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Let $b$ be continuous and increasing with $b(0) = 0$. Define $\Psi(t) = \int_0^t b(s)\;ds$ and $\Psi^*(s) = \sup_{r \in \mathbb{R}} (sr-\Psi(r))$.

(Note $\Psi^*(b(s)) + \Psi(s) = sb(s)$).

Let $v \in L^2(0,T;H^1_0(\Omega))$ with $\frac{\partial b(v)}{\partial t} \in L^2(0,T;H^{-1}(\Omega))$. Then $$\langle \frac{\partial b(v)}{\partial t}, v \rangle_{H^{-1}(\Omega), H^1(\Omega)} = \frac{d}{dt}\int_{\Omega}\Psi^*(b(v)).$$

This result follows by (the more general) Lemma 1.5 of the famous Quasilinear Elliptic-Parabolic Differential Equations paper by Alt and Luckhaus. The authors prove it by discretising in time.

Does anyone know an alternative proof (where no discretisation is used)? Using density of smooth functions does not help. Let us assume $b$ and $b^{-1}$ are differentiable if necessary.

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  • $\begingroup$ Why do you want to avoid time discretization? $\endgroup$ Jun 9, 2014 at 22:00
  • $\begingroup$ Fro smooth $b$ and $v$'s you have by standard properties of the Legendre transform $\Psi^*$ that $<\partial_t b(v)\cdot v>=\partial_t \Psi^*(v)$. Here the dot denotes the euclidean scalar product and this also holds if $b,v$ are vector-valued. Why doesn't approximation help? $\endgroup$ Jun 9, 2014 at 23:06
  • $\begingroup$ @leomonsaingeon Suppose I approximate $b(v)$ by functions $b_n \in L^2(0,T;H^1) \cap H^1(0,T;L^2)$ or even $C_c^\infty([0,T];H^1)$. I will get an expression involving $\int(b^{-1})'(b_n)b_n'b_n - \frac{d}{dt}\int\Psi(b^{-1}(b_n))$. I can't take the derivative inside the brackets because the integrand needs to be in $L^2(0,T;H^1) \cap H^1(0,T;H^{-1})$ at least. But the it's not (eg. can't show that the gradient is in $L^2(0,T;L^2)$ after using chain rule). $\endgroup$
    – LapLace
    Jun 10, 2014 at 9:37
  • $\begingroup$ ..I don't want to use density of functions defined on $Q=(0,T)\times\Omega$, I would rather use Bochner space. Maybe something like $C^1(0,T;C^1(\Omega))$ is dense in the desired space or something.. $\endgroup$
    – LapLace
    Jun 10, 2014 at 9:37
  • $\begingroup$ @MarkPeletier I don't like discretisation and I think it's not necessary for something simple like this identity $\endgroup$
    – LapLace
    Jun 10, 2014 at 9:38

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