Uniqueness and continuous dependence on the data of Heat equation.

Question

Let two smooth $v_1$ and $v_2$ both satisfy the system

$$\partial_t{v}-\Delta v=f \quad \text{in} \quad U \times (0,\infty), $$ $$v = g \quad \text{on} \quad \partial U \times (0,\infty),$$

for some fixed given smooth $f: \bar{U}\times (0,\infty) \rightarrow \mathbb{R}$ and $g: \partial U \times (0,\infty).$ $U$ is open, bounded and $U \subset \mathbb{R}^n.$ Show that $$\sup_{x \in U} |v_1(t, x) − v_2(t, x)| \rightarrow 0,$$ as $t \rightarrow \infty.$

This is my work:

Let $ u =v_1 -v_2,$ it is sufficient to prove $\sup_{x \in U} |u(x,t)| \rightarrow 0,$ as $t \rightarrow \infty. (1)$

$u$ obeys the system $$\partial_t{u}-\Delta u=0 \quad \text{in} \quad U \times (0,\infty), $$ $$u = 0 \quad \text{on} \quad \partial U \times (0,\infty).$$ Multiply both sides by $u.|u|^{2(m-1)},$ note that $\partial_t(|u|^{2m})=2m\partial_tu.u.|u|^{2(m-1)}$ then $$\dfrac{1}{2m}\partial_t\int_{U}|u|^{2m}dx=\int_{U}\Delta u.u.|u|^{2(m-1)}dx$$ Apply integration by part for the RHS, we get $$\dfrac{1}{2m}\partial_t\int_{U}|u|^{2m}dx=-(2m-1)\int_{U}|\nabla u|^2|u|^{2(m-1)}dx.$$ By the generalize Poincare's inequality, we obtain $$\partial_t\int_{U}|u|^{2m}dx \leq -2C\left(2-\dfrac{1}{m}\right)\int_{U}|u|^{2m}dx$$ or $$\partial_t\Vert u(t,\cdot)\Vert^{2m}_{L^{2m}(U)} \leq -2C\left(2-\dfrac{1}{m}\right)\Vert u(t,\cdot)\Vert^{2m}_{L^{2m}(U)}$$ $\Rightarrow 2m \Vert u(t,\cdot)\Vert^{2m-1}.\partial_t\Vert u(t,\cdot)\Vert_{L^{2m}(U)} \leq -2C\left(2-\dfrac{1}{m}\right)\Vert u(t,\cdot)\Vert^{2m}_{L^{2m}(U)}$

$\Rightarrow \partial_t\Vert u(t,\cdot)\Vert_{L^{2m}(U)} \leq -2\dfrac{C}{m}\left(2-\dfrac{1}{m}\right)\Vert u(t,\cdot)\Vert_{L^{2m}(U)}$

Applying Gronwall's inequality, we get

$\Vert u(t,\cdot)\Vert_{L^{2m}(U)} \leq e^{-2\frac{C}{m}\left(2-\frac{1}{m}\right)t}\Vert u(0,\cdot)\Vert_{L^{2m}(U)}.$

I am planing to let $m \rightarrow \infty$ to obtain $\Vert u(t,\cdot)\Vert_{L^{\infty}(U)}$ and let $t \rightarrow \infty$ then $e^{-2\frac{C}{m}\left(2-\frac{1}{m}\right)t} \rightarrow 0$ to obtain (1). But the problem is as $m \rightarrow \infty,$ $e^{-2\frac{C}{m}\left(2-\frac{1}{m}\right)t} \rightarrow 1 \neq 0.$

Am I on the right track or did I make some wrong steps?

Could you provide any ideas to improve my work?

Answer 1

Yes, as mentioned in the comments, the energy energy integral is useful here. Let $u$ be defined as in the question.

Define

$$E(t)=\int_{U}{u(t,x)}^2~\mathrm d^m x$$ Note $E$ is bounded below by $0$.

