#### 5.2.1 Cartesian coordinates

##### 5.2.1.1 [394] Rectangular membrane. Fixed on all edges, General solution

problem number 394

Solve $u_{tt} = c^4 \left ( u_{xx}+ u_{yy} \right )$

\begin {align*} 0 & <x<L\\ 0 & <y<H \end {align*}

Boundary conditions on $$x$$

\begin {align*} u\left ( 0,y,t\right ) & =0\\ u\left ( L,y,t\right ) & =0 \end {align*}

And boundary conditions on $$y$$\begin {align*} u\left ( x,0,t\right ) & =0\\ u\left ( x,H,t\right ) & =0 \end {align*}

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

Mathematica

Failed

Maple

$u \left (x , y , t\right ) = \Mapleoverset {\infty }{\Mapleunderset {m =1}{\sum }}\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\frac {4 \left (H L \left (\left (\int _{0}^{H}\left (\right ) \sin \left (\frac {\pi m y}{H}\right )d y \right )_{\mathit {AllSolutions}}\right ) \sin \left (\frac {\pi \sqrt {H^{2} n^{2}+L^{2} m^{2}}\, c t}{H L}\right )+\pi \sqrt {H^{2} n^{2}+L^{2} m^{2}}\, c \left (\left (\int _{0}^{H}\left (\right ) \sin \left (\frac {\pi m y}{H}\right )d y \right )_{\mathit {AllSolutions}}\right ) \cos \left (\frac {\pi \sqrt {H^{2} n^{2}+L^{2} m^{2}}\, c t}{H L}\right )\right ) \sin \left (\frac {\pi m y}{H}\right ) \sin \left (\frac {\pi n x}{L}\right )}{\sqrt {H^{2} n^{2}+L^{2} m^{2}}\, \pi H L c}$

Hand solution

Assuming $$u=X\left ( x\right ) Y\left ( y\right ) T\left ( t\right )$$ and substituting into the PDE gives\begin {align*} \frac {1}{c^{2}}T^{\prime \prime }XY & =X^{\prime \prime }YT+Y^{\prime \prime }XT\\ \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =\frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y} \end {align*}

Therefore\begin {align*} \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =-\lambda \\ \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y} & =-\lambda \end {align*}

The time ODE becomes$T^{\prime \prime }+c^{2}\lambda T=0$ And the space ODE is$\frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y}=-\lambda$ Separating this again gives$\frac {X^{\prime \prime }}{X}=-\lambda -\frac {Y^{\prime \prime }}{Y}$ Let the second separation variable be $$\mu$$. This gives two new ODE’s to solve\begin {align*} \frac {X^{\prime \prime }}{X} & =-\mu \\ -\lambda -\frac {Y^{\prime \prime }}{Y} & =-\mu \end {align*}

Or\begin {align*} X^{\prime \prime }+\mu X & =0\\ Y^{\prime \prime }+Y\left ( \lambda -\mu \right ) & =0 \end {align*}

Solving for $$X\left ( x\right )$$ ODE ﬁrst, and knowing that only $$\mu >0$$ will give non trivial solutions (from the nature of the boundary conditions), gives the solution as$X\left ( x\right ) =A\cos \left ( \sqrt {\mu }x\right ) +B\sin \left ( \sqrt {\mu }x\right )$ Applying B.C. at $$x=0$$ results in$0=A$ Therefore $$X\left ( x\right ) =B\sin \left ( \sqrt {\mu }x\right )$$. Applying the B.C. at $$x=L$$ gives$0=B\sin \left ( \sqrt {\mu }L\right )$ For non trivial solution\begin {align*} \sqrt {\mu }L & =n\pi \\ \mu & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}

Therefore the $$X_{n}\left ( x\right )$$ eigenfunctions are$X_{n}\left ( x\right ) =B_{n}\sin \left ( \frac {n\pi }{L}x\right ) \qquad n=1,2,3,\cdots$ Now, solving the $$Y\left ( y\right )$$ ODE above$Y^{\prime \prime }+Y\left ( \lambda -\left ( \frac {n\pi }{L}\right ) ^{2}\right ) =0$ The solution is$Y_{n}\left ( y\right ) =A\cos \left ( \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}y\right ) +B\sin \left ( \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}y\right )$ Applying ﬁrst B.C. gives $0=A$ Hence$Y_{n}\left ( y\right ) =B\sin \left ( \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}y\right )$ Applying the second B.C. gives$0=B\sin \left ( \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}H\right )$ For non trivial solution\begin {align*} \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}H & =m\pi \qquad m=1,2,3,\cdots \\ \lambda _{nm}-\left ( \frac {n\pi }{L}\right ) ^{2} & =\left ( \frac {m\pi }{H}\right ) ^{2}\\ \lambda _{nm} & =\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}

