#### 4.3.1 Cartesian coordinates

4.3.1.1 [327] All boundaries at zero

4.4.0.1 [330] In rectangle
4.4.0.2 [331] On whole plane

##### 4.3.1.1 [327] All boundaries at zero

problem number 327

Solve for $$u(x,y)$$ \begin{align*} \frac{u_{xx}}{A} + \frac{u_{xx}}{B} = -2 \theta \end{align*}

Where $$A,B,\theta$$ are constants, and the boundary conditions are \begin{align*} u(x, -b) &= 0\\ u(x, b) &= 0\\ u(-a, y) &= 0\\ u(a, y) &= 0 \end{align*}

Mathematica

Failed

Maple

sol=()

Hand solution

solve \begin{align*} \frac{u_{xx}}{A}+\frac{u_{yy}}{B} & =-2\theta \\ Bu_{xx}+Au_{yy} & =-2\theta AB\\ & =C \end{align*}

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

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{align*} u\left ( \tilde{x},0\right ) & =0\\ u\left ( \tilde{x},2b\right ) & =0\\ u\left ( 0,\tilde{y}\right ) & =0\\ u\left ( 2a,\tilde{y}\right ) & =0 \end{align*}

And the pde becomes $$Bu_{\tilde{x}\tilde{x}}+Au_{\tilde{y}\tilde{y}}=C$$. 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  $$Bu_{xx}+Au_{yy}=C$$ with BC\begin{align*} u\left ( x,0\right ) & =0\\ u\left ( x,2b\right ) & =0\\ u\left ( 0,y\right ) & =0\\ u\left ( 2a,y\right ) & =0 \end{align*}

Using eigenfunction expansion method. Let $$u\left ( x,y\right ) =\sum _{n=1}^{\infty }b_{n}\left ( y\right ) X_{n}\left ( x\right ) \tag{1}$$ 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 $$Bu_{xx}+Au_{yy}=C$$ gives$B\sum _{n=1}^{\infty }b_{n}\left ( y\right ) X_{n}^{\prime \prime }\left ( x\right ) +A\sum _{n=1}^{\infty }b_{n}^{\prime \prime }\left ( y\right ) X_{n}\left ( x\right ) =C$ Expanding $$C$$ (a constant) as Fourier sine series the above becomes$B\sum _{n=1}^{\infty }b_{n}\left ( y\right ) X_{n}^{\prime \prime }\left ( x\right ) +A\sum _{n=1}^{\infty }b_{n}^{\prime \prime }\left ( y\right ) X_{n}\left ( x\right ) =\sum _{n=1}^{\infty }q_{n}X_{n}\left ( x\right )$ But $$X_{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}X_{n}\left ( x\right )$$, hence the above becomes\begin{align} -B\sum _{n=1}^{\infty }\lambda _{n}b_{n}\left ( y\right ) X_{n}\left ( x\right ) +A\sum _{n=1}^{\infty }b_{n}^{\prime \prime }\left ( y\right ) X_{n}\left ( x\right ) & =\sum _{n=1}^{\infty }q_{n}X_{n}\left ( x\right ) \nonumber \\ Ab_{n}^{\prime \prime }\left ( y\right ) -B\lambda _{n}b_{n}\left ( y\right ) & =q_{n}\tag{1A} \end{align}

But \begin{align*} C & =\sum _{n=1}^{\infty }q_{n}X_{n}\left ( x\right ) \\ \int _{0}^{2a}CX_{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}}}\left ( \left ( -1\right ) ^{n}-1\right ) & =q_{n}a\\ q_{n} & =\frac{-C}{a\sqrt{\lambda _{n}}}\left ( \left ( -1\right ) ^{n}-1\right ) \end{align*}

Hence (1A) becomes$Ab_{n}^{\prime \prime }\left ( y\right ) -B\lambda _{n}b_{n}\left ( y\right ) =\frac{-C}{a\sqrt{\lambda _{n}}}\left ( \left ( -1\right ) ^{n}-1\right )$ This is standard second order linear ODE. The solution is$b_{n}\left ( y\right ) =D_{n}e^{\sqrt{\frac{B}{A}\lambda _{n}}y}+E_{n}e^{-\sqrt{\frac{B}{A}\lambda _{n}}y}+\frac{C}{aB\lambda _{n}^{\frac{3}{2}}}\left ( \left ( -1\right ) ^{n}-1\right )$ Using the above in (1) gives the solution$$u\left ( x,y\right ) =\sum _{n=1}^{\infty }\left ( D_{n}e^{\sqrt{\frac{B}{A}\lambda _{n}}y}+E_{n}e^{-\sqrt{\frac{B}{A}\lambda _{n}}y}+\frac{C}{aB\lambda _{n}^{\frac{3}{2}}}\left ( \left ( -1\right ) ^{n}-1\right ) \right ) X_{n}\left ( x\right ) \tag{1A}$$ We now need to ﬁnd $$D_{n},E_{n}$$.