Now, observe $$\dot E(t)=\int_U 2 ~u(t,x)~\partial_tu(t,x)~\mathrm d^m x \\ =2\int_U (u ~\Delta u)(t,x)\mathrm d^mx \\ =2\int_U \big(u ~\nabla\cdot( \nabla u)\big)(t,x)\mathrm d^mx$$

Recall the generalized integration by parts: $$\int_U \phi~\nabla\cdot v~\mathrm d\mu^m=\int_{\partial U}n\cdot \phi v~\mathrm d\mu^{n-1}-\int_{U}v\cdot \nabla\phi~\mathrm d\mu^m$$

Taking in our case $\phi=u$ and $v=\nabla u$, we get $$\dot E(t)=2\int_U \big(u ~\nabla\cdot( \nabla u)\big)(t,x)\mathrm d^mx \\ =2\int_{\partial U} \big( n\cdot (u\nabla u)\big)(t,x)\mathrm d^m x-2\int_U |\nabla u|^2(t,x)\mathrm d^m x$$ The first integral is zero due to the assumptions on the boundary data of $u$, and therefore we obtain $$\dot E(t)=-2\int_U|\nabla u|^2(t,x)\mathrm d^m x$$

Poincare's inequality implies

$$\dot E(t)=-2\int_U |\nabla u|^2\mathrm d^m x=2{\left\Vert\nabla u(t,\cdot)\right\Vert_2}^2\leq -2C {\Vert u(t,\cdot)\Vert_2}^2$$ Since ${\Vert u(t,\cdot)\Vert_2}^2=E$, we have $$\dot E\leq -2C E$$ Hence by Gronwall's inequality we get $E(t)\leq \mathrm e^{-2Ct}E(0)$ which implies $E\to 0 $ which implies ${\Vert u(t,\cdot)\Vert_2}\to 0$ as $t\to\infty$.

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So, we have shown that $\Vert u(t,\cdot)\Vert_2\to 0$, but this is not enough to show that $\Vert u(t,\cdot)\Vert_\infty\to 0$, as desired in the question. However, this is rectified using the strong maximum principle.

Define the parabolic cylinder and its boundary $$U(T):=(0,T]\times U \\ \Gamma(T)=\overline{U(T)}\setminus U(T)=(\{0\}\times \bar U)\cup ([0,T]\times\partial U)$$

THEOREM: Strong maximum principle for the heat equation. Assume $u$ is a classical solution of the heat equation in $U(T)$. Then, (i) $$\max_{\overline{U(T)}}u=\max_{\Gamma(T)}u$$ Furthermore (ii), if $U$ is connected and there exists a point $(t_0,x_0)\in U(T)$ such that $u(t_0,x_0)=\max_{\overline{U(T)}}u$, then $u$ is constant in $\overline{U(t_0)}$.

(For proof: See page 55 of Evans PDE book.)

The (ii) statement means that, if ever $u$ assumes its maximum inside $U$ at any positive time, then $u$ must be constant.

In our case, we know that $u=0$ on $\partial U$. That means that, aside from the trivial case $u\equiv 0$, we know that $$\operatorname{argmax}_{\overline{U(T)}}|u|\in \{0\}\times U$$ I.e, it must occur in the open domain $U$ at $t=0$. However, the (ii) statement of the above theorem now tells us that , aside from the trivial solution $u\equiv 0$, for all times $t>0$, $$\sup_U |u(t,\cdot)| < \sup_U |u(0,\cdot)|$$ Because otherwise , $u$ would be constant in the domain $U(t)$, and we know the only constant solution satisfying our BCs is the zero solution. So, we have shown that the function $$M(t)=\sup_{U}|u(t,\cdot)|$$ Is a decreasing function bounded below by zero, and hence $M\to M_\infty\in\mathbb R_{\geq 0}$.