Hence the $$Y_{nm}\left ( y\right )$$ solution is$Y_{nm}=B_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \qquad n=1,2,3,\cdots ,m=1,2,3,\cdots$ The time ode $$T\left ( t\right )$$ is now solved\begin {align*} T_{nm}^{\prime \prime }+c^{2}\lambda _{nm}T_{nm} & =0\\ T_{nm}\left ( t\right ) & =A_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) +B_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) \end {align*}

Combining all solutions, and merging all constants into two results in\begin {align} u_{nm}\left ( x,y,t\right ) & =X_{n}\left ( x\right ) Y_{nm}\left ( y\right ) T_{nm}\left ( t\right ) \nonumber \\ u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }X_{m}\left ( x\right ) Y_{mn}\left ( y\right ) T_{mn}\left ( t\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\lambda _{nm}}t\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \sin \left ( c\sqrt {\lambda _{nm}}t\right ) \tag {1} \end {align}

Initial conditions are now used to ﬁnd $$A_{nm},B_{nm}$$. At $$t=0$$\begin {align*} u\left ( x,y,0\right ) & =f\left ( x,y\right ) \\ \frac {\partial u}{\partial t}\left ( x,y,0\right ) & =g\left ( x,y\right ) \end {align*}

Applying ﬁrst initial condition to (1)  gives$f\left ( x,y\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right )$ Applying 2D orthogonality gives\begin {align*} \int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy & =A_{nm}\left ( \frac {L}{2}\right ) \left ( \frac {H}{2}\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}

Taking time derivative of (1) gives\begin {align*} \frac {\partial u}{\partial t}\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }-c\sqrt {\lambda _{nm}}A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \sin \left ( c\sqrt {\lambda _{nm}}t\right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}B_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\lambda _{nm}}t\right ) \end {align*}

At $$t=0$$ the above becomes$g\left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}B_{nm}\sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right )$ Applying 2D orthogonality gives\begin {align*} \int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy & =B_{nm}\left ( \frac {L}{2}\right ) \left ( \frac {H}{2}\right ) \\ B_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}

Summary of solution\begin {align*} u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) & +\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \sin \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy\\ B_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy\\ \lambda _{nm} & =\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2} \end {align*}

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##### 5.2.1.2 [395] Rectangular membrane. Fixed on all edges, zero velocity. Speciﬁc example

problem number 395

Solve $u_{tt} = c^4 \left ( u_{xx}+ u_{yy} \right )$

\begin {align*} 0 & <x<L\\ 0 & <y<H \end {align*}

Boundary conditions on $$x$$

\begin {align*} u\left ( 0,y,t\right ) & =0\\ u\left ( L,y,t\right ) & =0 \end {align*}

And boundary conditions on $$y$$\begin {align*} u\left ( x,0,t\right ) & =0\\ u\left ( x,H,t\right ) & =0 \end {align*}

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

Using $$L=1,H=2,c=\frac {1}{10},f(x,y)=x \cos (y),g(x,y)=0$$.

Mathematica

Failed

Maple

$u \left (x , y , t\right ) = \Mapleoverset {\infty }{\Mapleunderset {m =1}{\sum }}\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\frac {4 \left (-\left (-1\right )^{n}+\cos (2) \left (-1\right )^{m +n}\right ) m \cos \left (\frac {\pi \sqrt {m^{2}+4 n^{2}}\, t}{20}\right ) \sin \left (\pi n x \right ) \sin \left (\frac {\pi m y}{2}\right )}{\left (\pi ^{2} m^{2}-4\right ) n}$

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.1.1 on page 1833 as

\begin {align*} u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) & +\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \sin \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy\\ B_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy\\ \lambda _{nm} & =\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}

