5
$\begingroup$

Consider the elliptic problem $Lu = \exp(x)$ on $[0,1]$ with $Lu = -\frac{d^2u}{dx^2} + \frac{du}{dx}$ and boundary conditions $u(0) = 5$, $\frac{du}{dx}(1) + u(1) = 2$. Answer the following questions.

  1. Is $L$ a self-adjoint operator?
  2. Show that $(u, u_x) = \frac{1}{2}(u(1)^2 - u(0)^2)$
  3. Put the system in the form $a(w,u)=F(w)$ and give both $a(w, u)$ and $F(w)$ such that we successfully can show coercivity of the bilinear form.
  4. Show that the bilinear form is coercive.
  5. Let a linear function space in $C^2[0, 1]$ be spanned by $\{Q_1(x), Q_2(x), ..., Q_N(x)\}$ with $Q_i(0) = 0; i = 1, ...,N$. Give the linear system that arises from the Galerkin projection of the above problem on this space.
  • 5
    Questions 1, 2, and 4 are completely "definition" questions. Please at the very least try to do those yourself, or if you really cannot figure it out, edit the question to include the appropriate definitions and which parts of the definitions you do not know how to check.2011-09-18

1 Answers 1

4

My solution to 3 and 4.

I found convenient to solve 3 and 4 together, at one time. The solution is quite long, but something can be left out (I wrote it for sake of completeness).

A. Reduction to a standard BVP with Robin BCs.

There is a standard trick to reduce your inhomogeneous BCs to homogeneous ones, in order to obtain a BVP with classical Robin BCs.

First of all, you have to find a function $u_0\in W^{1,2}(0,1)$ which satisfies you BCs. It seems reasonable to look for such $u_0$ in the simple form $\alpha x+\beta$, with $\alpha ,\beta$ parameters to be determined imposing the BCs; it is easily seen that $u_0(x):= -3/2 x+5$ satisfies $u_0(0)=5$ and $u_0(1)+u_0^\prime (1)=2$.

Now, set $w=u-u_0$, in such a way that $w\in W^{1,2}(0,1)$ satisfies the Robin BCs: $w(0)=0 \qquad \text{and}\qquad w(1)+w^\prime (1)=0 \; ;$ hence $u$ solves the problem: $\begin{cases} -u^{\prime \prime} +u^\prime =e^x &\text{, in } ]0,1[ \\ u(0)=5 \\ u(1)+u^\prime (1)=2\end{cases}$ if and only if $w$ solves: $\tag{R} \begin{cases} -w^{\prime \prime} +w^\prime =e^x +\frac{3}{2} &\text{, in } ]0,1[ \\ w(0)=0 \\ w(1)+w^\prime (1)=0\end{cases}$ which is the aforementioned problem with standard Robin BCs.

B. Weak formulation of problem (R).

Problem (R) can be rewritten in weak form as follows. Let: $K:=\left\{ v\in W^{1,2}(0,1):\ v(0)=0\right\}\; .$ $K$ is the closed convex subset of $W^{1,2}(0,1)$ in which you expect to find the solution of (R), hence you can use any $v\in K$ as test function to recast (R) in weak form. Notice that the ODE in (R) is in fact equivalent to: $-\frac{\text{d}}{\text{d} x}\Big[ e^{-x}\ w^\prime (x)\Big] =1+\frac{3}{2}\ e^{-x}$ ($e^{-x}$ is an integrating factor), thus, after having multiplied both sides of (1) by $v\in K$, you can integrate to get: $-\int_0^1 \frac{\text{d}}{\text{d} x}\Big[ e^{-x}\ w^\prime (x)\Big]\ v(x)\ \text{d} x = \int_0^1 \left( 1+\frac{3}{2}\ e^{-x}\right) \ v(x)\ \text{d} x$ and a further integration by parts in the LHside yields: $\tag{1} \frac{1}{e}\ w(1)\ v (1) +\int_0^1 e^{-x}\ w^\prime (x)\ v^\prime (x)\ \text{d} x=\int_0^1 \left( 1+\frac{3}{2}\ e^{-x}\right)\ v(x)\ \text{d} x\; ;$ if you set: $B[w,v]:=\frac{1}{e}\ w(1)\ v (1) +\int_0^1 e^{-x}\ w^\prime (x)\ v^\prime (x)\ \text{d} x$ and $\phi (x):=1+3/2\ e^{-x}\in C^\infty ([0,1])\subseteq W^{1,2}(0,1)$, you rewrite (1) as: $\tag{W} B[w,v]=\langle \phi ,w\rangle$ which is the weak form of (R).

