For $1 \leq p < \infty$ and initial-data $u _ 0 \in L ^p ( \mathbb{R} ^d)$ due to P. Li (Uniqueness of $L^1$ solutions for the Laplace equation and the heat equation on Riemannian manifolds) there exists a unique solution $u$ of the heat equation $u_t - \Delta u = 0$ in $(0, \infty) \times \mathbb{ R }^d$ which satisfies \begin{equation*} u \in C ( [0 , \infty ) ; L^p ( \mathbb{ R }^d) ) \quad \text{and} \quad u ( 0 , \cdot ) = u_0. \end{equation*} This solution is given by $u ( t ) = e ^{t \Delta} u_0$, the convolution with the fundamental solution.

My question is, whether there is known literature for the uniqueness in the case of non-negative $u_0 \in L^\infty ( \mathbb{ R }^d )$ for the heat equation in $(0, \infty) \times \mathbb{ R }^d$, where we replace $u \in C ( [0 , \infty ) ; L^p ( \mathbb{ R }^d) )$ by $u \in L^\infty ( (0, \infty) \times \mathbb{ R }^d)$ and $u(t,x) \to u_0(x)$ for almost every $x \in \mathbb{ R }^d$ as $t \to 0$. Is weak star convergency of $u(t, \cdot)$ to $u_0$ in $L^\infty(\mathbb{R}^d)$ as $t \to 0$ sufficient?

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    $\begingroup$ I think there we need $u_0$ to be continuous. $\endgroup$ Commented Sep 4, 2020 at 9:37
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    $\begingroup$ I don't think that you actually need continuity of $u_0$ in the proof. You can probably modify the proof such that it works for $u_0 \in L^\infty$. The only thing that changes is that instead of continuity up to $t=0$ we get $u(t,x) \rightarrow u_0(x)$ a.e. as $t\rightarrow 0$. $\endgroup$
    – SC2020
    Commented Sep 4, 2020 at 9:51
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    $\begingroup$ Thank you, this might actually work! Theorem 3 should hold, the question is probably whether Theorem 4 also holds in our case by replacing max with $\lVert \cdot \rvert _ {L^\infty}$ $\endgroup$ Commented Sep 4, 2020 at 11:13
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    $\begingroup$ Yeah, I guess we get the following assertion: For $U$ connected and bounded, if there exists a point $(x_0,t_0) \in U_T$ such that $u(x_0, t_0) = \lVert u \rVert _ {L ^\infty (\overline{U_T})}$ then $u$ is constant a.e. in $\overline{U_{t_0}}$. $\endgroup$ Commented Sep 4, 2020 at 12:48
  • $\begingroup$ I posted an answer, does it work that way? We do not need any non-negativity of $u_0$ and $u$ $\endgroup$ Commented Sep 4, 2020 at 22:10

1 Answer 1


Let $u$ and $v$ be two solutions with the stated properties. Let $\eta$ be a standard mollifier on $\mathbb{R}^d$ and let $u_\varepsilon :=u \ast \eta_\epsilon$ and $v_\varepsilon := v \ast \eta_\epsilon$, where we only convolve with respect to the $x$-variable.

Claim: We have $u_\varepsilon, v_\varepsilon \in C ^\infty((0, \infty) \times \mathbb{R} ^d) \cap C ( [0, \infty) \times \mathbb{R} ^d)$, $\partial_t u_\varepsilon - \Delta u_\varepsilon = \partial_t v_\varepsilon - \Delta v_\varepsilon = 0$ and $u_\varepsilon(0) = v_\varepsilon(0) = u_0 \ast \eta_\epsilon \in C_b ( \mathbb{R} ^d)$.

Proof of the Claim: As $u$ and $v$ are bounded, we can pass the differentiation into the integrals and get \begin{equation*} \partial_t u_\varepsilon - \Delta u_\varepsilon = (u_t - \Delta u) \ast \eta_\epsilon = 0 \end{equation*} and similarly $\partial_t v_\varepsilon - \Delta v_\varepsilon = 0$. Furthermore, $u_\varepsilon, v_\varepsilon \in C ^\infty((0, \infty) \times \mathbb{R} ^d)$ and $u_0 \ast \eta_\epsilon \in C_b ( \mathbb{R} ^d)$. Thus, we are left to show continuity in $t = 0$. This follows by the weak$^*$ convergence: Let $x \in \mathbb{R} ^d$. Then \begin{align*} u_\varepsilon (t,x) - u_\varepsilon (0,x) & = \int_{ \mathbb{R} ^d }\! ( u(t,y) - u_0 (y) ) \eta_\epsilon (x - y) \, dy \\ & \to 0 \end{align*} as $\eta_\epsilon (x - \cdot) \in L^1(\mathbb{R} ^d)$ and $u(t,\cdot) \to u_0$ weak$^*$ in $L^\infty ( \mathbb{R}^d )$ as $t \to 0$. In the same fashion, we see that $v_\varepsilon$ is continuous in $t = 0$, which proofs the claim.

The heat equation for regular initial data in $C_b(\mathbb{R}^d)$ has a unique bounded classical solution, see Thm. 6 in Evans, Chapter 2.3. Hence, $u_\varepsilon = v_\varepsilon$ for all $\varepsilon > 0$. As $u_\varepsilon \to u$ and $v_\varepsilon \to v$ locally uniformly in $(0,\infty) \times \mathbb{R} ^d$ as $\varepsilon \downarrow 0$, we get $u = v$ in $(0,\infty) \times \mathbb{R} ^d$.

  • $\begingroup$ This was exactly what I needed for my research! Thank you! $\endgroup$ Commented Nov 30, 2023 at 14:10

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