I want to prove that the approximaton $$\int_0^1u^2(x,t) dx \leq \int_0^1u_0(x)dx$$ holds for $t>0$ where \begin{cases} u_t-u_{xx}=0,&\text{if $0<x<1, t>0$}\\ u(0,t)=u(1,t)=0, & \text{if $t>0$}\\u(x,0)=u_0(x), &\text{if $x\in(0,1)$} \end{cases} What I got so far:
From the the fact $$f(t)=\int_a^t f'(s)ds +f(a)$$ we obtain $$\int_0^tu^2(x,t)dx= \int ...+\int_o^tu_0^2(x)dx $$
Can anyone tell me what to fill in in "...". Then I can show that $... \leq 0$ and I am done. Thanks in advance.
Multiply the equation by $u$ and integrate with respect to $x$ on $[0,1]$: $$ \int_0^1u_t\,u-\int_0^1u_{xx}\,u=0. $$ Integrating the second integral by parts and using the boundary conditionswe get $$ \frac12\frac{d}{dt}\int_0^1u^2+\int_0^1(u_x)^2=0. $$ Since the second term is non-negative, we see that $$ \frac{d}{dt}\int_0^1u^2\le0. $$ It follows that $\int_0^1u^2$ is decreasing as a function of $t$.