证明下列函数也分别满足laplace方程函数(x,y)=f(x2-y2,2xy)

Question

证明下列函数也分别满足laplace方程函数(x,y)=f(x2-y2,2xy)

Answer 1

F(x,y)=f(u,v)=f(x²-y²,2xy)
∂F/∂x=∂f/∂u.∂u/∂x+∂f/∂v.∂v/∂x
=2x∂f/∂u+2y∂f/∂v
∂²F/∂x²=2∂f/∂u+2x（2x∂²f/∂u²+2y∂²f/∂u∂v）+4xy∂²f/∂u∂v+4y²∂²f/∂v²
=2∂f/∂u+4x²∂²f/∂u²+8xy∂²f/∂u∂v+4y²∂²f/∂v²
∂F/∂y=∂f/∂u.∂u/∂y+∂f/∂v.∂v/∂y
=-2y∂f/∂u+2x∂f/∂v
∂²F/∂y²=-2∂f/∂u-2y∂²f/∂u².（-2y）-2y∂²f/∂u∂v.2x+2x（∂²f/∂u∂v.（-2y）+∂²f/∂v².2x）
=-2∂f/∂u+4y²∂²f/∂u²-4xy∂²f/∂u∂v-4xy∂²f/∂u∂v+4x²∂²f/∂v²
=-2∂f/∂u+4y²∂²f/∂u²-8xy∂²f/∂u∂v+4x²∂²f/∂v²
相加：
∂²F/∂x²+∂²F/∂y²=4x²（∂²f/∂u²+∂²f/∂v²）+4y²（∂²f/∂u²+∂²f/∂v²）
如果f(u,v)满足拉普拉斯方程，则∂²f/∂u²+∂²f/∂v²=0
因此：
∂²F/∂x²+∂²F/∂y²=0

Answer 2

函数呢，列出来