Thread: Mixed derivative with change of variables

1. Mixed derivative with change of variables

Hello,

I'm wondering how I can show that for all functions which satisfy:
$\displaystyle \frac{\partial ^2 f}{\partial x^2} - \frac{d^2 f}{dy^2} = 0$
that the following is true:
$\displaystyle \frac{\partial ^2 f}{\partial s \partial t} = 0$

Where $\displaystyle f(x,y) = f(x(s,t),y(s,t)) = f(s,t)$

Thanks very much for your help
Jon

2. Originally Posted by jonmondalson
Hello,

I'm wondering how I can show that for all functions which satisfy:
$\displaystyle \frac{\partial ^2 f}{\partial x^2} - \frac{d^2 f}{dy^2} = 0$ I assume that those d's in the second term should be partials?
that the following is true:
$\displaystyle \frac{\partial ^2 f}{\partial s \partial t} = 0$

Where $\displaystyle f(x,y) = f(x(s,t),y(s,t)) = f(s,t)$
That result is not true. For example, if $\displaystyle f(x,y) = xy$, $\displaystyle x(s,t)=s$ and $\displaystyle y(s,t)=t$ then $\displaystyle \frac{\partial ^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial y^2} = 0$ but $\displaystyle \frac{\partial ^2 f}{\partial s \partial t} = 1$.

3. What is true is that if you let s= x+ y, t= s- y, then the equation becomes
$\displaystyle \frac{\partial^2 f}{\partial s\partial t}= 0$
and, from that, $\displaystyle f(s, t)= F(s)+ G(t)= F(x+y)+ G(x- y)$ where F and G are any differentiable functions of a single variable.

4. Originally Posted by HallsofIvy
What is true is that if you let s= x+ y, t= s- y, then the equation becomes
$\displaystyle \frac{\partial^2 f}{\partial s\partial t}= 0$
and, from that, $\displaystyle f(s, t)= F(s)+ G(t)= F(x+y)+ G(x- y)$ where F and G are any differentiable functions of a single variable.
Thanks bud, but can I ask, how do you calculate $\displaystyle \frac{\partial^2 f}{\partial s\partial t}$, for example when s= x+ y and t= s- y, as in your example?