#### 4.1.1 Cartesian coordinates

4.1.1.4 [284] Haberman 2.5.1 (a)
4.1.1.5 [285] Haberman 2.5.1 (b)
4.1.1.6 [286] Haberman 2.5.1 (c)
4.1.1.7 [287] Haberman 2.5.1 (d)
4.1.1.8 [288] Haberman 2.5.1 (e)
4.1.1.9 [289] Unit triangle B.C.
4.1.1.10 [290] Top edge at inﬁnity
4.1.1.11 [291] Top edge at inﬁnity
4.1.1.12 [292] Right edge at inﬁnity
4.1.1.13 [293] Right edge at inﬁnity
4.1.1.14 [294] Right edge at inﬁnity

4.1.1.16 [296] One side homogeneous
4.1.1.17 [297] In right half plane
4.1.1.18 [298] Right edge at inﬁnity

4.1.1.20 [300] Right half-plane

4.1.1.24 [304] Strip in upper half

##### 4.1.1.1 [281] Rectangle, 3 edges zero, buttom edge not

problem number 281

Solve Laplace PDE $\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ inside a rectangle $$0 \leq x \leq L, 0 \leq y \leq H$$, with following boundary conditions \begin{align*} u(x,0)&=f(x) \\ u(x,H)&=0 \\ u(0,y)&=0 \\ u(L,y)&=0 \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \underset{n=1}{\overset{\infty }{\sum }}\text{csch}\left (\frac{H n \pi }{L}\right ) \text{FourierSinCoefficient}\left [f(x),x,n,\text{FourierParameters}\to \left \{1,\frac{\pi }{L}\right \}\right ] \sin \left (\frac{n \pi x}{L}\right ) \sinh \left (\frac{n \pi (H-y)}{L}\right )\right \}\right \}$

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty } \left ( -2\,{\frac{1}{L}\sin \left ({\frac{n\pi \,x}{L}} \right ) \int _{0}^{L}\!\sin \left ({\frac{n\pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x \left ( -{{\rm e}^{-{\frac{n\pi \, \left ( -2\,H+y \right ) }{L}}}}+{{\rm e}^{{\frac{n\pi \,y}{L}}}} \right ) \left ({{\rm e}^{2\,{\frac{n\pi \,H}{L}}}}-1 \right ) ^{-1}} \right )$

Hand solution

Solve

\begin{align*} \nabla ^{2}u & =0\qquad \text{on a rectangle}\qquad R=\left \{ 0<x<a,0<y<b\right \} \\ u\left ( x,0\right ) & =f\left ( x\right ) \\ u\left ( x,H\right ) & =0\\ u\left ( 0,y\right ) & =0\\ u\left ( L,y\right ) & =0 \end{align*}

Solution

Let $$u\left ( x,y\right ) =X\left ( x\right ) Y\left ( x\right )$$. Substituting this into the PDE $$\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}=0$$ and simplifying gives$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}$ Each side depends on diﬀerent independent variable and they are equal, therefore they must be equal to same constant.$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}=\pm \lambda$ Since the boundary conditions along the $$x$$ direction are the homogeneous ones, $$-\lambda$$ is selected in the above.$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}=-\lambda$ Two ODE’s are obtained$$X^{\prime \prime }+\lambda X=0\tag{1}$$ With the boundary conditions\begin{align*} X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

And$$Y^{\prime \prime }-\lambda Y=0\tag{2}$$ With the boundary conditions\begin{align*} Y\left ( 0\right ) & =f\left ( x\right ) \\ Y\left ( H\right ) & =0 \end{align*}

Case $$\lambda <0$$

The solution to (1) is

$X=A\cosh \left ( \sqrt{\left \vert \lambda \right \vert }x\right ) +B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }x\right )$ At $$x=0$$, the above gives $$0=A$$. Hence $$X=B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }x\right )$$. At $$x=L$$ this gives $$X=B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }L\right )$$. But $$\sinh \left ( \sqrt{\left \vert \lambda \right \vert }L\right ) =0$$ only at $$0$$ and $$\sqrt{\left \vert \lambda \right \vert }L\neq 0$$, therefore $$B=0$$ and this leads to trivial solution. Hence $$\lambda <0$$ is not an eigenvalue.

Case $$\lambda =0$$

$X=Ax+B$ Hence at $$x=0$$ this gives $$0=B$$ and the solution becomes $$X=B$$. At $$x=L$$, $$B=0$$. Hence the trivial solution. $$\lambda =0$$ is not an eigenvalue.

Case $$\lambda >0$$

Solution is $X=A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right )$ At $$x=0$$ this gives $$0=A$$ and the solution becomes $$X=B\sin \left ( \sqrt{\lambda }x\right )$$. At $$x=L$$$0=B\sin \left ( \sqrt{\lambda }L\right )$ For non-trivial solution $$\sin \left ( \sqrt{\lambda }L\right ) =0$$ or $$\sqrt{\lambda }L=n\pi$$ where $$n=1,2,3,\cdots$$, therefore$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots$ Eigenfunctions are$$X_{n}\left ( x\right ) =B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \tag{3}$$ For the $$Y$$ ODE, the solution is$$Y_{n}=C_{n}\cosh \left ( \frac{n\pi }{L}y\right ) +D_{n}\sinh \left ( \frac{n\pi }{L}y\right ) \tag{4}$$ Applying B.C. at $$y=H$$ gives\begin{align*} 0 & =C_{n}\cosh \left ( \frac{n\pi }{L}H\right ) +D_{n}\sinh \left ( \frac{n\pi }{L}H\right ) \\ C_{n} & =-D_{n}\frac{\sinh \left ( \frac{n\pi }{L}H\right ) }{\cosh \left ( \frac{n\pi }{L}H\right ) }\\ & =-D_{n}\tanh \left ( \frac{n\pi }{L}H\right ) \end{align*}

Hence (4) becomes \begin{align*} Y_{n} & =-D_{n}\tanh \left ( \frac{n\pi }{L}H\right ) \cosh \left ( \frac{n\pi }{L}y\right ) +D_{n}\sinh \left ( \frac{n\pi }{L}y\right ) \\ & =D_{n}\left ( \sinh \left ( \frac{n\pi }{L}y\right ) -\tanh \left ( \frac{n\pi }{L}H\right ) \cosh \left ( \frac{n\pi }{L}y\right ) \right ) \end{align*}

Now the complete solution is produced\begin{align*} u_{n}\left ( x,y\right ) & =Y_{n}X_{n}\\ & =D_{n}\left ( \sinh \left ( \frac{n\pi }{L}y\right ) -\tanh \left ( \frac{n\pi }{L}H\right ) \cosh \left ( \frac{n\pi }{L}y\right ) \right ) B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Let $$D_{n}B_{n}=B_{n}$$ since a constant. (no need to make up a new symbol).$u_{n}\left ( x,y\right ) =B_{n}\left ( \sinh \left ( \frac{n\pi }{L}y\right ) -\tanh \left ( \frac{n\pi }{L}H\right ) \cosh \left ( \frac{n\pi }{L}y\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right )$ Sum of eigenfunctions is the solution, hence$$u\left ( x,y\right ) =\sum _{n=1}^{\infty }B_{n}\left ( \sinh \left ( \frac{n\pi }{L}y\right ) -\tanh \left ( \frac{n\pi }{L}H\right ) \cosh \left ( \frac{n\pi }{L}y\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right ) \tag{5}$$ The nonhomogeneous boundary condition is now resolved.  At $$y=0$$$u\left ( x,0\right ) =f\left ( x\right )$ Therefore (5) becomes$f\left ( x\right ) =\sum _{n=1}^{\infty }-B_{n}\tanh \left ( \frac{n\pi }{a}b\right ) \sin \left ( \frac{n\pi }{a}x\right )$ Multiplying both sides by $$\sin \left ( \frac{m\pi }{L}x\right )$$ and integrating gives\begin{align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx & =-\int _{0}^{a}\sin \left ( \frac{m\pi }{L}x\right ) \sum _{n=1}^{\infty }B_{n}\tanh \left ( \frac{n\pi }{L}H\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =-\sum _{n=1}^{\infty }B_{n}\tanh \left ( \frac{n\pi }{L}b\right ) \int _{0}^{L}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx\\ & =-B_{n}\tanh \left ( \frac{m\pi }{L}H\right ) \left ( \frac{L}{2}\right ) \end{align*}

Hence$B_{n}=-\frac{2}{L}\frac{\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx}{\tanh \left ( \frac{n\pi }{L}H\right ) }$ The solution (5) becomes\begin{align*} u\left ( x,y\right ) & =-\frac{2}{L}\sum _{n=1}^{\infty }\frac{\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx}{\tanh \left ( \frac{n\pi }{L}H\right ) }\left ( \sinh \left ( \frac{n\pi }{L}y\right ) -\tanh \left ( \frac{n\pi }{L}H\right ) \cosh \left ( \frac{n\pi }{L}y\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right ) \\ & =-\frac{2}{L}\sum _{n=1}^{\infty }\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\left ( \frac{\sinh \left ( \frac{n\pi }{L}y\right ) }{\tanh \left ( \frac{n\pi }{L}H\right ) }-\cosh \left ( \frac{n\pi }{L}y\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

________________________________________________________________________________________

##### 4.1.1.2 [282] Rectangle, 3 edges zero, right edge not

problem number 282

Solve Laplace PDE inside square $$\nabla ^2 u(x,y) = 0$$ with $$0 \leq x \leq 1, 0 \leq y \leq 1$$, with following boundary conditions \begin{align*} u(x,0)&=0 \\ u(x,1)&=0 \\ u(0,y)&=0 \\ u(1,y)&=y(1-y) \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \underset{n=1}{\overset{\infty }{\sum }}-\frac{4 \left (-1+(-1)^n\right ) \text{csch}(n \pi ) \sin (n \pi y) \sinh (n \pi x)}{n^3 \pi ^3}\right \}\right \}$

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }-4\,{\frac{ \left ( \left ( -1 \right ) ^{n}-1 \right ){{\rm e}^{\pi \,n}}\sin \left ( n\pi \,y \right ) \left ({{\rm e}^{n\pi \,x}}-{{\rm e}^{-n\pi \,x}} \right ) }{{n}^{3}{\pi }^{3} \left ({{\rm e}^{2\,\pi \,n}}-1 \right ) }}$

Hand solution

$$a$$ is used for the length of the $$x$$ dimension and $$b$$ for the length of the $$y$$ dimension.

