Is the Laplace Equation with Initial Conditions Ill-Posed?

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SUMMARY

The discussion centers on the Laplace equation with initial conditions defined by the partial differential equation u_tt + u_xx = 0 for -∞ < x < ∞ and t > 0, with initial conditions u(x,0) = 0 and u_t(x,0) = f_k(x) where f_k(x) = sin(kx). The unique solution is given by u(x,t) = (1/k)sin(kx)sin(kt). The problem is established as ill-posed because, although a unique solution exists, it does not depend continuously on the initial data, particularly as k approaches 0, leading to discontinuities in the solution.

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The laplace equation whit initial conditions
u_tt + u_xx = 0 -oo<x<oo , t>0
u(x,0)=0
u_t(x,0)=f_k(x)
where f_k(x)=sin(kx), has the unique solution
u(x,t)=(1/k)sin(kx)sin(kt)
Show that the problem is ill posed.


I know that the equations is elliptic so i tried first whit the maximum principle but
this Partial differential equation has no boundary condition so i can use that principle.
The Fourier method requires a periodic boundary condition, but again there is no boundary condition in this PDE.
I then tried the energy method and i get this:
d||u(*,t)||^2/dt=d/dt (int (u(x,t)^2)) after some work i get (2k/tan(kx))||u(*,t)||^2 but how does this show that the problem is ill posed? am i doing it right? if not the how do i do? thanks :D
 
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A solution is well-posed if it satisfies

1. A solution exists
2. The solution is unique
3. The solution depends continuously on the data, in some reasonable topology.Otherwise it's ill posed. They tell you you have a unique solution, so you want to show the solution is not continuously dependent on the data. So, if you had something like, 1/tan(x), if tanx = 0, you have a discontinuity, and that would be a nice place to start
 
Office_Shredder said:
A solution is well-posed if it satisfies

1. A solution exists
2. The solution is unique
3. The solution depends continuously on the data, in some reasonable topology.


Otherwise it's ill posed. They tell you you have a unique solution, so you want to show the solution is not continuously dependent on the data. So, if you had something like, 1/tan(x), if tanx = 0, you have a discontinuity, and that would be a nice place to start

thx for the fast answer :), but i don't really understand. I'm using the energy method but i don't know if it is right to use it because everything gets very complicated and strange, and i think also that i must show that u(x,0)=0
and ||u||=k||f|| somehow? or don't i need to do this, the expression i get doenst show this :(
 
You talk about using the "energy principle" and earlier about the "maximum principle". Further you say "i think also that i must show that u(x,0)=0" when you are TOLD that this is true! The problem only asks you to show that the problem is "ill posed". Do what office-shredder said: use the definition of "ill-posed"! You are told that the solution exists, you are told that the solution is unique, so there is only one thing left. As Office Shredder told you "you want to show the solution is not continuously dependent on the data." What happens to your soluition as k goes to 0? Does this problem have a solution if k= 0?
 

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