C. Existence and uniqueness of weak solution to (R); variational formulation of (R).

The bilinear form $B[w,v]$ is continuous, symmetric and coercive on $W^{1,2}$: in fact, continuity and simmetry are both obvious; on the other hand, in order to prove coercivity you have to use $e^{-x}\geq 1/e$ and the Poincaré inequality: $\lVert w\rVert_{W^{1,2}}\leq \gamma\ \lVert w^\prime \rVert_{L^2} \qquad \text{(} \gamma >0\text{)}$ (which does hold because $w(0)=0$; see [Brezis, Functional Analysis, Sobolev Spaces and PDEs, Prop. 8.13]), for these inequalities yield: $B[w,w]\geq \frac{1}{e}\ \lVert w^\prime \rVert_{L^2}^2\geq \frac{1}{e\ \gamma}\ \lVert w\rVert_{W^{1,2}}^2\; .$ Since $\phi \in W^{1,2}$, the functional $W^{1,2}\ni w\mapsto \langle \phi ,w\rangle \in \mathbb{R}$ is linear and continuous and Stampacchia theorem [Brezis, Thm. 5.6] applies to (W) providing a unique weak solution $w\in K$ to that problem. Such a solution also solves the variational problem: $\tag{V} \min \Big\{ B[v,v]-\langle \phi ,v\rangle,\ v\in K\Big\}$ (problem (V) actually is the variational form of (R)).

D. Regularity of the weak solution.

Here a sort of bootstrap argument is used... Maybe the proof can be shortened somehow.

Using Sobolev embedding [Brezis, Thm. 8.8], from $w\in W^{1,2}(0,1)$ you gain $w\in C([0,1])$. Now, multiply both sides of the ODE in (R) by $w$ and integrate over $[0,x]$. You find: $\int_0^x \phi (t)\ w(t)\ \text{d} t =\int_0^x \frac{\text{d}}{\text{d} t}\Big[ e^{-t} w^\prime (t)\Big]\ w(t)\ \text{d} t\; ;$ after an integration by parts in the RHside you get (if I'm not mistaken): $\tag{2} -e^{-x}\ w^\prime (x)\ w(x) +\int_0^x e^{-t}\ (w^\prime (t))^2\ \text{d} t=\int_0^x \phi (t)\ w(t)\ \text{d} t$ and after another integration by parts you have: $\tag{3} \frac{1}{2}\ e^{-x} w^2 (x) +\underbrace{\frac{1}{2} \int_0^x e^{-t} w^2(t)\ \text{d} t}_{A(x)}+\underbrace{\int_0^x \int_0^t e^{-\tau}\ (w^\prime (\tau))^2\ \text{d} \tau\ \text{d} t}_{B(x)}=\underbrace{\int_0^x \int_0^t \phi (\tau)\ w(\tau)\ \text{d} \tau\ \text{d} t}_{C(x)}\; .$ Functions $A$ and $C$ are obviously of class $C^1([0,1])$ (actually $C$ is of class $C^2$); because $w^\prime\in L^2$ and $\phi \in L^\infty$, you have $\phi\ (w^\prime)^2\in L^1$ hence $\int_0^t e^{-\tau}\ (w^\prime (\tau))^2\ \text{d} \tau$ is absolutely continuous and $B$ is of class $C^1([0,1])$; then from (3) $e^{-x}\ w^2(x) =2[C(x)-A(x)-B(x)]$ is of class $C^1([0,1])$ and $w\in C^1([0,1])$. But then from (2) you gain $w^\prime \in C^1([0,1])$, therefore $w\in C^2([0,1])$ and the weak solution $w$ is actually a classical solution of (R).


Coming to 5, I didn't catch what the problem is actually asking...

Let $X:=\text{span} \Big\{ Q_1,\ldots Q_N\Big\}$. Does the problem ask to find $\tilde{w}=\sum_{n=1}^N \alpha_n\ Q_n\in X$ s.t. $B[\tilde{w},\tilde{w}]=\langle \phi ,\tilde{w}\rangle$ via solving a linear system in $\alpha_1,\ldots ,\alpha_N$?

In such a case, I think there are some problems: in fact the bilinear form $B[\cdot, \cdot]$ contains the term $\tilde{w}^2(1)$, which seems to me a quadratic polynomial in $\alpha_1,\ldots ,\alpha_N$...

  • 0
    Thank you for your solution.2011-12-18