# Uniqueness of the heat equation for initial data in $L^\infty$

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?

• I think there we need $u_0$ to be continuous. Commented Sep 4, 2020 at 9:37
• 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 $t\rightarrow 0$. Commented Sep 4, 2020 at 9:51
• 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 $\lVert \cdot \rvert _ {L^\infty}$ Commented Sep 4, 2020 at 11:13
• 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 $\overline{U_{t_0}}$. Commented Sep 4, 2020 at 12:48
• I posted an answer, does it work that way? We do not need any non-negativity of $u_0$ and $u$ Commented Sep 4, 2020 at 22:10

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 $$(0,\infty) \times \mathbb{R} ^d$$.