \right ) ^{2}}}+t/2-1/2$

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##### 5.1.2.7 [371] B.C. at $$x=1$$

problem number 371

Solve $\frac{\partial ^2 u}{\partial t^2} = \frac{1}{4} \frac{\partial ^2 u}{\partial x^2}$ With initial conditions \begin{align*} u(x,0) &= e^{-x^2}\\ \frac{\partial u}{\partial t}(x,0) &= 0 \end{align*}

And Boundary conditions $$\frac{\partial u}{\partial x}(1,t)= 0$$

Mathematica

Failed

Maple

$u \left ( x,t \right ) =1/2\,\cases{{{\rm e}^{-1/4\, \left ( t+2\,x \right ) ^{2}}}+{{\rm e}^{-1/4\, \left ( t-2\,x \right ) ^{2}}}&t/2<-1+x\cr{{\rm e}^{-1/4\, \left ( t+2\,x \right ) ^{2}}}+{{\rm e}^{-1/4\, \left ( t-2\,x+4 \right ) ^{2}}}&-1+x<t/2\cr }$

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##### 5.1.2.8 [372] Left end free. zero initial velocity (general solution)

problem number 372

Solve for $$u(x,t)$$ with $$x>0,t>0$$ $u_{tt} = c^2 u_{xx}$ With initial conditions \begin{align*} u(x,0) &= f(x)\\ \frac{\partial u}{\partial t}(x,0) &= 0 \end{align*}

And boundary condition $$\frac{\partial u}{\partial x}(0,t) = 0$$.

Mathematica

Failed

Maple

$u \left ( x,t \right ) =1/2\,\cases{f \left ( ct+x \right ) +f \left ( -ct+x \right ) &ct<x\cr f \left ( ct+x \right ) +f \left ( ct-x \right ) &x<ct\cr }$

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##### 5.1.2.9 [373] Left end free. zero initial velocity (Special solution)

problem number 373

Solve for $$u(x,t)$$ with $$x>0,t>0$$ $u_{tt} = c^2 u_{xx}$ With initial conditions \begin{align*} u(x,0) &= f(x)\\ \frac{\partial u}{\partial t}(x,0) &= 0 \end{align*}

And boundary condition $$\frac{\partial u}{\partial x}(0,t) = 0$$ using \begin{align*} c &= 3\\ f(x) &=\left \{ \begin{array} [c]{cc}1 & 4\leq x\leq 5\\ 0 & \text{otherwise}\end{array} \right . \end{align*}

Mathematica

$\left \{\left \{u(x,t)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{1}{2} \left (\left (\begin{array}{cc} \{ & \begin{array}{cc} 1 & 4<x-3 t<5 \\ 0 & \text{True} \\\end{array} \\\end{array}\right )+\left (\begin{array}{cc} \{ & \begin{array}{cc} 1 & 4<3 t+x<5 \\ 0 & \text{True} \\\end{array} \\\end{array}\right )\right ) & x>3 t \\ \frac{1}{2} \left (\left (\begin{array}{cc} \{ & \begin{array}{cc} 1 & 4<3 t-x<5 \\ 0 & \text{True} \\\end{array} \\\end{array}\right )+\left (\begin{array}{cc} \{ & \begin{array}{cc} 1 & 4<3 t+x<5 \\ 0 & \text{True} \\\end{array} \\\end{array}\right )\right ) & x\leq 3 t \\\end{array} \\\end{array}\right \}\right \}$

Maple

$u \left ( x,t \right ) =1/2\,\cases{ \left ( \cases{1&4<6\,t+x\ and \ 6\,t+x<5\cr 0&otherwise\cr } \right ) \left ( 3\,t+x \right ) + \left ( \cases{1&4<x-6\,t\ and \ x-6\,t<5\cr 0&otherwise\cr } \right ) \left ( -3\,t+x \right ) &3\,t<x\cr \left ( \cases{1&4<6\,t+x\ and \ 6\,t+x<5\cr 0&otherwise\cr } \right ) \left ( 3\,t+x \right ) + \left ( \cases{1&4<x\ and \ x<5\cr 0&otherwise\cr } \right ) \left ( 3\,t-x \right ) &x<3\,t\cr }$

Hand solution

Solve $$\frac{\partial ^{2}u}{\partial t^{2}}=c^{2}\frac{\partial ^{2}u}{\partial x^{2}}$$ for $$x\geq 0,t\geq 0$$ with initial conditions $$u\left ( x,0\right ) =f\left ( x\right ) =\left \{ \begin{array} [c]{cc}1 & 4\leq x\leq 5\\ 0 & \text{otherwise}\end{array} \right .$$ and $$\frac{\partial u\left ( x,0\right ) }{\partial t}=0$$ and boundary condition $$\frac{\partial u\left ( 0,t\right ) }{\partial x}=0$$ (Free end)

