I've been set this question, however I haven't seen one like this and I'm not sure what to do. Thanks for any answer.
Consider the diffusion equation:
$\frac{∂u}{∂t} - D\frac{∂^2u}{∂x^2} = f(x,t), x∈[0,L],t∈[0,∞] $
subject to periodic boundary conditions:
$u(0,t) = u(L,t),$ $u_x(0,t) = u_x(L,t).$
Prove that there exists a constant $C>0$ such that if $u_1(x,t)$ and $u_2(x,t) $ are solutions corresponding to the intitial conditions $φ_1(x) $ and $φ_2(x) $ respectively, then:
$$\int_0^L [u_1(x,t) - u_2(x,t)]^2 dx \leqslant C \int_0^L [φ_1(x) - φ_2(x)]^2 dx$$
for all $t \geqslant 0$.
Take care to indicate clearly where and how you proof uses the boundary conditions. What is the smallest value of C for which is inequality can be guaranteed.
The question linked in the comments provided a method, but there wasn't a full answer. I'm writing one down here for reference.
First, set $v = u_1-u_2$. Since $u_1$ and $u_2$ both satisfy the same inhomogeneous equation, $v$ has to be a solution of the homogeneous equation (by linearity) and the same periodic B.C.s
\begin{cases} v_t - Dv_{xx} = 0 \\ v(0,t) = v(L,t) \\ v_x(0,t) = v_x(L,t) \\ v(x,0) = \varphi_1(x) - \varphi_2(x) = \psi(x) \end{cases}
Mutiplying the PDE by $v$ and integrating over $x\in(0,L)$
$$ \int_0^L v(x,t)v_t(x,t)\ dx = D\int_0^L v(x,t)v_{xx}(x,t)\ dx $$
Integration by parts on the RHS gives
$$ \int_0^L v(x,t)v_{xx}(x,t)\ dx = -\int_0^L (v_x(x,t))^2\ dx $$
Since $v(L,t)v_x(L,t) - v(0,t)v_x(0,t) = 0$ as per the B.C.s
The LHS is
$$ \int_0^L v(x,t)v_t(x,t)\ dx = \int_0^L \frac12 \frac{\partial}{\partial t}(v(x,t))^2 \ dx = \frac12 \frac{d}{dt} \left(\int_0^L (v(x,t))^2 \ dx\right) $$
by Leibniz's integral rule. Therefore
$$ \frac{d}{dt}\left(\int_0^L (v(x,t))^2\ dx\right) = -2D\int_0^L (v_x(x,t))^2 \ dx \le 0 $$
which means, if $g(t) = \int_0^L (v(x,t))^2 \ dx$ then $g(t)$ is non-increasing in $t \in (0,\infty)$. Hence, $g(t)$ attains its maximum at $t=0$, or
$$ \int_0^L (v(x,t))\ dx \le \int_0^L (v(x,0)^2\ dx = \int_0^L (\psi(x))^2 \ dx $$