11 Laplace PDE in Cartesian coordinates

 11.1 Laplace PDE inside rectangle (Haberman 2.5.1 (a))
 11.2 Laplace PDE inside rectangle (Haberman 2.5.1 (b))
 11.3 Laplace PDE inside rectangle (Haberman 2.5.1 (c))
 11.4 Laplace PDE inside rectangle (Haberman 2.5.1 (d))
 11.5 Laplace PDE inside rectangle (Haberman 2.5.1 (e))
 11.6 Laplace PDE inside rectangle, top/bottom edges non-zero
 11.7 Laplace PDE inside rectangle, top edge at infinity
 11.8 Laplace PDE in 2D Cartessian with boundary condition as Dirac function
 11.9 Laplace PDE in rectangle, one side homogeneous and 3 sides are not
 11.10 Laplace PDE example 18 from Maple help page
 11.11 Laplace PDE on rectangle with one edge at infinity
 11.12 Laplace PDE inside a disk, periodic boundary conditions
 11.13 Dirichlet problem for the Laplace equation in upper half plan
 11.14 Dirichlet problem for the Laplace equation in right half-plane:
 11.15 Dirichlet problem for the Laplace equation in the first quadrant
 11.16 Neumann problem for the Laplace equation in the upper half-plane
 11.17 Dirichlet problem for the Laplace equation in a rectangle
 11.18 Cartessian coordinates with boundary conditions on two sides only
 11.19 in Rectangle, right edge at infinity
 11.20 Laplace PDE inside quarter disk, Neumann BC at edge

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11.1 Laplace PDE inside rectangle (Haberman 2.5.1 (a))

problem number 83

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

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

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11.2 Laplace PDE inside rectangle (Haberman 2.5.1 (b))

problem number 84

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

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

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11.3 Laplace PDE inside rectangle (Haberman 2.5.1 (c))

problem number 85

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

\[ \text{DSolve}\left [\left \{u^{(0,2)}(x,y)+u^{(2,0)}(x,y)=0,\left \{u^{(1,0)}(0,y)=0,u(L,y)=g(y),u(x,0)=0,u(x,H)=0\right \}\right \},u(x,y),\{x,y\},\text{Assumptions}\to \{0\leq x\leq L\land 0\leq y\leq H\}\right ] \]

Maple

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

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11.4 Laplace PDE inside rectangle (Haberman 2.5.1 (d))

problem number 86

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

\[ \text{DSolve}\left [\left \{u^{(0,2)}(x,y)+u^{(2,0)}(x,y)=0,\left \{u(0,y)=0,u(L,y)=0,u^{(0,1)}(x,0)=0,u(x,H)=0\right \}\right \},u(x,y),\{x,y\},\text{Assumptions}\to \{0\leq x\leq L\land 0\leq y\leq H\}\right ] \]

Maple

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

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11.5 Laplace PDE inside rectangle (Haberman 2.5.1 (e))

problem number 87

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

\[ \text{DSolve}\left [\left \{u^{(0,2)}(x,y)+u^{(2,0)}(x,y)=0,\left \{u(0,y)=0,u(L,y)=0,u(x,0)-u^{(0,1)}(x,0)=0,u(x,H)=f(x)\right \}\right \},u(x,y),\{x,y\},\text{Assumptions}\to \{0\leq x\leq L\land 0\leq y\leq H\}\right ] \]

Maple

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

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\begin{equation} X^{\prime \prime }+\lambda X=0\tag{1} \end{equation} With the boundary conditions\begin{align*} X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

And\begin{equation} Y^{\prime \prime }-\lambda Y=0\tag{2} \end{equation} 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

\[ X=A\cosh \left ( \sqrt{\lambda }x\right ) +B\sinh \left ( \sqrt{\lambda }x\right ) \] At \(x=0\), the above gives \(0=A\). Hence \(X=B\sinh \left ( \sqrt{\lambda }x\right ) \). At \(x=L\) this gives \(X=B\sinh \left ( \sqrt{\lambda }L\right ) \). But \(\sinh \left ( \sqrt{\lambda }L\right ) =0\) only at \(0\) and \(\sqrt{\lambda }L\neq 0\), 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\begin{equation} X_{n}\left ( x\right ) =B_{n}\sin \left ( \frac{n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \tag{3} \end{equation} 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\begin{equation} 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} \end{equation} 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).