Solution

Let $$u\left ( x,y\right ) =X\left ( x\right ) Y\left ( x\right )$$. Substituting this into the PDE gives$X^{\prime \prime }Y+Y^{\prime \prime }X=0$ Dividing throughout by $$XY\neq 0\,$$ and simplifying gives$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}=\lambda$ This gives the eigenvalue ODE\begin{align} Y^{\prime \prime }+\lambda Y & =0\tag{1}\\ Y\left ( 0\right ) & =0\nonumber \\ Y\left ( b\right ) & =0\nonumber \end{align}

The solution to (1) gives the eigenvalues $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}$$ for $$n=1,2,3\cdots$$ and since $$L=b$$, this becomes $\lambda _{n}=\left ( \frac{n\pi }{b}\right ) ^{2}\qquad n=1,2,\cdots$ And the corresponding eigenfunction\begin{align*} Y_{n}\left ( y\right ) & =c_{n}\sin \left ( \sqrt{\lambda _{n}}y\right ) \\ & =c_{n}\sin \left ( \frac{n\pi }{b}y\right ) \end{align*}

Therefore the corresponding nonhomogeneous $$X\left ( x\right )$$ ODE\begin{align} X_{n}^{\prime \prime }-\lambda _{n}X_{n} & =0\tag{2}\\ X_{n}\left ( 0\right ) & =0\nonumber \\ X_{n}\left ( a\right ) & =y-y^{2}\nonumber \end{align}

The solution to (2), since $$\lambda _{n}$$ is positive is \begin{align*} X_{n}\left ( x\right ) & =A_{n}\cosh \left ( \sqrt{\lambda _{n}}x\right ) +B_{n}\sinh \left ( \sqrt{\lambda _{n}}x\right ) \\ & =A_{n}\cosh \left ( \frac{n\pi }{b}x\right ) +B_{n}\sinh \left ( \frac{n\pi }{b}x\right ) \end{align*}

Boundary conditions $$X\left ( 0\right ) =0$$ gives$0=A_{n}$ The solution (3) now simpliﬁes to$X_{n}\left ( x\right ) =B_{n}\sinh \left ( \frac{n\pi }{b}x\right )$ Hence the fundamental solution is\begin{align*} u_{n}\left ( x,y\right ) & =X_{n}Y_{n}\\ & =c_{n}\sinh \left ( \frac{n\pi }{b}x\right ) \sin \left ( \frac{n\pi }{b}y\right ) \end{align*}

Where the constants $$B_{n}$$ is merged with $$c_{n}$$. The solution is$$u\left ( x,y\right ) =\sum _{n=1}^{\infty }c_{n}\sinh \left ( \frac{n\pi }{b}x\right ) \sin \left ( \frac{n\pi }{b}y\right ) \tag{3}$$ $$c_{n}$$ is now found by applying the boundary condition at $$x=a$$. The above becomes$y-y^{2}=\sum _{n=1}^{\infty }c_{n}\sinh \left ( \frac{n\pi }{b}a\right ) \sin \left ( \frac{n\pi }{b}y\right )$ Multiplying both sides by $$\sin \left ( \frac{m\pi }{b}y\right )$$ and integrating gives$\int _{0}^{b}\left ( y-y^{2}\right ) \sin \left ( \frac{m\pi }{b}y\right ) dy=\sum _{n=1}^{\infty }c_{n}\sinh \left ( \frac{n\pi }{b}a\right ) \left ( \int _{0}^{b}\sin \left ( \frac{m\pi }{b}y\right ) \sin \left ( \frac{n\pi }{b}y\right ) dy\right )$ By orthogonality the above reduces to \begin{align*} \int _{0}^{b}\left ( y-y^{2}\right ) \sin \left ( \frac{m\pi }{b}y\right ) dy & =c_{n}\sinh \left ( \frac{m\pi }{b}a\right ) \int _{0}^{b}\sin ^{2}\left ( \frac{m\pi }{b}y\right ) dy\\ & =\frac{b}{2}c_{m}\sinh \left ( \frac{m\pi }{b}a\right ) \end{align*}

Therefore$c_{n}=\frac{2}{b}\frac{1}{\sinh \left ( \frac{m\pi }{b}a\right ) }\int _{0}^{b}\left ( y-y^{2}\right ) \sin \left ( \frac{n\pi }{b}y\right ) dy$ Now replacing $$a=1,b=1$$, the above becomes\begin{align*} c_{n} & =\frac{2}{\sinh \left ( n\pi \right ) }\int _{0}^{1}\left ( y-y^{2}\right ) \sin \left ( n\pi y\right ) dy\\ & =\frac{2}{\sinh \left ( n\pi \right ) }\left ( \frac{-2\left ( -1+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}}\right ) \\ & =\frac{-4}{\sinh \left ( n\pi \right ) }\frac{\left ( -1+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}} \end{align*}

Hence the solution (3) becomes$u\left ( x,y\right ) =\frac{-4}{\pi ^{3}}\sum _{n=1}^{\infty }\frac{\left ( -1+\left ( -1\right ) ^{n}\right ) }{n^{3}}\frac{\sinh \left ( n\pi x\right ) }{\sinh \left ( n\pi \right ) }\sin \left ( n\pi y\right )$ This is a 3D plot of the solution.

This is a contour plot

________________________________________________________________________________________

##### 4.1.1.3 [283] Rectangle, 3 edges zero, buttom edge has impulse

problem number 283

This is Problem 6.3.10 from Introduction to Partial Diﬀerential Equations by Peter Olver ISBN 9783319020983.

Solve

\begin{align*} \nabla ^{2}u & =0\qquad \text{on a rectangle}\qquad R=\left \{ 0<x<a,0<y<b\right \} \\ u\left ( x,0\right ) & =f\left ( x\right ) \\ u\left ( x,b\right ) & =0\\ u\left ( 0,y\right ) & =0\\ u\left ( a,y\right ) & =0 \end{align*}

When the boundary data $$f\left ( x\right ) =\delta \left ( x-\xi \right )$$ is a delta function at a point $$0<\xi <a$$.

Mathematica

$\left \{\left \{u(x,y)\to \underset{n=1}{\overset{\infty }{\sum }}\frac{2 \text{csch}\left (\frac{b n \pi }{a}\right ) \sin \left (\frac{n \pi x}{a}\right ) \sin \left (\frac{n \pi \zeta }{a}\right ) \sinh \left (\frac{n \pi (b-y)}{a}\right )}{a}\right \}\right \}$

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }2\,{\frac{1}{a}\sin \left ({\frac{n\pi \,x}{a}} \right ) \sin \left ({\frac{n\pi \,\zeta }{a}} \right ) \left ({{\rm e}^{{\frac{n\pi \, \left ( 2\,b-y \right ) }{a}}}}-{{\rm e}^{{\frac{n\pi \,y}{a}}}} \right ) \left ({{\rm e}^{2\,{\frac{n\pi \,b}{a}}}}-1 \right ) ^{-1}}$

Hand solution

Let $$u\left ( x,y\right ) =X\left ( x\right ) Y\left ( x\right )$$. Substituting this into the PDE $$\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}=0$$ and simplifying gives$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}$ Each side depends on diﬀerent independent variable and they are equal, therefore they must be equal to same constant.$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}=\pm \lambda$ Since the boundary conditions along the $$x$$ direction are the homogeneous ones, $$-\lambda$$ is selected in the above.

$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}=-\lambda$

Two ODE’s are obtained$$X^{\prime \prime }+\lambda X=0 \tag{1}$$ With the boundary conditions\begin{align*} X\left ( 0\right ) & =0\\ X\left ( a\right ) & =0 \end{align*}

And$$Y^{\prime \prime }-\lambda Y=0 \tag{2}$$ With the boundary conditions\begin{align*} Y\left ( 0\right ) & =f\left ( x\right ) \\ Y\left ( b\right ) & =0 \end{align*}

In all these cases $$\lambda$$ will turn out to be positive. This is shown below.

