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I tried finding the General solution of the PDE: $x^2u_{xx}-y^2u_{yy}=0$

I first tried reducing it to canonical form and then I got stuck.

Here was what I did:

I got the characteristic equation to be:

$$\frac{dy}{dx}=\pm\frac{y}{x}$$

Then solving further, I got:

$$\ln y = \ln x+C_1$$ and $$\ln y=-\ln x+C_2$$

So in order to reduce the PDE into its canonical form, I introduced the new functions: $\xi, \eta$

Such that: $$\xi=\ln y-\ln x$$ and $$\eta=\ln y+\ln x$$

Thus the function becomes: $$u=[\xi(x,y),\eta(x,y)]$$

So I got $u_{xx}$ & $u_{yy}$ in terms of $u_{\xi}, u_{\eta}, u_{\xi\eta}, u_{\xi\xi}, u_{\eta\eta}$and slotted it into the PDE and I got this:

$$u_{\xi}-2u_{\xi\eta}=0$$

Is that the right canonical form for the PDE? And if it, how can I solve further to get the General Solution

Obinoscopy
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  • See http://math.stackexchange.com/questions/1546800/solving-y2u-xx-x2u-yy-0?rq=1 – DrHAL Nov 07 '16 at 14:59
  • @DrHAL there's a difference between mine and the question in the link. The coefficients were interchanged. Besides, my question is how to proceed after getting the canonical form and also to verify if my canonical form is correct – Obinoscopy Nov 07 '16 at 15:24
  • @Obinoscopy : Can you see us your calculus for the characteristic equation supposed to be : $\frac{dy}{dx}=\pm\frac{y}{x}$ : I suspect a mistake. I think that one of your characteristic equation is correct, the other false. – JJacquelin Nov 08 '16 at 10:14
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    Yes $u_{\xi} - 2u_{\xi\eta}=0 $ is correct. – Mahmoud Hassan Nov 13 '16 at 20:01
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    To find the general solution See http://math.stackexchange.com/questions/2012080/general-solution-to-pde-after-transformation/2012362#2012362 – Mahmoud Hassan Nov 13 '16 at 20:02
  • @JJacquelin there's no mistake there. The post after yours has resolved my issue. – Obinoscopy Nov 14 '16 at 12:03
  • @Obinoscopy : Very well. Just for curiosity, what functions $u(x,y)$ you obtained as general solution of the PDE ? – JJacquelin Nov 14 '16 at 12:28
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    @JJacquelin I obtained $u(x, y)=xyF(\frac{y}{x})+G(xy) $ – Obinoscopy Nov 15 '16 at 06:37
  • I agree, that's also what I obtained. – JJacquelin Nov 15 '16 at 06:57

2 Answers2

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Let $u(x,y)=X(x)Y(y)$ ,

Then $x^2X''(x)Y(y)-y^2X(x)Y''(y)=0$

$x^2X''(x)Y(y)=y^2X(x)Y''(y)$

$\dfrac{x^2X''(x)}{X(x)}=\dfrac{y^2Y''(y)}{Y(y)}=\dfrac{4s^2-1}{4}$

$\begin{cases}x^2X''(x)-\dfrac{4s^2-1}{4}X(x)=0\\y^2Y''(y)-\dfrac{4s^2-1}{4}Y(y)=0\end{cases}$

$\begin{cases}X(x)=\begin{cases}c_1(s)x^{\frac{1}{2}+s}+c_2(s)x^{\frac{1}{2}-s}&\text{when}~s\neq0\\c_1\sqrt{x}\ln x+c_2\sqrt{x}&\text{when}~s=0\end{cases}\\Y(y)=\begin{cases}c_3(s)y^{\frac{1}{2}+s}+c_4(s)y^{\frac{1}{2}-s}&\text{when}~s\neq0\\c_3\sqrt{y}\ln y+c_4\sqrt{y}&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,y)=\int_sC_1(s)(xy)^{\frac{1}{2}+s}~ds+\int_sC_2(s)(xy)^\frac{1}{2}\left(\dfrac{x}{y}\right)^s~ds+\int_sC_3(s)(xy)^\frac{1}{2}\left(\dfrac{y}{x}\right)^s~ds+\int_sC_4(s)(xy)^{\frac{1}{2}-s}~ds$

or $\sum\limits_sC_1(s)(xy)^{\frac{1}{2}+s}+\sum\limits_sC_2(s)(xy)^\frac{1}{2}\left(\dfrac{x}{y}\right)^s+\sum\limits_sC_3(s)(xy)^\frac{1}{2}\left(\dfrac{y}{x}\right)^s+\sum\limits_sC_4(s)(xy)^{\frac{1}{2}-s}$

i.e. $u(x,y)=F(xy)+\sqrt{xy}~G\left(\dfrac{x}{y}\right)$

doraemonpaul
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  • After change $\quad \xi=\log(x),\quad \eta=\log(y)\quad$ we get equation with constant coeficients $$u_{\xi\xi}-u_\xi-u_{\eta\eta}+u_\eta=0$$
  • $${{D}_{\xi}^{2}}-{D_{\xi}}-{{D}_{\eta}^{2}}+{D_{\eta}}=\left( {D_{\xi}}-{D_{\eta}}\right) \, \left( {D_{\xi}}+{D_{\eta}}-1\right)$$
  • solution of $\;u_\xi-u_\eta=0\;$ is $\;u_1=f(\xi+\eta)$
  • solution of $\;u_\xi+u_\eta-u=0\;$ is $\;u_2=e^\eta g(\xi-\eta)$
  • $$u=u_1+u_2=f(\xi+\eta)+e^\eta g(\xi-\eta)\\ =f\left(\log(x)+\log(y)\right)+e^{\log(y)}g(\log(x)-\log(y))\\= f(\log(xy))+y\,g\left(\log(\frac{x}{y})\right)\\ =F(xy)+y\,G(\frac{x}{y}) $$
Aleksas Domarkas
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