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Let $u(t, x)$ be the unique solution of the heat equation on the unit interval with Dirichlet boundary conditions and initial data $u_0$: $$ \left\{ \begin{array}{l} \partial_t u(t, x) = \partial_x^2 u(t, x), \quad t > 0,\ x \in [0, 1] \\ u(t, 0) = u(t, 1) = 0, \quad t > 0 \\ u(0, x) = u_0(x), \quad x \in [0, 1] \end{array} \right. $$ The steady state solution of this problem is $u_s(x) = 0$.

Question: Assume that $|u_0^\prime(x)| \leq \delta$ on $[0, 1]$. Does it remain true for all later times, that is $|\partial_x u(t, x)| \leq \epsilon$ on $[0, 1]$?

I would guess yes. Intuitively, this means that the solution to the heat equation, started close to the steady state solution, remains close. The maximum principle seems to be useful, but I can not control the derivative on the boundary. An other relevant result is that both $\| u(t, \cdot) \|_{L^2}$ and $\| \partial_x u(t, \cdot) \|_{L^2}$ vanish as $t \to \infty$.

I would also like to know if this is true for

  • the heat equation on the $d$-dimensional unit box $[0, 1]^d$
  • the quasilinear heat equation on the unit interval: $$ \left\{ \begin{array}{l} \partial_t u(t, x) = \frac{d}{dx} g(\partial_x u(t, x)), \quad t > 0,\ x \in [0, 1] \\ u(t, 0) = u(t, 1) = 0, \quad t > 0 \\ u(0, x) = u_0(x), \quad x \in [0, 1] \end{array} \right. $$

I asked this question also here: https://math.stackexchange.com/q/2164593/420667

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Let $v:=\partial_xu$. It still satisfies the heat equation $\partial_tv=\partial_x^2v$, but with the Neumann boundary condition: $$\partial_xv(t,0)=\partial^2u(t,0)=\partial_tu(t,0)=\partial_t0=0,$$ and the same at $x=1$. Therefore you may apply the maximum principle: $$\sup_x|v(t,x)|\le\sup_x|v(0,x)|=\sup_x|u_0'(x)|\le\delta.$$

Edit. To see why the maximum principle holds true, let me remind that $v$ is ${\cal C}^\infty$ up to the boundary for $t>0$. Then extending $V$ by parity to $x\in(-1,1)$, then by $2$-periodicity, one obtains a solution $V$ of the heat equation in ${\mathbb R}^+\times{\mathbb R}$. This one satisfies the maximum principle, here $|V|\le\delta$, which gives the result.

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  • $\begingroup$ I don't understand the first inequality in the calculation with the maximum principle. Why aren't the values $v(t, 0)$, $v(t,1)$ appearing on the right hand side? $\endgroup$ – user105541 Mar 2 '17 at 15:31
  • $\begingroup$ @Alexander. See my edits. $\endgroup$ – Denis Serre Mar 2 '17 at 15:56
  • $\begingroup$ To summarize the idea for myself: There is no effect of the boundary values $v(t,0)$, $v(t,1)$, because the lateral boundary can be removed by extending the solution periodically. $\endgroup$ – user105541 Mar 3 '17 at 8:28
  • $\begingroup$ @Alexander. Yes. At least, this is the way I prented here. But the maximum principle for the solution of the heat equation with Nemann boundary condition is valid in every dimension, every domain, even if a reflexion is not available. $\endgroup$ – Denis Serre Mar 3 '17 at 9:55
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Fortunately your boundary conditions are very simple, such that you can express the analytical solution of your problem explicitly. In fact, by using the ansatz of separation of variables one derives that for dimension $d=1$ and sufficiently smooth initial function $u_0$ the unique solution of your problem can be derived by Fourier expansion and it is given by \begin{align}\tag{1}\label{1} u(x,t)=\sum_{n=1}^\infty B_n\sin(n\pi x)e^{-n^2\pi^2t}, \end{align} where \begin{align*} B_n:=2\int_0^1\sin(n\pi x)u_0(x)\,dx \end{align*} for $n\in\mathbb{N}.$

In order to verify $\partial_tu(x,t)=\partial_x^2u(x,t)$ for all $t>0$ and $x\in[0,1],$ one has to justify the identities \begin{align}\tag{2}\label{2} \partial_tu(x,t)=\sum_{n=1}^\infty\partial_t B_n\sin(n\pi x)e^{-n^2\pi^2t} \end{align} and \begin{align}\tag{3}\label{3} \partial_x^2u(x,t)=\sum_{n=1}^\infty\partial_x^2 B_n\sin(n\pi x)e^{-n^2\pi^2t} \end{align} for all $t>0$ and $x\in[0,1].$ In oder to prove e.g. identity \eqref{2}, it suffices to show that \eqref{1} converges uniformly, the series of derivatives $$\sum_{n=1}^\infty\partial_t B_n\sin(n\pi x)e^{-n^2\pi^2t}$$ converges uniformly and the derivatives $\partial_t B_n\sin(n\pi x)e^{-n^2\pi^2t}$ are continuous for all $n\in\mathbb{N}.$ Those uniform convergence results can easily be verified by using the Weierstraß M-test. Now, identity \eqref{3} can be shown in the exact same way.

Furthermore, it is straightforward to see that $u(0,t)=u(1,t)=0$ for all $t>0.$ Finally, it holds \begin{align*} u(x,0)&=\sum_{n=1}^\infty B_n\sin(n\pi x)\\ &=2\sum_{n=1}^\infty\sin(n\pi x)\int_0^1\sin(n\pi x)u_0(x)\,dx\\ &=u_0(x) \end{align*} for sufficiently smooth initial conditions $u_0(x)$ satisfying $u_0(0)=u_0(1)=0.$ Minimal regularity assumptions for the justification of this Fourier Sine expansion can be found in the literature.

Exactly as above one now derives \begin{align*} \partial_xu(x,t)&=\sum_{n=1}^\infty\partial_x B_n\sin(n\pi x)e^{-n^2\pi^2t}\\ &=\sum_{n=1}^\infty n\pi B_n\cos(n\pi x)e^{-n^2\pi^2t}, \end{align*} such that elementary estimates show \begin{align*} |\partial_x u(x,t)|\leq 2\pi\int_0^1|u_0(\tilde x)|\,d\tilde x\sum_{n=1}^\infty ne^{-n^2\pi^2t}, \end{align*} which vanishes uniformly in $x\in[0,1]$, as $t\to\infty.$ Furthermore, if $\int_0^1|u_0(\tilde x)|\,d\tilde x$ is chosen sufficiently small, it also holds $|\partial_xu(x,t)|\leq \varepsilon$ for $t>0,\,x\in[0,1]$ and a given $\varepsilon>0.$

The same arguments should also hold true for dimensions $d>1.$ In fact, two dimensional separation of variables yields the candidate \begin{align*} u(x,y,t)=\sum_{m,n=1}^\infty B_{m,n}\sin(m\pi x)\sin(n\pi y)e^{-(m^2+n^2)\pi^2t}, \end{align*} where \begin{align*} B_{m,n}:=4\int_0^1\int_0^1\sin(m\pi x)\sin(n\pi y)u_0(x,y)\,dxdy \end{align*} for $m,n\in\mathbb{N}.$ By repeating the above arguments you should therefore also be able to obtain your desired decay and smallness properties for $d=2.$ As far as i can judge, the extension to $d\in\mathbb{N}$ is now more or less straightforward.

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