# The chain rule for 2nd+ order partial differential equations

1. Oct 31, 2011

### Mr.Rockwater

1. The problem statement, all variables and given/known data

w= f(x,y)
x = u + v Verify that Wxx - Wyy = Wuv
y = u - v

2. Relevant equations

3. The attempt at a solution

I know how to find Wu or Wv but I have no idea on how to proceed to find the 2nd order derivative (or 3rd,4rth etc.. obviously). I assume this somehow uses the chain rule but I had no idea on how to apply it in this situation. Thank you!

2. Nov 1, 2011

### HallsofIvy

I presume that your "w" and "W" are the same.

Since W(x,y)= f(x,y), x=u+ v, and y= u- v,
$$W_u= \frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}= f_x(1)+ f_y(1)= f_x+ f_y$$

$$W_v= \frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}= f_x(1)+ f_y(-1)= f_x- f_y$$

Then $W_{uu}= (W_u)_u= (f_x+ f_y)_u$
$$= \left(\frac{\partial f_x}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial f_x}{\partial y}\frac{\partial y}{\partial u}\right)+ \left(\frac{\partial f_y}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial f_y}{\partial y}\frac{\partial y}{\partial u}\right)$$
$$= \left(f_{xx}(1)+ f_{xy}(1)\right)+\left(f_{xy}(1)+ f_{yy}(1)\right)= f_{xx}+ 2f_{xy}+ f_{yy}$$

Can you try the others, $W_{uv}$ and $W_{vv}$ yourself?

3. Nov 1, 2011

### Mr.Rockwater

Wow! Thanks for that helpful explanation! I get it now !