Let $u(t,x)$ be a solution to the heat equation $$\partial_t = \partial_{xx} \quad (t,x) \in [0,T) \times [-1,1]$$ subject to the initial/boundary conditions $$u(0,x) = f(x), \quad x \in [-1,1], \\ u(t,\pm 1) = g^{\pm}(t), \quad t \in [0,T),$$ with the usual compatibility conditions in corners: $f(\pm 1) = g^{\pm}(0)$. Suppose also that $f$ and $g$ are bounded and continuous. Then one can invoke the maximum principle or the energy method to prove that $u$ is the only solution.

What happens when $f$ or $g$ are unbounded? Say for instance, when $f(x) = \delta_0(x)$ (point mass at zero) and $g^\pm \equiv 0$? This problem has a solution that can be easily represented as a series.

How does one go about proving uniqueness in such a situation?

In fact, come to think of it, how does one prove the uniqueness of the fundamental solution $v(t,x) = \exp \{-x^2 / (4t)\}/ \sqrt{4 \pi t}$?

Is it some kind of weak uniqueness, where you show uniqueness of all classical solutions resulting from mollification of initial/boundary conditions? Is that the best one can do?

Any references would be deeply appreciated.

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    $\begingroup$ For the corresponding problem on Rn, uniqueness depends on the class in which you take the initial data: if the data grow too fast at infinity then uniqueness is lost. In your case this problem does not occur but still you should specify clearly the class of data to make the question meaningful $\endgroup$ – Piero D'Ancona Jan 18 at 10:21
  • $\begingroup$ @PieroD'Ancona thank you. I know uniqueness proofs when the data grows sub-exponentially at infinity. But what class does the Dirac delta fit in? It's a generalized function. $\endgroup$ – bm76 Jan 18 at 11:13
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    $\begingroup$ Let us assume for simplicity that $g^\pm=0$. Then we have a periodic continuation. Any periodic distribution is in some negative Sobolev space and therefore has a Fourier series. So we can separate variables and reduce the problem to ODEs. $\endgroup$ – Michael Renardy Jan 18 at 17:46
  • $\begingroup$ Michael: but in which sense do you recover the boundary condition? $\endgroup$ – Piero D'Ancona Jan 18 at 20:50
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    $\begingroup$ @PieroD'Ancona: Recovering the boundary condition really is no problem: The solution for $t>0$ is analytic. $\endgroup$ – Michael Renardy Jan 19 at 3:04

(Not sure if I understand the question correctly.)

If $p_{t,x}(y) = p(t, x, y)$ is the fundamental solution (a.k.a. the heat kernel), then $p_{t,x}$ converges as $t \to 0^+$ to the Dirac measure $\delta_0$ in the sense of weak* convergence of measures, and hence also in the sense of distributions.

For any initial value given by a distribution $f$ in $(-1,1)$ (say: compactly supported, but this can be extended slightly), $u(t, x) = \langle f, p_{t,x}$ makes sense. Then $u$ can be proved to solve the heat equation, and $u(t, \cdot)$ converges in the space of distributions to $f$ as $t \to 0^+$.

I do not know the literature well, but I would check any book on applications of the distribution theory to PDEs for more on that. For example, Section 16.7 in Vladimirov's Methods of the theory of generalized functions might be a good reference.


The fundamental solution of the heat equation is not unique: it is only unique modulo an entire solution of the heat equation, i.e. a solution of the heat equation which is analytic in the whole $\Bbb R_t\times \Bbb R_x \equiv\Bbb R^2$. For example we may consider the class of heat polynomials ([1], §1.4, pp. 17-18) $$ p_n(t,x)= n! \sum_{k=0}^{[n/2]} \frac{t^k}{k!}\frac{x^{n-2k}}{(n-2k)!} $$ and see that, if $v(t,x)$ is the standard fundamental solution of the heat equation recalled in the question, $$ \partial_t \big(p_n(t,x)+v(t,x)\big)-\partial_{xx} \big(p_n(t,x)+v(t,x)\big)=\delta(t,x)=\partial_t v(t,x) -\partial_{xx} v(t,x) $$ for all $n\in \Bbb N$, by linearity. However, this non uniqueness property is not a unique characteristic of the heat equation: the fundamental solution of every linear partial differential operators (if obviously existing) is always defined modulo a solution of the associated homogeneous partial differential equation. To obtain uniqueness, you should add other conditions which depend on the structure of the given differential operator.

Classically, uniqueness for the solutions to the Cauchy problem for the heat equations can be proved only assuming some restriction on the initial data and/or the non-homogeneous "forcing" term (if present): these restrictions are

  1. measurability of the data and
  2. spatial growth bounded by $e^{\varepsilon\Vert x\Vert^2}$ for an arbitrary $\varepsilon>0$ in any finite time interval $[0,T]$.

The theory is due to Andrei Tikhonov and is described, for example, in reference [1], §3.6, pp. 40-42 for the one spatial dimensional case and in the wonderful monograph of Vladimirov ([2] §16.7 pp. 228-229) indicated by Mateusz Kwaśnicki in his answer above.
Be it noted that, if the second condition is not fulfilled, then null solutions to the given Cauchy problem may occur ([1], §3.7, p. 42).


[1] John Rozier Cannon, The one-dimensional heat equation, Foreword by Felix E. Browder, (English) Encyclopedia of Mathematics and Its Applications, Vol. 23. Menlo Park, California etc.: Addison-Wesley Publishing Company; Cambridge etc.: Cambridge University Press, pp. XXV+483 (1984), ISBN: 0-201-13522-1, MR0747979, Zbl 0567.35001.

[2] Vassilij Sergeevič Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.

  • $\begingroup$ DanieleTampieri How about the Cauchy-Dirichlet problem on $(t,x) \in [0,T]\times [-1,1]$ with nice bounded lateral boundary conditions, but Dirac delta at zero $\delta_0(x)$ for the initial condition? Does this problem also have many solutions? $\endgroup$ – bm76 Feb 9 at 7:48
  • $\begingroup$ Hi @bm76: the answer is negative. The boundary condition impose the unique solvability of the problem: in the linear space of solution to the non-homogeneous equation which differ by a solution to the homogeneous one, imposing a boundary condition is equivalent to choose a single one. $\endgroup$ – Daniele Tampieri Feb 9 at 8:20
  • $\begingroup$ (Continued) @bm76, you can fin the details when $f\in L^2$ (thus not when $f=\delta_0(x)$) in the last chapter of another wonderful book by Vasslilij Vladimirov, *Equations of Mathematical Physics" (various editions by Marcel Dekker and MIR): perhaps you can adapt the method (which is basically a Fourier expansion in eigenvector of the heat operator) for the case when the initial condition is the Dirac $\delta_0(x)$ distribution $\endgroup$ – Daniele Tampieri Feb 9 at 8:23

I'm not a specialist in this area. So for beginning I recommend you to study the theory of equations with $L_1$ data in F.Petitta Not So Long Introduction to the Weak Theory of Parabolic Problems with Singular Data (Chapter 4). There is also an extensive literature on nonlinear equations with measure data, for example,

  1. P.Baras, M.Pierre Problems paraboliques semilineaires avec donnees measures. Applicable Analysis, 1984, V.18, 11l-149,

  2. L.Boccardo, T.Gallouet Non-linear Elliptic and Parabolic Equations Involving Measure Data. J. of Functional Analysis 1989, V.87, 149-169.


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