A Conversation about the Heat Equation in My TA Session — $L^2$ Convergence of Energy
Student: Can I discuss the last question of the midterm exam with you?
Me: Sure, I can listen to your approach of solving it.
Student: Let’s consider the heat equation defined $\mathbb{R}_+\times [0,1]$:
\[\begin{align*} \begin{cases} u_t-u_{xx}=0 &\text{in }\mathbb{R}_+\times(0,1)\\ u(0,x)=u_0(x) & x\in [0,1]\\ u_x(t,0)=u_x(t,1)=0 & t\in \mathbb{R}_+ \end{cases} \end{align*}\]We define $\mathcal{H}(t)=\int_0^1u(t,x)dx$, which is the energy at time $t$. First, we need to prove that the energy is conserved, meaning that $\mathcal{H}(t)$ is equal to some constant $H$.
Me: This shouldn’t be too difficult for you. For such kind of physics problems, you can exchange the order of differentiation and integration and use the given boundary conditions to find:
\[\begin{align*} \frac{d}{dt}\mathcal{H}(t)&=\frac{d}{dt}\int_0^1u(t,x)dx\\ &=\int_0^1u_t(t,x)dx\\ &=\int_0^1u_{xx}(t,x)dx\\ &=u_x(t,1)-u_x(t,0)\\ &=0 \end{align*}\]Then you will see that the derivative of the energy with respect to time is zero, so the energy is equal to some constant $H$.
Student: Yes I agree with you. Next we want to prove that $u(t,x)$ converges to $H$ in the $L^2$-sense as $t\rightarrow\infty$. Following the hint, we should first prove that
\[\begin{align*} E'(t)= -\int_0^1(v_x(t,x))^2dx \end{align*}\]where $v:=u-H$ and $E(t)=\int_0^1v^2(t,x)dx$.
Me: I guess you still apply the trick of exchanging the order of differentiation and integration.
Student: That’s correct. After exchanging we get
\[\begin{align*} E'(t)&=\frac{1}{2}\int_0^1\frac{\partial}{\partial t}(v^2(t,x))dx\\ &=\int_0^1v(t,x)v_t(t,x)dx\\ &=\int_0^1v(t,x)v_{xx}(t,x)dx \end{align*}\]We need to integrate by part:
\[\begin{align*} E'(t)&=vv_x|_0^1-\int_0^1(v_x(t,x))^2dx\\ &=-\int_0^1(v_x(t,x))^2dx \end{align*}\]By the definition of $v$, we can know that
\[\int_0^1v(t,x)dx=\int_0^1u(t,x)dx-H=\mathcal{H}(t)-H=0\]The problem then asks me to prove that $\forall t\geq 0$, $\exists x_0\in [0,1]$ such that $v(t,x_0)=0$.
Me: Well how do you prove that?
Student: I used the Mean Value Theorem. Define $F(s)=\int_0^sv(t,x)dx$. Since $F(1)=F(0)=0$，the Mean Value Theorem tells us that $\exists x_0$ such that
\[\begin{align*} F'(x_0)&=\frac{F(1)-F(0)}{1-0}=0\\ v(t,x_0)&=0 \end{align*}\]Me: I totally agree with you.
Student: Then we are asked to use the Cauchy-Schwarz inequality to prove that
\[E(t)\leq \frac{1}{2}\int_0^1(v_x(t,x))^2dx\]I don’t have time to solve this problem during the exam, but I do have a solution right now (Calculating on the blackboard).
\[\begin{align*} (v(t,x))^2&=(v(t,x)-v(t,x_0))^2\\ &=(\int_x^{x_0}v_x(t,x)dx)^2\\ &\leq (\int_0^{1}v_x(t,x)dx)^2\\ &\leq \int_0^1(v_x(t,x))^2dx\text{ (Cauchy Schwarz})\\ E(t)&=\frac{1}{2}\int_0^1(v(t,x)-v(t,x_0))^2dx\\ &\leq \frac{1}{2}\int_0^1(\int_0^1(v_x(t,x))^2dx)dx\\ &=\frac{1}{2}\int_0^1(v_x(t,x))^2dx \end{align*}\]Me: Nice calculations. Let’s review what we know about the function $E$ right now:
\[\begin{align} E'(t)&= -\int_0^1(v_x(t,x))^2dx\\ E(t)&\leq \frac{1}{2}\int_0^1(v_x(t,x))^2dx \end{align}\]Therefore, $E$ is a nonnegative decreasing function, so it has a limit at infinity.
Student: I believe that the limit of $E’(t)$ at infinity is $0$, so by $(1)$ we have
\[\lim_{t\rightarrow\infty}\int_0^1(v_x(t,x))^2dx=0\]We then can apply the squeeze theorem to $(2)$ to see that the limit of $E(t)$ at infinity is 0.
