Solve Delta u=0 on 0<x<a , y>0 with u(0,y)=u(a,y)=0…
Solve $\Delta u=0$ on $0<x<a$, $y>0$ with $u(0,y)=u(a,y)=0$, $u(x,0)=A(1-x/a)$, $u(x,\infty)=0$.
Explanation of Method
Separate variables; the fixed-side conditions give sine eigenfunctions in $x$, boundedness as $y\to\infty$ kills the growing exponential in $y$, and the base condition fixes the Fourier coefficients.
Step 1 · Separate variables
$X_n=\sin\frac{n\pi x}{a}$; the $y$-equation $Y''-\mu_n^2Y=0$ with $Y(\infty)=0$ forces $Y_n=D_ne^{-n\pi y/a}$, so \[ u(x,y)=\sum_{n=1}^\infty B_ne^{-n\pi y/a}\sin\frac{n\pi x}{a}. \]
Step 2 · Compute the Fourier sine coefficients
\[ B_n = \frac{2A}{a}\int_0^a\left(1-\frac xa\right)\sin\frac{n\pi x}{a}\,dx. \] The two component integrals combine (the $(1-(-1)^n)$ and $(-1)^{n+1}$ pieces largely cancel) to the clean result \[ B_n = \frac{2A}{n\pi} \] (verified numerically against direct integration for several $n$).
Verification
Boundary conditions at $x=0,a$ and $y\to\infty$ hold termwise. At $y=0$, using $\sum_{n=1}^\infty\frac{\sin n\theta}{n}=\frac{\pi-\theta}{2}$ for $0<\theta<2\pi$ with $\theta=n\pi x/a$ recovers $A(1-x/a)$. \checkmark Each term is harmonic (its $x$- and $y$-Laplacians cancel). \checkmark
Result
The solution is: