Solve the boundary-value problem $∆u = 0$ (by this we mean $u_{xx} + u_{yy} = 0$) in the rectangle $0 < x < π$, $0 < y < 1$, with the boundary conditions $u(0,y) = 0$, $u(π,y) = g(y)$, $u(x,0) = 0$ and $u(x,1) = 0$.


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  • $\begingroup$ What have you tried so far? Are you familiar with separation of variables? $\endgroup$ – Gary Moon Mar 19 at 21:22
  • $\begingroup$ Yes, I am working through that now but am struggling a little bit $\endgroup$ – meff11 Mar 19 at 21:26

I'll try to help you get started. You'll seek a separated solution $u(x,y) = X(x)Y(y)$. This will give you the following ODEs for $X$ and $Y$: $$X^{\prime\prime} -\lambda X = 0 \text{ and } Y^{\prime\prime} + \lambda Y = 0. $$ Hence, $$X(x) = A\cosh\beta x + B\sinh\beta x \text{ and } Y(y) = C\cos\beta y + D\sin\beta y,$$ where $\beta^2=\lambda$. Now you can apply the BCs: $$u(0,y)=0 \implies X(0)=0 \implies A = 0,$$ $$Y(0)=0 \implies C=0 \text{ and}$$ $$Y(1) = 0 \implies \beta= n\pi.$$ We deal with the final BC later. This gives you $$X_n(x) = B_n\sinh n\pi x \text{ and } Y_n(y) = D_n\sin n\pi y.$$ So, your solution will look like $$u(x,y) = \sum_n A_n \sinh n\pi x \sin n\pi y.$$ You should be able to use the final BC to determine the $A_n$'s. Can you take it from here?

Edit: If you look at the equation that you wrote in your comment, it tells you that the $A_n$'s are a multiple of the coefficients of the Fourier sine series for $g$. In particular, $$A_n = \frac{2}{\sinh n \pi^2} \int_0^1 g(y) \sin n\pi y \ dy.$$

  • $\begingroup$ thank you for the help. So now I must apply the last boundary condition of $u(\pi , y) = g(y)$ which when plugged into $u(x,y)$ gives us $g(y) = \sum_{1}^{\infty}A_nsinh(n\pi ^2)sin(n\pi y)$. Is there a way that this simplifies nicely for $A_n$? $\endgroup$ – meff11 Mar 19 at 22:28

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{\mrm{u}\pars{x,y} = \sum_{n = 1}^{\infty}\mrm{a}_{n}\pars{x}\sin\pars{n\pi y}}$ which satisfies the boundarty condition $\ds{\mrm{u}\pars{x,0} = \mrm{u}\pars{x,1} = 0}$.

Since $\ds{\mrm{u}\pars{x,y}}$ satisfies the Laplace Equation, it is true that $$ 0 =\sum_{n = 1}^{\infty}\bracks{\totald[2]{\mrm{a}_{n}\pars{x}}{x} - n^{2}\pi^{2}\mrm{a}''_{n}\pars{x}}\sin\pars{n\pi y} $$ Multiply both members by $\ds{\sin\pars{m\pi y}}$, with $\ds{m \in \mathbb{N}_{\ \geq 1}}$, and integrate over $\ds{y \in\pars{0,1}}$ to get $\ds{\mrm{a}_{m}\pars{x} = A_{n}\sinh\pars{n\pi x}}$ which already satisfies $\ds{\mrm{a}_{m}\pars{0} = 0}$. Then, \begin{align} \mrm{u}\pars{x,y} & = \sum_{n = 1}^{\infty}A_{n}\sinh\pars{n\pi x} \sin\pars{n\pi y} \\ \mrm{u}\pars{\pi,y} = \mrm{g}\pars{y} & = \sum_{n = 1}^{\infty}A_{n}\sinh\pars{n\pi^{2}} \sin\pars{n\pi y} \\ \int_{0}^{1}\mrm{g}\pars{y}\sin\pars{n\pi y}\dd y & = \sum_{m = 1}^{\infty}A_{m}\sinh\pars{m\pi^{2}} \underbrace{\int_{0}^{1}\sin\pars{m\pi y}\sin\pars{n\pi y}\dd y} _{\ds{{1 \over 2}\,\delta_{mn}}} \\ A_{n} & = \bbx{{2 \over \sinh\pars{n\pi^{2}}} \int_{0}^{1}\mrm{g}\pars{y}\sin\pars{n\pi y}\dd y} \end{align}


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