In this problem \begin {align*} L & =1\\ H & =2\\ c & =\frac {1}{10}\\ f\left ( x,y\right ) & =x\cos y\\ g\left ( x,y\right ) & =0 \end {align*}

Hence the solution becomes

\begin {align*} u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {m\pi }{2}y\right ) \cos \left ( \frac {1}{10}\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( n\pi x\right ) & +\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac {m\pi }{2}y\right ) \sin \left ( \frac {1}{10}\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( n\pi x\right ) \\ A_{nm} & =2\int _{0}^{1}\int _{0}^{2}x\cos y\sin \left ( n\pi x\right ) \sin \left ( \frac {m\pi }{2}y\right ) \ dxdy\\ B_{nm} & =0\\ \lambda _{nm} & =\left ( \frac {m\pi }{2}\right ) ^{2}+\left ( n\pi \right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}

But \begin {align*} A_{nm} & =2\int _{0}^{1}\int _{0}^{2}x\cos y\sin \left ( n\pi x\right ) \sin \left ( \frac {m\pi }{2}y\right ) \ dxdy\\ & =\frac {4\left ( -1\right ) ^{n}m\left ( -1+\left ( -1\right ) ^{m}\cos \left ( 2\right ) \right ) }{n\left ( m^{2}\pi ^{2}-4\right ) } \end {align*}

Hence the solution simpliﬁes to

\begin {align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }\frac {4\left ( -1\right ) ^{n}m\left ( -1+\left ( -1\right ) ^{m}\cos \left ( 2\right ) \right ) }{n\left ( m^{2}\pi ^{2}-4\right ) }\sin \left ( \frac {m\pi }{2}y\right ) \cos \left ( \frac {1}{10}\sqrt {\left ( \frac {m\pi }{2}\right ) ^{2}+\left ( n\pi \right ) ^{2}}t\right ) \right ) \sin \left ( n\pi x\right ) \\ & =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }\frac {4\left ( -1\right ) ^{n}m\left ( -1+\left ( -1\right ) ^{m}\cos \left ( 2\right ) \right ) }{n\left ( m^{2}\pi ^{2}-4\right ) }\cos \left ( \pi \frac {\sqrt {m^{2}+4n^{2}}}{20}t\right ) \sin \left ( \frac {m\pi }{2}y\right ) \right ) \sin \left ( n\pi x\right ) \end {align*}

Animation is below

Source code used for the above

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##### 5.2.1.3 [396] All 4 edges ﬁxed, zero initial velocity, Speciﬁc example

problem number 396

Solve for $$u(x,y,t)$$ with $$0<x<L$$ and $$0<y<H$$ and $$t>0$$.

Solve $u_{tt} = c^2 \nabla ^2 u(x,y)$ With boundary conditions \begin {align*} u(x,0,t) &=0 \\ u(0,y,t) &= 0 \\ u(L,y,t) &=0 \\ u(x,H,t) &= 0 \end {align*}

With initial conditions \begin {align*} u(x,y,0) &=3 f_1(x) f_2(y) \\ u_t(x,y,0) &= 0 \end {align*}

And $f_{1}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}x & & 0<x<\frac {L}{2}\\ L-x & & \frac {L}{2}<x<L \end {array} \right .$ Where $f_{2}\left ( y\right ) =\left \{ \begin {array} [c]{ccc}y & & 0<y<\frac {H}{2}\\ H-y & & \frac {H}{2}<y<H \end {array} \right .$ And $$L=2,H=3$$ and $$c=\frac {1}{3}$$.

Mathematica

$\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {n=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {288 \cos \left (\frac {1}{18} \pi t \sqrt {9 n^2+4 K[3]^2}\right ) \sin \left (\frac {n \pi }{2}\right ) \sin \left (\frac {n \pi x}{2}\right ) \sin \left (\frac {1}{2} \pi K[3]\right ) \sin \left (\frac {1}{3} \pi y K[3]\right )}{n^2 \pi ^4 K[3]^2} & (n|K[3])\in \mathbb {Z}\land n\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , y , t\right ) = \Mapleoverset {\infty }{\Mapleunderset {m =1}{\sum }}\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\frac {288 \cos \left (\frac {\pi \sqrt {4 m^{2}+9 n^{2}}\, t}{18}\right ) \sin \left (\frac {\pi m}{2}\right ) \sin \left (\frac {\pi n}{2}\right ) \sin \left (\frac {\pi m y}{3}\right ) \sin \left (\frac {\pi n x}{2}\right )}{\pi ^{4} m^{2} n^{2}}$