Case $$\lambda <0$$

The solution to (1) is

$X=A\cosh \left ( \sqrt{\left \vert \lambda \right \vert }x\right ) +B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }x\right )$ At $$x=0$$, the above gives $$0=A$$. Hence $$X=B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }x\right )$$. At $$x=a$$ this gives $$X=B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }a\right )$$. But $$\sinh \left ( \sqrt{\left \vert \lambda \right \vert }a\right ) =0$$ only at $$0$$ and $$\sqrt{\left \vert \lambda \right \vert }a\neq 0$$, therefore $$B=0$$ and this leads to trivial solution. Hence $$\lambda <0$$ is not an eigenvalue.

Case $$\lambda =0$$

$X=Ax+B$ Hence at $$x=0$$ this gives $$0=B$$ and the solution becomes $$X=B$$. At $$x=a$$, $$B=0$$. Hence the trivial solution. $$\lambda =0$$ is not an eigenvalue.

Case $$\lambda >0$$

Solution is $X=A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right )$ At $$x=0$$ this gives $$0=A$$ and the solution becomes $$X=B\sin \left ( \sqrt{\lambda }x\right )$$. At $$x=a$$ $0=B\sin \left ( \sqrt{\lambda }a\right )$ For non-trivial solution $$\sin \left ( \sqrt{\lambda }a\right ) =0$$ or $$\sqrt{\lambda }a=n\pi$$ where $$n=1,2,3,\cdots$$, therefore$\lambda _{n}=\left ( \frac{n\pi }{a}\right ) ^{2}\qquad n=1,2,3,\cdots$ Eigenfunctions are$$X_{n}\left ( x\right ) =B_{n}\sin \left ( \frac{n\pi }{a}x\right ) \qquad n=1,2,3,\cdots \tag{3}$$ For the $$Y$$ ODE, the solution is$$Y_{n}=C_{n}\cosh \left ( \frac{n\pi }{a}y\right ) +D_{n}\sinh \left ( \frac{n\pi }{a}y\right ) \tag{4}$$ Applying B.C. at $$y=b$$ gives\begin{align*} 0 & =C_{n}\cosh \left ( \frac{n\pi }{a}b\right ) +D_{n}\sinh \left ( \frac{n\pi }{a}b\right ) \\ C_{n} & =-D_{n}\frac{\sinh \left ( \frac{n\pi }{a}b\right ) }{\cosh \left ( \frac{n\pi }{a}b\right ) }\\ & =-D_{n}\tanh \left ( \frac{n\pi }{a}b\right ) \end{align*}

Hence (4) becomes \begin{align*} Y_{n} & =-D_{n}\tanh \left ( \frac{n\pi }{a}b\right ) \cosh \left ( \frac{n\pi }{a}y\right ) +D_{n}\sinh \left ( \frac{n\pi }{a}y\right ) \\ & =D_{n}\left ( \sinh \left ( \frac{n\pi }{a}y\right ) -\tanh \left ( \frac{n\pi }{a}b\right ) \cosh \left ( \frac{n\pi }{a}y\right ) \right ) \end{align*}

Now the complete solution is produced\begin{align*} u_{n}\left ( x,y\right ) & =Y_{n}X_{n}\\ & =D_{n}\left ( \sinh \left ( \frac{n\pi }{a}y\right ) -\tanh \left ( \frac{n\pi }{a}b\right ) \cosh \left ( \frac{n\pi }{a}y\right ) \right ) B_{n}\sin \left ( \frac{n\pi }{a}x\right ) \end{align*}

Let $$D_{n}B_{n}=B_{n}$$ since a constant. (no need to make up a new symbol).$u_{n}\left ( x,y\right ) =B_{n}\left ( \sinh \left ( \frac{n\pi }{a}y\right ) -\tanh \left ( \frac{n\pi }{a}b\right ) \cosh \left ( \frac{n\pi }{a}y\right ) \right ) \sin \left ( \frac{n\pi }{a}x\right )$ Sum of eigenfunctions is the solution, hence$$u\left ( x,y\right ) =\sum _{n=1}^{\infty }B_{n}\left ( \sinh \left ( \frac{n\pi }{a}y\right ) -\tanh \left ( \frac{n\pi }{a}b\right ) \cosh \left ( \frac{n\pi }{a}y\right ) \right ) \sin \left ( \frac{n\pi }{a}x\right ) \tag{5}$$ The nonhomogeneous boundary condition is now resolved.  At $$y=0$$$u\left ( x,0\right ) =f\left ( x\right ) =\delta \left ( x-\xi \right )$ Therefore (5) becomes$\delta \left ( x-\xi \right ) =\sum _{n=1}^{\infty }-B_{n}\tanh \left ( \frac{n\pi }{a}b\right ) \sin \left ( \frac{n\pi }{a}x\right )$ Multiplying both sides by $$\sin \left ( \frac{m\pi }{a}x\right )$$ and integrating gives\begin{align*} \int _{0}^{a}\delta \left ( x-\xi \right ) \sin \left ( \frac{m\pi }{a}x\right ) dx & =-\int _{0}^{a}\sin \left ( \frac{m\pi }{a}x\right ) \sum _{n=1}^{\infty }B_{n}\tanh \left ( \frac{n\pi }{a}b\right ) \sin \left ( \frac{n\pi }{a}x\right ) dx\\ & =-\sum _{n=1}^{\infty }B_{n}\tanh \left ( \frac{n\pi }{a}b\right ) \int _{0}^{a}\sin \left ( \frac{n\pi }{a}x\right ) \sin \left ( \frac{m\pi }{a}x\right ) dx\\ & =-B_{n}\tanh \left ( \frac{m\pi }{a}b\right ) \left ( \frac{a}{2}\right ) \end{align*}

Hence$B_{n}=-\frac{2}{a}\frac{\int _{0}^{a}\delta \left ( x-\xi \right ) \sin \left ( \frac{n\pi }{a}x\right ) dx}{\tanh \left ( \frac{n\pi }{a}b\right ) }$ But $$\int _{0}^{a}\delta \left ( x-\xi \right ) \sin \left ( \frac{m\pi }{L}x\right ) dx=\sin \left ( \frac{m\pi }{L}\xi \right )$$ by the property delta function. Therefore$B_{n}=-\frac{2}{a}\frac{\sin \left ( \frac{n\pi }{a}\xi \right ) }{\tanh \left ( \frac{n\pi }{a}b\right ) }$ This completes the solution. (4) becomes\begin{align*} u\left ( x,y\right ) & =-\frac{2}{a}\sum _{n=1}^{\infty }\frac{\sin \left ( \frac{n\pi }{a}\xi \right ) }{\tanh \left ( \frac{n\pi }{a}b\right ) }\left ( \sinh \left ( \frac{n\pi }{a}y\right ) -\tanh \left ( \frac{n\pi }{a}b\right ) \cosh \left ( \frac{n\pi }{a}y\right ) \right ) \sin \left ( \frac{n\pi }{a}x\right ) \\ & =-\frac{2}{a}\sum _{n=1}^{\infty }\sin \left ( \frac{n\pi }{a}\xi \right ) \sin \left ( \frac{n\pi }{a}x\right ) \left ( \frac{\sinh \left ( \frac{n\pi }{a}y\right ) }{\tanh \left ( \frac{n\pi }{a}b\right ) }-\cosh \left ( \frac{n\pi }{a}y\right ) \right ) \end{align*}

Looking at the solution above, it is composed of functions that are all diﬀerentiable. Hence the solution is inﬁnitely diﬀerentiable inside the rectangle.

Here is a plot of the above solution using $$a=\pi ,b=\frac{1}{2},\xi =1$$.

________________________________________________________________________________________

##### 4.1.1.4 [284] Haberman 2.5.1 (a)

problem number 284

This is problem 2.5.1 part (a) from Richard Haberman applied partial diﬀerential equations, 5th edition

Solve Laplace equation $\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ inside a rectangle $$0 \leq x \leq L, 0 \leq y \leq H$$, with following boundary conditions \begin{align*} \frac{\partial u}{\partial x}(0,y) &= 0 \\ \frac{\partial u}{\partial x}(L,y) &= 0 \\ u(x,0)&=0 \\ u(x,H)&=f(x) \\ \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \underset{n=1}{\overset{\infty }{\sum }}\frac{2 \cos \left (\frac{n \pi x}{L}\right ) \text{csch}\left (\frac{H n \pi }{L}\right ) \left (\int _0^L \cos \left (\frac{n \pi x}{L}\right ) f(x) \, dx\right ) \sinh \left (\frac{n \pi y}{L}\right )}{L}+\frac{y \int _0^L f(x) \, dx}{H L}\right \}\right \}$

Maple

$u \left ( x,y \right ) ={\frac{1}{H\,L} \left ( 4\,\sum _{n=1}^{\infty } \left ( 1/2\,\cos \left ({\frac{n\pi \,x}{L}} \right ) \int _{0}^{L}\!f \left ( x \right ) \cos \left ({\frac{n\pi \,x}{L}} \right ) \,{\rm d}x\sinh \left ({\frac{n\pi \,y}{L}} \right ) \left ( \sinh \left ({\frac{n\pi \,H}{L}} \right ) \right ) ^{-1} \right ) H+\int _{0}^{L}\!f \left ( x \right ) \,{\rm d}xy \right ) }$

