16 Poisson PDE in Cartesian coordinates

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16.1 Poisson equation in a rectangle, all boundaries are zero

problem number 124

Added March 13, 2019.

Solve for $u(x,y)$

$$\begin{array}{r}\frac{u_{xx}}{A}+\frac{u_{xx}}{B}=-2\theta \end{array}$$

Where $A,B,\theta$ are constants, and the boundary conditions are

$$\begin{array}{rl}u(x,-b)& =0\\ u(x,b)& =0\\ u(-a,y)& =0\\ u(a,y)& =0\end{array}$$

Mathematica ✗

ClearAll[a, b, A, B, theta, x, y, u]; 
 pde = D[u[x, y], {x, 2}]/A + D[u[x, y], {y, 2}]/B == -2*theta; 
 bc = {u[x, -b] == 0, u[x, b] == 0, u[-a, y] == 0, u[a, y] == 0}; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
 

$$\text{Failed}$$

Maple ✗

 
x:='x'; y:='y'; u:='u';A:='A';B:='B';theta:='theta';a:='a';b:='b'; 
pde:=diff(u(x,y),x$2)/A+diff(u(x,y),y$2)/B=-2*theta; 
bc:=u(x,-b)=0, 
    u(x,b)=0, 
    u(-a,y)=0, 
    u(a,y)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));
 

$$\text{ sol=() }$$

Hand solution

solve $$\begin{array}{rl}\frac{u_{xx}}{A}+\frac{u_{yy}}{B}& =-2\theta \\ B{u}_{xx}+A{u}_{yy}& =-2\theta AB\\ & =C\end{array}$$

Where $C=-2\theta AB$ is a new constant. With boundary conditions$$\begin{array}{rl}u(x,-b)& =0\\ u(x,b)& =0\\ u(-a,y)& =0\\ u(a,y)& =0\end{array}$$

To simplify solution, shift the rectangle so its lower left corner on the origin. Let $\tilde{x}=x+a$, and $\tilde{y}=y+b$. The boundary conditions becomes$$\begin{array}{rl}u(\tilde{x},0)& =0\\ u(\tilde{x},2b)& =0\\ u(0,\tilde{y})& =0\\ u(2a,\tilde{y})& =0\end{array}$$

And the pde becomes $B{u}_{\tilde{x}\tilde{x}}+A{u}_{\tilde{y}\tilde{y}}=C$. Instead of keep writing $\tilde{x},\tilde{y}$, will use $x,y$, but remember that these are shifted version. At the end, we shift back.

Hence the PDE to solve is  $B{u}_{xx}+A{u}_{yy}=C$ with BC$$\begin{array}{rl}u(x,0)& =0\\ u(x,2b)& =0\\ u(0,y)& =0\\ u(2a,y)& =0\end{array}$$

Using eigenfunction expansion method. Let $$\begin{array}{}\text{(1)}& u(x,y)=\sum _{n=1}^{\mathrm{\infty }}{b}_n\left(y\right)X_n\left(x\right)\end{array}$$ Where $X_n\left(x\right)$ is eigenfunctions for $X^{\prime \prime }\left(x\right)+{\lambda }_{n}X\left(x\right)=0$ with boundary conditions $X\left(0\right)=X\left(2a\right)=0$. This has eigenfunctions as $X_n\left(x\right)=\sin \left(\sqrt{{\lambda }_n}x\right)$ with eigenvalues ${\lambda }_n={\left(\frac{n\pi }{2a}\right)}^2$  for $n=1,2,\cdots$.

Substituting (1) into the PDE $B{u}_{xx}+A{u}_{yy}=C$ gives$$B\sum _{n=1}^{\mathrm{\infty }}{b}_n\left(y\right)X_n^{\prime \prime }\left(x\right)+A\sum _{n=1}^{\mathrm{\infty }}{b}_n^{\prime \prime }\left(y\right)X_n\left(x\right)=C$$ Expanding $C$ (a constant) as Fourier sine series the above becomes$$B\sum _{n=1}^{\mathrm{\infty }}{b}_n\left(y\right)X_n^{\prime \prime }\left(x\right)+A\sum _{n=1}^{\mathrm{\infty }}{b}_n^{\prime \prime }\left(y\right)X_n\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{q}_n{X}_n\left(x\right)$$ But $X_n^{\prime \prime }\left(x\right)=-{\lambda }_n{X}_n\left(x\right)$, hence the above becomes$$\begin{array}{rl}-B\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n{b}_n\left(y\right)X_n\left(x\right)+A\sum _{n=1}^{\mathrm{\infty }}{b}_n^{\prime \prime }\left(y\right)X_n\left(x\right)& =\sum _{n=1}^{\mathrm{\infty }}{q}_n{X}_n\left(x\right)\\ \text{(1A)}& A{b}_n^{\prime \prime }\left(y\right)-B{\lambda }_n{b}_n\left(y\right)& =q_n\end{array}$$

