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Let $S = [0,1]\times [0,T]$ for $0 < T \leq 1$. Let $X([0,T]) = \tilde{C}^{2, \alpha}(S\times [0,T])$ and $Y([0,T])=\tilde{C}^{0, \alpha}(S\times [0,T])$ be functions in the respective parabolic Hölder spaces.

Consider the PDE $$u_t = a(x,t)u_{xx} + b(x,t)u_x + c(x,t)u + f$$ $$u(\cdot, 0) \equiv 0$$ If $a, b, c, f \in Y([0,T])$, then we have the a-priori estimate $$\lVert u \rVert_{X([0,T])} \leq C(T)\lVert f \rVert_{Y([0,T])}$$ where $C(T)$ is a constant depending on $T$. This is a standard result.

Now define coefficients $a'(x,t) = \begin{cases}a(x,t) &\text{if $0 \leq t \leq T$}\\ a(x,T) &\text{if $T \leq t \leq 1$} \end{cases}$ and likewise for the other coefficients.

Then it holds that $$\lVert a'(x,t) \rVert_{Y([0,1])} \leq \lVert a(x,t) \rVert_{Y([0,T])}$$ and similarly for the others.

From this, $$\lVert u \rVert_{X([0,T])} \leq C\lVert f \rVert_{Y([0,T])}$$ where $C$ does not depend on $T$ anymore.

Is this a valid argument? Can someone elaborate a bit on this? This seems a very easy to get rid of dependence on $T$, so why are people not using it more often?


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@DavideGiraudo Suppose that it is. (It would be anyway I think since we're just taking supremums and we only made the domain bigger but not added any more values that the function can take). – TagWoh Aug 2 '12 at 20:23
@DavideGiraudo You're right. I should take $k=0$, so the coefficients need only be continuous. Thanks – TagWoh Aug 2 '12 at 21:00
@DavideGiraudo That is the point that I do not fully understand. I think, since we can bound the coefficients on $[0,1]$, we can perhaps use standard results to give that a-priori bound (the last displayed equation). PS: the constant $C$ corresponds to taking $T=1$. – TagWoh Aug 2 '12 at 21:41
@DavideGiraudo Sorry for confusion. This is something I saw in paper so it must be true. I am trying to understand it. – TagWoh Aug 2 '12 at 21:47
up vote 1 down vote accepted

The equation is the same for $t\leq T$, and the initial condition doesn't change, and using the a priori estimate: $$\lVert u\rVert_{Y([0,T])}\leq \rVert u'\rVert_{Y([0,1])}\leq C(1)\lVert f'\rVert_{Y([0,1])}.$$ But the supremum norm $f'$ is the same as the supremum norm of $f$, since they reach the same values, and the Hölder semi-norm is the same. Indeed, we have to make $|x-y|^{\alpha}$ smaller hence we should take them in $[0,T]$ (taking $y>T$ won't make the $\sup$ relevant).

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As for the question why people are not using it more often, it is kind of trivial because you restrict $T$ to be smaller than $1$. It essentially means that you replaced $C(T)$ by its maximum over $T\in(0,1]$.

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But there's nothing special about choosing 1 here, right? So one can choose a bigger time. – Court Aug 13 '12 at 11:46
@Court: Yes, with the constant increasing as you choose bigger time. In particular, one cannot choose $T=\infty$ because the constant $C(T)$ may blow up. – timur Aug 13 '12 at 14:33

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