________________________________________________________________________________________

##### 4.1.1.5 [285] Haberman 2.5.1 (b)

problem number 285

This is problem 2.5.1 part (b) from Richard Haberman applied partial diﬀerential equations, 5th edition

Solve Laplace equation $\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ inside a rectangle $$0 \leq x \leq L, 0 \leq y \leq H$$, with following boundary conditions \begin{align*} \frac{\partial u}{\partial x}(0,y) &= g(y) \\ \frac{\partial u}{\partial x}(L,y) &= 0 \\ u(x,0)&=0 \\ u(x,H)&=0 \\ \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \underset{n=1}{\overset{\infty }{\sum }}-\frac{2 \cosh \left (\frac{n \pi (L-x)}{H}\right ) \text{csch}\left (\frac{L n \pi }{H}\right ) \left (\int _0^H g(y) \sin \left (\frac{n \pi y}{H}\right ) \, dy\right ) \sin \left (\frac{n \pi y}{H}\right )}{n \pi }\right \}\right \}$

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty } \left ( -2\,{\frac{1}{n\pi }\sin \left ({\frac{n\pi \,y}{H}} \right ) \int _{0}^{H}\!\sin \left ({\frac{n\pi \,y}{H}} \right ) g \left ( y \right ) \,{\rm d}y \left ( \cosh \left ({\frac{\pi \,n \left ( -2\,L+x \right ) }{H}} \right ) -\sinh \left ({\frac{\pi \,n \left ( -2\,L+x \right ) }{H}} \right ) +\cosh \left ({\frac{n\pi \,x}{H}} \right ) +\sinh \left ({\frac{n\pi \,x}{H}} \right ) \right ) \left ( \cosh \left ( 2\,{\frac{n\pi \,L}{H}} \right ) +\sinh \left ( 2\,{\frac{n\pi \,L}{H}} \right ) -1 \right ) ^{-1}} \right )$

________________________________________________________________________________________

##### 4.1.1.6 [286] Haberman 2.5.1 (c)

problem number 286

This is problem 2.5.1 part (c) from Richard Haberman applied partial diﬀerential equations, 5th edition

Solve Laplace equation $\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ inside a rectangle $$0 \leq x \leq L, 0 \leq y \leq H$$, with following boundary conditions \begin{align*} \frac{\partial u}{\partial x}(0,y) &= 0 \\ u(L,y) &= g(y) \\ u(x,0)&=0 \\ u(x,H)&=0 \\ \end{align*}

Mathematica

Failed

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty } \left ( 4\,{\frac{1}{H}\sin \left ({\frac{n\pi \,y}{H}} \right ) \int _{0}^{H}\!\sin \left ({\frac{n\pi \,y}{H}} \right ) g \left ( y \right ) \,{\rm d}y \left ( \cosh \left ({\frac{n\pi \,L}{H}} \right ) +\sinh \left ({\frac{n\pi \,L}{H}} \right ) \right ) \cosh \left ({\frac{n\pi \,x}{H}} \right ) \left ( \cosh \left ( 2\,{\frac{n\pi \,L}{H}} \right ) +\sinh \left ( 2\,{\frac{n\pi \,L}{H}} \right ) +1 \right ) ^{-1}} \right )$

________________________________________________________________________________________

##### 4.1.1.7 [287] Haberman 2.5.1 (d)

problem number 287

This is problem 2.5.1 part (d) from Richard Haberman applied partial diﬀerential equations, 5th edition

Solve Laplace equation $\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$

inside a rectangle $$0 \leq x \leq L, 0 \leq y \leq H$$, with following boundary conditions

\begin{align*} u(0,y) &= g(y) \\ u(L,y) &= 0 \\ \frac{\partial u}{\partial y}u(x,0)&=0 \\ u(x,H)&=0 \\ \end{align*}

Mathematica

Failed

Maple

$u \left ( x,y \right ) =\sum _{n=0}^{\infty } \left ( 2\,{\frac{1}{H}\sin \left ( 1/2\,{\frac{\pi \, \left ( 2\,ny+H+y \right ) }{H}} \right ) \int _{0}^{H}\!\sin \left ( 1/2\,{\frac{\pi \, \left ( 2\,ny+H+y \right ) }{H}} \right ) g \left ( y \right ) \,{\rm d}y \left ( -\cosh \left ({\frac{\pi \, \left ( 1/2+n \right ) \left ( -2\,L+x \right ) }{H}} \right ) -\sinh \left ({\frac{\pi \, \left ( 1/2+n \right ) \left ( -2\,L+x \right ) }{H}} \right ) +\cosh \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{H}} \right ) -\sinh \left ( 1/2\,{\frac{ \left ( 1+2\,n \right ) \pi \,x}{H}} \right ) \right ) \left ( -\cosh \left ({\frac{ \left ( 1+2\,n \right ) \pi \,L}{H}} \right ) +\sinh \left ({\frac{ \left ( 1+2\,n \right ) \pi \,L}{H}} \right ) +1 \right ) ^{-1}} \right )$

________________________________________________________________________________________

##### 4.1.1.8 [288] Haberman 2.5.1 (e)

problem number 288

This is problem 2.5.1 part (e) from Richard Haberman applied partial diﬀerential equations, 5th edition

Solve Laplace equation $\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ inside a rectangle $$0 \leq x \leq L, 0 \leq y \leq H$$, with following boundary conditions \begin{align*} u(0,y) &= 0 \\ u(L,y) &= 0 \\ u(x,0) - \frac{\partial u}{\partial y}u(x,0)&=0 \\ u(x,H)&= f(x) \\ \end{align*}

Mathematica

Failed

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty } \left ( 4\,{\frac{1}{L} \left ( \cosh \left ({\frac{n\pi \,H}{L}} \right ) +\sinh \left ({\frac{n\pi \,H}{L}} \right ) \right ) \left ( \pi \,\cosh \left ({\frac{n\pi \,y}{L}} \right ) n+L\,\sinh \left ({\frac{n\pi \,y}{L}} \right ) \right ) \int _{0}^{L}\!\sin \left ({\frac{\pi \,nx}{L}} \right ) f \left ( x \right ) \,{\rm d}x\sin \left ({\frac{\pi \,nx}{L}} \right ) \left ( \left ( \pi \,n+L \right ) \cosh \left ( 2\,{\frac{n\pi \,H}{L}} \right ) + \left ( \pi \,n+L \right ) \sinh \left ( 2\,{\frac{n\pi \,H}{L}} \right ) +\pi \,n-L \right ) ^{-1}} \right )$

Hand solution

Let $$u\left ( x,y\right ) =X\left ( x\right ) Y\left ( x\right )$$. Substituting this into the PDE $$\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}=0$$ and simplifying gives$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}$ Each side depends on diﬀerent independent variable and they are equal, therefore they must be equal to same constant.$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}=\pm \lambda$ Since the boundary conditions along the $$x$$ direction are the homogeneous ones, $$-\lambda$$ is selected in the above. Two ODE’s (1,2) are obtained as follows$$X^{\prime \prime }+\lambda X=0 \tag{1}$$ With the boundary conditions\begin{align*} X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

And$$Y^{\prime \prime }-\lambda Y=0 \tag{2}$$ With the boundary conditions\begin{align*} Y\left ( 0\right ) & =Y^{\prime }\left ( 0\right ) \\ Y\left ( H\right ) & =f\left ( x\right ) \end{align*}

In all these cases $$\lambda$$ will turn out to be positive. This is shown for this problem only and not be repeated again.

Case $$\lambda <0$$

The solution to (1) us

$X=A\cosh \left ( \sqrt{\left \vert \lambda \right \vert }x\right ) +B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }x\right )$ At $$x=0$$, the above gives $$0=A$$. Hence $$X=B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }x\right )$$. At $$x=L$$ this gives $$X=B\sinh \left ( \sqrt{\left \vert \lambda \right \vert }L\right )$$. But $$\sinh \left ( \sqrt{\left \vert \lambda \right \vert }L\right ) =0$$ only at $$0$$ and $$\sqrt{\left \vert \lambda \right \vert }L\neq 0$$, therefore $$B=0$$ and this leads to trivial solution. Hence $$\lambda <0$$ is not an eigenvalue.

Case $$\lambda =0$$

$X=Ax+B$ Hence at $$x=0$$ this gives $$0=B$$ and the solution becomes $$X=B$$. At $$x=L$$, $$B=0$$. Hence the trivial solution. $$\lambda =0$$ is not an eigenvalue.

