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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?

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

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