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Fix $T > 0.$

Let $V \subset H \subset V^*$ be a Gelfand triple. Consider the linear parabolic PDE $$u_t - Au = f\quad\text{in $L^2(0,T;V^*)$}$$ $$u(0) = u_0$$ where $u_0 \in H$ and $f \in L^2(0,T;V^*)$ and $A$ is some elliptic smooth operator.

we know that this problem has a unique solution $$u \in L^2(0,T;V), u_t \in L^2(0,T;V^*)$$ by using a Galerkin method for example.

Questions:

  1. What exactly does it mean to say that we can extend $u$ to a global solution? I assume this means we can write $u \in L^2(0,\infty;V)$ and that $u$ solves the PDE I wrote above on $[0,\infty)$. How is $f$ extended from $[0,T]$ -- do we assume we are given such an extension.

  2. Under what conditions does one obtain a global solution?

(I tried all the other threads). Any reference to source that talks about this in detail would be appreciated too. Thanks.

Edit: This is confusing. Some papers

  • consider a PDE and say that "because we have existence of $u \in L^2(0,T;V)$ for any $T>0$, we have global existence".
  • other papers say solve the IVP, and then solve another IVP with $\tilde u(0) = u(T)$ and in this way extend the solution

Please someone give me authoritative reference on this topic.

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Define $u(x)=u_n(x)$ for $x\in[0,n]$, where $u_n$ is the unique solution in the interval [0,n]. I think that $u$ is a good candidate for a global solution. –  Tomás Jan 23 at 17:31
    
@Tomás That is what I was thinking, but see this: mathoverflow.net/questions/155475/… I kind of agree with what is written there.. this approach is just local in some way, so I think there is another way. –  michael_carbon Jan 23 at 18:10
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I think that if there is a global solution then, it must be $u$ because of uniqueness. –  Tomás Jan 23 at 18:51

2 Answers 2

  1. Global solution means to extend it from $[0,T]$ to $[0,\infty)$, provided of course that $f\in L^2_{\mathrm{loc}}([0,\infty),V^*)$.

  2. The solution is obtainable with standard semigroup methods (unless $f$ depends on $u$, in which case you need to develop a suitable energy method), i.e., $$ u(t)=\mathrm{e}^{tA}u_0+\int_0^t\mathrm{e}^{(t-s)A}f(s)\,ds. \tag{1} $$ If $A$ is an elliptic operator (i.e. $A=-\Delta$), then $\mathrm{e}^{tA}$ defines a strongly continuous semigroup which is even contractive: $\|\mathrm{e}^{tA}g\|_{V^*}\le\|g\|_{V^*}$, for $t>0$, which makes $(1)$ very easy to understand.

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Thanks. I am unfamiliar with semigroup method, please can you maybe say something using Galerkin method? –  michael_carbon Jan 23 at 21:34
    
In Galerkin formulation you get a system of ODEs with respect to the time dependent coefficients, and the solution can be extended to infinity provided that none of the coefficients blows up at a finite time. –  uvs Jan 24 at 11:15
    
@uvs In a paper I read, it says that global existence of a solution is true because we can get a solution $u \in L^2(0,T;H^1)$ for any $T>0$. I guess the authors are using bad terminology here. –  michael_carbon Jan 24 at 14:10
    
@uvs So then do we get $u_n \to u$ (weakly) in $L^2(0,\infty;H^1)$ and $u_n' \to u'$ in $L^2(0,\infty;H^{-1})$ weakly, where $u_n$ are the Galerkin approximations? –  michael_carbon Jan 24 at 14:39

Once you have a solution $u$ this solution will have a maximum lifespan $T^*$.

The statement "there exists a solution for every $T$" implies $T^*=\infty$:

Assume that the solution is not global, i.e. $T^*<\infty$. Consequently there is no solution for some $t>T^*$ (by the definition of $T^*$). This is a contradiction with thre previos statement "there exists a solution for every $T$".

So, from my viewpoint and if my reasoning is correct, the authors are NOT using "bad terminology", right?

Now, let me use the (onedimensional with domain the real line) heat equation as an example:

For the heat equation with an $H^s$ initial data, one gets the estimates $$ \|f(t)\|_{H^s}\leq \|f_0\|_{H^s}. $$ So, by standard arguments one gets a weak solution up to time $T$. If this weak solution is not global, then the solution should leave the space $H^s$ (Otherwise write $T^*$ for the maximum lifespan, assuming that the solution doesn't leave $H^s$, the function $f(T^*)\in H^s$. Now take $g_0=f(T^*)$. By the same argument before, you gave a solution $g$ at least for small time. This contradicts the definition of $T^*)$. As the bounds implies that the solution never leaves $H^s$ (the solution operator is a contraction), you get global solutions.

The idea of using the final data $u(T)$ as a new initial data $v(0)$ to extend the solution requires the time of existence of the new "step" $v$ to be larger (or at least, the same) as the previous step. Otherwise, one can have infinitely many "steps" $T_i$ in such a way the total lifespan $T^*=\sum T_i$ is finite.

I don't know if this clarify something...

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Thank you, very good answer. So if we suppose $u$ is a global solution as in $u \in L^2(0,T;V)$ for each $T>0,$ then $u:[0,\infty) \to V$ is well-defined but we cannot yet say $u \in L^2(0,\infty;V)$, we have to check that (need some estimates). Let me ask one question; in the previous answer, uvs (and elsewhere I have heard too) claimed that in the Galerkin approximation, if the coefficients don't blow up we can extend the solution to infinity. What does this mean? The Galerkin approximations satisfy $u_m \in L^2(0,\infty;V)$?? –  michael_carbon Jan 26 at 9:30
    
Well, as long as the galerkin coefficients exist, you will have a weak solution. However, notice that a solution such that $\limsup_{t\rightarrow\infty}|u(t)|_{V}=\infty$ is also a global solution that can't be in $L^2((0,\infty),V)$ –  guacho Jan 26 at 18:39

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