We consider the problem $$-u''+a(x)u=f, x\in (0,1)=\Omega , u(0)=\alpha,u(1)=\beta$$ with $f \in L^2(\Omega) , a(x) \geq a_0 > 0, a \in L^{\infty}(\Omega)$

Question 1: prove that the variational formulation associate to this problem admits a unique solution in an adequate Hilbert space $V$ (we can use for this proof the function $u_1(x)=\alpha +(\beta - \alpha) x$).

Question 2: prove that $$||u||_V \leq C ||f||_{L^2}$$

For the question 1, I consider $u=w + u_1$ with $w \in V = H^1(\Omega)$ and solution for the equation of the problem, and I prove that $w$ is unique in $H^1_0(\Omega)$ by Lax-Milgram, so the problem admits a unique solution $u$ in $H^1_0.$

My problem is in question 2, I don't know how we prove it. Thank's for the help.

  • $\begingroup$ Your second question is strange. If $u$ is a solution that belongs to $V$, then it is immediate that $\|u\|_V$ is finite. $\endgroup$ – Tomás Aug 10 '13 at 20:24
  • $\begingroup$ this is a mistake, sorry. We have to prouve that $||u||_V \leq C ||f||_{L^2}$$ $\endgroup$ – jijii Aug 10 '13 at 21:34
  • $\begingroup$ Where did you find this problem? $\endgroup$ – Tomás Aug 10 '13 at 23:13
  • $\begingroup$ this is an old exam, and i haven't a solution $\endgroup$ – jijii Aug 11 '13 at 15:53

I'll treat the case $\alpha = \beta = 0$ (so that I do not need to jump throught the $u_1$ hoops); from the bound required in part 2, I deduce that we assume $f \in L^2(\Omega)$.

  1. If you test your original equation with functions from $C^\infty_0(\Omega)$, you arrive at $$ \int_0^1 [ -u'' + au ]v = \int_0^1 fv$$ Integration by parts turns this into $$ u'(1)v(1) -u'(0)v(0)+ \int_0^1 [u'v' + au v] = \int_0^1 fv$$ where the first two terms cancel since we have $v(0) = v(1) = 0$. The final variational problem reads $$ \int_0^1 [u'v' + au v] = \int_0^1 fv \quad \forall v \in C^\infty_0(\Omega)$$ or more concisely $$ a(u,v) = \ell(v) $$ The right space to pose this problem over is $H^1_0(\Omega)$, which is a Hilbert space w.r.t. the scalar product $(v,w) = \int_0^1 uv + \int_0^1 u'v'$.

    The Lax-Milgram theorem can now be used to prove that the variational problem has exactly one solution in $H^1_0$. To that end, observe that $\ell$ is continuous; this is true since we assumed $f \in L^2(\Omega)$. We also need the bilinear form to be coercive and continuous with respect to the aforementioned scalar product: For ellipticity, we can use the fact that $a \ge a_0 > 0$, so that clearly (without loss of generality, $a_0 \le 1$), we have $a(v,v) \ge a_0 (v,v)$. We also have $a(v,w) \le \|a\|_\infty (v,w)$ and thus continuity.

    Note that because of Poincare's inequality, we could handle the case $a_0 = 0$ as well.

  2. Since $a(u,v) = \ell(v)$ holds for all $v \in C^\infty_0$ functions, it also holds for all $H^1_0$ functions by density. In particular, we can choose $v = u$ and obtain $$ a_0 \|u\|_{H^1}^2 = a_0 (u,u) \le a(u,u) = \ell(u) \le \|f\|_{L^2}\|u\|_{L^2} \le \|f\|_{L^2}\|u\|_{H^1} $$ where our earlier thoughts on the bilinear form and the Cauchy-Schwarz inequality were used. Dividing by $\|u\|_{H^1}$ (if $u = 0$, the claim would be trivial) yields $$ \|u\|_{H^1} \le \frac 1{a_0} \|f\|_{L^2} $$ as desired.


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