Case $$\lambda >0$$

Solution is $X=A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right )$ At $$x=0$$ this gives $$0=A$$ and the solution becomes $$X=B\sin \left ( \sqrt{\lambda }x\right )$$. At $$x=L$$ $0=B\sin \left ( \sqrt{\lambda }L\right )$ For non-trivial solution $$\sin \left ( \sqrt{\lambda }L\right ) =0$$ or $$\sqrt{\lambda }L=n\pi$$ where $$n=1,2,3,\cdots$$, therefore$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots$ Eigenfunctions are$$X_{n}\left ( x\right ) =B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \tag{3}$$ For the $$Y$$ ODE, the solution is\begin{align*} Y_{n} & =C_{n}\cosh \left ( \frac{n\pi }{L}y\right ) +D_{n}\sinh \left ( \frac{n\pi }{L}y\right ) \\ Y_{n}^{\prime } & =C_{n}\frac{n\pi }{L}\sinh \left ( \frac{n\pi }{L}y\right ) +D_{n}\frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}y\right ) \end{align*}

Applying B.C. at $$y=0$$ gives\begin{align*} Y\left ( 0\right ) & =Y^{\prime }\left ( 0\right ) \\ C_{n}\cosh \left ( 0\right ) & =D_{n}\frac{n\pi }{L}\cosh \left ( 0\right ) \\ C_{n} & =D_{n}\frac{n\pi }{L} \end{align*}

The eigenfunctions $$Y_{n}$$ are\begin{align*} Y_{n} & =D_{n}\frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}y\right ) +D_{n}\sinh \left ( \frac{n\pi }{L}y\right ) \\ & =D_{n}\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}y\right ) +\sinh \left ( \frac{n\pi }{L}y\right ) \right ) \end{align*}

Now the complete solution is produced\begin{align*} u_{n}\left ( x,y\right ) & =Y_{n}X_{n}\\ & =D_{n}\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}y\right ) +\sinh \left ( \frac{n\pi }{L}y\right ) \right ) B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Let $$D_{n}B_{n}=B_{n}$$ since a constant. (no need to make up a new symbol).$u_{n}\left ( x,y\right ) =B_{n}\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}y\right ) +\sinh \left ( \frac{n\pi }{L}y\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right )$ Sum of eigenfunctions is the solution, hence$u\left ( x,y\right ) =\sum _{n=1}^{\infty }B_{n}\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}y\right ) +\sinh \left ( \frac{n\pi }{L}y\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right )$ The nonhomogeneous boundary condition is now resolved.  At $$y=H$$$u\left ( x,H\right ) =f\left ( x\right )$ Therefore$f\left ( x\right ) =\sum _{n=1}^{\infty }B_{n}\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}H\right ) +\sinh \left ( \frac{n\pi }{L}H\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right )$ Multiplying both sides by $$\sin \left ( \frac{m\pi }{L}x\right )$$ and integrating gives\begin{align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx & =\int _{0}^{L}\sin \left ( \frac{m\pi }{L}x\right ) \sum _{n=1}^{\infty }B_{n}\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}H\right ) +\sinh \left ( \frac{n\pi }{L}H\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =\sum _{n=1}^{\infty }B_{n}\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}H\right ) +\sinh \left ( \frac{n\pi }{L}H\right ) \right ) \int _{0}^{L}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx\\ & =B_{m}\left ( \frac{m\pi }{L}\cosh \left ( \frac{m\pi }{L}H\right ) +\sinh \left ( \frac{m\pi }{L}H\right ) \right ) \frac{L}{2} \end{align*}

Hence$$B_{n}=\frac{2}{L}\frac{\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx}{\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}H\right ) +\sinh \left ( \frac{n\pi }{L}H\right ) \right ) } \tag{4}$$ This completes the solution. In summary$u\left ( x,y\right ) =\sum _{n=1}^{\infty }B_{n}\left ( \frac{n\pi }{L}\cosh \left ( \frac{n\pi }{L}y\right ) +\sinh \left ( \frac{n\pi }{L}y\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right )$ With $$B_{n}$$ given by (4).

________________________________________________________________________________________

##### 4.1.1.9 [289] Unit triangle B.C.

problem number 289

Taken from Mathematica DSolve help pages.

Solve Laplace equation $\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ inside a rectangle $$0 \leq x \leq 1, 0 \leq y \leq 2$$, with following boundary conditions \begin{align*} u(0,y) &= 0 \\ u(1,y) &= 0 \\ u(x,0) &= \text{UnitTriagle(2 x-1)} \\ u(x,2) &= \text{UnitTriagle(2 x-1)} \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \underset{n=1}{\overset{\infty }{\sum }}\frac{8 \text{csch}(2 n \pi ) \sin \left (\frac{n \pi }{2}\right ) \sin (n \pi x) (\sinh (n \pi (2-y))+\sinh (n \pi y))}{n^2 \pi ^2}\right \}\right \}$

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }8\,{\frac{\sin \left ( 1/2\,\pi \,n \right ){{\rm e}^{2\,\pi \,n}}\sin \left ( \pi \,nx \right ) \left ( -{{\rm e}^{\pi \,n \left ( y-2 \right ) }}+{{\rm e}^{-\pi \,n \left ( y-2 \right ) }}+{{\rm e}^{n\pi \,y}}-{{\rm e}^{-n\pi \,y}} \right ) }{{n}^{2}{\pi }^{2} \left ({{\rm e}^{4\,\pi \,n}}-1 \right ) }}$

________________________________________________________________________________________

##### 4.1.1.10 [290] Top edge at inﬁnity

problem number 290

Solve Laplace equation $u_{xx}+u_{yy} = 0$ Inside a rectangle $$0 \leq x \leq L, 0 \leq y \leq \infty$$, with following boundary conditions \begin{align*} u(0,y) &= A \\ u(L,y) &= 0 \\ u(x,0) &= 0 \end{align*}

Mathematica

Failed

Maple

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

Hand solution

Let $$u=U+v\tag{1}$$ Where $$U$$ satisﬁes $$\nabla ^{2}U=0$$ but with right edge boundary conditions zero, and $$v\left ( x\right )$$ satisﬁes the nonhomogeneous boundary conditions $$v\left ( 0\right ) =A,v\left ( L\right ) =0$$. This implies $v\left ( x\right ) =A\left ( 1-\frac{x}{L}\right )$ Hence $$u=U+A\left ( 1-\frac{x}{L}\right )$$. Substituting this back in $$\nabla ^{2}u=0$$ gives$\nabla ^{2}U=0$ But with boundary condition on right edge being zero now. Let $$U=X\left ( x\right ) Y\left ( x\right )$$. Substituting this in the above gives$\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=0$ We want the eigenvalue problem to be in the $$X$$ direction. Hence \begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

This has eigenvalues $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,\cdots$$ with eigenfunctions $$X_{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right )$$. The $$Y$$ ode is\begin{align*} Y_{n}^{\prime \prime }-\lambda _{n}Y_{n} & =0\\ Y_{n}\left ( 0\right ) & =0 \end{align*}

Since $$\lambda _{n}>0$$ then the solution is $$Y_{n}\left ( y\right ) =c_{1_{n}}e^{\sqrt{\lambda _{n}}y}+c_{2_{n}}e^{-\sqrt{\lambda _{n}}y}$$. Since $$Y_{n}\left ( y\right )$$ is bounded, then $$c_{1_{n}}=0$$ and the $$Y_{n}\left ( y\right ) =c_{2_{n}}e^{-\sqrt{\lambda _{n}}y}$$. Hence\begin{align*} U\left ( x,y\right ) & =\sum _{n=1}^{\infty }X_{n}\left ( x\right ) Y_{n}\left ( y\right ) \\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) e^{-\sqrt{\lambda _{n}}y}\\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{n\pi }{L}y} \end{align*}

Using the above in (1) gives the solution$$u\left ( x,y\right ) =A\left ( 1-\frac{x}{L}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{n\pi }{L}y}\tag{2}$$ At $$y=0$$ the above gives\begin{align*} 0 & =A\left ( 1-\frac{x}{L}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \\ A\left ( \frac{x}{L}-1\right ) & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Therefore $$B_{n}$$ are the Fourier sine coeﬃcients of $$\frac{A}{L}x$$\begin{align*} B_{n} & =\frac{2}{L}\int _{0}^{L}A\left ( \frac{x}{L}-1\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =\frac{2A}{L}\int _{0}^{L}\left ( \frac{x}{L}-1\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =-\frac{2A}{L}\frac{L}{n\pi }\\ & =-\frac{2A}{n\pi } \end{align*}

Hence the solution (2) becomes$u\left ( x,y\right ) =A\left ( 1-\frac{x}{L}\right ) -2\frac{A}{\pi }\sum _{n=1}^{\infty }\frac{1}{n}\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{n\pi }{L}y}$

________________________________________________________________________________________

##### 4.1.1.11 [291] Top edge at inﬁnity

problem number 291

Solve Laplace equation $u_{xx} + u_{yy} = 0$

Inside a rectangle $$0 \leq x \leq L, 0 \leq y \leq \infty$$, with following boundary conditions

\begin{align*} u(0,y) &= 0 \\ u(L,y) &= A \\ u(x,0) &= 0 \end{align*}

Mathematica

Failed

Maple

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

Hand solution

Let $$u=U+v\tag{1}$$ Where $$U$$ satisﬁes $$\nabla ^{2}U=0$$ but with right edge boundary conditions zero, and $$v\left ( x\right )$$ satisﬁes the nonhomogeneous boundary conditions $$v\left ( 0\right ) =0A,v\left ( L\right ) =A$$. This implies $v\left ( x\right ) =A\frac{x}{L}$ Hence $$u=U+\frac{A}{L}x$$. Substituting this back in $$\nabla ^{2}u=0$$ gives$\nabla ^{2}U=0$ But with boundary condition on right edge being zero now. Let $$U=X\left ( x\right ) Y\left ( x\right )$$. Substituting this in the above gives$\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=0$ We want the eigenvalue problem to be in the $$X$$ direction. Hence \begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

