Partial Differential problem

[54stickers]

If $z = f(x,y)$ and $x = e^{u}, y =e^{v}$ Prove:

$x^{2}\frac{\partial^{2}z}{\partial x^{2}} + y^{2}\frac{\partial^{2}z}{\partial y^{2}} + x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{\partial^{2}z}{\partial u^{2}} + \frac{\partial^{2}z}{\partial v^{2}}$

I used $u = ln(x), v = ln(y)$ and the following partial differential set ups:


$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x}$
$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}$
$\frac{\partial z}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u}$
$\frac{\partial z}{\partial u} = \frac{\partial f}{\partial y} \frac{\partial y}{\partial v}$

Then:

$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{1}{x}$
$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial v} \frac{1}{y}$
$\frac{\partial f}{\partial u} = \frac{\partial z}{\partial u} e^{u}$
$\frac{\partial f}{\partial v} = \frac{\partial z}{\partial v} e^{v}$

For the second derivatives:

$\frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} f}{\partial u^{2}} \frac{-1}{x^{2}}$
$\frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} f}{\partial v^{2}} \frac{-1}{y^{2}}$
$\frac{\partial^{2} z}{\partial u^{2}} = \frac{\partial^{2} f}{\partial x^{2}} e^{u}$
$\frac{\partial^{2} z}{\partial v^{2}} = \frac{\partial^{2} f}{\partial y^{2}} e^{v}$

I cannot get these terms to equal the statement that is to be proved. Am I doing something wrong?

Thanks!

 

[jvicens]

You are on the right track with your approach. However, there are a few mistakes in your calculations. Here is a step-by-step explanation of how to prove the given statement:

1. Start with the given equation: x^{2}\frac{\partial^{2}z}{\partial x^{2}} + y^{2}\frac{\partial^{2}z}{\partial y^{2}} + x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{\partial^{2}z}{\partial u^{2}} + \frac{\partial^{2}z}{\partial v^{2}}

2. Use the given substitutions: x = e^{u}, y = e^{v}. This will give us: e^{2u}\frac{\partial^{2}z}{\partial u^{2}} + e^{2v}\frac{\partial^{2}z}{\partial v^{2}} + e^{u} \frac{\partial z}{\partial u} + e^{v} \frac{\partial z}{\partial v} = \frac{\partial^{2}z}{\partial u^{2}} + \frac{\partial^{2}z}{\partial v^{2}}

3. Use the chain rule to find the partial derivatives with respect to u and v:

\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u} = e^{u}\frac{\partial z}{\partial x}

\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} = e^{v}\frac{\partial z}{\partial y}

4. Substitute these expressions into the equation from step 2: e^{2u}\frac{\partial^{2}z}{\partial u^{2}} + e^{2v}\frac{\partial^{2}z}{\partial v^{2}} + e^{u} \left(e^{u}\frac{\partial z}{\partial x}\right) + e^{