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Find general solution of the PDE $$x^2u_x^2+y^2u_y^2=u^2$$ Solution:

when we use transform $v=\ln u$ (or $u=e^v$), we get $x^2v_x^2+y^2v_y^2=1$.

And by the using separation of variables $v(x,y)=f(x)+g(y)$, I got the $x^2f'^2=\lambda^2$ and $g'^2=(1-\lambda^2)/y^2$.

And how can we solve the rest?

MrYouMath
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HD239
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2 Answers2

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Hint:

Let $u=e^v$ ,

Then $u_x=e^vv_x$

$u_y=e^vv_y$

$\therefore x^2e^{2v}v_x^2+y^2e^{2v}v_y^2=e^{2v}$

$x^2v_x^2+y^2v_y^2=1$

Let $\begin{cases}p=\ln x\\q=\ln y\end{cases}$ ,

Then $v_x=v_pp_x+v_qq_x=\dfrac{v_p}{x}$

$v_y=v_pp_y+v_qq_y=\dfrac{v_q}{y}$

$\therefore x^2\dfrac{v_p^2}{x^2}+y^2\dfrac{v_q^2}{y^2}=1$

$v_p^2+v_q^2=1$

doraemonpaul
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Why not use separation of variables from the beginning. Introduce $u=X(x)Y(y)$ to get

$$x^2X'^2Y^2+y^2X^2Y'^2=X^2Y^2 \implies \frac{x^2X'^2}{X^2}+\frac{y^2Y'^2}{Y^2}=1$$

It is clear that the equation can only be true if $\frac{x^2X'^2}{X^2}$ and $\frac{y^2Y'^2}{Y^2}$ are constants such that their sum adds up to 1. So we obtain two ordinary differential equations:

$$X'(x)^2=k^2\frac{X^2(x)}{x^2}$$ $$Y'(y)^2=(1-k^2)\frac{Y(y)^2}{y^2}$$

Now we take the square root of both expressions

$$X'(x)=\pm k\frac{X(x)}{x} \implies X(x)=c_1x^{\pm k}$$ $$Y'(y)=\pm \sqrt{1-k^2}\frac{Y(y)}{y} \implies Y(y)=c_2y^{\pm \sqrt{1-k^2}}.$$

Now, the solution by separation of variables is given by $$u(x,y)=c_1c_2x^{\pm k}y^{\pm \sqrt{1-k^2}}.$$

Note, that you cannot superimpose different solutions for $k$ as the differential equation is not linear.

EDIT: If you want to solve

$$f'(x)^2=\frac{\lambda^2}{x^{2}}\implies f'(x)=\pm\frac{\lambda}{x} \implies f(x)=\pm \lambda\ln x +c_1$$

$$g'(y)^2=\frac{1-\lambda^2}{y^2} \implies g'(y)=\pm\frac{\sqrt{1-\lambda^2}}{y} \implies g(y)=\pm\sqrt{1-\lambda^2}\ln y +c_2$$

You will see that $|\lambda|\leq 1$ is a side constraint such that the solutions stay real.

MrYouMath
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  • Thank you very much for your valuable solution. My teacher purposedly wanted me to use the solution method as I wrote above :) How can we apply the method as above? Do you have any idea? – HD239 Oct 05 '17 at 17:28
  • I have added an additional Edit at the end of my answer. – MrYouMath Oct 05 '17 at 17:31
  • Thank you :) Then $v=\pm(\lambda lnx+\sqrt{1-\lambda^2}lny)+c_3$ and $u=\pm(x^\lambda+y^{\sqrt{1-\lambda^2}}+c_4)$. Right? – HD239 Oct 05 '17 at 17:48
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    Almost,I would keep the plus minus for both expressions. – MrYouMath Oct 05 '17 at 19:18
  • $u(x,y)=\pm (x^{ \lambda}+y^{\sqrt{1-\lambda^2}}+c_4)$ is not equal to your first solution $u(x,y)=c_1c_2x^{\pm \lambda}y^{\pm \sqrt{1-\lambda^2}}.$ Do the two solutions really provide our PDE? I can' t test it in Maple. – HD239 Oct 06 '17 at 06:18