Can anyone help me prove the well-posedness of the following heat equation with Robin boundary condition?
$u_t(x,t)=u_{xx}(x,t)$
$u(0,t)=0$
$u_x(1,t)=-au(1,t)$
where $a>0$.
The existence of the solution may be simply obtained by separation of variable. Are there any good references on this problem?
On
your problem can be written under the form $$u'(t)=Au\\u(0)=u_0$$ where $$A:D(A) \subset L^2(0,1) \to L^2(0,1)$$ $$D(A)=\left\lbrace v\in H^1(0,1), v(0)=0, v_x(1)+av(1)=0\right\rbrace $$ It is straitforward to see that $(Au,u)_{L^2(0,1)} \ \leqslant 0$, and $I-A$ is maximal, ($A$ is the second derevative), therefore, by semigroup theory, there exists one solution in $C(0,T;L^2(0,1)$ (if your initial state is $L^2(0,1)$)
Your problem is missing an initial condition such as $u(x,0)=u_0(x)$, which is necessary. Assuming $u_1,u_2$ are two such solutions satisfying the same initial condition and conditions you have specified, then $v=u_1-u_2$ satisfies $$ v_t = v_{xx} \\ v(x,0)=0 \\ v(0,t)=0,\;\;v_x(1,t)=-av(1,t). $$ Then, \begin{align} &\frac{d}{dt}\int_{0}^{1}v(x,t)^2dx \\ &=2\int_0^1v(x,t)v_t(x,t)dx \\ &= 2\int_0^1v(x,t)v_{xx}(x,t)dx \\ &= 2v(x,t)v_{x}(x,t)|_{x=0}^{1}-2\int_0^1 v_x(x,t)^2dx \\ &= 2v(1,t)v_x(1,t)-2v(0,t)v_x(0,t)-2\int_0^1 v_x^2x \\ &= 2v(1,t)v_x(1,t)-2\int_0^1 v_x^2dx \\ &= -2av_x(1,t)^2-2\int_0^1 v_x^2 dx \le 0. \end{align} Because $\int_0^1v(x,t)^2dx = 0$ for $t=0$, the above forces $$\int_0^1v(x,t)^2dx=0,\;\;\; t \ge 0.$$
This is enough to give $v(x,t)=0$ for all $t\ge 0$. So $u_1=u_2$.