Does anyone know how to prove there exists at most one smooth solution $ u $ of the following problem for the beam equation?
$ u_{tt} + u_{xxxx} = 0 $ in $ (0,1) \times (0, T) $
$ u(0,t) = u(1,t) = u_x(0,t) = u_x(1,t) = 0 $ for all $ t \in (0,T) $
$ u(.,0) = g $ and $ u_t(.,0) = h $.
Thank you so much in advanced.
Let $u$ and $\tilde u$ be two solutions, and let $v = u - \tilde u$. Then $$ v_{tt} + v_{xxxx} = 0,\\ v(0,t) = v(1,t) = v_x(0,t) = v_x(1,t) = 0, \\v(\cdot,0) = v_t(\cdot,0) = 0 .$$ Multiply the first equation by $v_t$, and integrate w.r.t. $x$ from $x=0$ to $1$, and integrate by parts twice on one of the integrals (noting $v_t(0,t) = \cdots = v_{tx}(1,t) = 0$), and use the product rule for differentiation to see $(v_t^2)_t = 2v_{tt}v_t$ and $(v_{xx}^2)_t = 2v_{xxt}v_{xx}$, to conclude $$ \frac12 \frac{\partial}{\partial t}\left(\|v_t\|_2^2 + \|v_{xx}\|_2^2\right) = 0$$ where we define $\|f\|_2^2 = \int_0^1 |f(x)|^2 \, dx $. Hence $$ \|v_t\|_2^2 + \|v_{xx}\|_2^2 = \text{constant} .$$ Setting $t=0$, we see that this constant is zero. Hence for all $t>0$ $$ v_t = 0, v_{xx} = 0 .$$ Since $v_{xx} = 0$, from the boundary conditions we obtain $v = 0$.