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.1.8 on page 1874 as\begin {align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}

In this problem \begin {align*} L & =2\\ H & =3\\ c & =\frac {1}{3}\\ f\left ( x,y\right ) & =3f_{1}\left ( x\right ) f_{2}\left ( y\right ) \end {align*}

And$f_{1}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}x & & 0<x<\frac {L}{2}\\ L-x & & \frac {L}{2}<x<L \end {array} \right .$ And$f_{2}\left ( y\right ) =\left \{ \begin {array} [c]{ccc}y & & 0<y<\frac {H}{2}\\ H-y & & \frac {H}{2}<y<H \end {array} \right .$ This is animation of the above solution using these speciﬁc values for for $$40$$ seconds. (Animation will only show in the HTML version)

Source code used for the above

The following shows selected modes. For example, for $$n=1,m=1$$ the solution becomes$u\left ( x,y,t\right ) =A_{1,1}\sin \left ( \frac {\pi }{L}x\right ) \sin \left ( \frac {\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {\pi }{H}\right ) ^{2}+\left ( \frac {1\pi }{L}\right ) ^{2}}t\right )$ And for $$n=1,m=5$$ then the solution becomes$u\left ( x,y,t\right ) =A_{1,5}\sin \left ( \frac {\pi }{L}x\right ) \sin \left ( \frac {5\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {5\pi }{H}\right ) ^{2}+\left ( \frac {\pi }{L}\right ) ^{2}}t\right )$ And so on.

 $$n$$ $$m$$ animation $$1$$ $$1$$ $$1$$ $$3$$ $$1$$ $$5$$ $$3$$ $$1$$ $$3$$ $$5$$ $$5$$ $$1$$ $$5$$ $$3$$

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##### 5.2.1.4 [397] All 4 edges ﬁxed, zero initial velocity, Speciﬁc example, delta in center

problem number 397

Solve for $$u(x,y,t)$$ with $$0<x<L$$ and $$0<y<H$$ and $$t>0$$.

Solve $u_{tt} = c^2 \nabla ^2 u(x,y)$ With boundary conditions \begin {align*} u(x,0,t) &=0 \\ u(0,y,t) &= 0 \\ u(L,y,t) &=0 \\ u(x,H,t) &= 0 \end {align*}

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

And \begin {align*} L & =20\\ H & =30\\ c & =\frac {1}{3}\\ f\left ( x,y\right ) & =f_{1}\left ( x\right ) f_{2}\left ( y\right ) \end {align*}

Where $$f\left ( x,y\right )$$ is an approximation of delta in the middle of the membrane $f_{1}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}1 & & \frac {45}{100}L<x<\frac {55}{100}L\\ 0 & & \text {otherwise}\end {array} \right .$ And$f_{2}\left ( y\right ) =\left \{ \begin {array} [c]{ccc}1 & & \frac {45}{100}H<y<\frac {55}{100}H\\ 0 & & \text {otherwise}\end {array} \right .$

Mathematica

$\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {n=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 \left (\cos \left (\frac {9 n \pi }{20}\right )-\cos \left (\frac {11 n \pi }{20}\right )\right ) \left (\cos \left (\frac {9}{20} \pi K[3]\right )-\cos \left (\frac {11}{20} \pi K[3]\right )\right ) \cos \left (\frac {1}{180} \pi t \sqrt {9 n^2+4 K[3]^2}\right ) \sin \left (\frac {n \pi x}{20}\right ) \sin \left (\frac {1}{30} \pi y K[3]\right )}{n \pi ^2 K[3]} & (n|K[3])\in \mathbb {Z}\land n\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , y , t\right ) = \Mapleoverset {\infty }{\Mapleunderset {m =1}{\sum }}\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\frac {16 \left (-2 \cos \left (\frac {\pi n}{5}\right )+2 \cos \left (\frac {\pi n}{10}\right )-1\right ) \left (-2 \cos \left (\frac {\pi n}{10}\right )+2 \cos \left (\frac {\pi n}{20}\right )-1\right ) \left (2 \cos \left (\frac {\pi n}{10}\right )+2 \cos \left (\frac {\pi n}{20}\right )+1\right ) \left (\cos \left (\frac {9 \pi m}{20}\right )-\cos \left (\frac {11 \pi m}{20}\right )\right ) \cos \left (\frac {\pi n}{20}\right ) \cos \left (\frac {\pi \sqrt {4 m^{2}+9 n^{2}}\, t}{180}\right ) \left (\sin ^{2}\left (\frac {\pi n}{20}\right )\right ) \sin \left (\frac {\pi m y}{30}\right ) \sin \left (\frac {\pi n x}{20}\right )}{\pi ^{2} m n}$