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11.6 Laplace PDE inside rectangle, top/bottom edges non-zero

problem number 88

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

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

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11.7 Laplace PDE inside rectangle, top edge at infinity

problem number 89

Added December 20, 2018.

Example 21, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

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

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

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

Mathematica

\[ \text{DSolve}\left [\left \{u^{(0,2)}(x,y)+u^{(2,0)}(x,y)=0,\{A=u(0,y),u(L,y)=0,u(x,0)=0\}\right \},u(x,y),\{x,y\},\text{Assumptions}\to \{x>0,y>0,L>0\}\right ] \] May be there is workaround.

Maple

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

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11.8 Laplace PDE in 2D Cartessian with boundary condition as Dirac function

problem number 90

Added December 20, 2018

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

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

\[ u \left ( x,y \right ) =1/2\,{\frac{\int _{-\infty }^{\infty }\!{{\rm e}^{-sy+isx}}\,{\rm d}s}{\pi }} \]

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11.9 Laplace PDE in rectangle, one side homogeneous and 3 sides are not

problem number 91

Added December 20, 2018

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

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(0,y)&=0 \ u(\pi ,y) &= \sinh (\pi ) \cos (y) \ u(x,0) &= \sin (x) \ u(x,\pi ) &= -\sinh (x) \end{align*}

Mathematica

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

Maple

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

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11.10 Laplace PDE example 18 from Maple help page

problem number 92

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

With boundary conditions

\begin{align*} u(0,y) &= \frac{\sin y}{y} \end{align*}

Mathematica

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

Maple

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

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11.11 Laplace PDE on rectangle with one edge at infinity

problem number 93

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

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

\[ \text{DSolve}\left [\left \{u^{(0,2)}(x,y)+u^{(2,0)}(x,y)=0,\{u(x,0)=0,u(x,a)=0,u(0,y)=\sin (y),u(\infty ,y)=0\}\right \},u(x,y),\{x,y\},\text{Assumptions}\to a>0\right ] \]

Maple

\[ u \left ( x,y \right ) =\sum _{n=1}^{\infty }2\,{\frac{1}{a}\cases{1/2\,\pi \,n&$a=\pi \,n$\cr -{\frac{\pi \, \left ( -1 \right ) ^{n}\sin \left ( a \right ) na}{{\pi }^{2}{n}^{2}-{a}^{2}}}&otherwise\cr }{{\rm e}^{-{\frac{n\pi \,x}{a}}}}\sin \left ({\frac{\pi \,ny}{a}} \right ) } \]

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11.12 Laplace PDE inside a disk, periodic boundary conditions

problem number 94

Solve Laplace equation in polar coordinates inside a disk

Solve for \(u\left ( r,\theta \right ) \)

\begin{align*} \frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r} +\frac{1}{r^{2}}\frac{\partial ^{2}u}{\partial \theta ^{2}} & =0\\ 0 & \leq r\leq a\\ 0 & <\theta \leq 2\pi \end{align*}

Boundary conditions

\begin{align*} u\left ( a,\theta \right ) & =f\left ( \theta \right ) \\ \left \vert u\left ( 0,\theta \right ) \right \vert & <\infty \\ u\left ( r,0\right ) & =u\left ( r,2\pi \right ) \\ \frac{\partial u}{\partial \theta }\left ( r,0\right ) & =\frac{\partial u}{\partial \theta }\left ( r,2\pi \right ) \end{align*}

Mathematica

\[ \left \{\left \{u(r,\theta )\to \underset{n=1}{\overset{\infty }{\sum }}r^n \left (\frac{\cos (n \theta ) \left (\int _{-\pi }^{\pi } \cos (n \theta ) f(\theta ) \, d\theta \right ) a^{-n}}{\pi }+\frac{\left (\int _{-\pi }^{\pi } f(\theta ) \sin (n \theta ) \, d\theta \right ) \sin (n \theta ) a^{-n}}{\pi }\right )+\frac{\int _{-\pi }^{\pi } f(\theta ) \, d\theta }{2 \pi }\right \}\right \} \]

