Answer to Question #135652 in Differential Equations for Ritika Lad

Question

The general solution u(x,t)=(A sin pcx +B cos pcx )(C sin pct+D cos pct ) with u(0,t)=0 and u(l,t)=0 gives..........

Answer 1

Given the Dirichlet problem with the initial Dirichlet boundary conditions

$u(0,t)=0=u(L,t)$

Then the problem has at most one solution.

To solve this,

Suppose that the problem has two solutions $u_1$ and $u_2$ . Setting $u=u_1−u_2$ , the function $u$ solves

$$\begin{cases} u_t-ku_{xx}=0 & x\in [0,L], t>0\\ u(0,t)=0, u(L,t)=0 & t \geq0 \\ u(x,0)=0 & x\in [0,L] \end{cases}$$

Define the energy

$$E(t)=\frac12 \intop_0^L((A\ sin\ pcx+ B\ cos\ pcx)(C\ sin\ pct+ D\ cos\ pct))^2dx$$

Differentiating with respect tot, we get

$E'(t)=\frac12 \intop_0^L(C\ sin\ pct+ D\ cos\ pct)(A\ sin\ pcx+ B\ cos\ pcx)(C\ sin\ pct+ D\ cos\ pct)=\intop _0^k u(x,t)u_{xx}(x,t)dx$

Integrating by part and using the boundary conditions, we find that

$E'(t) \intop_0^k u(x,t)v_{xx}(x,t)dx=[ku(x,t)u_x(x,t)]_0^L-\intop_0^Lku_x(x,t)^2dx \leq0$

Hence the energy $E$ is decreasing. Since $E(0) = 0$ and $E(t)≥0$ , we get that $E≡0$ . his implies that $u(0,t)≡0$ for every $t$ . Consequently,$u_1(x,t) =u_2(x,t)$ for all $x∈[0,L]$ and $t≥0$ .