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I have been given the second order equation $x^2 u_{xx} - y^2 u_{yy} = 0 , V={(x,y) \in \mathbb{R} : x > 0,y > 0}.$

I have been asked to reduce to canonical form as well as obtain the general solution of the equation. I can reach the canonical form of the equation but I am unsure how to achieve the general solution. The canonical form I obtained was:

$4x^2u_{ηε} - \frac{2x}{y}u_ε = 0 $

with constraints as $ ε = \frac{x}{y} , η = xy$

Any help or advice on how to obtain the general solution would be appreciated.

Dylan
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C. Jones
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2 Answers2

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$\displaystyle \epsilon = \frac{x}{y} , \eta = xy \rightarrow x^2=\epsilon \eta$

Then $$ 4x^2u_{\eta \epsilon} - \frac{2x}{y}u_{\epsilon} = 4\eta \epsilon u_{\eta \epsilon} - 2\epsilon u_{\epsilon} =0 \rightarrow u_{\eta \epsilon} - \frac{u_{\epsilon}}{2\eta}=0 $$

Now, write $ \displaystyle v=u_{\epsilon}$ , then you have to solve the ODE $\displaystyle v_{\eta}- \frac{v}{2\eta}=0$

idk
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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 $\;u_2=e^\eta g(\xi-\eta)$
  • General solution of $x^2 u_{xx} - y^2 u_{yy} = 0\;$ is $$u=u_1+u_2=f(\xi+\eta)+e^\eta g(\xi-\eta)\\ =f\left(\log(x)+\log(y)\right)+e^{\log(y)}g(\log(x)-\log(y))\\= f(\log(xy))+y\,g\left(\log(\frac{x}{y})\right)\\ =F(xy)+y\,G(\frac{x}{y}) $$
Aleksas Domarkas
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