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I am looking for results that show smooth dependence of a solution to a parabolic equation, from the initial data.

More specifically I have the following problem:

CONSIDER spaces $P:=\mathbb{R}^k$ ("parameter space"), $\Omega:=\mathbb{R}^n$ ("actual space") and $T=[0,t_0]$ ("time").

Consider also $u_0,f_0\in C^{\infty}(P\times \Omega)$, and $L$ a linear elliptic operator on $\Omega$, whose components depend smoothly on $P$. For every $p\in P$, there is a solution $u_p\in C^{\infty}(\Omega\times T)$ of the PDE

${\partial\over \partial t}u_p+L_p u_p=f_p \qquad on\;\{p\}\times\Omega\times T$

$u_p=(u_0)_p\qquad on\; \{p\}\times\Omega\times\{0\}$

QUESTION: is the function $\overline{u}(p,x,t):=u_p(x,t)$ a function in $C^{\infty}(P\times\Omega\times T)$?

The problem comes from the following, geometric situation: I have a space $P\times \Omega$ with a metric on it, and $L$ is just the Laplacian $\Delta$ of $P\times \Omega$, restricted to the levels $\{p\}\times \Omega$.

Any reference will be very welcome. Thanks in advance.

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You differentiate (wrt $p$) the equation and derive an equation for the $p$-derivatives of $u$. This will be a parabolic equation so you are fine. –  timur Jul 19 '12 at 13:29
Can you expand a little bit on that? In particular: when you differentiate wrt $p$, how does the term $(L_pu_p)'$ behave? It feels like it becomes $(L_p)'u_p+ L_p(u_p')$ (here by "u'" i mean differentiation wrt $P$ of course). But then i don't see how the resulting equation is still parabolic... –  Marco Radeschi Jul 19 '12 at 16:04
Yes, then you move the term $(L_p)'u_p$ to the right hand side and treat is as a source term. –  timur Jul 19 '12 at 16:36

1 Answer 1

If you wish to consider continuous dependence for your problem, you must further specify the type of solutions you are interested in (i.e. solutions that satisfy a particular bound), for otherwise non-uniqueness of solutions can be an issue. See Non-uniqueness of solutions of the heat equation for some additional information.

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