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To show $M_\infty=0$ is a little bit more difficult. But essentially, the idea is, the only way for $\Vert u(t,\cdot)\Vert _2\to 0$ (as already shown) while maintaining $\Vert u(t,\cdot)\Vert_\infty \to M_\infty >0$ would be if the family of functions $u(t,\cdot)$ "clustered" around some finite collection of points $\{x_0,x_1,...x_{N-1}\}$, i.e $u(t,x)\to 0$ as $t\to\infty$ for all $x\in U\setminus \{x_0,...,x_{N-1}\}$ but with $u(t,x)\to L\leq M_\infty$ for all $x\in \{x_0,...,x_{N-1}\}$.

However, such "clustering" is impossible, as it would violate the smoothing properties of the heat equation. To make the proof easier, assume that only one "cluster point" $x_0$ exists satisfying $\Vert u(t,x_0) \Vert_\infty\to M_\infty>0$. Since we already know that ($\star$) $u(t,x)\to 0$ as $t\to\infty$ $\forall x\neq x_0$, this means that, given $t$ large enough and $\epsilon$ small enough, we can make the derivative estimate

$(\star) ~:~ \text{proof needed!}$ $$\left|\frac{u(t,x_0+\epsilon\upsilon)-u(t,x_0)}{\epsilon}\right| \\ (\upsilon \in \mathbb R^m , |\upsilon|=1)$$ Arbitrarily large. But, by the mean value theorem, we know that $\exists x^*$ on the line segment connecting $x_0$ and $x_0+\epsilon\upsilon$ such that $$\upsilon \cdot \nabla u(t,x^*)=\frac{u(t,x_0+\epsilon\upsilon)-u(t,x_0)}{\epsilon}$$ Which implies $$|\upsilon \cdot \nabla u(t,x^*)|=|\nabla u(t,x^*)|=\left|\frac{u(t,x_0+\epsilon\upsilon)-u(t,x_0)}{\epsilon}\right|$$ And since we know we can make the RHS arbitrarily large, that means we can make $|\nabla u(t,x^*)|$ arbitrarily large as long as we choose a point $x^*$ close enough to $x_0$ and a time $t$ large enough. However, we know from theorem 9 on page 61 of Evans PDE that $\exists c\in\mathbb R_+$ such that $$\max_{C(t,x;r/2)}|\nabla u|\leq \frac{c}{r^{m+3}}\Vert u \Vert_{L^1\big(C(t,x;r)\big)}$$ Where $C(t,x;r)$ is the cylinder $$C(t,x;r)=\{(s,y)\in\mathbb R_{\geq 0}\times\mathbb R^m : |x-y|\leq r ~\text{and}~t-r^2\leq s\leq t\}$$

But, since $L^2$ convergence implies $L^1$ convergence we know that $\Vert u \Vert_{L^1\big(C(t,x;r)\big)}\to 0$ as $t\to\infty$, and thus our ability to make $|\nabla u|$ arbitrarily large would violate our initial assumptions. Therefore, it is not possible for the families of functions $u(t,\cdot)$ to "cluster" around a point $x_0$ and therefore the only possible value for $M_\infty$ is $0$, in other words,

$$\boxed{\Vert u(t,\cdot)\Vert_\infty\to 0~~\text{as}~t\to\infty}$$

As desired. $\blacksquare$.

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Addendum: Details.

The key point is to show that $\Vert u \Vert_{L^1\big(C^1(t,x;r)\big)}\to 0$ as $t\to\infty$. We already know that $E(t)={\Vert u(t,\cdot) \Vert_2}^2\leq \mathrm e^{-2Ct}E(0)$ from Gronwall's inequality. This means that, for any $0<T<t$, we have $$\sqrt{\int_{t-T}^{t}{E(t')}^2\mathrm dt}=\Vert u \Vert_{L^2\big(U(t)\setminus U(t-T)\big)}\to 0 ~~\text{as}~t\to\infty$$ But, since $C(t,x;\sqrt{T})\subset \big(U(t)\setminus U(t-T)\big)$ which implies $\Vert u \Vert_{L^2\big(C(t,x;\sqrt{T})\big)}\to 0$ as $t\to\infty$ for all positive $T$. But, since $L^2$ convergence implies $L^1$ convergence, we get $\Vert u \Vert_{L^1\big(C(t,x;r)\big)}\to 0 $ as $t\to\infty$ for any $x\in U$ and suitable $r>0$.