This has eigenvalues $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,\cdots$$ with eigenfunctions $$X_{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right )$$. The $$Y$$ ode is\begin{align*} Y_{n}^{\prime \prime }-\lambda _{n}Y_{n} & =0\\ Y_{n}\left ( 0\right ) & =0 \end{align*}

Since $$\lambda _{n}>0$$ then the solution is $$Y_{n}\left ( y\right ) =c_{1_{n}}e^{\sqrt{\lambda _{n}}y}+c_{2_{n}}e^{-\sqrt{\lambda _{n}}y}$$. Since $$Y_{n}\left ( y\right )$$ is bounded, then $$c_{1_{n}}=0$$ and the $$Y_{n}\left ( y\right ) =c_{2_{n}}e^{-\sqrt{\lambda _{n}}y}$$. Hence\begin{align*} U\left ( x,y\right ) & =\sum _{n=1}^{\infty }X_{n}\left ( x\right ) Y_{n}\left ( y\right ) \\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) e^{-\sqrt{\lambda _{n}}y}\\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{n\pi }{L}y} \end{align*}

Using the above in (1) gives the solution$$u\left ( x,y\right ) =\frac{A}{L}x+\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{n\pi }{L}y}\tag{2}$$ At $$y=0$$ the above gives\begin{align*} 0 & =\frac{A}{L}x+\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \\ -\frac{A}{L}x & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Therefore $$B_{n}$$ are the Fourier sine coeﬃcients of $$-\frac{A}{L}x$$\begin{align*} B_{n} & =-\frac{2}{L}\int _{0}^{L}\frac{A}{L}x\sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =-\frac{2A}{L^{2}}\int _{0}^{L}x\sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =-\frac{2A}{L^{2}}\frac{\left ( -1\right ) ^{n+1}L^{2}}{n\pi }\\ & =\frac{2A}{n\pi }\left ( -1\right ) ^{n} \end{align*}

Hence the solution (2) becomes$u\left ( x,y\right ) =\frac{A}{L}x+\frac{2A}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{n\pi }{L}y}$

________________________________________________________________________________________

##### 4.1.1.12 [292] Right edge at inﬁnity

problem number 292

Solve Laplace equation $u_{xx}+u_{yy}= 0$ Inside a rectangle $$0 \leq y \leq L, 0 \leq x \leq \infty$$, with following boundary conditions \begin{align*} u(0,y) &= 0 \\ u(x,0) &= A \\ u(x,L) &= 0 \end{align*}

Mathematica

Failed

Maple

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

Hand solution

Let $$u\left ( x,y\right ) =U\left ( x,y\right ) +v\left ( y\right ) \tag{1}$$ Where $$U$$ satisﬁes $$\nabla ^{2}U=0$$ but with bottom edge boundary conditions zero, and $$v\left ( y\right )$$ satisﬁes the nonhomogeneous boundary conditions $$v\left ( 0\right ) =A,v\left ( L\right ) =0$$. This implies $v\left ( y\right ) =A\left ( 1-\frac{y}{L}\right )$ Substituting (1) back in $$\nabla ^{2}u=0$$ results in$\nabla ^{2}U=0$ But with boundary condition on bottom edge as $$U=0$$. Now we can use separation of variables. Let $$U=X\left ( x\right ) Y\left ( x\right )$$. Substituting this in the above gives$\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=0$ We want the eigenvalue problem to be in the $$Y$$ direction. Hence$\frac{Y^{\prime \prime }}{Y}=-\frac{X^{\prime \prime }}{X}=-\lambda$ Therefore the eigenvalue problem is\begin{align*} Y^{\prime \prime }+\lambda Y & =0\\ Y\left ( 0\right ) & =0\\ Y\left ( L\right ) & =0 \end{align*}

This has eigenvalues $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,\cdots$$ with eigenfunctions $$Y_{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}y\right )$$. The $$X$$ ode is\begin{align*} X_{n}^{\prime \prime }-\lambda _{n}X_{n} & =0\\ X_{n}\left ( 0\right ) & =0 \end{align*}

Since $$\lambda _{n}>0$$ then the solution is $$X_{n}\left ( y\right ) =c_{1_{n}}e^{\sqrt{\lambda _{n}}x}+c_{2_{n}}e^{-\sqrt{\lambda _{n}}x}$$. Since $$X_{n}\left ( x\right )$$ is bounded, then $$c_{1_{n}}=0$$ and the $$X_{n}\left ( x\right ) =c_{2_{n}}e^{-\sqrt{\lambda _{n}}x}$$. Hence by superposition the solution is\begin{align*} U\left ( x,y\right ) & =\sum _{n=1}^{\infty }X_{n}\left ( x\right ) Y_{n}\left ( y\right ) \\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt{\lambda _{n}}y\right ) e^{-\sqrt{\lambda _{n}}x}\\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}y\right ) e^{-\frac{n\pi }{L}x} \end{align*}

Substituting the above in (1) gives$$u\left ( x,y\right ) =A\left ( 1-\frac{y}{L}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}y\right ) e^{-\frac{n\pi }{L}x}\tag{2}$$ At $$x=0$$ the above gives\begin{align*} 0 & =A\left ( 1-\frac{y}{L}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}y\right ) \\ A\left ( \frac{y}{L}-1\right ) & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac{n\pi }{L}y\right ) \end{align*}

Therefore $$B_{n}$$ are the Fourier sine coeﬃcients of $$A\left ( \frac{y}{L}-1\right )$$\begin{align*} B_{n} & =\frac{2}{L}\int _{0}^{L}A\left ( \frac{y}{L}-1\right ) \sin \left ( \frac{n\pi }{L}y\right ) dy\\ & =\frac{2A}{L}\int _{0}^{L}\left ( \frac{y}{L}-1\right ) \sin \left ( \frac{n\pi }{L}y\right ) dy\\ & =-\frac{2A}{L}\frac{L}{n\pi }\\ & =-\frac{2A}{n\pi } \end{align*}

Hence the solution (2) becomes$u\left ( x,y\right ) =A\left ( 1-\frac{y}{L}\right ) -\frac{2A}{\pi }\sum _{n=1}^{\infty }\frac{1}{n}\sin \left ( \frac{n\pi }{L}y\right ) e^{-\frac{n\pi }{L}x}$

________________________________________________________________________________________

##### 4.1.1.13 [293] Right edge at inﬁnity

problem number 293

Solve Laplace equation $u_{xx}+u_{yy} = 0$ Inside a rectangle $$0 \leq y \leq L, 0 \leq x \leq \infty$$, with following boundary conditions \begin{align*} u(0,y) &= 0 \\ u(x,L) &= e^{-x} \\ u(x,0) &= 0 \end{align*}

Mathematica

Failed

Maple

$u \left ( x,y \right ) ={\frac{1}{L} \left ( \sum _{n=1}^{\infty }-{\frac{1}{\pi \,n \left ( \pi \,n+L \right ) }\sin \left ({\frac{n\pi \,y}{L}} \right ) \left ({\it \_C5} \left ( n \right ) \left ( -{\pi }^{2}{n}^{2}+{L}^{2} \right ){{\rm e}^{-{\frac{\pi \,nx}{L}}}}+ \left ( \left ( -2\,\pi \,n-4\,L \right ) \left ( -1 \right ) ^{n}+{\it \_C5} \left ( n \right ) \left ( \pi \,n+L \right ) ^{2} \right ){{\rm e}^{{\frac{\pi \,nx}{L}}}}-2\, \left ( - \left ( -1 \right ) ^{n}{{\rm e}^{-x}}+{\it \_C5} \left ( n \right ) \left ( \pi \,n+L \right ) \right ) L \right ) }L+y{{\rm e}^{-x}} \right ) }$

Hand solution

Let $$u=X\left ( x\right ) Y\left ( x\right )$$. Substituting this in $$\nabla ^{2}u=0$$$\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=0$ We want the eigenvalue problem to be in the $$X$$ direction. Hence$\frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}=-\lambda$ Therefore the eigenvalue problem is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ \left \vert X\left ( x\right ) \right \vert & <\infty \end{align*}

case $$\lambda <0$$

Solution is $$X\left ( x\right ) =c_{1}\cosh \left ( \sqrt{-\lambda }x\right ) +c_{2}\sinh \left ( \sqrt{-\lambda }x\right )$$. Since $$X\left ( 0\right ) =0$$ then $$c_{1}=0$$. Solution becomes $$X\left ( x\right ) =c_{2}\sinh \left ( \sqrt{-\lambda }x\right )$$. Since $$\sinh$$ is not bounded on $$x>0$$ as $$x\rightarrow \infty$$ then $$c_{2}=0$$. Therefore $$\lambda <0$$ is not eigenvalue.

case $$\lambda =0$$

Solution is $$X\left ( x\right ) =c_{1}x+c_{2}$$. At $$x=0$$ this gives $$c_{2}=0$$.  Hence solution is $$X\left ( x\right ) =c_{1}x$$. This is bounded as $$x\rightarrow \infty$$ only when $$c_{1}=0$$. Therefore $$\lambda =0$$ is not eigenvalue.