Maple

\[ u \left ( r,\theta \right ) =1/2\,{\frac{1}{\pi } \left ( 2\,\sum _{n=1}^{\infty } \left ({\frac{\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \sin \left ( n\theta \right ) \,{\rm d}\theta \sin \left ( n\theta \right ) +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \cos \left ( n\theta \right ) \,{\rm d}\theta \cos \left ( n\theta \right ) }{\pi } \left ({\frac{a}{r}} \right ) ^{-n}} \right ) \pi +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \,{\rm d}\theta \right ) } \]

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11.13 Dirichlet problem for the Laplace equation in upper half plan

problem number 95

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

Mathematica

\[ \left \{\left \{u(x,y)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{\begin{array}{cc} \{ & \begin{array}{cc} \tan ^{-1}\left (\frac{\frac{1}{2}-x}{y}\right )+\tan ^{-1}\left (\frac{x+\frac{1}{2}}{y}\right ) & y>0\lor x>\frac{1}{2}\lor x<-\frac{1}{2} \\ 0 & \text{True} \\\end{array} \\\end{array}}{\pi } & y\geq 0 \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \} \]

Maple

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

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11.14 Dirichlet problem for the Laplace equation in right half-plane:

problem number 96

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 \(u(0,y)=\sinc (y)\).

Mathematica

\[ \left \{\left \{u(x,y)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{x+(x \cos (y)-y \sin (y)) (\sinh (x)-\cosh (x))}{x^2+y^2} & x\geq 0 \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \} \]

Maple

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

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11.15 Dirichlet problem for the Laplace equation in the first quadrant

problem number 97

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 \(u(0,y)=\sinc (y)\).

Mathematica

\[ \left \{\left \{u(x,y)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{2 \left (\frac{3 \left (y \left (3 \pi (x+1) \left (x^4-4 x^3+2 \left (y^2+12\right ) x^2-4 \left (y^2+10\right ) x+y^4-16 y^2+100\right )+x \left (\left (6 \tan ^{-1}(3)-\log (10)\right ) x^4+2 \left (\left (6 \tan ^{-1}(3)-\log (10)\right ) y^2+10 \left (6 \tan ^{-1}(3)+\log (10)\right )\right ) x^2-20 y^2 \left (6 \tan ^{-1}(3)+\log (10)\right )+260 \log (10)+y^4 \left (6 \tan ^{-1}(3)-\log (10)\right )+360 \tan ^{-1}(3)\right )+x \left (x^4+2 \left (y^2-10\right ) x^2+y^4+20 y^2-260\right ) \log \left (x^2+y^2\right )\right )-\left (x^6+\left (y^2+6\right ) x^4-\left (y^4-92 y^2+60\right ) x^2-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{3 \left (x^6+\left (y^2+5\right ) x^4-\left (y^4+46 y^2-35\right ) x^2-y^6+5 y^4-35 y^2+343\right ) \tan ^{-1}\left (\frac{x}{y}\right )+x \left (2 \pi \left (\sqrt{3} y^5-9 y^4+14 \sqrt{3} y^3-6 y^2-35 \sqrt{3} y+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 )+147\right )+y \left (\left (4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )-3 \log (7)\right ) x^4+2 \left (y^2 \left (4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )-3 \log (7)\right )-7 \left (4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )+\log (343)\right )\right ) x^2+14 y^2 \left (4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )+\log (343)\right )+189 \log (7)+y^4 \left (4 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )-3 \log (7)\right )-140 \sqrt{3} \tan ^{-1}\left (\frac{2}{\sqrt{3}}\right )\right )+3 y \left (x^4+2 \left (y^2+7\right ) x^2+y^4-14 y^2-63\right ) \log \left (x^2+y^2\right )\right )}{\left (x^4-8 x^3+2 \left (y^2+15\right ) x^2-8 \left (y^2+7\right ) x+y^4+2 y^2+49\right ) \left (x^4+8 x^3+2 \left (y^2+15\right ) x^2+8 \left (y^2+7\right ) x+y^4+2 y^2+49\right )}\right )}{3 \pi } & x\geq 0\land y\geq 0 \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \} \]

