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Let $\boldsymbol{u}\in L^2(0,T;H_0^1(U))$, $\boldsymbol{u}'\in L^2(0,T;H^{-1}(U))$ and $\boldsymbol{f}\in L^2(0,T;L^2(U))$. Suppose that $$ \int_0^T \langle \boldsymbol{u}',\boldsymbol{v}\rangle + B[\boldsymbol{u},\boldsymbol{v};t]\,dt = \int_0^T (\boldsymbol{f},\boldsymbol{v})\,dt, $$ for all $\boldsymbol{v}\in L^2(0,T;H_0^1(U))$.

Question:

How to derive that $$ \langle \boldsymbol{u}', v\rangle + B[\boldsymbol{u}, v;t] = (\boldsymbol{f},v) $$ for each $v\in H_0^1(U)$ and a.e. $0\leq t\leq T$.

This is an assertion in Evans' PDE book (section 7.1, Theorem 3) but I can't figure out why. Any hints are appreciated!

Peter Phipps
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Stephen
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1 Answers1

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Let $w \in H^1_0(U)$ and $\varphi \in C^\infty_c((0,T))$ be arbitrary and consider $v \in L^2(0,T;H^1_0(U))$ given by the product $v(x,t) = w(x) \varphi(t)$. Using this $v$ as a test function, we get $$ \int_0^T [\langle \boldsymbol{u}',\boldsymbol{w}\rangle + B[\boldsymbol{u},\boldsymbol{w};t] - (\boldsymbol{f},\boldsymbol{w})]\varphi(t)\,dt = 0. $$ Since this holds for all such $\varphi$, we conclude that the map it's integrated against must vanish a.e. in time. That is, we learn that $$ \langle \boldsymbol{u}',\boldsymbol{w}\rangle + B[\boldsymbol{u},\boldsymbol{w};t] - (\boldsymbol{f},\boldsymbol{w}) =0 $$ for each such $w$ and a.e. $t \in (0,T)$. This is what you want upon swapping notation $w \mapsto v$.

Glitch
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