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I want to solve exercise 4 in chapter 7.5 Problems in Evans book, i.e.

$u_k$ converges weakly to $u$ in $L^2(0,T;H^1_0(U))$ and $u_k'$ converges weakly to $v$ in $L^2(0,T;H^{-1}(U))$, then $v=u'$. There is a hint: let $\phi\in C_c^1(0,T)$ and $w\in H^1_0(U)$. Then $\int_0^T \langle u_k',\phi w\rangle dt = -\int_0^T \langle u_k,\phi ' w\rangle dt$. How should I use this hint? My plan was to use the fundamental lemma of calculus of variations. For that, I wanted to show,

$$\int_0^T (u'(t)-v(t))\phi(t) dt=0$$

for all $\phi \in C^\infty_c((0,T))$. By definition of weak derivative and weak convergence I can write

$$\int_0^T u_k'(t)\phi(t) dt= -\int_0^T u_k(t) \phi'(t) dt$$ The LHS converges to $\int_0^T v(t)\phi(t)dt$ and the RHS to $-\int_0^T u(t)\phi'(t)dt $. Therefore

$$\int_0^T v(t)\phi(t)+u(t)\phi'(t)dt=0$$

I guess the hint motivates something similar. But how could I proceed and prove the statement? Or is my start completely wrong?

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1 Answer 1

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I think some care needs be taken at the point, where you write

[...] The LHS converges to $\int_0^T v(t) \phi(t)dt$ and the RHS to $-\int_0^T u(t) \phi'(t)dt$. Therefore [...]

This is true, but actually not given in the exercise statement, and I believe its deduction is the main part of the problem. Otherwise you'd already be done, having shown that $\int_0^T u(t)\phi'(t)\, dt = -\int_0^T v(t)\phi(t)\, dt$ for all $\phi\in C_c^1(0,T)$. A function $v$ (by definition) is a weak derivative of $u$ if this holds.

Maybe we should first recall what it means for a sequence $w_n$ to converge to $w$ weakly in $L^2(0,T; X^\ast)$, where $X$ is some Banach space. By definition (at least if I'm not misunderstanding something myself) this means that

$$\int_0^T \langle w_n, g\rangle \, dt \to \int_0^T \langle w,g\rangle\, dt \qquad \forall \, g \in L^2(0,T; X)$$

where $\langle\; , \;\rangle $ is the natural pairing $X^\ast \times X \to \mathbb R$.

Now following the hint: If we set $X = H^1_0(U)$, $X^\ast= H^{-1}(U)$ and $g = \phi(t) w$ for $\phi\in C_c^1(0,T)$, $w\in H^1_0(U)$, then $g, g'\in L^2(0,T;H^1_0(U))$. So by assumption on $u$ and $v$:

\begin{align} \int_0^T \langle u_n, \phi' w\rangle \, dt &\to \int_0^T \langle u,\phi' w\rangle\, dt \\ \int_0^T \langle u_n', \phi w\rangle \, dt &\to \int_0^T \langle v,\phi w\rangle\, dt \end{align}

We can rewrite $$\int_0^T \langle u, \phi' w\rangle \, dt = \int_0^T \phi' \langle u, w\rangle \, dt = \int_0^T \langle u\phi' , w\rangle \, dt = \left\langle \left(\int_0^T u(t)\phi'(t) \, dt\right) , w\right \rangle$$

where I have made use of Fubini in the last equality. A similar equality is true for $v$. Using this together with $\int_0^T \langle u_n, \phi' w\rangle \, dt = -\int_0^T \langle u_n', \phi w\rangle \, dt$, we obtain

$$\left\langle \left(\int_0^T u(t)\phi'(t) \, dt\right) , w\right \rangle = \left\langle \left(\int_0^T -v(t)\phi(t) \, dt\right) , w\right \rangle \qquad \forall\, w\in H^1_0(U)$$

Therefore $\int_0^T u(t)\phi'(t) \, dt = -\int_0^T v(t)\phi(t) \, dt$ for all $\phi \in C_c^1(0,T)$. So $v$ is a weak derivative of $u$.

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Note: I'm just learning about this stuff myself, so please be critical! I'm not to be trusted. ;-) –  Sam Jun 24 '12 at 14:33
    
Just a small comment after a while: maybe it is a little bit petty. If I'm not wrong it would be better to write $\int_0^T\langle \phi^'w,u_n\rangle dt\to\int_0^T\langle \phi^'w,u\rangle dt$. Since weak convergence of $u_n\to u$ in $L^2(0,T;H^1_0)$ means $\int_0^T\langle \phi,u_n\rangle dt \to \int_0^T \langle \phi,u\rangle dt$ for all $\phi \in H^{-1}$. Of course we naturally associate $\phi^'w$ with an element in $H^{-1}$ an everything works exactly in the same way. –  math Aug 13 '12 at 7:32

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