# Partial Differential Equation, help with initial conditions

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.

• Did you solve the PDE? Sep 27, 2018 at 11:27

## 1 Answer

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 ($$y=x+c_1$$)

• thank you!! this is really helpful! Sep 27, 2018 at 19:18