I was doing a problem in the book A Collection of Problems on MATHEMATICAL PHYSICS by B. M. BUDAK, A. A. SAMARSKII and A. N. TIKHONO of the form $$x^2 u_{xx} - y^2 u_{yy} = 0$$

In the answer section it only said $ ε = \frac{y}{x} , η = xy$ and the general solution $u(x,y)=F(xy)+{x}~G\left(\dfrac{x}{y}\right)$

On my attempts i got the conical form to be $$u_{\eta \epsilon} - \frac{u_{\epsilon}}{2\eta}=0 $$

and the general solution to be

Looking at posts on this question i have seen people also get $u(x,y)=F(xy)+\sqrt{xy}~G\left(\dfrac{x}{y}\right)$ as well as $u(x,y)=F(xy)+{xy}~G\left(\dfrac{x}{y}\right)$ I feel confused about this question as i have seen 3 different answer and no method. So i wish to ask if my answer is correct and if not how do i arrive at the correct answer?


I'm not an expert on PDEs so can't help you with the general method. However the question of the different answers is not too hard to resolve.

Remember that $F$ and $G$ will be arbitrary functions of a single variable (well, almost arbitrary - there will be certain differentiability conditions). Your answer can be written as $$\eqalign{u(x,y) &=F(xy)+\sqrt{xy}G\Bigl(\frac xy\Bigr)\cr &=F(xy)+x\sqrt{\frac yx}G\Bigl(\frac xy\Bigr)\cr &=F(xy)+xH\Bigl(\frac xy\Bigr)\cr}$$ where $$H(t)=\frac1{\sqrt t}G(t)\ ,$$ so it is really the same as the book answer. For the final answer $$u(x,y)=F(xy)+xyG\Bigl(\frac xy\Bigr)$$ you can carefully differentiate using the chain rule, product rule and quotient rule: if I have done this right you end up with $$x^2u_{xx}-y^2u_{yy}=2x^2G'\Bigl(\frac xy\Bigr)\ ;$$ this is not (usually) $0$, so the answer is wrong. You can use a similar method to check that your answer and the book answer are correct.

  • 1
    $\begingroup$ Your answer is the same as the book answer for $x,y > 0$. You'll run into some trouble outside that quadrant. Note that if you change the signs of both $x$ and $y$ you change the sign of $x H(x/y)$, but not of $\sqrt{xy} G(x/y)$. $\endgroup$ – Robert Israel Nov 6 '17 at 6:07

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