How to perform a this type of variable change $(X,Y) = (x+y, x-y)$ for $\frac{\partial^2u}{\partial x^2} - \frac{\partial ^2 u}{\partial y^2}=0 $

Question

I've been asked to solve this PDE: $$\frac{\partial ^2 u}{\partial x^2} - \frac{\partial ^2 u}{\partial y^2}=0 $$ by using this change of variables $(X,Y) = (x+y, x-y)$. And I am not sure how to do it as I am unsure how the function for the variable change would look. Do I put $$u(x,y) = f(x+y,x-y)$$ or do I put it differently? My question is how exactly would the function $f$ look like?

Answer 1

You have $u(x,y) = f(X,Y) = f(x+y, x-y)$, then you use chain rule: $$ \frac{\partial u}{\partial x} = \frac{\partial f}{\partial X}\frac{\partial X}{\partial x} + \frac{\partial f}{\partial Y}\frac{\partial Y}{\partial x}=\frac{\partial f}{\partial X} + \frac{\partial f}{\partial Y},\\ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial X}\left(\frac{\partial f}{\partial X} + \frac{\partial f}{\partial Y}\right)\frac{\partial X}{\partial x} + \frac{\partial}{\partial Y}\left(\frac{\partial f}{\partial X} + \frac{\partial f}{\partial Y}\right)\frac{\partial Y}{\partial x} = \frac{\partial^2 f}{\partial X^2} + 2\frac{\partial^2 f}{\partial X\partial Y} + \frac{\partial^2 f}{\partial Y^2} $$

Apply the same logic to $\partial/\partial y$.

Answer 2

You can keep

$$u(x,y) = f(x+y,x-y)=f(X,Y)$$

where $X=X(x,y)=x+y$ and $Y=Y(x,y)=x-y$. Then

$$X_x = 1, ~~X_y=1, ~~Y_x=1, Y_y=-1$$

so by the chain rule

$$u_x=f_X X_x + f_Y Y_x = f_X + f_Y$$

$$u_y=f_X X_y + f_Y Y_y = f_X - f_Y$$

$$u_{xx}=\left(f_X + f_Y\right)_x=f_{XX}X_x+f_{XY}Y_x+f_{XY}X_x+f_{YY}Y_x=f_{XX}+2f_{XY}+f_{YY}$$

$$u_{yy}=\left(f_X -f_Y\right)_y=f_{XX}X_y+f_{XY}Y_y-f_{XY}X_y-f_{YY}Y_y=f_{XX}-2f_{XY}+f_{YY}$$