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\noindent{\bf 269C\ \ \ HW\#2, due on Monday, April 24}

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\noindent{\bf [1]} Find the linear basis functions for the triangle $K$ with vertices at 
$(0,0)$, $(h,0)$ and $(0,h)$. Show that the corresponding element stiffness
matrix is given by 
$$
\left[
\begin{array}{ccc}
1 & -\frac{1}{2} & -\frac{1}{2} \\
-\frac{1}{2} & \frac{1}{2} & 0\\
-\frac{1}{2} & 0 & \frac{1}{2}
\end{array}
\right]
$$
Using this result, show that the linear system (1.25) of Example 1.1. has the
stated form (pages 30-31) (there is a typo in the textbook regarding the matrix)
 
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\noindent{\bf [2]} Let $V$ be a Hilbert space and $L:V\rightarrow R$ be a linear form. Show that $L$ is bounded if and only if $L$ is continuous. Then directly apply this result to a bilinear form $a:V\times V \rightarrow R$. 

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\noindent{\bf [3]} Let $V$ be a Hilbert space and the (nonlinear) operator $A:V\rightarrow V$, satisfying 

(i) there is $M\geq 0$ s.t. $\forall$ $u,v\in V$, $\|Au-Av\|\leq M\|u-v\|$ 

(ii) there is $\alpha >0$ s.t. $\forall$ $u,v\in V$, $\langle Au-Av,u-v\rangle \geq \alpha \|u-v\|^2$ 

Show that the nonlinear equation $Au=f$ has a unique solution (for some $f\in V$), using the Banach fixed point theorem and the same technique for proving the Lax-Milgram theorem (introduce the function $g_{\lambda}$). 

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\noindent{\bf [4]} Let $V$ be a complex Hilbert space, $a:V\times V\rightarrow C$ a sesquilinear form, $L:V\rightarrow C$ an anti-linear form. Assume that $a$ is Hermitian, thus $a(v,u)=\overline{a(u,v)}$, $\forall u,v\in V$, and that $a(v,v)\geq 0$. Consider the problems 

$$\mbox{(V)} \mbox{ Find } u\in V \mbox{ s.t. } a(u,v)=L(v), \ \forall v\in V,$$

$$\mbox{(M)}  \mbox{ Find } u\in V \mbox{ s.t. } J(u)=\inf_{v\in V}J(v),$$
with $J:V\rightarrow R$ defined by $J(v)=\frac{1}{2}a(v,v)-\mbox{Re}L(v)$. 

Show that $u\in V$ is solution of (V) iff $u\in V$ is solution of (M). 

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\noindent $\bullet$ {\bf Note:} Over the complex space $C$, with the above notations, we say that $a:V\times V\rightarrow C$ is a sesquilinear form if $a$ is linear in the first argument (under the usual addition and scalar multiplication) and antilinear in the second argument (usual addition but $a(u,\lambda v)=\bar{\lambda}a(u,v)$). Note that if $\mbox{Re} a(v,v)\geq \alpha \|v\|^2$, together with $a$ and $L$ bounded, then the Lax-Milgarm Lemma holds in the complex case with the same proof. 

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