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Let $X\subseteq C(\mathbb{R}\times [0,\infty))$ be the collection of all solutions to the IV heat equation $$ \partial_t u(t,x) = \partial^2 u(t,x) \qquad u(0,x)=p(0,x), $$ for some fixed $p\in C^2(\mathbb{R}\times [0,\infty);(0,\infty))$. By standard methods one finds a candidate solution of the form $$ u=K\star p(0,\cdot); $$ where $K$ is the heat Kernel. However, since the domain is unbounded the the maximum principle fails and there may be more than one such solution.

Therefore, is there a solution $\hat{u}$ to the IV heat equation which minimizes the metric-like projection: $$ \sup_{(x,t) \in [-M,M]^d\times [0,\infty)} \| u(t,x)- p(t,x) \|, $$ where $M>0$ and where $$ \sup_{(x,t) \in [-M,M]^d\times [0,\infty)} \| p(t,x) \|<\infty $$

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    $\begingroup$ If there is any such solution, it is automatically a minimizer. Indeed: the difference of any two solutions is a solution of the heat equation with zero initial data, and hence it is either unbounded or identically zero. $\endgroup$
    – Mateusz Kwaśnicki
    Nov 26, 2019 at 10:56
  • $\begingroup$ But p does not solve the heat equation so the difference $u-p$ is not a solution to the heat equatoin, in general. $\endgroup$
    – ABIM
    Nov 26, 2019 at 12:36
  • $\begingroup$ I mean: if $u_1$ and $u_2$ are solutions and both $u_1-p$ and $u_2-p$ are bounded, then $u_1-u_2$ is a bounded solution with zero initial data, and hence $u_1=u_2$. $\endgroup$
    – Mateusz Kwaśnicki
    Nov 26, 2019 at 12:38
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    $\begingroup$ If you want to project, you should minimize instead of maximize, right? To clarify: $p(t,x)$ is given and you want to find $u\in X$ which minimizes the distance $\|u-p\|$? (Which norm, by the way…) $\endgroup$
    – Dirk
    Nov 26, 2019 at 15:59
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    $\begingroup$ @N00ber: Your edit changes nothing: every "unusual" solution will grow even faster. See this article by Tychonoff, and this answer by George Lowther. $\endgroup$
    – Mateusz Kwaśnicki
    Nov 27, 2019 at 12:46

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