Answer to Question #210053 in Calculus for Faith

Question

Laplace's equation holds that Uxx+Uyy=0,verify that the second derivative of the following equations are laphace's equation.

1.u=in√x²+y².

2.u=x²-y².

Answer 1

Part 1

$u=\ln \sqrt{x²+y²}$

$u=\frac{1}{2} \ln (x²+y²) \implies e^{2u}= x^2+y^2$

Differentiating partially with respect to x and y

$2e^{2u}.\frac{∂u}{∂x}=2x \implies U_x= xe^{-2u}$

$\therefore U_{xx}=e^{-2u}-2xe^{-2u}. U_x= e^{-2u}[1-2xU_x]= e^{-2u}[1-2x^2e^{-2u}]$

Similary

$2e^{2u}. U_y=2y \implies U_y=ye^{-2u}$

$\therefore U_{yy}=e^{-2u}-2ye^{-2u}. U_y= e^{-2u}[1-2yU_y]= e^{-2u}[1-2y^2e^{-2u}]$

$U_{xx}+U_{yy}=e^{-2u}[2-2e^{-2u}(x^2+y^2)]$

$U_{xx}+U_{yy}=e^{-2u}[2-2e^{-2u}*e^{2u}]$

$U_{xx}+U_{yy}=e^{-2u}[2-2]$

$U_{xx}+U_{yy}=0$

Thus $u=\ln \sqrt{x²+y²}$ is the laphace's equation.

Part 2

$u=x²-y²$

$U_x=2x; U_{xx}= 2$

$U_y=-2y; U_{yy}= -2$

$U_{xx}+U_{yy}=2+(-2)$

$U_{xx}+U_{yy}=0$

Thus $u={x²-y²}$ is the laphace's equation.