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## HW 12 Mathematics 503, analytical part, July 26, 2007

June 12, 2014

### 1 Problem

Let be the space of continuously diﬀerentiable functions such that . On this space consider the functional

where and are given constants.

(a) Show that achieves a minimum at iﬀ

for all

(b)Show that if is twice continuously diﬀerentiable and satisﬁes this optimality condition then satisﬁes the diﬀerential equation

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

### 2 Solution

#### 2.1 part(a)

.

Let , and let . The set is called the set of admissible functions (from which the function which will minimize the functional will be found), and is called the set of admissible directions. We require also that where 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: show that

if then is minimum

Consider

Where since and the other functions are squared. Hence this implies from the above that if then is a minimizer of . In other words

(This is because any change from along any of the admissible directions will result in a functional which is larger than it was at ).

Now to show the backward case:

If then is not a minimum.

Assume for some we have

Where .Hence we have

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

Hence is not a minimizer of . So no matter which we hope it is our minimum, we can ﬁnd an admissible direction such that if move very slightly away from this in this admissible direction, we ﬁnd that is smaller (this will always be the case if )

#### 2.2 Part (b)

Given , and for all admissible directions . Show that satisﬁes the diﬀerential equation

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

Since Now substitute the above back into and take as common factor, we obtain

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

or

with

#### 2.3 Part (c)

Given and  show that

Let Now

But since at both ends. Hence the above becomes

But since satisﬁes the diﬀerential equation. Hence the above becomes

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