Me: I don’t think your argument is correct. A nonnegative decreasing function does not need to have vanishing derivative at infinity. Consider the following example. Let’s $h$ be a bump function supported in $[0,1]$. $h$ satisfies
\[\begin{align*} h(x)\begin{cases} >0, &x\in (0,1)\\ =0, &\text{otherwise} \end{cases} \end{align*}\]Furthermore, $h(1/2)=1$. Define the following two functions:
\[\begin{align*} g(x)&=\sum_{n=1}^\infty h(n^2(x-n))\chi_{[n,n+1]}\\ f(x)&=(\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1h(t)dt)-\int_0^xg(x)dx \end{align*}\]Any finite interval $[a,b]$ can be covered by finitely many intervals of the form $[n,n+1]$, so $g$ is locally the sum of finitely many smooth functions，which implies that both $g$ and $f$ are smooth. $f’=-g\leq 0$, so $f$ is decreasing。$f\geq 0$ because
\[\begin{align*} \int_0^xg(x)dx&\leq \int_0^mg(x)dx\text{ ($m$ is an integer larger than $x$)}\\ &=\int_0^m\sum_{n=1}^\infty h(n^2(x-n))\chi_{[n,n+1]}dx\\ &=\int_0^m\sum_{n=1}^{m-1} h(n^2(x-n))\chi_{[n,n+1]}dx\\ &\leq \sum_{n=1}^{m-1}\int_{n}^{\frac{1}{n^2}+n}h(n^2(x-n))dx\\ &\leq \sum_{n=1}^{m-1}\frac{1}{n^2}\int_0^1h(u)du\text{ (substitution $u=n^2(x-n)$)}\\ &\leq \sum_{n=1}^{\infty}\frac{1}{n^2}\int_0^1h(u)du<\infty \end{align*}\]Consider the following two sequences:${n+\frac{1}{n^2}}, {n+\frac{1}{2n^2}}$.
\[\begin{align*} g(n+\frac{1}{n^2})&=h(n^2\frac{1}{n^2})=h(1)=0\\ g(n+\frac{1}{2n^2})&=h(n^2\frac{1}{2n^2})=h(\frac{1}{2})=1 \end{align*}\]We can see that $g$ is oscillating, so $f’=-g$ does not have a limit at infinity. Let’s reflect on the construction of this counterexample. Intuitively, we might expect a non-negative decreasing function to look something like this:
We may believe that in order for the function to remain non-negative and approach its limit at infinity, the rate of change of the function at infinity cannot be too large. However, we can actually have a function that changes very little for most of the time and changes a lot at some points, resulting in an overall small rate of change:
The red arrows in the above picture points at the moments that the function has a large rate of change. This helps us understand the construction of the above counterexample: it is simply the antiderivative of a non-negative function that oscillates at infinity.
Now let’s make a digression. The example above suggests that the existence of $\lim_{x\rightarrow \infty}f(x)$ does not guarantee that $\lim_{x\rightarrow \infty}f’(x)=0$ because $\lim_{x\rightarrow \infty}f’(x)$ may not even exist. However, what if we know that $\lim_{x\rightarrow \infty}f’(x)$ exists? In this case, do we have the equality $\lim_{x\rightarrow \infty}f’(x)=0$? The answer is Yes, and in fact we only need a weak assumption: the existence of $\lim_{x\rightarrow \infty}f(x)+f’(x)$.
Theorem (Hardy): If both $\lim_{x\rightarrow \infty}f(x)$ and $\lim_{x\rightarrow \infty}f(x)+f’(x)$ exists, then $\lim_{x\rightarrow \infty}f’(x)=0$.
Proof: We elegantly apply L’Hôpital’s rule:
\[\begin{align*} \lim_{x\rightarrow \infty}f(x)&=\lim_{x\rightarrow \infty}\frac{e^xf(x)}{e^x}\\ &=\lim_{x\rightarrow \infty}\frac{e^xf(x)+e^xf'(x)}{e^x}\text{ (L'Hôpital's rule)}\\ &=\lim_{x\rightarrow \infty}f(x)+f’(x) \end{align*}\]from which we easily deduce that $\lim_{x\rightarrow \infty}f’(x)=0$. $\square$
Let’s go back to the original problem，which is to prove that $\lim_{t\rightarrow\infty}E(t)=0$.
Student: Here’s another argument. Suppose $\lim_{t\rightarrow\infty}E(t)=c>0$. Then there exists a large enough real number $M$ such that $\forall t>M$,
\[\begin{align*} |E(t)-c|&<\frac{c}{2}\\ \frac{c}{2}&<E(t)\leq \frac{1}{2}\int_0^1(v_x(t,x))^2dx\text{ (by $(2)$)}\\ c&\leq \int_0^1(v_x(t,x))^2dx\\ -c&\geq -\int_0^1(v_x(t,x))^2dx=E'(t) \end{align*}\]Let $T$ be a real number larger than $M$. We have
\[\begin{align*} |E(T+1)-E(T)|&=|\int_T^{T+1}E'(t)dt|\\ &= -\int_T^{T+1}E'(t)dt\\ &\geq c \end{align*}\]Apply the triangle inequality to get
\[\begin{align*} c&\leq |E(T+1)-E(T)|\\ &\leq |E(T+1)-c|+|E(T)-c| \end{align*}\]Either $\lvert E(T+1)-c\rvert \geq c/2$ or $\lvert E(T)-c\rvert\geq c/2$. In both cases, we obtain a contradiction.
Reference:
- https://math.stackexchange.com/questions/42277/proving-that-lim-limits-x-to-inftyfx-0-when-lim-limits-x-to-inftyf?noredirect=1&lq=1
- https://math.stackexchange.com/questions/788813/the-limit-of-the-derivative-of-an-increasing-and-bounded-function-is-always-0/788818#788818