1
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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 $L^2(0,1)$)

$\endgroup$
0
$\begingroup$

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 $u_1=u_2$.

$\endgroup$
3
  • $\begingroup$ One might say that the initial condition is just arbitrary if none is chosen..I.E $u(x,0)= g(x)$. Even if it isn’t stated, it is somewhat implied, no? $\endgroup$
    – DaveNine
    Commented Dec 8, 2018 at 20:26
  • $\begingroup$ @DaveNine : This question is all about having a well-posed problem. So I don't think skipping a discussion of conditions that lead to "well-posedness" is called for in this case. $\endgroup$ Commented Dec 8, 2018 at 21:04
  • $\begingroup$ @DisintegratingByParts : Thank you. Does the uniqueness of the solution imply well-posedness of the problem? $\endgroup$
    – D. Zh
    Commented Dec 9, 2018 at 18:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .