\begin{titlepage} \begin{center} %\vspace*{1.29375in} \vspace*{2in} {\Large Computing Assignment, Part 1}\\ \vspace{.1in} Mathematics 503, Mathematical Modeling\\ \vspace{1.5in} California State University, Fullerton\\ \vspace{.1in} July 26 2007.\\ %\vspace{1.334375in} \vspace{1in} BY\\ \vspace{.1in} Nasser Abbasi\footnote{Student, Applied Mathematics Department, California State University, Fullerton.}\\ \end{center} \end{titlepage}

problem:

Let $V$ be the space of continuously differentiable functions $y$ such that MATH. On this space consider the functional

MATH

where $q>0$ and $f$ are given constants.

(a) Show that $J$ achieves a minimum at $y$ iff

MATH

for all MATH

(b)Show that if $y\in V$ is twice continuously differentiable and satisfies this optimality condition then $y$ satisfies the differential equation MATH

(c)Conversely show that if $y\in V$ is twice continuously differentiable, and satisfies the differential equation above, then $y$ satisfies the optimality condition of part (a)




answer:




part(a). Let MATH, and let MATH. The set $A$ is called the set of admissible functions (from which the function which will minimize the functional will be found), and $A_{y}$ is called the set of admissible directions. We require also that $y+t\phi \in A$ where $t$ is some small scalar.

Since this is an IFF, then we need to show the forward and the backward case. Start with the forwards case:

Given: MATHshow that

if MATH then $J\left( y\right) $ is minimum

Consider MATH

Where MATH $\geq 0$ since $q>0$ and the other functions are squared. Hence this implies from the above that if MATH then $y$ is a minimizer of $J\left( y\right) $. In other words MATH

(This is because any change from $y$ along any of the admissible directions $\phi $ will result in a functional MATH which is larger than it was at $J\left( y\right) $).

Now to show the backward case:

If MATH then $J\left( y\right) $ is not a minimum.

Assume for some $\phi =\phi _{0}$ we have MATH

MATH

Where MATH .Hence we have

MATH

Now, no matter how large $Q$ is, we can make $t$ small enough so that $tQ$ is smaller than the absolute value of MATH. $\ $But since MATH, then MATH will be a negative quantity. Hence, since $t\times $negative quantity is also negative quantity, then we conclude that MATH

Hence

$y$ is not a minimizer of $J\left( y\right) $
. So no matter which $y$ we hope it is our minimum, we can find an admissible direction $\phi _{0}$ such that if move very slightly away from this $y$ in this admissible direction, we find that MATH is smaller (this will always be the case if MATH)

Part (b)

Given $y\in C^{2}[0,1]$, and MATH for all admissible directions $\phi $. Show that $y$ satisfies the differential equation MATH

Since MATH, in other wordsMATH. Then now we do integration by parts.

MATH

Since MATHNow substitute the above back into MATH and take $\phi $ as common factor, we obtain

MATH

Now we apply the fundamental theory of variational calculus (which Lemma 3.13 is special case) and argue that since $\phi $ is arbitrary admissible direction, then for the above to be zero for every $\phi $, we must have MATH

or

MATH

with MATH

Part (c)

Given $y\in C^{2}[0,1]$ and MATH show that MATH

Let $\phi \in A_{d}.$ Now MATH

But MATH since $\phi =0$ at both ends. Hence the above becomes

MATH

But MATH since $y$ satisfies the differential equation. Hence the above becomes

MATH

Which is the optimality condition (weak form) of part (a) we are asked to show.