Case $$n$$ even

When $$n$$ is even $$\left ( \left ( -1\right ) ^{n}-1\right ) =0$$ and the solution (1A) becomes$u\left ( x,y\right ) =\sum _{n=1}^{\infty }\left ( D_{n}e^{\sqrt{\frac{B}{A}\lambda _{n}}y}+E_{n}e^{-\sqrt{\frac{B}{A}\lambda _{n}}y}\right ) X_{n}\left ( x\right )$ At $$y=0$$ the above gives$0=\sum _{n=1}^{\infty }\left ( D_{n}+E_{n}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$ Therefore$$D_{n}+E_{n}=0\tag{2}$$ And at $$y=2b$$$0=\sum _{n=1}^{\infty }\left ( D_{n}e^{\sqrt{\frac{B}{A}\lambda _{n}}2b}+E_{n}e^{-\sqrt{\frac{B}{A}\lambda _{n}}2b}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$ Therefore$$D_{n}e^{\sqrt{\frac{B}{A}\lambda _{n}}2b}+E_{n}e^{-\sqrt{\frac{B}{A}\lambda _{n}}2b}=0\tag{3}$$ From (2,3) we see that $$D_{n}=E_{n}=0$$, Hence $$u\left ( x,y\right ) =0$$ when $$n$$ even.

Case $$n$$ odd

When $$n$$ is odd $$\left ( \left ( -1\right ) ^{n}-1\right ) =-2$$ and the solution (1A) becomes$u\left ( x,y\right ) =\sum _{n=1}^{\infty }\left ( D_{n}e^{\sqrt{\frac{B}{A}\lambda _{n}}y}+E_{n}e^{-\sqrt{\frac{B}{A}\lambda _{n}}y}-\frac{2C}{aB\lambda _{n}^{\frac{3}{2}}}\right ) X_{n}\left ( x\right )$ At $$y=0$$ the above gives$0=\sum _{n=1}^{\infty }\left ( D_{n}+E_{n}-\frac{2C}{aB\lambda _{n}^{\frac{3}{2}}}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$ Therefore$$D_{n}+E_{n}-\frac{2C}{aB\lambda _{n}^{\frac{3}{2}}}=0\tag{4}$$ And at $$y=2b$$$0=\sum _{n=1}^{\infty }\left ( 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}}}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$ Therefore$$D_{n}e^{\sqrt{\frac{B}{A}\lambda _{n}}2b}+E_{n}e^{-\sqrt{\frac{B}{A}\lambda _{n}}2b}-\frac{2C}{aB\lambda _{n}^{\frac{3}{2}}}=0\tag{5}$$ Solving (4,5) for $$D_{n},E_{n}$$ gives\begin{align*} 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{align*}

Therefore the ﬁnal solution from (1A) becomes\begin{align*} u\left ( x,y\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( 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}}}\right ) X_{n}\left ( x\right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \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}}}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}

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$u\left ( x,y\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \left ( \frac{2C}{aB\lambda _{n}^{\frac{3}{2}}}\frac{1}{1+e^{\sqrt{\frac{B}{A}\lambda _{n}}2b}}\right ) e^{\sqrt{\frac{B}{A}\lambda _{n}}\left ( y+b\right ) }+\left ( \frac{2C}{aB\lambda _{n}^{\frac{3}{2}}}\frac{e^{\sqrt{\frac{B}{A}\lambda _{n}}2b}}{1+e^{\sqrt{\frac{B}{A}\lambda _{n}}2b}}\right ) e\left ( ^{-\sqrt{\frac{B}{A}\lambda _{n}}y+b}\right ) -\frac{2C}{aB\lambda _{n}^{\frac{3}{2}}}\right ) \sin \left ( \sqrt{\lambda _{n}}\left ( x+a\right ) \right )$ Where $$C=-2\theta AB$$, hence\begin{align*} u\left ( x,y\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \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}}\left ( y+b\right ) }+\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}}\left ( y+b\right ) }+\frac{4\theta AB}{aB\lambda _{n}^{\frac{3}{2}}}\right ) \sin \left ( \sqrt{\lambda _{n}}\left ( x+a\right ) \right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \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}}\left ( y+b\right ) }+\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}}\left ( y+b\right ) }+\frac{4\theta A}{a\lambda _{n}^{\frac{3}{2}}}\right ) \sin \left ( \sqrt{\lambda _{n}}\left ( x+a\right ) \right ) \end{align*}

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##### 4.3.1.2 [328] Dirichlet problem in a rectangle

problem number 328

Taken from Mathematica DSolve help pages.

