can someone help me?, I don't know how to proceed in the last boundary condition $u_{y}(x,1)=x(1-x)\ $
$u_{xx}+u_{yy}=0\ $, $\ 0<x<1,\ 0<y<1$
$u(0,y)=0$
$u(1,y)=0$
$u_{y}(x,0)=0$
$u_{y}(x,1)=x(1-x)\ $
Now, I obtain this solution for the first three conditions:
$$ A_o/2+\sum_{n=1}^{\infty} A_n \cdot \cosh(kx) \cdot \cos(ky) $$
k=n$\pi$
it's that correct? how I can proceed in the last boundary condition? any help would be appreciated!
The separation of variables equation is $$ -\frac{X''}{X} = \lambda = \frac{Y''}{Y} $$ In order to vanish in $x$ and $0$ and $1$, the separated solutions in $X$ must be $$ X_n(x) = \sin(n\pi x),\;\;\; n=1,2,3,\cdots. $$ This determines $\lambda=n^2\pi^2$; because the corresponding solution $Y_n$ must satisfy $Y_n'(0)=0$, then $$ Y_{n}(y)=\cosh(n\pi y). $$ The general solution is $$ u(x,y) = \sum_{n=1}^{\infty}A_n\sin(n\pi x)\cosh(n\pi y), $$ where $A_n$ are constants determined by $$ u_{y}(x,1)=x(1-x) = \sum_{n=1}^{\infty}A_nn\pi\sin(n\pi x)\sinh(n\pi). $$ Using the orthogonality of the $X_n$, $$ \int_{0}^{1}x(1-x)\sin(n\pi x)dx=A_n\sinh(n\pi)\int_{0}^{1}\sin^2(n\pi x)dx \\ A_n = \frac{\int_{0}^{1}x(1-x)\sin(n\pi x)dx}{\sinh(n\pi)\int_{0}^{1}\sin^2(n\pi x)dx}. $$