Consider the initial value problem $$ \partial_t u = \partial_{xx} u$$ $$ u(0,x) = u_0(x)$$ for the heat equation in one dimension, where $u_0: {\bf R} \to {\bf R}$ is a smooth initial datum and $u: [0,+\infty) \times {\bf R} \to {\bf R}$ is the smooth solution. Under reasonable growth conditions on $u_0$ and $u$ (e.g. $u_0$ and $u$ are at most polynomial growth), there is a unique solution $u$ to this problem given by the classical formula

$$ u(t,x) = \frac{1}{\sqrt{4\pi t}} \int_{\bf R} e^{-|x-y|^2/4t} u_0(y)\ dy.$$

However, as is well known, once one allows $u_0$ or $u$ to grow sufficiently rapidly at infinity, then smooth solutions to the heat equation are no longer unique, as demonstrated first by Tychonoff in 1935.

My question then concerns the corresponding question for existence : does there exist smooth initial data $u_0$ for which there are no global smooth solutions $u$ to the initial value problem for the heat equation?

One obvious candidate for such "bad" data would be a backwards heat kernel, such as $$ u_0(x) = e^{|x|^2/4}.$$ One can verify that $$ u(t,x) = \frac{1}{(1-t)^{1/2}} e^{|x|^2/4(1-t)}$$ is a smooth solution to the initial value problem for the heat equation with initial datum $u_0$ up to time $t=1$, at which point it blows up (rather dramatically). However, this does not fully solve the problem due to the aforementioned lack of uniqueness; just because this particular solution $u$ blows up, there could be some other more exotic smooth solution with the same data which somehow manages to retain its smoothness beyond the time $t=1$. This seems highly unlikely to me, but I was not able to demonstrate such an "unconditional non-existence" result - the absence of any growth hypotheses at infinity seems to destroy most methods of controlling solutions, and could potentially create some strange scenario in which one could continually keep singularities from forming by pumping in infinite quantities of energy from spatial infinity in just the right manner. But perhaps there is some literature on this problem?

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    $\begingroup$ Yes you can match any boundary conditions by, as you put it, pumping in infinite quantities of energy from spatial infinity. At least, I'm very confident that this is the case, and have a method of doing this in mind. I'll just have to write it out first though... $\endgroup$ Aug 5, 2011 at 22:13

1 Answer 1


It is true that for any initial datum $u_0\in C^\infty(\mathbb{R})$ there exists a solution $u\in C^\infty(\mathbb{R}^+\times\mathbb{R})$ to the heat equation with initial condition $u(0,x)=u_0(x)$. As you point out, this will not be unique.

I can give a method of constructing such solutions now. The idea is to show that we can write $u=\sum_{n=1}^\infty f_n$ where $f_n$ are carefully constructed solutions chosen such that the partial sums $\sum_{n=0}^mf_n(0,x)$ eventually agree with $u_0(x)$ any bounded subset of the reals, and for which $f_n$ tends to zero arbitrarily quickly in the compact-open topology. First a bit of notation. I use $\mathbb{R}^+=[0,\infty)$ for the nonnegative reals. For a space $X$ then $C_0(X)$, $C^\infty(X)$, $C^\infty_0(X)$, and $C^\infty_K(X)$, represent the continuous real-valued functions on $X$ which are respectively vanishing at infinity, smooth, smooth and vanishing at infinity, and smooth with compact support. Let $(P_t)_{t\geq 0}$ be the kernels $$ \begin{align} &P_t\colon C_0(\mathbb{R})\to C_0(\mathbb{R}),\\\\ &P_tu(x)=\frac{1}{\sqrt{4\pi t}}\int_\mathbb{R}e^{-(x-y)^2/4t}u(y)\,dy \end{align} $$ for $t > 0$, and $P_0u=u$. This is the Markov transition function for Brownian motion (more precisely, for standard Brownian motion scaled by $\sqrt{2}$, because of the normalization used here). For $u\in C_K^\infty(\mathbb{R})$, then $f(t,x)=P_tu(x)$ is in $C_0^\infty(\mathbb{R}^+\times\mathbb{R})$, and is a solution to the heat equation with initial condition $f(0,x)=u(x)$, agreeing with the classical solution stated in the question. I'll also consider initial conditions $u\in C^\infty_K((a,\infty))$ (for $a\in\mathbb{R}$) by setting $u(x)\equiv0$ for all $x\le a$. The first step in the construction is to find initial conditions supported in $(a,\infty)$ so that $P_tu(0)$ approximates any given continuous function of time that we like.

