calculus of variations type problem

[david....@gmail.com]

Hi,

I have recently been thinking about an optimization problem to which I
can't find an easy solution. The problem is: find a C^3 function f :
[0,T] -> R which minimizes

\int_0^T (d^3f/dt^3)^2 dt

subject to the constraint

\sum_i exp(-f(t_i)) c_i

where the sum is over a finite set, and c_i are arbitrary real numbers.
obviously the problem is quite specific and can be translated into
various equivalent forms. However I haven't figured out a way to make
it amenable to the techniques I learnt as an undergrad, so if anyone
could advise me on an approach, that would be great.

Thanks,
David

[Rod]

My first guess would be to replace the sum with an integral involving dirac
deltas.
\int G_i(t)exp(-f(t))

where G_i(t) = sum_i c_i * delta(t-t_i)

<david....@gmail.com> wrote in message
news:1140391131.2...@g43g2000cwa.googlegroups.com...

[David Ziskind]

<david....@gmail.com> wrote in message
news:1140391131.2...@g43g2000cwa.googlegroups.com:
> ... The problem is: find a C^3 function f : [0,T] -> R which

> minimizes
>\int_0^T (d^3f/dt^3)^2 dt
>subject to the constraint

>\sum_i exp(-f(t_i)) c_i [etc]

(This post will look best if viewed in a monospace font.)

As suggested by Rod in the previous reply, this problem can be
transformed to an isoperimetric problem by use of the Dirac delta
function.

In the theorem following, I stands for the identity function, Int
abbreviates "integral taken from 0 to 1", and (Ex) reads as "there
exists an x such that". PiF denotes the partial with respect to the
i th argument of F. Generally, upper-case letters denote functions
and lower-case, numbers. The use of 0 and 1 for the boundary
arguments make the presentation slightly easier to read through.

The following result can be established:

"Isoperimetric variational problem with higher derivatives"

If [ Int(K(I,F*,F*',F*'',F*''')
= min{v:((EF) ( v = Int{K(I,F,F',F'',F''')}
and 0 = Int{J(I,F,F',F'',F''')}
and (F0,F'0,F''0)=(F*0,F*'0,F*''0)
and (F1,F'1,F''1)=(F*1,F*'1,F*''1)}
and 0 = Int(J(I,F*,F*',F*'',F*''')]
then
(Eu)(EL)[ L = K + u*J
and u is a constant
and (P2L)(I,F,F',F'',F''')
- [(P3L)(I,F,F',F'',F''')]'
+ [(P4L)(I,F,F',F'',F''')]''
- [(P5L)(I,F,F',F'',F''')]''' = 0]

(In this theorem, expressions such as K(I,F,F',F'',F''') mean the
function x->K(x,Fx,F'x,F''x,F'''x).)

-----------------------------------------------------

Now, let x->d(x) denote the Dirac delta function, and U denote the
unit step function. (U has no connection with u above.)

Impose the condition that all x_i are interior to [0,1].

Now, the above theorem is applicable to the problem posed in the OP
since by letting

J(v,w,x,y,z) =def Sum{i: d(v-x_i)*H(F(v))*c_i} - e

the constraint equation recited in the theorem which is

0 = Int{s=0 to 1: J(s,Fs,F's,F''s,F'''s)} ,

by the above def for J becomes

0 = Int{s=0 to 1: Sum{i: d(s-x_i)*H(F(s))*c_i} - e}}

which evaluates to

0 = Sum{i: H(F(x_i))*c_i} - e

which is the constraint posed in the OP. (I have made two
adjustments here: first, the exp function in the OP is replaced by
generic H, and second, I have assumed that the OP intended that the
ordinary sum involving F(x_i) should equate to some prescribed value
e.)

The associated Lagrangian (ie, basic + multiplier*constraint) is:

L(x,Fx,F'x,F''x,F'''x) =
(F'''x)^2 + u*Sum{i: d(x-x_i)*H(F(x))*c_i} - u*e

Technically, in order to compute the partials, one should write out
L(v,w,x,y,z) but for clarity, I am following the practice of using
x,Fx,F'x,... etc as though they were dummies.

From this point forward, the minimizing function, ie F*, will be
written as F. There should be no confusion as the former use of F
as a comparison function is no longer needed.

Computing the four relevant partials of L gives the Euler-Lagrange
equation:

u*Sum{i: d(x-x_i)*H'(F(x))*c_i} - 2*(F^[6])(x) = 0

Integrate over x, from 0 to x*, and relabel x* to x. Obtain:

u*Sum{i: U(x-x_i)*H'(F(x_i))*c_i} - 2*(F^[5])(x) + 2*(F^[5])(0) = 0

Perform five more integrations and obtain the primitive which is an
equation involving:
-F(x),
-six constants of integration: F0, F'0, F''0,...,F'''''0, and
-six "ramp" functions: U(x), U_1(x),U_2(x),...,U_5(x).
U_1(x) = x for x>0, = 0 otherwise. U_2(x) = x^2/2 for x>0,
= 0 otherwise. The higher U_i's have parallel definitions.

The first three constants of integration are prescribed data. The
last three are obtained by evaluating the equations for Fx, F'x, and
F''x at x=1 and using the prescribed right-hand boundary values.

This leaves determination of the Lagrangian multiplier u. One takes
the constraint equation, in this case
0 = Sum{i: H(F(x_i))*c_i} - e and substitutes in for the various
F(x_i) by evaluating F(x) at x=x_i. The result is an equation
involving u (since F(x) involves u). Solve this equation for u and
substitute it into the equation for F(x). This completes the
problem of calculating F. As always, additional analysis will be
required if it is desired to show that F is in fact a minimizing
function rather than merely an extremal.

--------------------------

If the problem posed is one where one of the boundary value
prescriptions is omitted, say F''(1), then the isoperimetric
approach is unnecessary and the basic Euler-Lagrange equation can be
employed. That is, the constant of integration for the missing
boundary value can be used to satisfy the constraint equation.

For a good compilation of many tools and results for variational
theory, see Chapter 11 of Korn and Korn, "Mathematical Handbook for
Scientists and Engineers", Dover, 2000, ISBN 0-486-41147-8.

David Ziskind
zis...@ntplx.net

[Thomas Mautsch]

In news:<1140391131.2...@g43g2000cwa.googlegroups.com>
schrieb david....@gmail.com <david....@gmail.com>:

I may be mistaken, but I think that the minimum value should be zero,
taken on for an appropriate choice of a constant,
or at most linear function for f.