case $$\lambda >0$$

Let $$\lambda =\alpha ^{2},\alpha >0$$. Then solution is $$X\left ( x\right ) =c_{1}\cos \left ( \alpha x\right ) +c_{2}\sin \left ( \alpha x\right )$$. At $$x=0$$ this results in $$0=c_{1}$$. Hence the eigenvalues are $$\lambda =\alpha ^{2}$$ for all real positive real numbers and eigenfunctions are $X_{\alpha }\left ( x\right ) =\sin \left ( \alpha x\right )$ For the $$Y$$ ode, \begin{align*} Y^{\prime \prime }-\alpha ^{2}Y & =0\\ Y\left ( 0\right ) & =0 \end{align*}

The solution is $$Y_{\alpha }\left ( y\right ) =c_{1}e^{\alpha y}+c_{2}e^{-\alpha y}$$. Since $$Y\left ( 0\right ) =0$$ then $$c_{2}=-c_{1}$$ and the solution becomes $$Y_{\alpha }\left ( y\right ) =c_{1}\left ( e^{\alpha y}-e^{-\alpha y}\right ) =c_{1}\sinh \left ( \alpha y\right )$$. Hence the solution is generalized linear combination of $$Y\left ( y\right ) X\left ( x\right )$$ given by Fourier integral (since eigenvalues are continuous now and not discrete)\begin{align} u\left ( x,y\right ) & =\int _{0}^{\infty }A\left ( \alpha \right ) Y_{\alpha }\left ( y\right ) X_{\alpha }\left ( x\right ) d\alpha \nonumber \\ & =\int _{0}^{\infty }A\left ( \alpha \right ) \sinh \left ( \alpha y\right ) \sin \left ( \alpha x\right ) d\alpha \tag{1} \end{align}

When $$y=L$$, then above becomes$e^{-x}=\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \sinh \left ( \alpha L\right ) \right ) \sin \left ( \alpha x\right ) d\alpha$ Hence the coeﬃcient$$\ A\left ( \alpha \right ) \sinh \left ( \alpha L\right )$$ is given by\begin{align*} A\left ( \alpha \right ) \sinh \left ( \alpha L\right ) & =\frac{2}{\pi }\int _{0}^{\infty }e^{-x}\sin \left ( \alpha x\right ) dx\\ & =\frac{2}{\pi }\frac{\alpha }{1+\alpha ^{2}} \end{align*}

Therefore $$A\left ( \alpha \right ) =\frac{2}{\pi \sinh \left ( \alpha L\right ) }\frac{\alpha }{1+\alpha }$$. The solution (1) becomes$u\left ( x,y\right ) =\frac{2}{\pi }\int _{0}^{\infty }\frac{\alpha \sinh \left ( \alpha y\right ) \sin \left ( \alpha x\right ) }{\left ( 1+\alpha ^{2}\right ) \sinh \left ( \alpha L\right ) }d\alpha$

________________________________________________________________________________________

##### 4.1.1.14 [294] Right edge at inﬁnity

problem number 294

Second midterm exam problem, Math 4567, UMN. Spring 2019.

Solve Laplace equation $u_{xx}+u_{yy} = 0$ Inside a rectangle $$0 \leq y \leq 1, 0 \leq x \leq \infty$$, with following boundary conditions \begin{align*} u(0,y) &= 0 \\ u(x,1) &= f(x) \\ u(x,0) &= 0 \end{align*}

Mathematica

Failed

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }{\frac{\sin \left ( n\pi \,y \right ) \left ({\it \_C5} \left ( n \right ){{\rm e}^{-\pi \,nx}}\pi \,n-{{\rm e}^{\pi \,nx}} \left ( \pi \,n{\it \_C5} \left ( n \right ) -2\,f \left ( 0 \right ) \left ( -1 \right ) ^{n} \right ) \right ) }{\pi \,n}}+\int _{0}^{x}\!\sum _{n=1}^{\infty }{\frac{\sin \left ( n\pi \,y \right ) \left ( 2\, \left ({\frac{{\rm d}^{2}}{{\rm d}{\tau }^{2}}}f \left ( \tau \right ) \right ) \left ( -{{\rm e}^{\pi \, \left ( -x+\tau \right ) }} \right ) ^{-n}+\pi \,n{\it \_C5} \left ( n \right ) \left ({{\rm e}^{n\pi \, \left ( -x+\tau \right ) }}-{{\rm e}^{n\pi \, \left ( x-\tau \right ) }} \right ) \right ) }{\pi \,n}}\,{\rm d}\tau +yf \left ( x \right )$

________________________________________________________________________________________

##### 4.1.1.15 [295] Laplace PDE in 2D Cartessian with boundary condition as Dirac function

problem number 295

Solve Laplace equation for $$u(x,y)$$ $\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$ With boundary condition \begin{align*} u(x,0) &= \delta (x) \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{y}{\pi \left (x^2+y^2\right )} & y\geq 0 \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \}$

Maple

$u \left ( x,y \right ) =1/2\,{\frac{\int _{-\infty }^{\infty }\!{{\rm e}^{s \left ( ix-y \right ) }}\,{\rm d}s}{\pi }}$

________________________________________________________________________________________

##### 4.1.1.16 [296] One side homogeneous

problem number 296

Solve Laplace equation for $$u(x,y)$$ $u_{xx}+u_{yy} = 0$ With boundary condition \begin{align*} u(0,y)&=0 \\ u(\pi ,y) &= \sinh (\pi ) \cos (y) \\ u(x,0) &= \sin (x) \\ u(x,\pi ) &= -\sinh (x) \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \underset{K[1]=1}{\overset{\infty }{\sum }}\text{csch}(\pi K[1]) \left (\delta (K[1]-1) \sin (x K[1]) \sinh ((\pi -y) K[1])+\frac{2 K[1] \sinh (\pi ) \left (\left (1+(-1)^{K[1]}\right ) \left (K[1]^2+1\right ) \sin (y K[1]) \sinh (x K[1])+(-1)^{K[1]} \left (K[1]^2-1\right ) \sin (x K[1]) \sinh (y K[1])\right )}{\pi \left (K[1]^4-1\right )}\right )\right \}\right \}$

Maple

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

________________________________________________________________________________________

##### 4.1.1.17 [297] In right half plane

problem number 297

PDE example 18 from Maple help page

see march_20_2019_11_pm.tex for start of solution. Not completed yet

Solve Laplace equation $u_{xx} + u_{yy} = 0$ With boundary conditions \begin{align*} u(0,y) &= \frac{\sin y}{y} \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \frac{-2 y \sinh (x) \sin (y)-2 x \cosh (x) \cos (y)+x}{x^2+y^2}\right \}\right \}$

Maple

$u \left ( x,y \right ) ={\frac{\sin \left ( ix-y \right ) +{\it \_F2} \left ( y-ix \right ) \left ( y-ix \right ) + \left ( ix-y \right ){\it \_F2} \left ( y+ix \right ) }{ix-y}}$

________________________________________________________________________________________

##### 4.1.1.18 [298] Right edge at inﬁnity

problem number 298

Solve Laplace equation $u_{xx} + u_{yy} =0$ With boundary conditions \begin{align*} u(0,y) &= \sin y \\ u(x,0) &= 0 \\ u(x,a) &= 0 \\ u(\infty ,y) &= 0 \end{align*}

Mathematica

Failed

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }{\frac{1}{\pi \,n+a} \left ( -2\,\pi \,{{\rm e}^{{\frac{\pi \,nx}{a}}}}\cases{-1&a=\pi \,n\cr -{\frac{\sin \left ( a \right ) \left ( -1 \right ) ^{n}}{-\pi \,n+a}}&otherwise\cr }n+{\it \_C5} \left ( n \right ) \left ( \pi \,n+a \right ) \left ({{\rm e}^{-{\frac{\pi \,nx}{a}}}}-{{\rm e}^{{\frac{\pi \,nx}{a}}}} \right ) \right ) \sin \left ({\frac{n\pi \,y}{a}} \right ) }$

________________________________________________________________________________________

##### 4.1.1.19 [299] Dirichlet problem Upper half

problem number 299

Taken from Mathematica DSolve help pages

Solve for $$u( x,y)$$ \begin{align*} u_{xx}+ y_{yy} =0 \end{align*}

Boundary conditions $$u(x,0)=1$$ for $$-\frac{1}{2}\leq x \leq \frac{1}{2}$$ and $$x=0$$ otherwise. This is called UnitBox in Mathematica.

Mathematica

$\left \{\left \{u(x,y)\to \frac{\tan ^{-1}\left (\frac{\frac{1}{2}-x}{y}\right )+\tan ^{-1}\left (\frac{x+\frac{1}{2}}{y}\right )}{\pi }\right \}\right \}$

Maple

$u \left ( x,y \right ) =-{\it \_F2} \left ( ix-y \right ) +\cases{0&x+iy<-1/2\cr 1&x+iy\leq 1/2\cr 0&1/2<x+iy\cr }+{\it \_F2} \left ( y+ix \right )$

________________________________________________________________________________________

##### 4.1.1.20 [300] Right half-plane

problem number 300

Taken from Mathematica DSolve help pages

Solve for $$u( x,y)$$ \begin{align*} u_{xx}+ y_{yy} =0 \end{align*}

Boundary conditions $$u(0,y)=\sinc (y)$$.

