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Show that the initial value problem $u_t+u_x=0$, with $u=x$ on $x^2+t^2=1$ has no solution.

However, if the initial data are given only over the semicircle that lies in the half-plane $x+t\le0$, the solution exists but is not differentiable along the characteristics that issue from the two end points of the circle.

I know how to graphically explain this as, if the condition was a circle, the circle would intersect the characteristic lines twice. This would give two solutions to the pde which is a contradiction as the solutions must be unique. Thus restricting the boundary to a semi-circle would solve this problem. However, I am struggling to represent this mathematically. Any help would be greatly appreciated.

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  • $\begingroup$ Did you solve the PDE? $\endgroup$ Sep 27, 2018 at 11:27

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With the full circle you get contradictory conditions on the same characteristic (e.g.: on $x=y$ must be $u(x,x)=\sqrt 2/2$ and $u(x,x)=-\sqrt 2/2$, obviously it's impossible), so it has not solutions (not "the solution has to be unique")

But you don't fix the problem giving values for a half circle. In this case occurs that the solution is not unique! With these boundary conditions $u$ is not determined along, e.g. $y=x+3$ as it doesn't cross that half circle. So said, the curve is too short to give value for $u$ along every characteristic curve ($y=x+c_1$)

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  • $\begingroup$ thank you!! this is really helpful! $\endgroup$
    – Shaun
    Sep 27, 2018 at 19:18

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