For the Airy equation, which is a third-order linear partial differential equation given by: $$ \partial_{t}u+\partial_{xxx}u=0.$$
Suppose furthermore that we consider solutios of Airy equation which are spatially periodic $$u(x+L,t)=u(x,t)$$ for a given period $L$ and for $(t,x)$ in $[0, \infty) \times \mathbb{R}.$ Show that there is at most one such solution u corresponding to a given smooth, periodic initial condition $u_0: \mathbb{R} \rightarrow \mathbb{R}$ without using separation variables.
This is my effort:
Suppose $u, v$ both satisfy Airy equation, then $w=u-v$ obeys $$\partial_t w=\partial_{xxx}w $$ $$w(x+L,t)=w(x,t)$$ $$w(x,0)=0$$
Note that $\partial_t w.w=\dfrac{1}{2}\partial_tw^2$ then $$\int_{\mathbb{R}}\partial_t w.w dx=\dfrac{1}{2}\int_{\mathbb{R}}\partial_t w^2dx.$$
On the other hand, $$\int_{\mathbb{R}}\partial_{xxx} w.w dx=[w_{xx}.w]\Big|^{+\infty}_{-\infty}-\int_{\mathbb{R}}w_{xx} w_xdx.$$
Do $w_{xx}.w=0$ on the boundary? And am I on the right track?
You are close. Notice that these functions may not be integrable on the entire line, so we will consider these integrated quantities on just a single period, say $[0,L]$. By periodicity, if we find that $w=0$ on $[0,L]$, then $w=0$ for all $x\in\mathbb{R}$.
We constrauct $u$, $v$, and $w$ as in your post, then we have $$ \begin{aligned} \frac{\partial}{\partial t} \frac{1}{2}\int_0^L w^2~\mathrm{d}x &= \int_0^L ww_t~\mathrm{d}x \\ &=\int_0^L w w_{xxx}~\mathrm{d}x \\ &=\left.[w w_{xx}]\right|_0^L-\int_0^L w_x w_{xx}~\mathrm{d}x \\ &=\left(w(L)w_{xx}(L) - w(0)w_{xx}(0)\right) - \frac{1}{2}\left(w_x^2(L) - w_x^2(0)\right), \end{aligned} $$ where the last term comes from noticing that $2w_xw_{xx} = \frac{\partial}{\partial x}(w_x^2)$. Now since $w$ is smooth and periodic, all of $w$'s derivatives are also periodic with period $L$, so this expression is zero. We then have
$$ \frac{\partial}{\partial t} \frac{1}{2}\int_0^L w^2~\mathrm{d}x= 0, $$ so $\int_0^L w^2~\mathrm{d}x$ is constant in time. However, since $w_0=0$, we find that $\int_0^L w^2~\mathrm{d}x=0$ for all time, so we have that $w=0$ on $[0,L]$ for all time. since $w$ is periodic with period $L$, we find that $w$ is identicially zero for all $x$ and $t$.