Solving Laplace's equation with Dirichlet and periodic boundary conditions

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!

2 Answers

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?

• 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... Commented Oct 13, 2022 at 19:17
• 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, $$B_n = \frac{1}{1-e^{4\pi n}}, \,C_n = \frac{1}{1-e^{-4\pi n}}.$$ Commented Oct 13, 2022 at 21:10

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.

• 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? Commented Oct 13, 2022 at 15:41
• @PhilipWinchester : Thank you. I don't post much here anymore. I think I have fixed my errors. Commented Jun 25, 2023 at 22:54