Solve for $$u\left ( x,y\right )$$ \begin{align*} u_{xx}+ u_{yy} & = 6x - 6y \end{align*}

Boundary conditions \begin{align*} 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 \end{align*}

Mathematica

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

Maple

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

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##### 4.3.1.3 [329] Poisson PDE in whole 2D plane

problem number 329

Solve Poisson PDE $\frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=6y$

Mathematica

Failed

Maple

$u \left ( x,y \right ) ={\it \_F1} \left ( y-ix \right ) +{\it \_F2} \left ( y+ix \right ) +3\,{x}^{2}y$

Hand solution

Solve Poisson PDE $\frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=6y$

The solution is $$f=f_{h}+f_{p}\tag{1}$$ Where $$f_{h}$$ is the homogenous solution to Laplace $$\frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=0$$ and $$f_{p}$$ is a particular solution. The homogeneous solution is easily found as follows. Let $f_{h}=F\left ( mx+y\right )$ Then $$\frac{\partial f_{h}}{\partial x}=mF^{\prime }$$ and $$\frac{\partial ^{2}f_{h}}{\partial x^{2}}=m^{2}F^{\prime \prime }$$ and $$\frac{\partial f_{h}}{\partial y}=F^{\prime }$$ and $$\frac{\partial ^{2}f_{h}}{\partial y^{2}}=F^{\prime \prime }$$. Substituting these into $$\frac{\partial ^{2}f_{h}}{\partial x^{2}}+\frac{\partial ^{2}f_{h}}{\partial y^{2}}=0$$ gives\begin{align*} m^{2}F^{\prime \prime }+F^{\prime \prime } & =0\\ m^{2}+1 & =0\\ m & =\pm i \end{align*}

Since we assumed $$f_{h}=F\left ( mx+y\right )$$, then the homogeneous solution is sum of two arbitrary functions (one for each root of $$m$$)$f_{h}=F_{1}\left ( ix+y\right ) +F_{2}\left ( -ix+y\right )$ To ﬁnd particular solution, let $$f_{p}=Ax^{n}y^{m}\tag{2}$$ Hence $$\frac{\partial f_{p}}{\partial x}=Anx^{n-1}y^{m}$$ and $$\frac{\partial ^{2}f_{p}}{\partial x^{2}}=An\left ( n-1\right ) x^{n-2}y^{m}$$ and $$\frac{\partial f_{p}}{\partial y}=Ax^{n}my^{m-1}$$ and $$\frac{\partial ^{2}f_{p}}{\partial y^{2}}=Ax^{n}m\left ( m-1\right ) y^{m-2}$$. Substituting these into $$\frac{\partial ^{2}f_{p}}{\partial x^{2}}+\frac{\partial ^{2}f_{p}}{\partial y^{2}}=6y$$ gives\begin{align*} An\left ( n-1\right ) x^{n-2}y^{m}+Ax^{n}m\left ( m-1\right ) y^{m-2} & =6y\\ y^{m}\left ( An\left ( n-1\right ) x^{n-2}+Ax^{n}m\left ( m-1\right ) y^{-2}\right ) & =6y \end{align*}

Comparing terms, then $$m=1$$ and $An\left ( n-1\right ) x^{n-2}+Ax^{n}m\left ( m-1\right ) y^{-2}=6$ Since $$m=1$$ then the above simpliﬁes to$An\left ( n-1\right ) x^{n-2}=6$ Since there is no $$x$$ in RHS, then $$n-2=0$$ or $$n=2$$ and the above becomes\begin{align*} 2A\left ( 2-1\right ) & =6\\ A & =3 \end{align*}

Hence (2) becomes$f_{p}=Ax^{n}y^{m}=3x^{2}y$ And the complete solution (1) is\begin{align*} f\left ( x,y\right ) & =f_{h}+f_{p}\\ & =F_{1}\left ( ix+y\right ) +F_{2}\left ( -ix+y\right ) +3x^{2}y \end{align*}