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.1.8 on page 1874 as\begin {align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}

In this problem \begin {align*} L & =20\\ H & =30\\ c & =\frac {1}{3}\\ f\left ( x,y\right ) & =f_{1}\left ( x\right ) f_{2}\left ( y\right ) \end {align*}

Where $$f\left ( x,y\right )$$ is an approximation of delta in the middle of the membrane $f_{1}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}1 & & \frac {45}{100}L<x<\frac {55}{100}L\\ 0 & & \text {otherwise}\end {array} \right .$ And$f_{2}\left ( y\right ) =\left \{ \begin {array} [c]{ccc}1 & & \frac {45}{100}H<y<\frac {55}{100}H\\ 0 & & \text {otherwise}\end {array} \right .$ This is animation of the above solution using these speciﬁc values for for $$40$$ seconds. (Animation will only show in the HTML version)

Source code used for the above

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##### 5.2.1.5 [398] All 4 edges ﬁxed

problem number 398

Taken from Mathematica helps pages on DSolve

Solve for $$u(x,y,t)$$ with $$0<x<1$$ and $$0<y<2$$ and $$t>0$$.

Solve $\frac {\partial ^2 u}{\partial t^2} = \frac {\partial ^2 u}{\partial x^2}+ \frac {\partial ^2 u}{\partial y^2}$ With boundary conditions \begin {align*} u(x,0,t) &=0 \\ u(0,y,t) &= 0 \\ u(1,y,t) &=0 \\ u(x,2,t) &= 0 \end {align*}

With initial conditions \begin {align*} u(x,y,0) &=\frac {1}{10} (x-x^2)(2 y-y^2) \\ \frac {\partial u}{\partial t}(x,y,0) &= 0 \end {align*}

Mathematica

$\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {n=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {32 \left (-1+(-1)^n\right ) \left (-1+(-1)^{K[3]}\right ) \cos \left (\frac {1}{2} \pi t \sqrt {4 n^2+K[3]^2}\right ) \sin (n \pi x) \sin \left (\frac {1}{2} \pi y K[3]\right )}{5 n^3 \pi ^6 K[3]^3} & K[3]\in \mathbb {Z}\land n\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , y , t\right ) = \Mapleoverset {\infty }{\Mapleunderset {m =1}{\sum }}\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}-\frac {32 \left (\left (-1\right )^{m}+\left (-1\right )^{n}-\left (-1\right )^{m +n}-1\right ) \cos \left (\frac {\pi \sqrt {m^{2}+4 n^{2}}\, t}{2}\right ) \sin \left (\pi n x \right ) \sin \left (\frac {\pi m y}{2}\right )}{5 \pi ^{6} m^{3} n^{3}}$

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##### 5.2.1.6 [399] All edges ﬁxed (Haberman 8.5.5 (a)

problem number 399

This is problem 8.5.5 part(a) from Richard Haberman applied partial diﬀerential equations 5th edition.

Solve the initial value problem for membrane with time-dependent forcing and ﬁxed boundaries $$u=0$$. $u_{tt} = c^2 \nabla ^2 u + Q(x,y,t)$ If the memberane is rectangle $$(0<x<L,0<y<H)$$. With initial conditions \begin {align*} u(x,y,0) &=f(x,y) \\ \frac {\partial u}{\partial t}(x,y,0) &= 0 \end {align*}

See my HW9, Math 322, UW Madison.

