5
$\begingroup$

I need a check on the following problem, which is an application of the Lax-Milgram lemma. $$ \begin{cases} \dfrac{\partial u}{\partial t} -\dfrac{\partial }{\partial x} \Bigl( \alpha \dfrac{\partial u}{\partial x}\Bigr) - \beta u =0 \\ u(x,0)=u_0(x), & x \in [0,1] \\ u = \eta, & x=0,\; t>0 \\ \alpha \dfrac{\partial u}{\partial x} + \gamma u =0, & x=1,\; t>0 \end{cases} $$ where $\alpha(x), u_0(x)$ are given functions, $\beta >0$ and $\eta,\gamma \in \mathbb{R}$.

Question. Prove existence and uniqueness of the weak solution, giving some suitable assumptions on $\alpha(x)$,$\gamma, \eta$.


Here's my attempt:

As test functions, I choose $v \in V=H_{\Gamma_d}^1$, where $\Gamma_d={0}$,i.e. I use tes functions which are $0$ where I have the Dirichlet data. By multiplying, with standard arguments one obtains the following bilinear form $$a(u,v)=\gamma u(1)v(1) + \int_0^1 \alpha u' v' dx - \beta \int_0^1 u v dx$$

To show existence and uniqueness, I need to show

  • $a(v,v) + \lambda \geq \alpha \|v\|_V^2$ (weakly coercive)
  • $a(u,v) \leq M \|u\|_V \|v\|_V$

Weakly coercive: Assume $0<\alpha_0 < \alpha(x) < \alpha_1$ $$a(v,v) + \beta \|v\|_V^2 \geq \frac{\alpha_0}{1+C_p^2} \|v\|_V^2 + \gamma v(1)^2$$

where $C_p$ is the Poincarè constant.
If $\gamma >0$, then: $$a(v,v) + \beta \|v\|_V^2 \geq \frac{\alpha_0}{1+C_p^2} \|v\|_V^2$$ i.e. $a(\cdot, \cdot)$ is weakly coercive.

Continuity:

$$|a(u,v)| \leq \alpha_1 \|u\|_V \|v\|_V + \beta \|u\|_V \|v\|_V + \gamma u(1)v(1)$$

Now, since I want a bound with the $H^1$ norm, I note that $$|u(1)| \leq \int_0^1 |u'(s)|ds + \eta$$ and therefore $$|u(1)| \leq \|u\|_V + \eta$$ Also, $v(1) = \int_0^1 v'(s)ds$ implies $|v(1)| \leq \|v\|_V$

Hence, $$|a(u,v)| \leq (\alpha_1 + \beta) \|u\|_V \|v\|_V+\gamma \|u\|_V \|v\|_V + \gamma \eta \|v\|_V $$

Now, if $\eta <0$, then $$|a(u,v)| \leq (\alpha_1+\beta+\gamma)\|u\|_V \|v\|_V$$

$\endgroup$
1
  • $\begingroup$ My comment may be misplaced, but I have some queries about your "weak coercivity". In Babuska's generalisation of Lax - Milgram, coercivity is replaced with inf-sup stability. Is your weak coercivity really equivalent to this, especially considering your test and trial spaces are not equal? And for continuity, I would think that $\|u \|_V$ should be replaced by a norm on your trial space as $u\notin V$. $\endgroup$ – epiliam Sep 10 '20 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.