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Problem:

Try to simplify $$x^2\frac{\partial^2w}{\partial x^2}+y^2\frac{\partial^2w}{\partial y^2}+z^2\frac{\partial^2w}{\partial z^2}+yz\frac{\partial^2w}{\partial y\partial z}+zx\frac{\partial^2w}{\partial z\partial x}+xy\frac{\partial^2w}{\partial x\partial y}=0$$ with transform $x=uv, y=vt, z=tu$, where $w$ is well-behaved enough.


Source: Григорий Михайлович Фихтенгольц


I found the method on the book (differentiating the equations in the transform to solve out $\partial w/\partial x$, etc) not easy to solve this problem, with a horrendous calculation. I wonder whether there's a sensible way to deal with the problem, without too pain and loss of rigor, for example, $\vartheta_s=s\dfrac\partial{\partial s}$?

Any help? Thanks!

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Are you sure there are no factors of $2$ in front of the last three terms? If they were there, the expression would be $(x,y,z)H(x,y,z)^T=0$ where $H$ is the Hessian of $w$. In other words, the restriction to any line through the origin is linear. –  user31373 Jun 30 '12 at 1:29
    
Perhaps you mean $z = tu$? Surely you don't want both $y=vt$ and $z=tv$. –  Robert Israel Jun 30 '12 at 1:32
    
@RobertIsrael Sorry, $z=tu$. –  Frank Science Jun 30 '12 at 4:44
    
@LeonidKovalev No $2$ stated. Robert Israel's answer is right, but I'm looking for a method not an answer. –  Frank Science Jun 30 '12 at 5:14
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1 Answer

up vote 0 down vote accepted

Assuming you meant $z=tu$, here's how Maple does it:

W:= w(x,y,z);

Q:= x^2*diff(W,x,x) + y^2*diff(W,y,y) + z^2*diff(W,z,z) + x*y*diff(W,x,y) + y*z*diff(W,y,z) + x*z*diff(W,x,z);

simplify(PDEtools:-dchange({x=u*v,y=v*t,z=t*u},Q,[u,v,t]));

$$\frac{1}{2} \left( {\frac {\partial ^{2}}{\partial {u}^{2}}}w \left( u,v,t \right) \right) {u}^{2}+\frac12 \left( {\frac {\partial ^{2}}{ \partial {v}^{2}}}w \left( u,v,t \right) \right) {v}^{2}+\frac12 {t}^{2} {\frac {\partial ^{2}}{\partial {t}^{2}}}w \left( u,v,t \right) $$

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I know the answer because it's an example in that book. I'm looking for a sensible way, not just working blindly. –  Frank Science Jun 30 '12 at 4:46
    
In this day and age, resorting to a computer algebra system is a sensible way to tackle something like this. –  Robert Israel Jul 1 '12 at 6:16
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