Mathematica

$\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}2 \sqrt {\frac {1}{H}} \sqrt {\frac {1}{L}} \left (\int _0^t \frac {2 \left (\int _0^L\int _0^HQ(x,y,K[4]) \sin \left (\frac {\pi x K[1]}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right )dydx\right ) \sin \left (c \pi \sqrt {\frac {K[1]^2}{L^2}+\frac {K[3]^2}{H^2}} (t-K[4])\right )}{c \pi \sqrt {\frac {H K[1]^2}{L}+\frac {L K[3]^2}{H}}} \, dK[4]+\frac {2 \cos \left (c \pi t \sqrt {\frac {K[1]^2}{L^2}+\frac {K[3]^2}{H^2}}\right ) \int _0^L\int _0^Hf(x,y) \sin \left (\frac {\pi x K[1]}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right )dydx}{\sqrt {H L}}\right ) \sin \left (\frac {\pi x K[1]}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right ) & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

sol=()

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##### 5.2.1.7 [400] 2 edgs ﬁxed, 2 free, zero initial velocity

problem number 400

Taken from Maple PDE help pages. This wave PDE inside square with free to move on left edge and right edge, and top and bottom edges are ﬁxed. It has zero initial velocity, but given a non-zero initial position. Where $$0<x<\pi$$ and $$0<y<\pi$$ and $$t>0$$.

Solve $u_{tt} = \frac {1}{4} \left ( \frac {\partial ^2 u}{\partial x^2}+ \frac {\partial ^2 u}{\partial y^2} \right )$ With boundary conditions \begin {align*} \frac {\partial u}{\partial x}u(0,y,t) &= 0 \\ \frac {\partial u}{\partial x}u(\pi ,y,t) &= 0 \\ u(x,0,t) &= 0\\ u(x,\pi ,0) &=0 \end {align*}

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

Mathematica

$\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}-\frac {2 \left (-1+(-1)^{K[3]}\right ) \cos \left (\frac {1}{2} t K[3]\right ) \sin (y K[3])}{K[3]^3}+\underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}-\frac {8 \left (-1+(-1)^{K[1]}\right ) \left (-1+(-1)^{K[3]}\right ) \cos (x K[1]) \cos \left (\frac {1}{2} t \sqrt {K[1]^2+K[3]^2}\right ) \sin (y K[3])}{\pi ^2 K[1]^2 K[3]^3} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , y , t\right ) = -2 \left (\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\frac {\left (\left (-1\right )^{n}-1\right ) \cos \left (\frac {n t}{2}\right ) \sin \left (n y \right )}{n^{3}}\right )+\left (\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\Mapleoverset {\infty }{\Mapleunderset {m =1}{\sum }}\frac {8 \left (\left (-1\right )^{m}+\left (-1\right )^{n}-\left (-1\right )^{m +n}-1\right ) \cos \left (m x \right ) \cos \left (\frac {\sqrt {m^{2}+n^{2}}\, t}{2}\right ) \sin \left (n y \right )}{\pi ^{2} m^{2} n^{3}}\right )$

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##### 5.2.1.8 [401] All 4 edges ﬁxed, zero initial velocity, general solution

problem number 401

Solve for $$u(x,y,t)$$ with $$0<x<L$$ and $$0<y<H$$ and $$t>0$$.

Solve $u_{tt} = c^2 \nabla ^2 u(x,y)$ With boundary conditions \begin {align*} u(x,0,t) &=0 \\ u(0,y,t) &= 0 \\ u(L,y,t) &=0 \\ u(x,H,t) &= 0 \end {align*}

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

Mathematica

$\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {n=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 \cos \left (\pi t \sqrt {c^2 \left (\frac {n^2}{L^2}+\frac {K[3]^2}{H^2}\right )}\right ) \left (\int _0^L\int _0^Hf(x,y) \sin \left (\frac {n \pi x}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right )dydx\right ) \sin \left (\frac {n \pi x}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right )}{H L} & (n|K[3])\in \mathbb {Z}\land n\geq 1\land K[3]\geq 1\land c^2 H^2 L^2 \left (H^2 n^2+L^2 K[3]^2\right )>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

time expired

Hand solution

Solve for $$u\left ( r,\theta ,t\right )$$$u_{tt}=c^{2}\nabla ^{2}u\left ( x,y\right )$ With boundary conditions such that all edges are ﬁxed, and initial conditions $$u\left ( x,y,0\right ) =f\left ( x,y\right )$$ and initial velocity $$g\left ( x,y\right ) =0.$$

