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I wish to find the function, $T(x,y)$, on the unit square that satisfies Laplace's equation,

$$ \nabla^2 T = 0, \text{ in } \Omega = [0,1]^2, $$

with Dirichlet boundary conditions in $y$ and periodic in $x$, i.e.

\begin{align} T(x,0) &= 1 + A\sin{\left( n2\pi x \right)}, \\ T(x,1) &= 0, \\ T(0,y) &= T(1,y) \end{align}

What I have tried so far

First Attempt

I attempted to solve the above using separation of variables. I.e., by setting $T(x,y) = X(x)Y(y)$, which (I think) translates the boundary conditions to \begin{align} Y(0) &= 1, \\ Y(1) & = 0, \\ X(x) &= 1 + A\sin{\left( n2\pi x \right)} \end{align} and reduces the problem to the ODEs \begin{align} \frac{d^2Y}{dy^2} = \lambda^2 Y, \quad \frac{d^2X}{dx^2} = -\lambda^2 X. \end{align} It is easy enough to find a solution for $Y$, but for $X$, $X(x) = 1 + A\sin{\left( n2\pi x \right)}$ is clearly not consistent with $\frac{d^2X}{dx^2} = -\lambda^2 X$

Second Attempt

I set $T(x,y) = \tilde{T}(x,y) + \phi(x,y)$. I let $\phi$ deal with the inhomogeneous boundary conditions, i.e. set $$ \phi = \left(1 + A\sin{\left( n2\pi x \right)} \right)\left(1-y \right). $$ In this set-up, the problem for $\tilde{T}$ is $$ \nabla^2 \tilde{T} = A\left(2\pi n \right)^2\sin{\left( n2\pi x \right)}(1-y), \text{ in } \Omega = [0,1]^2, $$ with periodic boundary conditions in $x$ and homogeneous boundary conditions in $y$. Perhaps this is an easier problem to solve?

Is my problem ill-posed? Have I made a mistake above? Or is there another approach I should take? Grateful for any input and guidance!

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2 Answers 2

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Hint: Try expressing the solution in a Fourier series in the $x$ variable:

$$T(x,y)=\sum_{n=-\infty}^\infty T_n(y)e^{i2\pi n x}$$

This automatically guarantees the periodicity condition $T(0,y)=T(1,y)$. Now apply the Laplace operator and derive an equation for the $T_n$. They should be completely determined by your other boundary conditions. Can you finish from here?

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  • $\begingroup$ Thanks for the answer! But with this, I think you stumble upon a similar problem as the one mentioned in my comment to Disintegrating By Parts answer. For the $T_l$'s ($T_n$ in your answer), we require $$ T_l''(y) = (2\pi l)^2 T_l(y)\, \rightarrow \, T_l(y) = B_l e^{2\pi l} + C_l e^{-2\pi l}. $$ Consider now the $l=0$ case. At $y=1$, $T(x,1) = 0$, so $$B_0 + C_0 = 0.$$ But at $y = 0$, $T(x,0) = 1 + A\sin{\left(2\pi nx \right) }$, so $$ B_0 + C_0 = 1. $$ A contradiction! That said, please correct me if I'm wrong... $\endgroup$ Commented Oct 13, 2022 at 19:17
  • $\begingroup$ Got it. We simply set $$ T_0(y) = (1-y). $$ Then, we arrive at the solution $$ T(x,y) = (1-y) + A\sin{\left(2\pi n x \right)}\left(B_n e^{2\pi ny} + C_n e^{-2\pi ny} \right), $$ where, $$ B_n = \frac{1}{1-e^{4\pi n}}, \,C_n = \frac{1}{1-e^{-4\pi n}}. $$ $\endgroup$ Commented Oct 13, 2022 at 21:10
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This problem is built for separation of variables. Assume solutions of the form $X(x)Y(x)$ and separate variables: $$ X''(x)Y(y)+X(x)Y''(y) = 0 \\ \frac{X''(x)}{X(x)} = -\frac{Y''(x)}{Y(y)} \\ \frac{X''(x)}{X(x)}=\lambda,\;\;\lambda=-\frac{Y''(y)}{Y(y)} $$ The functions $X$, $Y$ must satisfy these homogeneous conditions: $$ X(0)=X(1),\;\; Y(1)=0. $$ The periodic solutions in $x$ have the form \begin{align} X_0(x) &= B_0 \\ X_n(x) &= B_n\cos(2n\pi x)+C_n\sin(2n\pi x), \;\; n=1,2,3,\cdots \end{align} These dictate the permissible values of the separation parameter $\lambda$: $$ \lambda_n = -4n^2\pi^2,\;\; n=0,1,2,3,\cdots. $$ For $n=0$, the $\sin(4n\pi x)$ term drops out. Because $Y(1)=0$ must hold, the corresponding solutions in $y$ are scalar multiples of $$ Y_n(y)=\sinh(2n\pi (y-1)). $$ Therefore, the general solution is $$ T(x,y)=B_0+\\ \sum_{l=1}^{\infty}\{B_l\cos(2l\pi x)+C_l\sin(2l\pi x)\}\sinh(2l\pi(y-1)) $$ The constants $B_l,C_l$ are determined by the required edge condition through a Fourier series in $x$: $$ 1+A\sin(2n\pi x)=T(x,0) \\ =B_0-\sum_{l=1}^{\infty}\{B_l\cos(2l\pi x)+C_l\sin(2l\pi x)\}\sinh(2l\pi) $$This condition is a straightforward Fourier series equality which is solved by matching Fourier coefficients on the left with those on the right.

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  • $\begingroup$ Thanks for the answer! Given $T(x,1) = 0$, we have $B_0 = 0$. Since $T$ is periodic in $x \in \left[0,1 \right]$, we have $B_l=C_l=0$ when $l$ is odd. So we are left with equating Fourier coefficients for $$1 + A\sin{\left( 2n\pi x\right)} = -\sum_{l=1}^\infty \{B_{2l} \cos{\left(2l\pi x \right)}+C_{2l} \sin{\left(2l\pi x \right)}\} \sinh{\left( 2l\pi \right)}.$$ We have that $$B_{2l} \propto \int_0^1 \left[1 + A\sin{\left( 2n\pi x \right)}\right] \cos{\left(2l\pi x \right)}\, dx,$$ and similarly for $C_{2l}$. Meaning all coefficients apart from $C_{2n}$ are $0$. This can't be right? $\endgroup$ Commented Oct 13, 2022 at 15:41
  • $\begingroup$ @PhilipWinchester : Thank you. I don't post much here anymore. I think I have fixed my errors. $\endgroup$ Commented Jun 25, 2023 at 22:54

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