The general solution by D’Almbert’s is given by$u\left ( x,t\right ) =\frac{1}{2}\left \{ \begin{array} [c]{ccc}f\left ( x+ct\right ) +f\left ( x-ct\right ) & & x>ct\geq 0\\ f\left ( x+ct\right ) +f\left ( ct-x\right ) & & x\leq ct \end{array} \right .$ But $$f\left ( x\right )$$ is deﬁned for $$4\leq x\leq 5$$ only, hence the solution becomes$u\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}\left \{ \begin{array} [c]{ccc}\frac{1}{2}f\left ( x+ct\right ) & & 4\leq x\leq 5\\ 0 & & \text{otherwise}\end{array} \right . +\left \{ \begin{array} [c]{ccc}\frac{1}{2}f\left ( x-ct\right ) & & 4\leq x\leq 5\\ 0 & & \text{otherwise}\end{array} \right . & & x>ct\geq 0\\ & & \\ \left \{ \begin{array} [c]{ccc}\frac{1}{2}f\left ( x+ct\right ) & & 4\leq x\leq 5\\ 0 & & \text{otherwise}\end{array} \right . +\left \{ \begin{array} [c]{ccc}\frac{1}{2}f\left ( ct-x\right ) & & 4\leq x\leq 5\\ 0 & & \text{otherwise}\end{array} \right . & & 0<x\leq ct \end{array} \right .$ Animation is below

Source code used for the above

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##### 5.1.2.10 [374] Left end ﬁxed. zero initial velocity (Special solution)

problem number 374

Solve for $$u(x,t)$$ with $$x>0,t>0$$ $u_{tt} = c^2 u_{xx}$ With initial conditions \begin{align*} u(x,0) &= f(x)\\ \frac{\partial u}{\partial t}(x,0) &= 0 \end{align*}

And boundary condition $$u(0,t) = 0$$ using \begin{align*} c &= 3\\ f(x) &=\left \{ \begin{array} [c]{cc}1 & 4\leq x\leq 5\\ 0 & \text{otherwise}\end{array} \right . \end{align*}

Mathematica

$\left \{\left \{u(x,t)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{1}{2} \left (\left (\begin{array}{cc} \{ & \begin{array}{cc} 1 & 4<x-3 t<5 \\ 0 & \text{True} \\\end{array} \\\end{array}\right )+\left (\begin{array}{cc} \{ & \begin{array}{cc} 1 & 4<3 t+x<5 \\ 0 & \text{True} \\\end{array} \\\end{array}\right )\right ) & x>3 t \\ \frac{1}{2} \left (\left (\begin{array}{cc} \{ & \begin{array}{cc} 1 & 4<3 t+x<5 \\ 0 & \text{True} \\\end{array} \\\end{array}\right )-\left (\begin{array}{cc} \{ & \begin{array}{cc} 1 & 4<3 t-x<5 \\ 0 & \text{True} \\\end{array} \\\end{array}\right )\right ) & x\leq 3 t \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \}$

Maple

$u \left ( x,t \right ) =1/2\,\cases{ \left ( \cases{1&4<6\,t+x\ and \ 6\,t+x<5\cr 0&otherwise\cr } \right ) \left ( 3\,t+x \right ) - \left ( \cases{1&4<x\ and \ x<5\cr 0&otherwise\cr } \right ) \left ( 3\,t-x \right ) &x<3\,t\cr \left ( \cases{1&4<6\,t+x\ and \ 6\,t+x<5\cr 0&otherwise\cr } \right ) \left ( 3\,t+x \right ) + \left ( \cases{1&4<x-6\,t\ and \ x-6\,t<5\cr 0&otherwise\cr } \right ) \left ( x-3\,t \right ) &3\,t<x\cr }$

Hand solution

Solve $$\frac{\partial ^{2}u}{\partial t^{2}}=c^{2}\frac{\partial ^{2}u}{\partial x^{2}}$$ for $$x\geq 0,t\geq 0$$ with initial conditions $$u\left ( x,0\right ) =f\left ( x\right ) =\left \{ \begin{array} [c]{cc}1 & 4\leq x\leq 5\\ 0 & \text{otherwise}\end{array} \right .$$ and $$\frac{\partial u\left ( x,0\right ) }{\partial t}=0$$ and boundary condition $$u\left ( 0,t\right ) =0$$ (Fixed end)