Let $$u=X\left ( x\right ) Y\left ( y\right ) T\left ( t\right )$$. Substituting into the above PDE gives\begin {align*} \frac {1}{c^{2}}T^{\prime \prime }XY & =X^{\prime \prime }YT+Y^{\prime \prime }XT\\ \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =\frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y} \end {align*}

Hence\begin {align*} \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =-\lambda \\ \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y} & =-\lambda \end {align*}

The time ODE becomes$T^{\prime \prime }+c^{2}\lambda T=0$ And the space ODE is\begin {align*} \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y} & =-\lambda \\ \frac {X^{\prime \prime }}{X} & =-\lambda -\frac {Y^{\prime \prime }}{Y} \end {align*}

Using a new separation variable $$\mu$$ gives the following two ODE’s\begin {align*} \frac {X^{\prime \prime }}{X} & =-\mu \\ -\lambda -\frac {Y^{\prime \prime }}{Y} & =-\mu \end {align*}

Or\begin {align*} X^{\prime \prime }+\mu X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end {align*}

And\begin {align*} Y^{\prime \prime }+Y\left ( \lambda -\mu \right ) & =0\\ Y\left ( 0\right ) & =0\\ Y\left ( H\right ) & =0 \end {align*}

Solving ﬁrst for the $$X\left ( x\right )$$ ODE, and knowing that $$\mu$$ must be positive only here from the nature of the boundary conditions gives$X=A\cos \left ( \sqrt {\mu }x\right ) +B\sin \left ( \sqrt {\mu }x\right )$ Applying B.C. at $$x=0$$$0=A$ Hence solution becomes $$X\left ( x\right ) =B\sin \left ( \sqrt {\mu }x\right )$$. Applying the B.C. at $$x=L$$ gives$0=B\sin \left ( \sqrt {\mu }L\right )$ Non trivial solution requires that\begin {align*} \sqrt {\mu }L & =n\pi \qquad n=1,2,3,\cdots \\ \mu _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2} \end {align*}

Therefore the eigenfunctions $$X_{n}\left ( x\right )$$ are$X_{n}\left ( x\right ) =\sin \left ( \frac {n\pi }{L}x\right ) \qquad n=1,2,3,\cdots$ Solving the $$Y\left ( y\right )$$ ODE$Y_{n}^{\prime \prime }+\left ( \lambda -\left ( \frac {n\pi }{L}\right ) ^{2}\right ) Y_{n}=0\qquad n=1,2,3,\cdots$ The nature of the boundary conditions on $$Y\left ( y\right )$$ suggests that $$\left ( \lambda -\left ( \frac {n\pi }{L}\right ) ^{2}\right )$$ must be positive (if $$\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}=0$$ or $$\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}<0$$, trivial solutions result).

Hence the solution for $$Y_{n}\left ( y\right )$$ becomes$Y_{n}\left ( y\right ) =A\cos \left ( \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}y\right ) +B\sin \left ( \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}y\right )$ Applying ﬁrst B.C. $$Y\left ( 0\right ) =0$$ gives$0=A$ The solution becomes$Y_{n}\left ( y\right ) =B_{n}\sin \left ( \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}y\right )$ Applying second B.C. $$Y\left ( H\right ) =0$$ gives$0=B\sin \left ( \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}H\right )$ Non trivial solution requires that\begin {align*} \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}H & =m\pi \qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \\ \lambda _{nm}-\left ( \frac {n\pi }{L}\right ) ^{2} & =\left ( \frac {m\pi }{H}\right ) ^{2}\\ \lambda _{nm} & =\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2} \end {align*}

Hence the $$Y_{nm}\left ( y\right )$$ eigenfunctions are$Y_{nm}\left ( y\right ) =\sin \left ( \frac {m\pi }{H}y\right ) \qquad n=1,2,3,\cdots ,m=1,2,3,\cdots$ Now the time $$T\left ( t\right )$$ ode is solved, and since $$\lambda _{nm}$$ is positive, then \begin {align*} T_{nm}^{\prime \prime }+c^{2}\lambda _{nm}T_{nm} & =0\\ T_{nm}\left ( t\right ) & =A_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) +B_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) \\ & =A_{nm}\cos \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) +B_{nm}\sin \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) \end {align*}

