Consider the following PDE for $u(x, y)$

$u_x + 2√yu_y = 0$

which is defined for $y>0$

Solve the PDE in the domain $Ω$ = {$(x, y) : x ∈ R, y > 0$} subject to the condition $u(x, 0) = h(x)$ for some given function $h$

Ok, so I have found the characteristic curves to be described by

$x −√y = c_1$

The problem I have is that the initial condition is a function h(x), usually its defined as being equal to something like $x^2$ or $0$ etc. Im probably overthinking this, but how does one tackle a problem like this?

  • $\begingroup$ I played around and got a solution of $u(x,y) = h(x-√y)$ $\endgroup$ – Clovers Dec 3 '16 at 1:22
  • 1
    $\begingroup$ You got it! Just posted a solution (I didn't see your comment). $\endgroup$ – User8128 Dec 3 '16 at 1:28

I'll write up a full solution.

The characteristics originating from points $(x,y) = (\xi, 0)$ and parameterized by $s$ are given by \begin{align*} \dot x &= 1, \,\,\,\,\,\,\,\,\,\, x(0) = \xi,\\ \dot y &= 2\sqrt y, \,\, y(0) = 0,\\ \dot z &=0, \,\,\,\,\,\,\,\,\,\, z(0) = h(\xi), \end{align*} where $z(s ;\xi) = u(x(s;\xi),y(s;\xi))$. Solving gives $x = s + \xi$, $y = s^2$, $z = h(\xi)$. Solving for $\xi$, we see $\xi = x - \sqrt y$. Then $$u(x,y) = z(s(x,y);\xi(x,y)) = h(x-\sqrt y).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.