Hadamard and well-posed problems.

[Kizaru]

Hadamard and "well-posed" problems.

I can't really find much clarification on Hadamard's definition of a well-posed problem.
My confusion comes from knowing exactly what is meant by the second and third properties:
2) The solution is unique
3) The solution depends continuously on the data, in some reasonable topology. http://en.wikipedia.org/wiki/Well-posed_problem"

For example, a solution would exist for a PDE y * u_xx = x * u_yy with u(0,y) = 2 and u(x,0) = 2 that is simply a constant. But is the solution, u=2, considered unique? Does the solution depend on the data?

Does the solution for u(x,y) fail property 3 because it no longer satisfies the PDE if the initial conditions are changed?

Edit: Maybe this belongs in the Differential Equations forum. Sorry. Can this be moved? Thanks.

 

[HallsofIvy]

I can't really find much clarification on Hadamard's definition of a well-posed problem.
My confusion comes from knowing exactly what is meant by the second and third properties:
2) The solution is unique
3) The solution depends continuously on the data, in some reasonable topology. http://en.wikipedia.org/wiki/Well-posed_problem"

For example, a solution would exist for a PDE y * u_xx = x * u_yy with u(0,y) = 2 and u(x,0) = 2 that is simply a constant. But is the solution, u=2, considered unique?

Is there any other function of x and y that will satisfy those conditions? If not, then it is unique.

Does the solution depend on the data?

Well suppose the problem were exactly the same differential equation but u(0,y)= 3 and u(x,0)= 3. Then u(x,y)= 3 is a solution. Since changing the "data" changes the solution, yes, the solution depends on the data.

Does the solution for u(x,y) fail property 3 because it no longer satisfies the PDE if the initial conditions are changed?

No, property 3 does not say that the solution must be independent of the initial conditions, it say it must depend on them continuously. suppose the intial conditions were u(x, 0)= 2+\delta, u(0, y)= 2+ \delta. Then [/itex]u(x,y)= 2+ \delta[/ itex] is obviously a solution. And, as \delta goes to 0, that solution goes to u(x,y)= 2. In that situation, the solution depends continuously on the initial values.

Edit: Maybe this belongs in the Differential Equations forum. Sorry. Can this be moved? Thanks.

Okay, I will move it.

 

[Kizaru]

Thank you so much! It's much clearer than my professor's explanation :)