For $f \in C^0([0,1])$, I have the following partial differential equation:
$$u''(x) = f(x)$$ in $\Omega = (0,1)$ $$u'(0) = u'(1) = 0$$
Why is this equation not well posed in $H^1(\Omega)$?
--> Is it because of the missing boundary conditions?
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That problem can be interpreted in a weak formulation, that is, a solution to that equation is a $u\in H^1(\Omega)$ such that $$ \int_\Omega - u'\phi' =\int_{\Omega} f\phi,\qquad \forall \phi \in H^1(\Omega).$$ Thus, there are no regularity problems. The lack of well-posedness comes from a much more trivial issue; there is no uniqueness, because if $u$ is a solution, then so is $u+C$ for any constant $C\in\mathbb R$.
See the book of Brezis, section 8.4.
Giuseppe Negro
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Set $v=u'$ then you are trying to solve the problem $$ v'=f(x)\\ v(0)=v(1)=0 $$ But $$v(1)-v(0)=\int_0^1v'(s)\,ds=\int_0^1f(s)\,ds$$ so the equality of the boundary values is only possible if $f$ has mean zero.
In general you only get one boundary condition for a first-order ODE. As solutions, if any exist at all, are in $C^2$, $$ u(t)=u(0)+\int_0^t(t-s)f(s)\,ds, $$ the discussion of $H^1$ properties is a side-issue.
Lutz Lehmann
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Good answer. You are right that a "weak formulation" is not really important since here the problem is 1-dimensional, so it can be recast as an integral equation. – Giuseppe Negro Oct 25 '18 at 13:06
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I see how regularity is a non-issue since the solution will be in $C^2$ anyway. Deleted the silly answer ;) – Miguel Oct 25 '18 at 16:09
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Thank you for this answer! What if $f(x) = (x- \frac{1}{2})$ and $u(0) = 0$ --> what basis for a finite dimensional subspace $S \subset S^1_G := { \phi \in C^0([0,1]) | \forall r \in G: \phi|_r \in \mathbb P_1 }$ would you choose to approximate the solution $u$ numerically? – StMan Oct 25 '18 at 19:59