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Uniqueness of strongly continuous mild heat solutions

Statement

Assume Countable Choice. Let $n \geq 1$ , $1 \leq p < \infty$ , $T > 0$ , $u \in C ([0, T]; L^{p} (R^{n}))$ , $u (0) = f$ , and $u (t) = H_{t - s} u (s)$ for every $0 < s < t \leq T$ . Then $u (t) = H_{t} f$ for all $0 \leq t \leq T$ . Hence the heat evolution is the unique solution in this class. This is uniqueness for the semigroup relation, not an unrestricted classical uniqueness assertion or a claim of strong continuity on all $L^{\infty}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq p < \infty$ , $T > 0$ , $u \in C ([0, T]; L^{p} (R^{n}))$ with $u (0) = f$ and $u (t) = H_{t - s} u (s)$ for all $0 < s < t \leq T$ , and $t \in (0, T]$ with $s \in (0, t)$ .

[A1]

Countable Choice is the hypothesis carried by the evolution suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $1 \leq p < \infty$ the heat evolution satisfies the semigroup law $H_{t + s} = H_{t} H_{s}$ on $L^{p}$ for all $s, t \geq 0$ with $H_{0}$ the identity, the contraction bound $∥ H_{r} f ∥_{p} \leq ∥ f ∥_{p}$ , and strong continuity at zero, $∥ H_{r} f - f ∥_{p} \to 0$ as $r ↓ 0^{+}$ , for every $f \in L^{p}$ (The heat Cauchy problem for $L^{p}$ data).

[F2]

If $(K_{ε})$ is an $L^{1}$ approximate identity and $f \in L^{p}$ , $1 \leq p < \infty$ , then $∥ f * K_{ε} - f ∥_{p} \to 0$ (Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ ); this is the mechanism behind the strong continuity recorded in [F1].

Proof

technique · direct

1.1A1F1given

Fix $t \in (0, T]$ and $0 < s < t$ . By the assumed semigroup relation at time $t$ and at time $s$ , and by linearity of $H_{t - s}$ and the semigroup law of [F1], $u (t) - H_{t} f = H_{t - s} u (s) - H_{t - s} H_{s} f = H_{t - s} (u (s) - H_{s} f)$ .

2.1step 1.1F1given

Norm bound: applying the contraction clause of [F1] to the last expression and then the triangle inequality gives $∥ u (t) - H_{t} f ∥_{p} \leq ∥ u (s) - H_{s} f ∥_{p} \leq ∥ u (s) - f ∥_{p} + ∥ f - H_{s} f ∥_{p}$ .

3.1F1F2step 2.1givenalgebra

Limit: the continuity of $u$ at $0$ in the $L^{p}$ norm gives $∥ u (s) - f ∥_{p} = ∥ u (s) - u (0) ∥_{p} \to 0$ as $s ↓ 0^{+}$ , and the strong continuity clause of [F1] gives $∥ H_{s} f - f ∥_{p} \to 0$ ; hence the right-hand side of step 2.1 tends to $0$ , so the nonnegative number $∥ u (t) - H_{t} f ∥_{p}$ is $0$ and $u (t) = H_{t} f$ in $L^{p}$ for every $t \in (0, T]$ . At $t = 0$ the identity $u (0) = f = H_{0} f$ is the hypothesis and the identity case of [F1].

4.1step 3.1F1given

Existence in the same class: the map $t \mapsto H_{t} f$ lies in $C ([0, T]; L^{p})$ : at $t = 0$ this is the strong continuity of [F1], and for $t > 0$ and $h > 0$ the semigroup law and contraction give $∥ H_{t + h} f - H_{t} f ∥_{p} \leq ∥ H_{h} f - f ∥_{p}$ , while for $h < 0$ with $t + h \geq 0$ they give $∥ H_{t + h} f - H_{t} f ∥_{p} \leq ∥ H_{- h} f - f ∥_{p}$ , and both bounds tend to $0$ as $∣ h ∣ \to 0$ .

5.1step 3.1step 4.1given∎

Steps 1.1, 2.1 and 3.1 show that any $u$ in the stated class coincides with $t \mapsto H_{t} f$ on $[0, T]$ , and step 4.1 shows that $t \mapsto H_{t} f$ itself lies in that class, so the heat evolution is the unique solution of the semigroup relation with the given initial datum in $C ([0, T]; L^{p})$ .

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Sources

John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis) (standard reference, not scraped)
Jared Speck, MIT 18.152 Introduction to Partial Differential Equations, Class Meeting #5: The Fundamental Solution for the Heat Equation (Fall 2011) (standard reference, not scraped)