Change of variables in differential equation

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Homework Statement



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

[tex](xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz.[/tex]


Homework Equations


chain rule: [tex]
\frac{dw}{dx} = \frac{\partial w}{\partial u} \frac{\partial du}{\partial dx} + \frac{\partial w}{\partial v} \frac{\partial dv}{\partial dx}
[/tex]


The Attempt at a Solution


Using the chain rule and the product rule:

[tex]\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}[/tex]

and a similar expression for [tex]\frac{\partial w}{\partial y}[/tex].
On the other hand

[tex]w=xy-z[/tex], so

[tex]\frac{\partial w}{\partial x}=y-\frac{\partial z}{\partial x}[/tex]

(and similar for [tex]\frac{\partial w}{\partial y}[/tex]), so

[tex]\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}[/tex] and therefore:

[tex]\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}[/tex] and

[tex]\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}.[/tex]

After multiplying the given equation by

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

and writing

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

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

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

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

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

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

I was thinking of a way to change [tex](x,y,z)[/tex] to [tex](u,v,w)[/tex] 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?
 

Answers and Replies

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

[tex]
(1-x^2-y^2-z^2-2xyz)=(xy+z)z+(1-y^2)(-1)+x(x+yz)
[/tex]
looks similar to
[tex]
(xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz.
[/tex] but I don't have other ideas.
 
Last edited:

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