Mathematica

$\left \{\left \{u(x,y)\to \frac{-2 y \sinh (x) \sin (y)-2 x \cosh (x) \cos (y)+x}{x^2+y^2}\right \}\right \}$

Maple

$u \left ( x,y \right ) ={\frac{\sin \left ( ix-y \right ) +{\it \_F2} \left ( y-ix \right ) \left ( y-ix \right ) + \left ( ix-y \right ){\it \_F2} \left ( y+ix \right ) }{ix-y}}$

________________________________________________________________________________________

problem number 301

Taken from Mathematica DSolve help pages

Solve for $$u( x,y)$$ \begin{align*} u_{xx}+ y_{yy} =0 \end{align*}

Boundary conditions \begin{align*} u(x,0) &= - \frac{1}{(x-2)^2+3}\\ u(0,y) &= \frac{1}{(y-3)^2+1}\\ \end{align*}

Mathematica

$\left \{\left \{u(x,y)\to \frac{2 \left (\frac{3 \left (y \left (x \left (x^4+2 x^2 \left (y^2-10\right )+y^4+20 y^2-260\right ) \log \left (x^2+y^2\right )+x \left (x^4 \left (6 \tan ^{-1}(3)-\log (10)\right )+2 x^2 \left (y^2 \left (6 \tan ^{-1}(3)-\log (10)\right )+10 \left (\log (10)+6 \tan ^{-1}(3)\right )\right )+y^4 \left (6 \tan ^{-1}(3)-\log (10)\right )-20 y^2 \left (\log (10)+6 \tan ^{-1}(3)\right )+260 \log (10)+360 \tan ^{-1}(3)\right )+3 \pi (x+1) \left (x^4-4 x^3+2 x^2 \left (y^2+12\right )-4 x \left (y^2+10\right )+y^4-16 y^2+100\right )\right )-\left (x^6+x^4 \left (y^2+6\right )-x^2 \left (y^4-92 y^2+60\right )-y^6+6 y^4+60 y^2-1000\right ) \tan ^{-1}\left (\frac{y}{x}\right )\right )}{\left (x^2-2 x+y^2-6 y+10\right ) \left (x^2+2 x+y^2-6 y+10\right ) \left (x^2-2 x+y^2+6 y+10\right ) \left (x^2+2 x+y^2+6 y+10\right )}-\frac{x \left (3 y \left (x^4+2 x^2 \left (y^2+7\right )+y^4-14 y^2-63\right ) \log \left (x^2+y^2\right )+y \left (x^4 \left (4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )-3 \log (7)\right )+2 x^2 \left (y^2 \left (4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )-3 \log (7)\right )-7 \left (\log (343)+4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )\right )\right )+y^4 \left (4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )-3 \log (7)\right )+14 y^2 \left (\log (343)+4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )\right )+189 \log (7)-140 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )\right )+2 \pi \left (x^4 \left (\sqrt{3} y+3\right )+2 x^2 \left (\sqrt{3} y^3-3 y^2-7 \sqrt{3} y-3\right )+\sqrt{3} y^5-9 y^4+14 \sqrt{3} y^3-6 y^2-35 \sqrt{3} y+147\right )\right )+3 \left (x^6+x^4 \left (y^2+5\right )-x^2 \left (y^4+46 y^2-35\right )-y^6+5 y^4-35 y^2+343\right ) \tan ^{-1}\left (\frac{x}{y}\right )}{\left (x^4-8 x^3+2 x^2 \left (y^2+15\right )-8 x \left (y^2+7\right )+y^4+2 y^2+49\right ) \left (x^4+8 x^3+2 x^2 \left (y^2+15\right )+8 x \left (y^2+7\right )+y^4+2 y^2+49\right )}\right )}{3 \pi }\right \}\right \}$

Maple

sol=()

________________________________________________________________________________________

##### 4.1.1.22 [302] Neumann problem upper half-plane

problem number 302

Taken from Mathematica DSolve help pages

Solve for $$u( x,y)$$ \begin{align*} \nabla ^2 u(x,y)=0 \end{align*}

Boundary conditions $$\frac{\partial u}{\partial y}(x,0)=\text{UnitBox[x]}$$ where $$\text{UnitBox[x]}$$ is $$1$$ for $$-\frac{1}{2}\leq x \leq \frac{1}{2}$$ and $$0$$ otherwise. This is called UnitBox in Mathematica.

Mathematica

$\left \{\left \{u(x,y)\to \frac{-2 x \log \left (4 x^2-4 x+4 y^2+1\right )+\log \left (4 x^2-4 x+4 y^2+1\right )+2 x \log \left (4 x^2+4 x+4 y^2+1\right )+\log \left (4 x^2+4 x+4 y^2+1\right )-4 y \tan ^{-1}\left (\frac{x-\frac{1}{2}}{y}\right )+4 y \tan ^{-1}\left (\frac{x+\frac{1}{2}}{y}\right )-4-2 \log (4)}{4 \pi }\right \}\right \}$

Maple

$u \left ( x,y \right ) ={\frac{-i/2}{\pi } \left ( \int _{-\infty }^{\infty }\!{\frac{{{\rm e}^{1/2\,s \left ( 2\,ix-i-2\,y \right ) }}}{{s}^{2}}}\,{\rm d}s-\int _{-\infty }^{\infty }\!{\frac{{{\rm e}^{1/2\,s \left ( 2\,ix+i-2\,y \right ) }}}{{s}^{2}}}\,{\rm d}s \right ) }$ used convert(sol,Int).

________________________________________________________________________________________

##### 4.1.1.23 [303] Dirichlet problem in a rectangle

problem number 303

Taken from Mathematica DSolve help pages

Solve for $$u( x,y)$$ \begin{align*} \frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}=0 \end{align*}

Boundary conditions $$u(x, 0) = x^2 (1 - x), u(x, 2) = 0, u(0, y) = 0, u(1, y) = 0$$.

Mathematica

$\left \{\left \{u(x,y)\to \underset{n=1}{\overset{\infty }{\sum }}-\frac{4 \left (1+2 (-1)^n\right ) \text{csch}(2 n \pi ) \sin (n \pi x) \sinh (n \pi (2-y))}{n^3 \pi ^3}\right \}\right \}$

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }8\,{\frac{ \left ( -{{\rm e}^{-\pi \,n \left ( y-4 \right ) }}+{{\rm e}^{n\pi \,y}} \right ) \left ( \left ( -1 \right ) ^{n}+1/2 \right ) \sin \left ( \pi \,nx \right ) }{{n}^{3}{\pi }^{3} \left ({{\rm e}^{4\,\pi \,n}}-1 \right ) }}$

________________________________________________________________________________________

##### 4.1.1.24 [304] Strip in upper half

problem number 304

Solve for $$u( x,y)$$ \begin{align*} \frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}=0 \end{align*}

Boundary conditions \begin{align*} u(x, 0) &= 0 \\ u(x, b) &= h(x) \end{align*}

Mathematica

Failed

Maple

$u \left ( x,y \right ) =1/2\,{\frac{1}{\pi \,b} \left ( \int _{-\infty }^{\infty }\!-{\frac{\int _{-\infty }^{\infty }\!h \left ( x \right ){{\rm e}^{-ixs}}\,{\rm d}x{{\rm e}^{s \left ( ix+b-y \right ) }}}{{{\rm e}^{2\,sb}}-1}}\,{\rm d}sb+\int _{-\infty }^{\infty }\!-{\frac{\int _{-\infty }^{\infty }\!h \left ( x \right ){{\rm e}^{-ixs}}\,{\rm d}x{{\rm e}^{s \left ( ix+2\,b \right ) }}}{{{\rm e}^{2\,sb}}-1}}\,{\rm d}sy-\int _{-\infty }^{\infty }\!-{\frac{\int _{-\infty }^{\infty }\!h \left ( x \right ){{\rm e}^{-ixs}}\,{\rm d}x{{\rm e}^{s \left ( ix+b+y \right ) }}}{{{\rm e}^{2\,sb}}-1}}\,{\rm d}sb-y \left ( -2\,\pi \,h \left ( x \right ) +\int _{-\infty }^{\infty }\!-{\frac{\int _{-\infty }^{\infty }\!h \left ( x \right ){{\rm e}^{-ixs}}\,{\rm d}x{{\rm e}^{isx}}}{{{\rm e}^{2\,sb}}-1}}\,{\rm d}s \right ) \right ) }$

________________________________________________________________________________________

##### 4.1.1.25 [305] in Rectangle, right edge at inﬁnity

problem number 305

Solve for $$u( x,y)$$ \begin{align*} \frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}=0 \end{align*}

Boundary conditions \begin{align*} u(x, 0) &= 0 \\ u(x, a) &= 0\\ u(0,y) &= \sin (y) \\ u(\infty ,y) &=0 \end{align*}

Mathematica

Failed

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }{\frac{1}{\pi \,n+a} \left ( -2\,\pi \,{{\rm e}^{{\frac{\pi \,nx}{a}}}}\cases{-1&a=\pi \,n\cr -{\frac{ \left ( -1 \right ) ^{n}\sin \left ( a \right ) }{-\pi \,n+a}}&otherwise\cr }n+{\it \_C5} \left ( n \right ) \left ( \pi \,n+a \right ) \left ({{\rm e}^{-{\frac{\pi \,nx}{a}}}}-{{\rm e}^{{\frac{\pi \,nx}{a}}}} \right ) \right ) \sin \left ({\frac{n\pi \,y}{a}} \right ) }$

________________________________________________________________________________________