# Thread: 2nd order partial differentiation

1. ## 2nd order partial differentiation

Hi, revising this subject i'm having problems with questions of this nature

2. Originally Posted by pkr
Hi, revising this subject i'm having problems with questions of this nature

I will assume the equation is, $\displaystyle z_{xx} - yz_{yy} = \tfrac{1}{2}z_y$ for $\displaystyle (x,y) \in (-\infty,\infty)\times (0,\infty)= D$.

Define $\displaystyle \bold{g}: D\to \mathbb{R}^2$ by $\displaystyle \bold{g}(x,y) = (u,v)$ where $\displaystyle u=x-2\sqrt{y}$ and $\displaystyle v=x + 2\sqrt{y}$.

Now we will define a new function (of two variable) $\displaystyle w$ in the following way: $\displaystyle w(a,b) = u(c,d)$ where $\displaystyle (c,d)$ are the unique points such that $\displaystyle \bold{g}(c,d)=(a,b)$. In other words, we define $\displaystyle w$ to be evaluated at a point $\displaystyle (a,b)$ to be $\displaystyle u$ evaluated at $\displaystyle (c,d)$ where $\displaystyle (c,d)$ is the (unique point in $\displaystyle D$) that is mapped into $\displaystyle (a,b)$ under the transformation that you gave.

It should be clear that $\displaystyle w(u,v) = z(x,y)$ i.e. $\displaystyle (w\circ \bold{g})(x,y) = z(x,y)$ (function composition) for each point $\displaystyle (x,y)\in D$.

Now we are going to use the chain rule. That is where I am going to be a little informal with notation because it gets cleaner to write this way and it perhaps be easier for you to follow.

$\displaystyle z_x = w_u\cdot u_x + w_v \cdot v_x$
$\displaystyle z_y = w_u \cdot u_y + w_v \cdot v_y$

Let us analyze what the first equation means (the second one is similar). It is saying that (by chain rule) we get the expression on the right hand side. When we write $\displaystyle w_u$ we mean the partial derivative with respect to the first coordinate evaluated at $\displaystyle u$ (that is at $\displaystyle x-2\sqrt{y}$) multiplied by $\displaystyle u_x$ - the partial derivative of the function $\displaystyle u=x-2\sqrt{y}$ with respect to the first coordinate. Similarly, $\displaystyle w_v$ means the partial derivative with respect to the second coordinate evaluated at $\displaystyle v$ (that is $\displaystyle x+2\sqrt{y}$).

Therefore, we get upon writing everything out,
$\displaystyle z_x = w_u + w_v$
$\displaystyle z_y = -y^{-1/2}w_u + y^{-1/2}w_v$

3. Hey pkr, can you post a plot for this? I think it would be cool. Just make up any old (reasonable) IBVP on it.
Can you do the second partials? What's it saying? What's it doing?
Why are so much phenomena in our world describable by second order PDEs?
Sumptin's up I tell you what.

Here's one of them:

$\displaystyle \frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left[\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}\right]$

$\displaystyle =\frac{\partial^2 w}{\partial u^2}\frac{\partial u}{\partial x}+\frac{\partial^2 w}{\partial v\partial u}\frac{\partial v}{\partial x}+\frac{\partial^2 w}{\partial v^2}\frac{\partial v}{\partial x}+\frac{\partial^2 w}{\partial u\partial v}\frac{\partial u}{\partial x}$

Pretty sure that's right. Double check it.
I'll assume the mixed partials will drop out when everything is simplified, or else everything else will drop out and you're left with just one mixed partial. I may try to solve it (with just any old IBVP) but I'm slow so don't wait on me.

4. Yeah i've figured mine out now, using the chain rule it simplifys down easy, had a few terms involving y left but it cancelled due to what was on the RHS of the equation, havent got time now but i'll post my answer soon.