Answer to Question #138692 in Differential Equations for Nikhil Singh

Question

Using U= 1/2x^2 and V= 1/2y^2, solve the partial differential equation
1/x^2∂^2/∂x^2 - 1/x^3∂z/∂x= 1/y^2∂^2z/∂y^2- 1/y^3∂z/∂y

Answer 1

$\frac{1}{x^2}\frac{∂^2z}{∂x^2} - \frac{1}{x^3}\frac{∂z}{∂x}= \frac{1}{y^2}\frac{∂^2z}{∂y^2}- \frac{1}{y^3}\frac{∂z}{∂y}$

Putting, $X = \frac{1}{2}x^2, Y = \frac{1}{2}y^2$

then, $xdx = dX, ydy = dY$

so,

$\frac{\partial z}{\partial X} =\frac{\partial z}{\partial x}\frac{\partial x}{\partial X} = \frac{1}{x}\frac{\partial z}{\partial x}$

and $\frac{\partial^2 z}{\partial X^2} = \frac{\partial }{\partial X}(\frac{\partial z}{\partial X}) = \frac{\partial }{\partial x}(\frac{1 }{ x}\frac{\partial z}{\partial x})\frac{\partial x}{\partial X}$

$= -\frac{1}{x^3}\frac{\partial z}{\partial x} + \frac{1}{x^2}\frac{\partial^2 z}{\partial x^2}$

Similarly, $\frac{\partial^2 z}{\partial Y^2}= -\frac{1}{y^3}\frac{\partial z}{\partial y} + \frac{1}{y^2}\frac{\partial^2 z}{\partial y^2}$

This transformation will lead the equation,

$\frac{\partial^2 z}{\partial X^2} - \frac{\partial^2 z}{\partial Y^2} = 0$

$(D^2-D^{'2})z = 0$ where

$$D^2 = \frac{\partial^2 }{\partial X^2}, D^{'2} = \frac{\partial^2 }{\partial Y^2}$$

$(D +D')(D -D')z = 0$

solution of the equation,

$z = \phi_1(Y-X)+\phi_2(Y+X)$

$z = f_1(y^2-x^2) + f_2(y^2+x^2)$