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For any $n > 0$, standard theory implies there is a unique $u_n \in L^2(0,n;V)$ with $u_n' \in L^2(0,n;V^*)$ such that $$u_n' + Au_n = f\quad\text{as an equality in $L^2(0,n;V^*)$}$$ $$u_n(0) = u_0$$ where $f \in L^2(0,n;V^*)$, $u_0 \in H$ and $A$ is some smooth elliptic operator. Here $V \subset H \subset V^*$ is Gelfand triple.

If $n > m$, then we see that $u_n|_{[0,m]} = u_m$ by uniqueness of solutions.

Now define $v(t) = u_n(t)$ if $t \leq n$.

Then isn't $v$ in some sense a global solution of the PDE (assuming we have $f \in L^2(0,\infty;V^*)$)?

My question is this is sense of a global solution useful or not; what is the usual sense of a global solution? Because I have read many times "if the norm of the solution stays bounded then we can extend the solution globally" but never got any details.

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  • $\begingroup$ @AthanagorWurlitzer No I don't assume there is a global solution. Since we have a unique solution of the PDE for each time interval $[0,T]$, it follows that there is a solution $u_{T_1}$ to the PDE on the time interval $[0,T_1]$ where $T_1 > T.$ Then the restriction of $u_{T_1}$ to the interval $[0,T]$ solves the PDE on the interval $[0,T]$. $T_1$ was arbitrary so we can make it as big as we want. $\endgroup$
    – michael_carbon
    Commented Jan 23, 2014 at 11:14
  • $\begingroup$ So your question then is whether $u\in L^2(0,\infty;V)$? In general no, even for ODEs! $\endgroup$
    – Michael Renardy
    Commented Jan 23, 2014 at 11:23
  • $\begingroup$ @AthanagorWurlitzer Please see my edited question. Sorry for initial bad phrasing. $\endgroup$
    – michael_carbon
    Commented Jan 23, 2014 at 14:15
  • $\begingroup$ @MichaelRenardy Please see the edited question, I clarify what I meant to say. $\endgroup$
    – michael_carbon
    Commented Jan 23, 2014 at 14:16
  • $\begingroup$ (I deleted my earlier comments to clarify the presentation). No, it is not a global solution, it is still a local solution in time. Local means valid on bounded subsets of the domain of interest, which is exactly what your solution is. It does not matter whether $n$ is large or not, it isn't infinite. $\endgroup$
    – username
    Commented Jan 23, 2014 at 14:37

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Take for example the function $u(x)=sin(\pi x)$, which is in $H^1_0(0,1)$, and satisfies
$$ u_t -u^{\prime\prime}=f,\quad u_0=sin(\pi x), \mbox{ on } L^2(0,T;H^{-1}(\Omega)) $$ with $f=\pi^2\sin(\pi x)$. The $L^2(0,T;H^1_0(\Omega))$ norm of $u$ is $$ \|u\|_{L^2(0,T;H^1_0(\Omega))}=\int_0^T \int_0^1 \pi^2 \cos^2(\pi x)dxdt= \frac{T \pi^2}{2}. $$ It isn't in $L^2(0,\infty;H^1_0(\Omega))$.

This answer corresponded to a previous version of the question. The key point that you don't pay attention to in your question is that the norm stays bounded! Your question in the end isn't about linear PDE, it is simply: Let $f$ is locally integrable and there is uniform upper bound on its integral which does not depend on the subdomain, is the function integrable, and the answer is yes, but that's more for stack exchange.

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  • $\begingroup$ Thanks for answering. If you (or anyone else) cares to answer I've opened a thread of SE math.stackexchange.com/questions/648887/…). $\endgroup$
    – michael_carbon
    Commented Jan 24, 2014 at 13:12
  • $\begingroup$ So if I understand you right, it is enough that the data $f$ be uniformly bounded independent of $T$. $\endgroup$
    – michael_carbon
    Commented Jan 24, 2014 at 13:14
  • $\begingroup$ @michael_carbon You SE thread is still about PDE, but I don't think this is what is stopping you. What you want is a proof of the integrability statement I stated, which you should ask on SE. 'My' f is your v, of course. $\endgroup$
    – username
    Commented Jan 24, 2014 at 13:37
  • $\begingroup$ I think I finally understand, let me just go back to the PDE I have: it is true that for each $T$ there is a unique solution on $[0,T]$. Defining $v$ the way I did in the OP, we just need to check that $v \in L^2(0,\infty;V)$, which you say holds if the integrability statement you stated is true. I was reading some papers and got confused; I guess in nonlinear PDEs it is in general NOT immediate that "for each $T$ there is a solution on $[0,T]$"; rather there is some $T$ (not given a priori) on which the solution exists. then the task is to show that this $T$ can be extended... $\endgroup$
    – michael_carbon
    Commented Jan 24, 2014 at 15:59
  • $\begingroup$ ..But we get this for the PDE in the OP easily because of the simple nature of the PDE if I am right. $\endgroup$
    – michael_carbon
    Commented Jan 24, 2014 at 15:59

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