Answer to Question #285662 in Complex Analysis for Sga

Question

show that u e^-2xy sin x2 y2 is harmonic

Answer 1

$u=e^{-2xy}\sin(x^2-y^2)$

$\dfrac{\partial u}{\partial x}=-2ye^{-2xy}\sin(x^2-y^2)+2xe^{-2xy}\cos(x^2-y^2)$

$\dfrac{\partial^2 u}{\partial x^2}=4y^2e^{-2xy}\sin(x^2-y^2)-4xye^{-2xy}\cos(x^2-y^2)$

$+2e^{-2xy}\cos(x^2-y^2)-4xye^{-2xy}\cos(x^2-y^2)$

$-4x^2e^{-2xy}\sin(x^2-y^2)$

$\dfrac{\partial u}{\partial y}=-2xe^{-2xy}\sin(x^2-y^2)-2ye^{-2xy}\cos(x^2-y^2)$

$\dfrac{\partial^2 u}{\partial y^2}=4x^2e^{-2xy}\sin(x^2-y^2)+4xye^{-2xy}\cos(x^2-y^2)$

$-2e^{-2xy}\cos(x^2-y^2)+4xye^{-2xy}\cos(x^2-y^2)$

$-4y^2e^{-2xy}\sin(x^2-y^2)$

$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=4y^2e^{-2xy}\sin(x^2-y^2)$

$-8xye^{-2xy}\cos(x^2-y^2)+2e^{-2xy}\cos(x^2-y^2)$

$-4x^2e^{-2xy}\sin(x^2-y^2)+4x^2e^{-2xy}\sin(x^2-y^2)$

$+8xye^{-2xy}\cos(x^2-y^2)-2e^{-2xy}\cos(x^2-y^2)$

$-4y^2e^{-2xy}\sin(x^2-y^2)=0$

Hence

$\nabla^2u=\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=0, x, y\in \R$

Therefore the function $u=e^{-2xy}\sin(x^2-y^2)$ is harmonic.