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)
.
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
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:
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 )
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
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.