The following is a local copy of a reply the author posted to sci.math on February 25, 2006. Subject line of the thread is "calculus of variations type problem", and the start date was February 19, 2006. For general remarks concerning these local copies, see the home page of this web site.
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In the post starting the thread, the OP wrote (in part):
> ... 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]
As suggested by another contributor in a 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
http://davidziskind.org/
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Rev. 12/15/11