Heat equation inequality

Question

Let $u \in C^{1}(0,\infty) \times C^{1}[0,1]$ be a solution to heat equation (1) with inital boundaries (2),(3)

(1) $\partial_tu(t,x)-\partial_{xx}u(t,x)=0$, for all $(t,x) \in[0,\infty)\times[0,1]$

(2) $u(t,0)=u(t,1)=0,$ for $t \in[0,\infty]$

(3) $u(0,x)=u_{0}(x),$ for $x\in[0,1]$

Show that there exists constants $C_{1},C_{2}$ such that $$ \int_{0}^{1}u^2(t,x)dx \leq C_{1}e^{-tC_2}\;\text{ for all }t\geq0 $$

Thoughts
I want to get this into a form where I can apply Wirtinger's inequality for functions: I tried by multiplying the through by $u(t,x)$ and intergrating but I couldn't get it out that way, so I'm not sure how to solve this.

Answer 1

Let's assume your solution has the regularity required by the equation. A way to get an estimate of this form is as follows: Consider a shifted multiple of the fundamental solution $G$, $w(x,t)=MG(x,t+1)$. For large enough $M$, $w(x,0)=MG(x,1)\ge u_0$. These shifted and rescaled Gaussians are solutions to your equation in all of $[0,\infty)\times\mathbb{R}$. Therefore they are solutions on $[0,\infty)\times[0,1]$ (with boundary data equal $w(t,0)>0$ and $w(t,1)>0$). By the parabolic comparison principle, the solution to your equation will be below $w$ at all times, $u(x,t)\le w(x,t)$ and, for $x\in[0,1]$, $w(x,t)\le Ct^{-1/2}$ where $C=M/{\sqrt{4\pi}}$.

As a consequence, $$ \int\limits_0^1 u^2(x,t)\, dx\le C^2t^{-1} $$

Answer 2

As mentioned in the comnents, I don't feel totally clear on the differentiability conditions stipulated for $u(x, t)$, so I assume $u(x, t) \in C^2([0, 1] \times [0, \infty))$; this seems to suffice for the present application.

We first develop a couple of useful identities; as usual

$u_x = \dfrac{\partial u}{\partial x}, \tag 1$

and so forth.

First,

$(u^2)_{xx} = (2uu_x)_x = 2(uu_x)_x = 2(u_x^2 + uu_{xx}) = 2u_x^2 + 2uu_{xx}; \tag 2$

second, by virtue of $u_t = u_{xx}$,

$(u^2)_t = 2uu_t = 2uu_{xx} = (u^2)_{xx} - 2u_x^2 = 2(uu_x)_x - 2u_x^2; \tag 3$

we have

$\dfrac{d}{dt} \displaystyle \int_0^1 u^2(x, t) \; dx = \int_0^1 (u^2)_t \; dx$ $= \displaystyle \int_0^1 (2(uu_x)_x - 2u_x^2) \; dx = 2\int_0^1 (uu_x)_x \; dx - 2 \int_0^1 u_x^2 \; dx; \tag 4$

we next take note of the fact that

$\displaystyle \int_0^1 (uu_x)_x \; dx = u(1, t)u_x(1, t) - u(1, t) u_x(0, t) = 0, \tag 5$

which follows from boundary condition (2); thus,

$\dfrac{d}{dt} \displaystyle \int_0^1 u^2(x, t) \; dx = \displaystyle \int_0^1 (u^2)_t \; dx = -2 \int_0^1 u_x^2 \; dx. \tag 6$

We are now in position to apply the Wirtinger inequality to the integral on the right of this equation and find

$\displaystyle \pi^2 \int_0^1 u^2 \; dx \le \int_0^1 u_x^2 \; dx, \tag 7$

which we multiply by $-2$:

$-2 \displaystyle \int_0^1 u_x^2 \; dx \le -2\pi^2 \int_0^1 u^2 \; dx; \tag 8$

in combination with (6) this yields

$\dfrac{d}{dt} \displaystyle \int_0^1 u^2\; dx \le -2\pi^2 \int_0^1 u^2 \; dx, \tag 9$

to which we may apply Gronwall:

$\displaystyle \int_0^1 u^2(x, t) \; dx \le \left ( \int_0^1 u^2(x, 0) \; dx \right ) e^{-2\pi^2 t} = \left ( \int_0^1 u_0^2 \; dx \right ) e^{-2\pi^2 t} , \tag{10}$

and arrive at the desired result. $OE\Delta$.