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I have to solve this equation :

$$ \frac{\partial u}{\partial x} \cdot \frac{\partial u}{\partial y} = xy $$

with initial condition $u(x,0) = x$.

I know it is easy with separation of variables, but I need to do it with the method of characteristics and I can't figure it out.

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    $\begingroup$ Are you sure it is not u_x + u_y = xy?, method of characteristics would be an unusual choice otherwise. $\endgroup$ – Gregory Dec 9 '14 at 21:04
  • $\begingroup$ Yes, this really is the question. And that is why I find it particularly diffucult to come up with an answer. $\endgroup$ – AppMathPhys Dec 10 '14 at 8:22
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Let $\begin{cases}p=x^2\\q=y^2\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=2x\dfrac{\partial u}{\partial p}$

$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=2y\dfrac{\partial u}{\partial q}$

$\therefore2xu_p2yu_q=xy$ with $u(p,0)=\pm\sqrt p$

$u_pu_q=\dfrac{1}{4}$ with $u(p,0)=\pm\sqrt p$

$u_q-\dfrac{1}{4u_p}=0$ with $u(p,0)=\pm\sqrt p$

$u_{pq}+\dfrac{u_{pp}}{4u_p^2}=0$ with $u(p,0)=\pm\sqrt p$

Let $v=u_p$ ,

Then $v_q+\dfrac{v_p}{4v^2}=0$ with $v(p,0)=\pm\dfrac{1}{2\sqrt p}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dq}{dt}=1$ , letting $q(0)=0$ , we have $q=t$

$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$

$\dfrac{dp}{dt}=\dfrac{1}{4v^2}=\dfrac{1}{4v_0^2}$ , letting $p(0)=f(v_0)$ , we have $p=\dfrac{t}{4v_0^2}+f(v_0)=\dfrac{q}{4v^2}+f(v)$ , i.e. $v=F\left(p-\dfrac{q}{4v^2}\right)$

$v(p,0)=\pm\dfrac{1}{2\sqrt p}$ :

$F(p)=\pm\dfrac{1}{2\sqrt p}$

$\therefore v=\pm\dfrac{1}{2\sqrt{p-\dfrac{q}{4v^2}}}$

$v^2=\dfrac{1}{4\left(p-\dfrac{q}{4v^2}\right)}$

$v^2=\dfrac{v^2}{4pv^2-q}$

$4pv^4-qv^2=v^2$

$4pv^4-(q+1)v^2=0$

$(4pv^2-(q+1))v^2=0$

$4pv^2-(q+1)=0$ or $v^2=0$ (reject)

$v^2=\dfrac{q+1}{4p}$

$v=\pm\dfrac{\sqrt{q+1}}{2\sqrt p}$

$u_p=\pm\dfrac{\sqrt{q+1}}{2\sqrt p}$

$u(p,q)=\pm\sqrt{p(q+1)}+g(q)$

$u_q=\pm\dfrac{\sqrt p}{2\sqrt{q+1}}+g_q(q)$

$\therefore\left(\pm\dfrac{\sqrt{q+1}}{2\sqrt p}\right)\left(\pm\dfrac{\sqrt p}{2\sqrt{q+1}}+g_q(q)\right)=\dfrac{1}{4}$

$\dfrac{1}{4}\pm\dfrac{\sqrt{q+1}}{2\sqrt p}g_q(q)=\dfrac{1}{4}$

$g_q(q)=0$

$g(q)=C$

$\therefore u(p,q)=\pm\sqrt{p(q+1)}+C$

$u(p,0)=\pm\sqrt p$ :

$C=0$

$\therefore u(p,q)=\pm\sqrt{p(q+1)}$

$u(x,y)=x\sqrt{y^2+1}$

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  • $\begingroup$ Thank you this is a very good and detailed answer. $\endgroup$ – AppMathPhys Dec 15 '14 at 14:28

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