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>



Heat equation close to the steady state
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.


