Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $u \in L^p(0,T;V)$ denote the solution to the parabolic PDE $$u_t + \Delta u = f\qquad\text{a.e $t \in [0,T]$}$$ where $u_t \in L^q(0,T;V^*).$ We have the usual assumptions on $V \subset H \subset V^*$ and $H$.

Now this PDE has a unique solution, i.e. $u$ is unique.

$T$ is arbitrary, so we have a set of solutions $u_n \in L^p(0,n;V)$ that solve the PDE a.e in $[0,n]$.

Now i read:

Define $$v(t) = u_n(t) \qquad \text{if $t \leq n$}$$ Then $v \in L^p(0,\infty;V)$ solves the PDE on $[0,\infty).$

Is this obvious that it solves the PDE on the real line? furthermore, to show that $v$ has finite norm in $L^p(0,\infty;V)$ am I right that we need a uniform bound independent of $n$ on $u_n$. These a priori estimates come from considering Galerkin method for example, but I thought that the constant depends on the end time, which in this cases happens to be $n$. So I am not sure how to show it has finite norm.

share|cite|improve this question
Where have you found this proof? – Tomás Sep 24 '13 at 15:03
@Tomás It is in book by Boyer, Fabrie "Mathematical Tools for the Study of the Incompressible Navier-Stokes Equations and Related Models", page 361 (Globality of weak solutions). – BigUser Sep 24 '13 at 15:18
The problem there is a particular one. Why do you think that it can be generalized in such way? – Tomás Sep 24 '13 at 15:42
@Tomas I also saw it here:'s_book_hyperref.pdf See the last paragraph of the proof of Theorem 11.2 on page 82. Tbh I do not understand that proof; I think there are several typos. But that is a separate thread. – BigUser Sep 24 '13 at 15:46
Well, so I think it is better to concentrate on the Navier Stokes equation. What do you think? – Tomás Sep 24 '13 at 15:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.