1) For any $a > 0$ and $h\in C_0((0,\infty])$, there exists a sequence $u_1,u_2,\ldots\in C^\infty_K((a,\infty))$ such that $\sqrt{t}P_tu_n(0)$ converges uniformly to $h(t)$ (over $t > 0$) as $n\to\infty$.

Consider the closure, $V$, in $C_0((0,\infty])$ (under the uniform norm) of the space of functions $t\mapsto\sqrt{4\pi t}P_tu(0)$ for $u\in C^\infty_K((a,\infty))$. Note that the limit as $t\to\infty$ does always exist and is just the integral of $u$. Then $V$ is a closed linear subspace of $C_0((0,\infty])$. Consider a sequence $u_n\in C_0((a,\infty))$ tending to the delta function $\delta_b$ at a point $b > a$, in the sense that $u_n$ all have support in the same compact set, and converge in distribution to $\delta_b$. Then, from the expression defining $P_t$, $\sqrt{4\pi t}P_tu_n(0)$ converges uniformly over $t > 0$ to $\exp(-b^2/4t)$. So, the function $t\mapsto\exp(-b^2/4t)$ is in $V$. As the set of functions of the form $t\mapsto\exp(-b^2/4t)$ for $b > a$ is closed under multiplication and separates points, the locally compact version of the Stone-Weierstrass theorem says that $V=C_0((0,\infty])$. Then, (1) follows.

2) For any $u\in C^\infty_K((0,\infty))$ and $a,T > 0$, there exists a sequence $u_1,u_2,\ldots\in C^\infty_0((a,\infty))$ such that $f_n(t,x)\equiv P_t(u+u_n)(x)$ converges uniformly to zero (along with all its partial derivatives to all orders) over $t\in[0,T]$ and $x\le0$.

Choosing $0 < \epsilon < a$ so that the support of $u$ is contained in $(\epsilon,\infty)$, (1) implies that we can choose $u_n\in C^\infty_K((a,\infty)$ so that $\sqrt{t}P_tu_n(\epsilon)$ converges uniformly to $-\sqrt{t}P_tu(\epsilon)$ over $t\ge0$ as $n\to\infty$. Then, $f_n(t,x)\equiv P_t(u+u_n)(x)$ is a bounded solution to the heat equation with boundary conditions $f_n(0,x)=0$ for $x\le\epsilon$ and $f_n(t,\epsilon)=P_t(u+u_n)(\epsilon)$. It is then standard that the solution is given by an integral over the boundary, $$ f_n(t,x)=\int_0^t\frac{\epsilon-x}{\sqrt{4\pi (t-s)^3}}e^{-(\epsilon-x)^2/4(t-s)}\sqrt{s}P_s(u_n+u)(\epsilon)\frac{ds}{\sqrt{s}} $$ for $x\le0$. Differentiating this wrt $x$ and $t$ an arbitrary number of times, and using dominated convergence as $n\to\infty$, it follows that $f_n(t,x)$ (and all its partial derivatives) converge uniformly to zero over $x\le0$ and $t\in[0,T]$ as $n\to\infty$.

3) Suppose that $T > 0$, $0 < a < b$ and $u\in C^\infty_K(\mathbb{R})$ has support contained in $(-\infty,-a)\cup(a,\infty)$. Then, there exists a sequence $u_n\in C^\infty_K(\mathbb{R})$ with supports in $(-\infty,-b)\cup(b,\infty)$ such that $P_t(u+u_n)(x)$ (and all its partial derivatives) tends to 0 uniformly over $\vert x\vert\le a$ and $t\in[0,T]$.