Answer 2

To fix the gap $(\star)$ in my previous answer.

In my other answer, I claimed that $\Vert u(t,\cdot)\Vert_2\to 0$ as $t\to\infty$ implied that $u(t,x)\to 0$ for almost all $x\in U$.

Note that the restriction to "almost all" is necessary, because $L^p$ convergence does not imply everywhere pointwise convergence, not even for a sequence of continuous functions. However, we can fix this, with the help of the following theorem:

THEOREM: Rapid convergence in measure implies pointwise convergence almost everywhere.

Let $(X,\Sigma,\mu)$ be a measure space and let $(f_n)_{n\in\mathbb N}$ be a sequence of measurable functions which converges in measure to the measurable function $f$. Then, as shown by John Dawkins HERE as a consequence of the Borel-Cantelli lemma, if $$\sum_{n=1}^\infty \mu\big(\{x\in X:|f_n(x)-f(x)|>\epsilon\}\big)<\infty \\ \forall \epsilon>0$$ Then, $f_n\overset{\text{p.w}}{\longrightarrow}f$ almost everywhere as $n\to\infty$.

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In our case, we can consider the measure space $(U,\mathscr B(U),\mu^m)$ where $\mathscr B(U)$ are the Borel subsets of $U$ and $\mu^m$ is the $m$ dimensional standard Euclidean measure. Since $L^p$ convergence implies convergence in measure, we know that $u_k\overset{\mu^m}{\longrightarrow} 0$ as $k\to\infty$ where $u_k=u(t_k,\cdot)$ is the function $u$ evaluated at time $t=t_k$, and $(t_k)_{k\in\mathbb N}$ is any strictly increasing sequence of positive real numbers.

First, observe that $$\text{IF}~~{\Vert u(t,\cdot )\Vert^2_{L^2(U)}}=\int_U {u(t,x)}^2\mathrm d^m x<\delta \\ \text{THEN}~~\mu^m\big(x\in U:{|u(t,x)|}^2>\epsilon\big)< \frac{\delta}{\epsilon} \\ \text{equivalently,}~~~\mu^m\big(x\in U:|u(t,x)|>\epsilon\big)< \frac{\delta}{\epsilon^2}$$

I.e, if ${\Vert u(t,\cdot )\Vert^2_{L^2(U)}}$ is less than $\delta$, then ${u(t,\cdot)}^2$ can only take values of $\epsilon$ or greater on a set $V\subset U$ with measure $\mu^m(V)$ no larger than $\delta/\epsilon$.

Now, recall our Gronwall inequality: $$\int_U {u(t,x)}^2\mathrm d^m x={\Vert u(t,\cdot )\Vert^2_{L^2(U)}}=E(t)\leq \mathrm e^{-2c_1 t}E(0)$$ Putting the above two together, we get

$${\Vert u(t,\cdot )\Vert^2_{L^2(U)}}<\mathrm e^{-2c_1 t}E(0)\implies \mu^m\big(x\in U:|u(t,x)|>\epsilon\big)< \frac{\mathrm e^{-2c_1 t}E(0)}{\epsilon^2}$$ And therefore, letting $t=t_k$,

$$\sum_{k=1}^\infty \mu^m(x\in U:|u_k(x)|>\epsilon) < \frac{E(0)}{\epsilon^2}\sum_{k=1}^\infty \mathrm e^{-2c_1k}<\infty$$

Thus $u_k\overset{\text{p.w}}{\longrightarrow} 0$ almost everywhere as $k\to\infty$. Since this holds for any sequence $(t_k)_{k\in\infty}$ we conclude that $u(t,x)\overset{\text{p.w}}{\longrightarrow} 0$ as $t\to \infty$ for almost all $x\in U$. My other answer uses this as a starting point to show that this "almost all" is in fact, all.