# Trouble with integral and derivatives

1. Jan 24, 2014

### brambrambram

I have to integrate

-partial^2f/partialx^2 -partial^2f/partialy^2

in the variable x-y

How to do this?

2. Jan 24, 2014

### HallsofIvy

Staff Emeritus
The 'chain rule'. If u= x- y and v= x+ y then $u_x= 1$, $u_v= -1$, $v_x= 1$ and $v_y= 1$.

So $f_x= f_u(u_x)+ f_v(v_x)= f_u+ f_v$ and then $f_{xx}= (f_u+ f_v)_x= (f_u+ f_v)_u+ (f_u+ f_v)v= f_{uu}+ 2f_{uv}+ f_vv$

Similarly $f_y= f_u(u_y)+ f_v(v_y)= -f_u+ f_v$ and then $f_{yy}= (-f_u+ f_v)_y= -(-f_u+ f_v)_u+ (-f_u+ f_v)_v= f_{uu}- 2f_{uv}+ f_{vv}$.

So $-f_{xx}- f_{yy}= -(f_{xx}+ f_{yy})= -(2f_{uu}+ 2f_{vv})= -2(f_{uu}+ f_{vv})$.

For x and y independent variables, u and v are independent so the integral of that, with respect to x- y= u is $-2f_u+ \Phi(v)$, where $\Phi(v)$ is an arbitrary function of v= x+ y.

(Did your problem really have "-" on both derivatives? With the derivative on only on variable, $u_{xx}- u_{yy}$, this is related to the "wave equation" and subtracting rather than adding gives $4f_{uv}$ and integration with respect to x- y= u gives $4f_v$, or four times the derivative with respect to x+y.)