The general solution by D’Almbert’s is given by$u\left ( x,t\right ) =\frac{1}{2}\left \{ \begin{array} [c]{ccc}f\left ( x+ct\right ) +f\left ( x-ct\right ) & & x>ct\geq 0\\ f\left ( x+ct\right ) -f\left ( ct-x\right ) & & x\leq ct \end{array} \right .$ But $$f\left ( x\right )$$ is deﬁned for $$4\leq x\leq 5$$ only, hence the solution becomes$u\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}\left \{ \begin{array} [c]{ccc}\frac{1}{2}f\left ( x+ct\right ) & & 4\leq x\leq 5\\ 0 & & \text{otherwise}\end{array} \right . +\left \{ \begin{array} [c]{ccc}\frac{1}{2}f\left ( x-ct\right ) & & 4\leq x\leq 5\\ 0 & & \text{otherwise}\end{array} \right . & & x>ct\geq 0\\ & & \\ \left \{ \begin{array} [c]{ccc}\frac{1}{2}f\left ( x+ct\right ) & & 4\leq x\leq 5\\ 0 & & \text{otherwise}\end{array} \right . +\left \{ \begin{array} [c]{ccc}-\frac{1}{2}f\left ( ct-x\right ) & & 4\leq x\leq 5\\ 0 & & \text{otherwise}\end{array} \right . & & 0<x\leq ct \end{array} \right .$ Animation is below

Source code used for the above

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##### 5.1.2.11 [375] Left end free. zero initial position (general solution)

problem number 375

Solve for $$u(x,t)$$ with $$x>0,t>0$$ $u_{tt} = c^2 u_{xx}$ With initial conditions \begin{align*} u(x,0) &= 0\\ \frac{\partial u}{\partial t}(x,0) &= g(x) \end{align*}

And boundary condition $$\frac{\partial u}{\partial x}(0,t) = 0$$.

Mathematica

Failed

Maple

$u \left ( x,t \right ) =1/2\,{\frac{\cases{\int _{-ct+x}^{ct+x}\!g \left ( \zeta \right ) \,{\rm d}\zeta& ct<x\cr \int _{0}^{ct+x}\!g \left ( \zeta \right ) \,{\rm d}\zeta +\int _{0}^{ct-x}\!g \left ( \zeta \right ) \,{\rm d}\zeta& x<ct\cr }}{c}}$

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##### 5.1.2.12 [376] Left end free. Non zero initial position and velocity (general solution)

problem number 376

Solve for $$u(x,t)$$ with $$x>0,t>0$$ $u_{tt} = c^2 u_{xx}$ With initial conditions \begin{align*} u(x,0) &= f(x)\\ \frac{\partial u}{\partial t}(x,0) &= g(x) \end{align*}

And boundary condition $$\frac{\partial u}{\partial x}(0,t) = 0$$.

Mathematica

Failed

Maple

$u \left ( x,t \right ) =1/2\,\cases{f \left ( ct+x \right ) +f \left ( -ct+x \right ) +{\frac{\int _{-ct+x}^{ct+x}\!g \left ( \zeta \right ) \,{\rm d}\zeta }{c}}&ct<x\cr{\frac{f \left ( ct-x \right ) c+f \left ( ct+x \right ) c+\int _{0}^{ct+x}\!g \left ( \zeta \right ) \,{\rm d}\zeta +\int _{0}^{ct-x}\!g \left ( \zeta \right ) \,{\rm d}\zeta }{c}}&x<ct\cr }$

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##### 5.1.2.13 [377] Left end free with source

problem number 377

Left end free with initial position and velocity given.

Solve for $$u(x,t)$$ with $$x>0,t>0$$ $\frac{\partial ^2 u}{\partial t^2} = 9 \frac{\partial ^2 u}{\partial x^2} + f(x,t)$ With initial conditions \begin{align*} u(x,0) &= 0\\ \frac{\partial u}{\partial t}(x,0) &= x^3 \end{align*}

And boundary condition $$\frac{\partial u}{\partial x}(0,t) = 0$$.

Mathematica

$\left \{\left \{u(x,t)\to 3 c_1 \delta (3 t-x)+\frac{1}{12} (x-3 t)^4 \theta \left (t-\frac{x}{3}\right )+9 t^3 x+t x^3\right \}\right \}$

Maple

$u \left ( x,t \right ) =\cases{9\,{t}^{3}x+t{x}^{3}&3\,t<x\cr{\frac{27\,{t}^{4}}{4}}+9/2\,{t}^{2}{x}^{2}+1/12\,{x}^{4}&x<3\,t\cr }$

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