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.