PDE - Energy - Wave Equation

Question

I dont know how solve a) and b), I'm read the book of Walter Strauss, but I have a lot of doubts, for the c) first, I tried estimate $(k(t)e^{-2at})'$ and integrate the inequal, but not unwind... :(

Let $a\in \mathbb{R}$ and consider the solution of the equation

$$u_{tt}+au_{t}-u_{xx} =0\;,\;t>0\;,\;x\in(0,1) \\ u(0,t)=u(1,t)=0,\; t>0 \\ u(x,0)= \varphi(x)\;,u_t(x,0)=\psi(x)\; ,\;x \in (0,1) $$

Define the energy associated by

$$E(t):= \frac{1}{2}\int_0^1(u_t^2(x,t)+u_x^2(x,t))dx. $$

a) show that $$E(t)=-ak(t)+E(0)$$ where $k(t)=\int_0^t\int_0^1u_t^2(x,s)dxds$

b) show that $$k'(t)\leq -2ak(t)+2E(0)$$

c) shows the estimated decay for $k$: $$k(t)\leq\frac{E(0)}{a}(1-e^{-2at})$$

Answer 1

This solves (a) and should get you started for the other questions. Note that $u_{tt}=u_{xx}-au_t$ hence $$E'(t)=\int_0^1 (u_tu_{tt}+u_xu_{xt})\mathrm dx=\int_0^1 (u_tu_{xx}+u_xu_{xt})\mathrm dx-a\int_0^1 (u_t)^2\mathrm dx.$$ The first term on the RHS is $$\int_0^1 (u_tu_x)_x\mathrm dx=u_t(1,t)u_x(1,t)-u_t(0,t)u_x(0,t)=0,$$ since $u(0,t)$ and $u(1,t)$ do not depend on $t$ hence $u_t(0,t)=u_t(1,t)=0$. Integrating $E'(t)$ from $0$ to $t$ yields the desired formula since the remaining term on the RHS is $-ak'(t)$.

Answer 2

Now I know the itens b) and c),

b) show that $$k'(t)\leq -2ak(t)+2E(0)$$

$k'(t)=\int_0^1u_t^2(x,s)dx$

$$2E(t) -\int_0^1u_x^2(x,t)) = \int_0^1(u_t^2(x,t)dx = k'(t)$$

and $\;2E(t) -\int_0^1u_x^2(x,t))\leq 2E(t)=(2-ak(t)+E(0)) $ because$\;\; \int_0^1u_x^2(x,t))\gt0$

finally $$k'(t)\leq \;-2ak(t)+2E(0)$$

Answer 3

c) shows the estimated decay for $k$: $$k(t)\leq\frac{E(0)}{a}(1-e^{-2at})$$

Solution:

Consider the integral factor $e^{\int 2a dt} = e^{2at}$

so $$k'(t)e^{2at} +2ak(t)e^{2at}\leq2E(0)e^{2at} \\ \implies[k(t)e^{2at}]_t\leq 2E(0)e^{2at} \\ \implies k(t)e^{2at}\leq [2E(0)e^{2as}]_0^t \\ \implies k(t)e^{2at}\leq \frac{E(0)}{a}(e^{2at}-1) \\ \implies k(t)\leq \frac{E(0)}{a}(1 - e^{-2at}) $$