Combining all solution , and merging all constants into two results in

\begin {align} u_{nm}\left ( x,y,t\right ) & =X_{n}\left ( x\right ) Y_{nm}\left ( y\right ) T_{nm}\left ( t\right ) \nonumber \\ u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }X_{n}\left ( x\right ) Y_{nm}\left ( y\right ) T_{nm}\left ( t\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \sin \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) \tag {1} \end {align}

Initial conditions are now used to ﬁnd $$A_{nm},B_{nm}$$. At $$t=0$$

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

Applying ﬁrst initial condition to (1)  gives$f\left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right )$ Applying 2D orthogonality gives\begin {align*} \int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy & =A_{nm}\left ( \frac {L}{2}\right ) \left ( \frac {H}{2}\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}

Taking time derivative of (1) gives\begin {align*} \frac {\partial u}{\partial t}\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }-c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \sin \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}B_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) \end {align*}

AT $$t=0$$ the above becomes$\int _{0}^{H}g\left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}B_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right )$ Applying 2D orthogonality gives$\int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy=B_{nm}\left ( \frac {L}{2}\right ) \left ( \frac {H}{2}\right )$ But the initial velocity $$g\left ( x,y\right ) =0$$. Hence $$B_{nm}=0$$ for all $$n,m$$.

Summary of solution\begin {align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}

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##### 5.2.1.9 [402] With damping

problem number 402

Taken from Maple PDE help pages. This wave PDE inside square with damping present.

Membrane is free to move on the right edge and also on top edge. But ﬁxed at left edge and bottom edge.

It has zero initial position, but given a non-zero initial velocity. Where $$0<x<1$$ and $$0<y<1$$ and $$t>0$$. Solve $u_{tt} + \frac {1}{10} u_t = \frac {1}{4} \nabla ^2 u(x,y)$ With boundary conditions \begin {align*} u(0,y,t) &=0\\ \frac {\partial u}{\partial x}u(1,y,t) &= 0 \\ u(x,0,t) &=0 \\ \frac {\partial u}{\partial y}u(x,1,t) &= 0 \end {align*}

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

Mathematica

$\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {5120 e^{-t/20} \sin \left (\frac {1}{2} \pi x (2 K[1]-1)\right ) \sin \left (\frac {1}{2} \pi y (2 K[3]-1)\right ) \sin \left (\frac {1}{20} t \sqrt {50 \pi ^2 \left (2 K[1]^2-2 K[1]+2 K[3]^2-2 K[3]+1\right )-1}\right )}{\pi ^6 (2 K[1]-1)^3 (2 K[3]-1)^3 \sqrt {50 \pi ^2 \left (2 K[1]^2-2 K[1]+2 K[3]^2-2 K[3]+1\right )-1}} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , y , t\right ) = \Mapleoverset {\infty }{\Mapleunderset {m =0}{\sum }}\Mapleoverset {\infty }{\Mapleunderset {n =0}{\sum }}\frac {5120 \,{\mathrm e}^{-\frac {t}{20}} \sin \left (\frac {\sqrt {-1+\left (100 m^{2}+100 n^{2}+100 m +100 n +50\right ) \pi ^{2}}\, t}{20}\right ) \sin \left (\frac {\left (2 n +1\right ) \pi x}{2}\right ) \sin \left (\frac {\left (2 m +1\right ) \pi y}{2}\right )}{\sqrt {-1+\left (100 m^{2}+100 n^{2}+100 m +100 n +50\right ) \pi ^{2}}\, \pi ^{6} \left (2 m +1\right )^{3} \left (2 n +1\right )^{3}}$

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##### 5.2.1.10 [403] On the whole plane

problem number 403

From Mathematica DSolve help pages.

Hyperbolic partial diﬀerential equation with non-rational coeﬃcients.

Solve for $$u(x,y)$$ $u_{xx} -2 \sin x u_{x y} -\cos ^2 x u_{y y} -\cos x u_y=0$

Mathematica

$\{\{u(x,y)\to c_1(x-\cos (x)+y)+c_2(-x-\cos (x)+y)\}\}$

Maple

sol=()

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