# Change of variables in differential equation

1. Jan 31, 2010

### michalpp

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

I have to transform the following equation using variables $$(u,v,w(u,v))=(yz-x,xz-y, xy-z)$$:

$$(xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz.$$

2. Relevant equations
chain rule: $$\frac{dw}{dx} = \frac{\partial w}{\partial u} \frac{\partial du}{\partial dx} + \frac{\partial w}{\partial v} \frac{\partial dv}{\partial dx}$$

3. The attempt at a solution
Using the chain rule and the product rule:

$$\frac{\partial w}{\partial x}=\frac{\partial z}{\partial x}(y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v})-\frac{\partial w}{\partial u}+z\frac{\partial w}{\partial v}$$

and a similar expression for $$\frac{\partial w}{\partial y}$$.
On the other hand

$$w=xy-z$$, so

$$\frac{\partial w}{\partial x}=y-\frac{\partial z}{\partial x}$$

(and similar for $$\frac{\partial w}{\partial y}$$), so

$$\frac{\partial z}{\partial x}(y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v})-\frac{\partial w}{\partial u}+z\frac{\partial w}{\partial v}=y-\frac{\partial z}{\partial x}$$ and therefore:

$$\frac{\partial z}{\partial x}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v}$$ and

$$\frac{\partial z}{\partial y}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=x+\frac{\partial w}{\partial v}-z\frac{\partial w}{\partial u}.$$

After multiplying the given equation by

$$[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]$$

and writing

$$\frac{\partial z}{\partial x}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]$$ as $$y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v}$$ I get

$$\frac{\partial w}{\partial v}(1-x^2-y^2-z^2-2xyz)=0.$$

If
$$[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]\neq 0$$

this equation is equivalent to the given one. But what if $$[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=0$$?.

I'm kind of stuck at this point, cause in the first case

$$\frac{\partial w}{\partial v}=0$$ or $$(1-x^2-y^2-z^2-2xyz)=0$$.

I was thinking of a way to change $$(x,y,z)$$ to $$(u,v,w)$$ in the last equation, but without success. I also have no idea what to do in the second case. So what should I do now? Is this a correct method to solve a problem like this?

2. Jan 31, 2010

### michalpp

I've found the answer to this problem in a book by B.P. Demidowich - it says
$$\frac{\partial w}{\partial v}=0$$
So this means that I need a proof that
$$(1-x^2-y^2-z^2-2xyz)\neq 0$$, or do I still miss something?

$$(1-x^2-y^2-z^2-2xyz)=(xy+z)z+(1-y^2)(-1)+x(x+yz)$$
looks similar to
$$(xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz.$$ but I don't have other ideas.

Last edited: Jan 31, 2010