find the general solution of Ux = 0, where U := U(x,y), and Explain why when it is subject to the initial condition U(x,0) = Uo(x) is not well-posed

Question

anonymous · Answer

@Zarkon can you help me please with this question?

anonymous · Answer

I try to solve it but I got stuck; here is what i did:
 I used the method of characteristic ; 
dx/ds=1 so x=s+c and we usually use the initial condition x(0)=x' so c=x' 
and dy/ds=0 so y=0 and by the initial condition y(0)=0 so c=0 so y=0,
 now u (x,y)= u (x,0) = uo (x')= uo (x - s),
 how to find s here in order to get the solution u (x,y) ???

kainui · Answer

Does Ux=0 mean $U*x = 0$ or $\frac{\partial U}{\partial x} = 0$?

kainui · Answer

I guess I'm just confused since the second case looks like how you'd see a problem you'd attack with method of characteristics but uhh, It just seems like it would be unnecessary to solve that way since you can immediately integrate it.

$$\frac{\partial U}{\partial x} = 0$$
Since it's 0 it must be a function of y, since it doesn't depend on x. Normally this would be a constant if it was single variable though.
$$U(x,y) = f(y)$$

kainui · Answer

Yeah this must be it, because this makes sense why they'd say $$U(x,0)=U_0(x)$$ is not well posed since
$$U_0(x)=f(y)$$
So the initial conditions in this sense would mean that $f(y)=C$ must be a constant since x and y are independent variables but equal everywhere and some condition like that would be meaningless haha.

anonymous · Answer

if it  is not well posed means there is either no solution
or the solution is not unique!

anonymous · Answer

i don't know i'm still a little confuse!

anonymous · Answer

why is it not well posed then? it is because the solutions depend on y so it is not unique?

kainui · Answer

Well by solving the differential equation we have:

$$U(x,y)=f(y)$$

So now imagine I give you some condition that $U(x,0)=x^3$ as a specific case. Plugging in gives us:
$$x^3=f(0)$$

But f(0) is a constant and doesn't depend on x! So we have a contradiction.

kainui · Answer

So to specifically answer your question: "if it  is not well posed means there is either no solution
or the solution is not unique!"

This means given an arbitrary $U(x,0)$ there is no solution.

kainui · Answer

There's only a special case of when this works, and that's only when the initial function of x is equal to a constant $U(x,0)=C$ (so really it's not even a function of x hah)

anonymous · Answer

thanks