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Joonas Ilmavirta
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For $p > 1$, consider the equation $$\langle u_t, v \rangle + \int_\Omega |\nabla u|^{p-2}\nabla u \nabla v = \langle f, v \rangle$$ $$u(0) = u_0$$ $$u|_{\partial\Omega} =0$$ for all $v \in W^{1,p}(\Omega)$, where $u_0 \in W_0^{1,p}\cap L^2$ and $f=f(x) \in W^{-1,q}(\Omega)$. This has a solution $u \in L^p(0,T;W^{1,p}_0)$ with $u_t \in L^q(0,T;W^{-1,q})$

If I assume additional smoothness on the data, can I get $u_t$ in a nicer space? Eg. $u_t \in L^s(0,T;L^r)$ for positive $s$ and $r$? I would like $u_t -\Delta_pu =f$ to hold pointwise a.e., basically.

Eg. take $p=q=2$. Then if $f \in L^2(0,T;L^2)$ and $u_0 \in H^1$ then $u_t \in L^2(0,T;L^2)$.

For $p > 1$, consider the equation $$\langle u_t, v \rangle + \int_\Omega |\nabla u|^{p-2}\nabla u \nabla v = \langle f, v \rangle$$ $$u(0) = u_0$$ $$u|_{\partial\Omega} =0$$ for all $v \in W^{1,p}(\Omega)$, where $u_0 \in W_0^{1,p}\cap L^2$ and $f=f(x) \in W^{-1,q}(\Omega)$. This has a solution $u \in L^p(0,T;W^{1,p}_0)$ with $u_t \in L^q(0,T;W^{-1,q})$

If I assume additional smoothness on the data, can I get $u_t$ in a nicer space? Eg. $u_t \in L^s(0,T;L^r)$ for positive $s$ and $r$?

Eg. take $p=q=2$. Then if $f \in L^2(0,T;L^2)$ and $u_0 \in H^1$ then $u_t \in L^2(0,T;L^2)$.

For $p > 1$, consider the equation $$\langle u_t, v \rangle + \int_\Omega |\nabla u|^{p-2}\nabla u \nabla v = \langle f, v \rangle$$ $$u(0) = u_0$$ $$u|_{\partial\Omega} =0$$ for all $v \in W^{1,p}(\Omega)$, where $u_0 \in W_0^{1,p}\cap L^2$ and $f=f(x) \in W^{-1,q}(\Omega)$. This has a solution $u \in L^p(0,T;W^{1,p}_0)$ with $u_t \in L^q(0,T;W^{-1,q})$

If I assume additional smoothness on the data, can I get $u_t$ in a nicer space? Eg. $u_t \in L^s(0,T;L^r)$ for positive $s$ and $r$? I would like $u_t -\Delta_pu =f$ to hold pointwise a.e., basically.

Eg. take $p=q=2$. Then if $f \in L^2(0,T;L^2)$ and $u_0 \in H^1$ then $u_t \in L^2(0,T;L^2)$.

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Strong solution to $u_t - \Delta_p u = f$

For $p > 1$, consider the equation $$\langle u_t, v \rangle + \int_\Omega |\nabla u|^{p-2}\nabla u \nabla v = \langle f, v \rangle$$ $$u(0) = u_0$$ $$u|_{\partial\Omega} =0$$ for all $v \in W^{1,p}(\Omega)$, where $u_0 \in W_0^{1,p}\cap L^2$ and $f=f(x) \in W^{-1,q}(\Omega)$. This has a solution $u \in L^p(0,T;W^{1,p}_0)$ with $u_t \in L^q(0,T;W^{-1,q})$

If I assume additional smoothness on the data, can I get $u_t$ in a nicer space? Eg. $u_t \in L^s(0,T;L^r)$ for positive $s$ and $r$?

Eg. take $p=q=2$. Then if $f \in L^2(0,T;L^2)$ and $u_0 \in H^1$ then $u_t \in L^2(0,T;L^2)$.