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For following PDE how can I show $||u(t,*)||_{L^2([0,l])} \le a e^{-bt}$ without solving it, where $a>0$, and $b>0$ are constants.

let, $l>0$, $S = (0,\infty)\times (0,l)$ and $u(t,x) \in C^{1,2}(\bar S)$

\begin{align*} u_t - u_{xx} = 0 & \; ; (t,x) \in S \\ u(x,0)=\frac{x(l-x)}{l^2} &\; ; 0\le x\le l\\ u_x(0, t) = u_x(l , t)=0&\; ; t\in (0,\infty) \end{align*}

The step by step hint is given here but I can't follow after first step. i.e. don't know how to show $$\frac{d}{dt} ||u(t,*)||_{L^2([0,l])}^2 = -2||\partial_xu(t,*)||_{L^2([0,l])}^2$$

and afterwards. Thank you for your time!!

ADDED::
How to show that
1. $||u(t,*)||_{C^0([0,l])} \le \sqrt l ||u_x(t,*)||_{L^2([0,l])}$ and
2. $ ||u(t,*)||_{L^2([0,l])}^2\le l^2 ||u_x(t,*)||_{L^2([0,l])}^2$

For (1), how can I show using fundamental theorem of Calculus and Cauchy Inequality $$||u(t,*)||_{C^0([0,l])} = \sup |u(t,x)|_{x\in(0,l)}\le \sqrt l \left( \int_0^l |u_x(t,x)|^2 dx\right)^{1/2}$$

Monkey D. Luffy
  • 1,632
  • Multiply both sides of the heat equation by $u$ and integrate over $[0,l]$. When integrating $u_{xx} u$ you can integrate by parts. (Why do the boundary terms vanish?) – Willie Wong Jul 17 '13 at 12:55

1 Answers1

1

$$\frac{d}{dt} ||u(t,*)||_{L^2([0,l])}^2 =\frac{d}{dt} \int_0^l u(t,x)^2 dx=2 \int_0^l u(t,x)u_t(t,x) $$(using equation and then integration by parts with boundary conditions) $$ 2 \int_0^l u(t,x)u_{xx}(t,x)dx= 2u(t,x)u_{x}(t,x)|_{x=0}^{x=l}-2\int_0^l u_{x}(t,x)u_{x}(t,x)dx$$ $$=-2||\partial_xu(t,*)||_{L^2([0,l])}^2$$

Edit Assuming that you talk about problem IV from your pdf, we note that

$$u(t, 0) = u(t , l)=0$$ (and not $u_x(t, 0) = u_x(t , l)=0$).

Let's show that $$||u(t,*)||_{C^0([0,l])} \le \sqrt l ||u_x(t,*)||_{L^2([0,l])}.$$

We write

$$\sup_y |u(t,y)|=\sup_y |u(t,y)|=\sup_y \left|\int_0^y u_x(t,s)ds\right|\le \sup_y \int_0^y 1\cdot|u_x(t,s)|ds$$ $$\le \sup_y \sqrt{\int_0^y |u_x(t,s)|^2ds \int_0^y 1^2ds}\le\sqrt l ||u_x(t,*)||_{L^2([0,l])}.$$

From that we can conlcude that

$$\int_0^l |u(t,x)|^2dx\le ||u(t,*)||^2_{C^0([0,l])}\int_0^ldx\le l^2 ||u_x(t,*)||^2_{L^2([0,l])}.$$

TZakrevskiy
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