Maple

\[ \text{ sol=() } \]

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11.16 Neumann problem for the Laplace equation in the upper half-plane

problem number 98

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 \(\frac{u}{y}(x,0)=\text{UnitBox[x]}\) 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

\[ \left \{\left \{u(x,y)\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{-4 y \tan ^{-1}\left (\frac{x-\frac{1}{2}}{y}\right )+4 y \tan ^{-1}\left (\frac{x+\frac{1}{2}}{y}\right )-2 x \log \left (4 x^2-4 x+4 y^2+1\right )+\log \left (4 x^2-4 x+4 y^2+1\right )+2 x \log \left (4 x^2+4 x+4 y^2+1\right )+\log \left (4 x^2+4 x+4 y^2+1\right )-2 \log (4)-4}{4 \pi } & y\geq 0 \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \} \]

Maple

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

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11.17 Dirichlet problem for the Laplace equation in a rectangle

problem number 99

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

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

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11.18 Cartessian coordinates with boundary conditions on two sides only

problem number 100

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

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

\[ \text{DSolve}\left [\left \{u^{(0,2)}(x,y)+u^{(2,0)}(x,y)=0,\{u(x,0)=0,u(x,b)=h(x)\}\right \},u(x,y),\{x,y\}\right ] \]

Maple

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

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11.19 in Rectangle, right edge at infinity

problem number 101

Added December 20, 2018.

Example 23, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

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

\[ \text{DSolve}\left [\left \{u^{(0,2)}(x,y)+u^{(2,0)}(x,y)=0,\{u(x,0)=0,u(x,a)=0,u(0,y)=\sin (y),u(\infty ,y)=0\}\right \},u(x,y),\{x,y\},\text{Assumptions}\to a>0\right ] \]

Maple

\[ u \left ( x,y \right ) =\sum _{n=1}^{\infty }2\,{\frac{1}{a}\cases{1/2\,\pi \,n&$a=\pi \,n$\cr -{\frac{\pi \, \left ( -1 \right ) ^{n}\sin \left ( a \right ) na}{{\pi }^{2}{n}^{2}-{a}^{2}}}&otherwise\cr }{{\rm e}^{-{\frac{n\pi \,x}{a}}}}\sin \left ({\frac{\pi \,ny}{a}} \right ) } \]

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11.20 Laplace PDE inside quarter disk, Neumann BC at edge

problem number 102

Added December 20, 2018.

Example 20, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve Laplace equation in polar coordinates inside quarter disk with \(0<r<1\) and \(0<\theta <\frac{\pi }{2}\)

Solve for \(u\left ( r,\theta \right ) \)

\begin{align*} \frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r} +\frac{1}{r^{2}}\frac{\partial ^{2}u}{\partial \theta ^{2}} & =0\\ \end{align*}

Boundary conditions

\begin{align*} u(r,0) &= 0 \\ u(r,\frac{\pi }{2}) &=0 \ \frac{\partial u}{\partial r}(1,\theta ) &= f(\theta ) \end{align*}

Mathematica

\[ \text{DSolve}\left [\left \{\frac{u^{(0,2)}(r,\theta )}{r^2}+\frac{u^{(1,0)}(r,\theta )}{r}+u^{(2,0)}(r,\theta )=0,\left \{u^{(1,0)}(1,\theta )=f(\theta )\right \},\left \{u(r,0)=0,u\left (r,\frac{\pi }{2}\right )=0\right \}\right \},u(r,\theta ),\{r,\theta \},\text{Assumptions}\to \left \{r>0,r<1,\theta >0,\theta <\frac{\pi }{2}\right \}\right ] \]

Maple

\[ u \left ( r,\theta \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac{\int _{0}^{\pi /2}\!f \left ( \theta \right ) \sin \left ( 2\,n\theta \right ) \,{\rm d}\theta{r}^{2\,n}\sin \left ( 2\,n\theta \right ) }{\pi \,n}} \right ) \]