But $$\begin{array}{rl}C& =\sum _{n=1}^{\mathrm{\infty }}{q}_n{X}_n\left(x\right)\\ {\int }_0^{2a}C{X}_n\left(x\right)dx& =q_n{\int }_0^{2a}{X}_n^2\left(x\right)dx\\ {\int }_0^{2a}C\sin \left(\sqrt{{\lambda }_n}x\right)dx& =q_n{\int }_0^{2a}{\sin }^2\left(\sqrt{{\lambda }_n}x\right)dx\\ \frac{-C}{\sqrt{{\lambda }_n}}({(-1)}^n-1)& =q_{n}a\\ q_n& =\frac{-C}{a\sqrt{{\lambda }_n}}({(-1)}^n-1)\end{array}$$

Hence (1A) becomes$$A{b}_n^{\prime \prime }\left(y\right)-B{\lambda }_n{b}_n\left(y\right)=\frac{-C}{a\sqrt{{\lambda }_n}}({(-1)}^n-1)$$ This is standard second order linear ODE. The solution is$$b_n\left(y\right)=D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}y}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}y}+\frac{C}{aB{\lambda }_n^{\frac{3}{2}}}({(-1)}^n-1)$$ Using the above in (1) gives the solution$$\begin{array}{}\text{(1A)}& u(x,y)=\sum _{n=1}^{\mathrm{\infty }}(D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}y}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}y}+\frac{C}{aB{\lambda }_n^{\frac{3}{2}}}({(-1)}^n-1))X_n\left(x\right)\end{array}$$ We now need to find $D_n,E_n$.

Case $n$ even

When $n$ is even $({(-1)}^n-1)=0$ and the solution (1A) becomes$$u(x,y)=\sum _{n=1}^{\mathrm{\infty }}(D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}y}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}y})X_n\left(x\right)$$ At $y=0$ the above gives$$0=\sum _{n=1}^{\mathrm{\infty }}(D_n+E_n)\sin \left(\sqrt{{\lambda }_n}x\right)$$ Therefore$$\begin{array}{}\text{(2)}& D_n+E_n=0\end{array}$$ And at $y=2b$$$0=\sum _{n=1}^{\mathrm{\infty }}(D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}2b}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}2b})\sin \left(\sqrt{{\lambda }_n}x\right)$$ Therefore$$\begin{array}{}\text{(3)}& D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}2b}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}2b}=0\end{array}$$ From (2,3) we see that $D_n=E_n=0$, Hence $u(x,y)=0$ when $n$ even.

Case $n$ odd

When $n$ is odd $({(-1)}^n-1)=-2$ and the solution (1A) becomes$$u(x,y)=\sum _{n=1}^{\mathrm{\infty }}(D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}y}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}y}-\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}})X_n\left(x\right)$$ At $y=0$ the above gives$$0=\sum _{n=1}^{\mathrm{\infty }}(D_n+E_n-\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}})\sin \left(\sqrt{{\lambda }_n}x\right)$$ Therefore$$\begin{array}{}\text{(4)}& D_n+E_n-\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}}=0\end{array}$$ And at $y=2b$$$0=\sum _{n=1}^{\mathrm{\infty }}(D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}2b}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}2b}-\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}})\sin \left(\sqrt{{\lambda }_n}x\right)$$ Therefore$$\begin{array}{}\text{(5)}& D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}2b}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}2b}-\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}}=0\end{array}$$ Solving (4,5) for $D_n,E_n$ gives$$\begin{array}{rl}{D}_n& =\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}}\frac{1}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\\ E_n& =\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}}\frac{e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\end{array}$$

