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\def\rit{ \hbox{\it I\hskip -2pt R} }

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{\bf Math 273b}: Calculus of Variations 

{\bf Homework \#3}

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\noindent{\bf [1]} Consider the 1D length functional minimization problem
$$\min_u F(u)=\int_{0}^1 L(u'(x))dx, \mbox{ or }\min_{u}\int_0^1 \sqrt{1+(u'(x))^2}dx,                                       
$$
for twice differentiable functions $u:[0,1]\rightarrow \rit$ with boundary conditions $u(0)=0$, $u(1)=1$.

(a) Show that the functional $u\mapsto F(u)$ is convex.

(b) Formally compute the Gateaux-differential and then obtain the Euler-Lagrange equation associated with the minimization. Solve the partial differential equation and thus obtain the unique solution of the minimization. 

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\noindent{\bf [2]} Let $A:\rit^n\rightarrow\rit^n$ be a linear self-adjoint operator, $b\in \rit^n$, and consider the quadratic function for $x\in \rit^n$
$$x\mapsto q(x):= \langle Ax,x\rangle -2\langle b,x\rangle. $$

Show that the three statements

(i) $\inf\{q(x):\ x\in \rit^n\} > -\infty $

(ii) $A\geq O$ and $b\in \mbox{Im} A$.

(iii) the problem $\inf\{q(x):\ x\in \rit^n\} > -\infty $ has a solution

\noindent are equivalent. When they hold, characterize the set of minimum points of $q$, in terms of the pseudo-inverse of $A$.

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\noindent{{\bf [3]} Computation of the Euler-Lagrange equation. 

(a) Consider the minimization problem
$$\inf_{u}F(u)=\int_{x_0}^{x_1} L(x,u(x),u'(x))dx,$$
with $u(x_0)=u_0$, $u(x_1)=u_1$ given constants, and $L$ a sufficiently smooth
function. Obtain formally the Euler-Lagrange equation of the minimization problem that is satisfied by a smooth optimal $u$. 

Hint: Consider smooth test functions $v$, such that $v(x_0)=v(x_1)=0$. Since $u$ is a minimizer, we must 
have $F(u)\leq F(u+\varepsilon v)$ for all such sufficiently smooth functions $v$ and every 
real $\epsilon$. Apply integration by parts to obtain the desired result. You should obtain a 
second-order differential equation.

(b) Let now $u(x,t)$ be a smooth solution of the time-dependent partial differential equation (PDE)
$$\frac{\partial u}{\partial t}=\frac{\partial}{\partial x} L_{u'}(x,u,u')-L_u(x,u,u'),$$
with $u(x,0)=u_0(x)$ on $(x_0,x_1)$ and $u(x_0,t)=U_0$, $u(x_1,t)=U_1$ for $t\geq 0$. 
Show that the function $E(t)=F(u(\cdot,t))$ is decreasing in time, where
$F(u)=\int_{x_0}^{x_1}L(x,u,u')dx.$

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\noindent{\bf Optional problems}

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\noindent{\bf [1]} Consider the minimization problem
$$\inf_{u}F(u)=\int_{x_0}^{x_1} L(x,u(x),u'(x),u''(x))dx,$$ 
with $u(x_0)=u_0$, $u(x_1)=u_1$, $u'(x_0)=U_0$, $u'(x_1)=U_1$ given, and $L$ a sufficiently smooth
function. As in the previous problem, derive the equation satisfied by a smooth optimal $u$. Choose test functions $v$ in $C^{\infty}[x_0,x_1]$ that satisfy $v(x_0)=v(x_1)=v'(x_0)=v'(x_1)=0$. 
(you should obtain a fourth-order differential equation).

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\noindent{\bf [2]} Consider the minimization problem in two dimensions $(x,y)$,
$$\inf_{u}E(u)=\int_{\Omega}L(x,y,u,u_x,u_y)dxdy,\ \ \ u=g\mbox{ on }\partial\Omega,$$
where $g$ is a given function on the boundary $\partial\Omega$, with $\Omega$ a bounded and open 
region in the plane. Assume that the integrand $L$ is differentiable.

(i) Show that a sufficiently smooth minimizer $u$ formally satisfies the Euler-Lagrange equation
$$\frac{\partial}{\partial x} L_{u_x}(P)+\frac{\partial}{\partial y}L_{u_y}(P)-L_u(P)=0$$
on $\Omega$, where $P=(x,y,u(x,y),u_x(x,y),u_y(x,y))$.

(ii) Apply the above result to the case when $L(x,y,u_x,u_y)=u_x^2+u_y^2-2fu$.

Hint for (i): consider another test function $v$, such that $v=0$ on $\partial \Omega$. Since $u$ is a 
minimizer, we must have $E(u)\leq E(u+\varepsilon v)$ for all such sufficiently smooth functions $v$ and 
all real $\epsilon$. Apply integration by parts to obtain the desired result. Here, $(u_x,u_y)=\nabla u$.

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\noindent{\bf Notes:} 

$\bullet$ {\bf Pseudo-Inverse.} If $A$ is a symmetric (or self-adjoint) linear operator on $X$, then 
$\mbox{Im} A^{\perp}=\mbox{Ker}A$. Let $p_{\mbox{Im}A}$ be the operator of orthogonal projection onto $\mbox{Im}A$. For given $y\in X$, there is a unique $x=x(y)$ in $\mbox{Im}A$ such that $Ax=p_{\mbox{Im}A}y$. Forthermore, the mapping $y\mapsto x(y)$ is linear. This mapping is called the pseudo-inverse, or generalized inverse of $A$. 

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$\bullet$ {\bf Integration by Parts Formula.} Let $\Omega$ be an open and bounded subset of $R^d$, with Lipschitz-continuous (or sufficiently smooth) boundary $\partial\Omega$. Let $\vec{n}=(n_1,n_2,...,n_d)$ be the exterior unit normal to $\partial\Omega$. 
 Let me recall the following fundamental Green's formula, or integration by parts formula: given two functions $u,v$ (with $u$, $v$, and all their 1st order partial derivatives belonging to $L^2(\Omega)$, or $u,v\in H^1(\Omega)$), then 
$$\int_{\Omega}u v_{x_i} dx=-\int_{\Omega} u_{x_i} v dx +\int_{\partial\Omega} u v n_i dS.$$

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