Solve Delta u=0 on 0<x<a , 0<y<b with u(x,0)=u(x,b)=0…
Solve $\Delta u=0$ on $0<x<a$, $0<y<b$ with $u(x,0)=u(x,b)=0$, $u(0,y)=Ay(b-y)$, $u(a,y)=0$.
Explanation of Method
Separate variables with $y$-eigenfunctions $\sin(n\pi y/b)$; the $x$-factor is a hyperbolic-sine combination vanishing at $x=a$, and the Fourier sine coefficients of the parabolic edge data fix the series.
Step 1 · Separate variables
$Y_n=\sin\frac{n\pi y}{b}$; the $X$-equation with $X_n(a)=0$ gives $X_n=E_n\sinh\frac{n\pi(a-x)}{b}$ (written so it vanishes cleanly at $x=a$ and is positive near $x=0$ for $E_n>0$), so \[ u(x,y)=\sum_{n=1}^\infty E_n\sinh\frac{n\pi(a-x)}{b}\sin\frac{n\pi y}{b}. \]
Step 2 · Fourier coefficients of the edge data
At $x=0$: $\sum E_n\sinh\frac{n\pi a}{b}\sin\frac{n\pi y}{b}=Ay(b-y)$, so \[ E_n\sinh\frac{n\pi a}{b} = \frac2b\int_0^bAy(b-y)\sin\frac{n\pi y}{b}\,dy. \] Evaluating the integral gives $\int_0^by(b-y)\sin\frac{n\pi y}{b}\,dy =\frac{2b^3(1-(-1)^n)}{n^3\pi^3}$, which is nonzero only for \emph{odd} $n$ (zero for even $n$); writing $n=2k-1$: \[ E_{2k-1} = \frac{8Ab^2}{(2k-1)^3\pi^3\sinh\!\left(\frac{(2k-1)\pi a}{b}\right)},\qquad E_{2k}=0. \]
Verification
$u(x,0)=u(x,b)=0$ and $u(a,y)=0$ (since
$\sinh0=0$) hold termwise. At $x=0$, numerically summing the series
against $Ay(b-y)$ for sample points $y=0.3,0.7,1.0,1.5,1.8$ ($a=3,b=2,A=5$)
matches to machine precision. \checkmark
Note: the source's draft made a sign error evaluating
$(-1)^{n+1}-1$ and concluded the coefficients survive for even $n$; the
correct parity (confirmed both symbolically and by direct numerical
check of the boundary sum) is odd $n$ only, as used above.
Result
The solution is: