Boundary value problem with nonhomogeneous conditions

Question

I am asked to solve the following boundary value problem $$u_{xx}+u_{yy}=0, 0<x<1, 0<y<1 (1')$$ $$u(0,y)=0, u(1,y)= \sin{(\pi y)} \cos{(\pi y)}, 0<y<1$$ $$u(x,0)=u(x,1)=0, 0<x<1$$

I have done the following:

$$u(x,y)=v(x,y)+s(x)$$

$$v_{xx}+v_{yy}+s''(x)=0, 0<x<1, 0<y<1$$ $$v(0,y)+s(0)=0, 0<y<1$$ $$v(1,y)+s(1)=\sin{( \pi y)} \cos{( \pi y)}, 0<y<1$$ $$v(x,0)+s(x)=v(x,1)+s(x)=0, 0<x<1$$

So we have the following problems:

$$s''(x)=0, 0<x<1$$ $$s(0)=0, s(1)= \sin{( \pi y)} \cos{( \pi y)}$$ $$\text{ and }$$ $$v_{xx}+v_{yy}=0, 0<x<1, 0<y<1$$ $$v(0,y)=0, v(1,y)=0, 0<y<1$$ $$v(x,0)=v(x,1)=-s(x), 0<x<1$$

Is this correct so far??

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ It's always convenient to use first the homogeneous condition: $$ {\rm u}\pars{x,y} = \sum_{n = 1}^{\infty}{\rm A}_{n}\pars{x}\sin\pars{n\pi y} \quad\imp\quad \sum_{n = 1}^{\infty}\bracks{% {\rm A}_{n}''\pars{x} - \pars{n\pi}^{2}{\rm A}_{n}\pars{x}}\sin\pars{n\pi y} = 0 $$ Then, $\ds{{\rm A}_{n}''\pars{x} - \pars{n\pi}^{2}{\rm A}_{n}\pars{x} = 0\quad\imp\quad {\rm A}_{n}\pars{x} \equiv a_{n}\sinh\pars{n\pi x} + b_{n}\cosh\pars{n\pi x}}$

$$ {\rm u}\pars{x,y} = \sum_{n = 1}^{\infty}\bracks{% a_{n}\sinh\pars{n\pi x} + b_{n}\cosh\pars{n\pi x}}\sin\pars{n\pi y} $$

$$ \begin{array}{l} {\rm u}\pars{0,y} = 0 =\sum_{n = 1}^{\infty}b_{n}\sin\pars{n\pi y} \\ {\rm u}\pars{1,y} = \sin\pars{\pi y}\cos\pars{\pi y} =\sum_{n = 1}^{\infty}\bracks{a_{n}\sinh\pars{n\pi} + b_{n}\cosh\pars{n\pi}}\sin\pars{n\pi y} \end{array} $$

Multiply both members of these equations by $\ds{\sin\pars{m\pi y}}$ and integrate over $\ds{\pars{0,1}}$. We'll get $\ds{b_{n} = 0\ \forall n = 1,2,\ldots}$ and \begin{align} &\overbrace{\int_{0}^{1}\sin\pars{m\pi y}\sin\pars{\pi y}\cos\pars{\pi y}\,\dd y} ^{\ds{{1 \over 4}\,\delta_{m,2}}}\ =\sum_{n = 1}^{\infty} a_{n}\sinh\pars{n\pi}\ \overbrace{\int_{0}^{1}\sin\pars{m\pi y}\sin\pars{n\pi y}\,\dd y} ^{\ds{\half\,\delta_{mn}}} \\[3mm]&\imp\quad a_{m} = {1 \over 2\sinh\pars{2\pi}}\,\delta_{m,2} \end{align}

$$\color{#00f}{\large% {\rm u}\pars{x,y} = {\sinh\pars{2\pi x}\sin\pars{2\pi y} \over 2\sinh\pars{2\pi}}} $$

Indeed, the solution was pretty obvious at the very beginning since $\ds{\sin\pars{\pi y}\cos\pars{\pi y} = \half\,\sin\pars{2\pi y}}$.

Answer 2

Let $u(x,y)=X(x)Y(y)$ and separate to obtain an $X$ problem and a $Y$ problem. The associated $y$ boundary conditions give you a pair of parallel sides with homogeneous BCs, so use that as your eigenvalue problem.

Once you have the $\lambda_n$ and $Y_n(y)$, find the general solution to the $X$ problem, $X_n(x)$.

Superimpose to get $u(x,y)=\sum Y_n(y)X_n(x)$ where $X_n(x)$ involves two families of constants. Use the $X$ boundary conditions (one of which is nonhomogeneous) to determine those two families of constants.