Show u=e^(-y)sin x solves Delta u=0 on the unit square with…

Show $u=e^{-y}\sin x$ solves $\Delta u=0$ on the unit square with $u(0,y)=0$, $u(1,y)=e^{-y}\sin1$, $u(x,0)=\sin x$, $u(x,1)=e^{-1}\sin x$.

Step 1 · Check the PDE

\[ u_{xx}=-e^{-y}\sin x,\qquad u_{yy}=e^{-y}\sin x,\qquad u_{xx}+u_{yy}=0. \]

Step 2 · Check all four edges

$u(0,y)=e^{-y}\sin0=0$; $u(1,y)=e^{-y}\sin1$; $u(x,0)=\sin x$; $u(x,1)=e^{-1}\sin x$ — all match the stated conditions directly.

Remark

$u={\rm Im}(e^{i(x+iy)}\cdot i)$... more directly, $u=e^{-y}\sin x$ is the imaginary part of the analytic function $-ie^{iz}$ (with $z=x+iy$), confirming harmonicity since real/imaginary parts of analytic functions are automatically harmonic.