Trying to solve Laplace's equation within a semi infinite strip. I have to use separation of variables, and I know how to solve the problem but I am having trouble setting up the boundary conditions.

Want to solve $\displaystyle \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$, for the semi infinite region $0 \leq x \leq 4$, $y\geq 0$.

The boundary conditions are $u(0,y) = u(4,y) = T$, where $T$ is constant, and $u(x,0) = 0$.

I know you start by assuming that $u(x,t) = H(x) \phi(y)$, and then differentiate and substitute into the PDE.

I get that $\displaystyle -\frac{H''(x)}{H(x)} = \frac{\phi''(x)}{\phi(x)} = -\lambda$.

Because the boundary conditions are non zero on the vertical sides of the plate, I don't know how to get specific boundary values for $H(x)$. I know that $\phi(0) = 0$, and that $\phi(y)$ is finite as $y\rightarrow \infty$ (we have to look for a bounded solution).

So, if someone would give me a bit of help in how to transform the boundary conditions in terms of $H(x)$ and $\phi(x)$, that would be greatly appreciate.


You want homogenous conditions $u(0,y)=0=u(4,y)$ instead, which can achieved by subtracting $T$ from the $u$ in order to obtain a new equation for $v=u-T$ $$ v_{xx}+v_{yy} = 0, \;\; v(0,y)=0=v(4,y),\;\; v(x,0)=-T. $$ Now when you separate variables $v(x,y)=X(x)Y(y)$, you obtain $$ -\frac{X''}{X} = \lambda = \frac{Y''}{Y}. $$ The negative can go in either place, but I know how it works out, so I chose the above so that $\lambda$ is positive instead of negative. Then $X$ must satisfy $$ X''+\lambda X = 0,\;\; X(0)=X(4)=0. $$ The solutions are $$ \lambda_n=\frac{n^2\pi^2}{16},\;\; X_n(x)=\sin\left(\frac{n\pi}{4}x\right) $$ The corresponding solutions in $Y$ are $$ Y_n = A_n\exp\left(-\frac{n\pi}{4}y\right). $$ (I've excluded solutions with positive exponent in order to have a bounded solution.) The constants $A_n$ must be chosen so that $$ v(x,y) = \sum_{n=1}^{\infty}A_n\sin\left(\frac{n\pi}{4}x\right)\exp\left(-\frac{n\pi}{4}y\right),\;\;\; v(x,0)=-T. \\ \implies \sum_{n=1}^{\infty}A_n\sin\left(\frac{n\pi}{4}x\right)=-T \\ $$ Multiplying both sides by $\sin\left(\frac{n\pi}{4}x\right)$, integrating in $x$ over $[0,4]$, and using the orthogonality of the eigenfunctions determines the constants $A_n$ by the equations $$ A_n\int_{0}^{4}\sin^{2}\left(\frac{n\pi}{4}x\right)dx=-T\int_{0}^{4}\sin\left(\frac{n\pi}{4}x\right)dx. $$ The solution of the original problem is $$ u(x,y) = T+v(x,y) = T+\sum_{n=1}^{\infty}A_n\sin\left(\frac{n\pi}{4}x\right)\exp\left(-\frac{n\pi}{4}y\right). $$


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