Set $u^+(x)=1_{\{x > 0\}}u(x)$ and $u^-(x)=1_{\{x < 0\}}u(x)$. Applying (2) to $u^+(x+a)$, there exists a sequence $u^+_n\in C^\infty_K(\mathbb{R})$ with supports in $(b,\infty)$ such that $P_t(u^++u^+_n)(x)$ tends to 0 uniformly over $x\le a$ and $t\in[0,T]$. Applying the same argument to $u^-(-x-a)$, there exists $u^-_n\in C^\infty_K(\mathbb{R})$ with support in $(-\infty,-b)$ such that $P_t(u^-+u^-_n)(x)$ tends to zero uniformly over $x\ge -a$ and $t\in[0,T]$. A sequence satisfying the statement of (3) is $u_n=u^+_n+u^-_n$.

4) If $u\in C^\infty(\mathbb{R})$ then there exists $f\in C^\infty(\mathbb{R}^+\times\mathbb{R})$ solving the heat equation with initial condition $f(0,x)=u(x)$.

We can inductively choose a sequence $u_n\in C^\infty_K(\mathbb{R})$ such that $\sum_{m=1}^nu_m(x)=u(x)$ for $\vert x\vert\le n+1$ and $n\ge1$. Let $u_1=u$ on $[-2,2]$ and then, for each $n\ge2$, apply the following step.

  • As $\tilde u=u-\sum_{m=1}^{n-1}u_m$ is zero on $[-n,n]$, it has support in $(-\infty,-a)\cup(a,\infty)$ for $a=n-1/2$. Choosing $b=n+1$, by (3), there exists $v\in C^\infty_K(\mathbb{R})$ with support in $(-\infty,-b)\cup(b,\infty)$ such that $P_t(\tilde u +v)(x)$ along with all its partial derivatives up to order $n$ are bounded by $2^{-n}$ over $\vert x\vert\le n-1/2$ and $t\le n$. Take $u_n=\tilde u + v$.

Setting $f_n(t,x)=P_tu_n(x)$, then $f_n$ are smooth functions satisfying the heat equation and the initial conditions $\sum_{m=1}^nf_m(0,x)=u(x)$ for $\vert x\vert\le n+1$. Also, by the choice of $u_n$, for $n > 1$ then $f_n$ together with all its derivatives up to order $n$ is bounded by $2^{-n}$ on $[0,n]\times[1-n,n-1]$. As $\sum_n2^{-n} < \infty$, the limit $f=\sum_nf_n$ exists and is smooth with all partial derivatives commuting with the summation. Then $f$ has the required properties.

  • $\begingroup$ hmm, I think there's still some issues with (1) here. I missed out a factor of $1/\sqrt{4\pi t}$, but it should be fixable... $\endgroup$ Aug 6, 2011 at 2:27
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    $\begingroup$ Ah, a nice "pyramid scheme" to recursively build a solution - I like it! I think (1) can be patched by dividing the solution by $\sqrt{t}$ first (one can see from the fundamental solution that this keeps the solution continuous on the interval where the initial data vanishes). One may also have to cap t to be bounded for (1)-(3), but that doesn't really damage (4), as one just needs control on $P_t(\tilde u+v)(x)$ for $t \leq n$, say. Incidentally it seems this gives a new proof of nonuniqueness for the heat equation... $\endgroup$
    – Terry Tao
    Aug 6, 2011 at 3:12
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    $\begingroup$ Ah, you patched (1) already :-). I just wanted to add that the Hahn-Banach theorem can be used to dualise (3), and it seems to say something like a (tempered distributional) solution to an inhomogenous heat equation with compactly supported forcing term and zero initial data cannot be compactly supported at any future time, which is a sort of "parabolic unique continuation" result. $\endgroup$
    – Terry Tao
    Aug 6, 2011 at 3:21
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    $\begingroup$ Thanks for the comments! Regarding the point in your first comment, I think it is possible to build on this method to show that uniqueness fails in the worst possible way. Consider $u\in C^\infty(\mathbb{R})$ and any solution $f\in C^\infty((0,\infty)\times\mathbb{R})$ to the heat equation. It is not assumed that $f(t,x)$ satisfies any particular initial condition, or even that the limit exists as $t\to0$. Then, there will exist a sequence $f_n$ of smooth solutions to the heat equation with initial condition $u$ so that, restricting to $t > 0$, $f_n$ tends to $f$ in the compact-open topology. $\endgroup$ Aug 6, 2011 at 11:04
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    $\begingroup$ Very interesting riff, @George L.! $\endgroup$ Aug 6, 2011 at 22:44

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