Therefore the final solution from (1A) becomes$$\begin{array}{rl}u(x,y)& =\sum _{n=1,3,5,\cdots }^{\mathrm{\infty }}(D_n{e}^{\sqrt{\frac{B}{A}{\lambda }_n}y}+E_n{e}^{-\sqrt{\frac{B}{A}{\lambda }_n}y}-\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}})X_n\left(x\right)\\ & =\sum _{n=1,3,5,\cdots }^{\mathrm{\infty }}(\left(\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}}\frac{1}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\right)e^{\sqrt{\frac{B}{A}{\lambda }_n}y}+\left(\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}}\frac{e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\right)e^{-\sqrt{\frac{B}{A}{\lambda }_n}y}-\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}})\sin \left(\sqrt{{\lambda }_n}x\right)\end{array}$$

Where ${\lambda }_n={\left(\frac{n\pi }{2a}\right)}^2$. Switching back to original coordinates using $\tilde{x}=x+a$, and $\tilde{y}=y+b$, then the above is$$u(x,y)=\sum _{n=1,3,5,\cdots }^{\mathrm{\infty }}(\left(\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}}\frac{1}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\right)e^{\sqrt{\frac{B}{A}{\lambda }_n}(y+b)}+\left(\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}}\frac{e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\right)e\left({}^{-\sqrt{\frac{B}{A}{\lambda }_n}y+b}\right)-\frac{2C}{aB{\lambda }_n^{\frac{3}{2}}})\sin \left(\sqrt{{\lambda }_n}(x+a)\right)$$ Where $C=-2\theta AB$, hence$$\begin{array}{rl}u(x,y)& =\sum _{n=1,3,5,\cdots }^{\mathrm{\infty }}(\left(\frac{-4\theta AB}{aB{\lambda }_n^{\frac{3}{2}}}\frac{1}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\right)e^{\sqrt{\frac{B}{A}{\lambda }_n}(y+b)}+\left(\frac{-4\theta AB}{aB{\lambda }_n^{\frac{3}{2}}}\frac{e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\right)e^{-\sqrt{\frac{B}{A}{\lambda }_n}(y+b)}+\frac{4\theta AB}{aB{\lambda }_n^{\frac{3}{2}}})\sin \left(\sqrt{{\lambda }_n}(x+a)\right)\\ & =\sum _{n=1,3,5,\cdots }^{\mathrm{\infty }}(\left(\frac{-4\theta A}{a{\lambda }_n^{\frac{3}{2}}}\frac{1}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\right)e^{\sqrt{\frac{B}{A}{\lambda }_n}(y+b)}+\left(\frac{-4\theta A}{a{\lambda }_n^{\frac{3}{2}}}\frac{e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}{1+e^{\sqrt{\frac{B}{A}{\lambda }_n}2b}}\right)e^{-\sqrt{\frac{B}{A}{\lambda }_n}(y+b)}+\frac{4\theta A}{a{\lambda }_n^{\frac{3}{2}}})\sin \left(\sqrt{{\lambda }_n}(x+a)\right)\end{array}$$

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16.2 Dirichlet problem for the Poisson equation in a rectangle

problem number 125

Taken from Mathematica DSolve help pages.

Solve for $u(x,y)$

$$\begin{array}{rl}\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}& =6x-6y\end{array}$$

Boundary conditions

$$\begin{array}{rl}u(x,0)& =1+11x+x^3\\ u(x,2)& =-7+11x+x^3\\ u(0,y)& =1-y^3\\ u(4,y)& =109-y^3\end{array}$$

Mathematica ✓

ClearAll[u, x, y]; 
 pde = Laplacian[u[x, y], {x, y}] == 6*x - 6*y; 
 bc = {u[x, 0] == 1 + 11*x + x^3, u[x, 2] == -7 + 11*x + x^3, u[0, y] == 1 - y^3, u[4, y] == 109 - y^3}; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
 

$$\left\{\{u(x,y)\to x^3+11x-y^3+1\}\right\}$$

Maple ✓

 
x:='x'; y:='y'; u:='u'; 
pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=6*x-6*y; 
bc:=u(x,0)=1+11*x+x^3, 
    u(x,2)=-7+11*x+x^3, 
    u(0,y)=1-y^3, 
    u(4,y)=109-y^3; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));
 

$$u(x,y)=x^3-y^3+11x+1$$