Average over a line integral
David Ziskind
<7dj1ra$rm9$1...@sloth.swcp.com> regarding the problem of maximizing the ratio
of two functionals (average value of f along path).
Specifically, the problem is to determine the R1->R1 function y which
relatively maximizes the functional E in the unconstrained endpoint sense,
where E is defined by:
E(z) =def Int[f(xt,zt)*((x't)^2+(z't)^2))^(1/2) dt]
/ Int[((x't)^2+(z't)^2))^(1/2) dt]
-Int indicates: integral from 0 to 1.
-z is a dummy variable, and represents an R1->R1 function. z is used in
place of y to avoid confusion with the maximizing element.
-x is a given R1->R1 function. f is a given R2->R1 function.
-y is an "unknown", to be determined by the maximization requirement.
-Both y(0) and y(1) are unconstrained (and should be allowed to vary so as
to fulfill the maximization requirement).
We can obtain a necessary condition for the maximization requirement by
following the usual Euler-Lagrange strategy. In this case, because of the
unconstrained endpoints, the problem is translated into a related problem
of maximizing another functional (say "e") with the relation E(z) = e(z,
z(0), z(1)). The original problem is then addressed by maximizing e over
the appropriate set of triples (z, a, b), with z in R1->R1, a in R1, and b
in R1.
In setting up the functional e, the dependency on z is "direct", and the
dependency on a (and b) is "indirect", i.e., a (and b) enter through terms
such as z(a,b,t). However, the distinction between "direct" versus
"indirect" does not change the strategy or mechanics of computing
stationary (aka critical) values. Thus, in summary, the program is:
- compute the variational derivative w/r to z and set it equal to zero,
- compute the ordinary derivative w/r to a and set it equal to zero,
- compute the ordinary derivative w/r to b and set it equal to zero.
It can be shown via the calculus that each of these three derivatives must
be zero when (y, a, b) is a maximizing triple.
Finally, as the functional to be maximized is the ratio of two integrals,
rather than a single integral is in the classic situation, it will be found
that this leads to an integro-differential equation as the necessary
condition (versus the classic Euler-Lagrange ODE), and also, that the
necessary condition can be efficiently stated using a Lagrange multiplier.
To formulate the necessary condition, we introduce the following
intermediates:
(Kf)(g,h) =def f(x(t),g)*[(x't)^2+h^2]^(1/2)
-K has built-in, or concealed dependency on x(t), and x'(t). When one
encounters K below the occurrence of x(t) and x'(t) is always the same, and
thus for brevity, these arguments are not shown explicitly.
-Kf has functional dependency on f. The definition of K shows f as an
argument because we will also have need to write expressions such as
(Ku)(g,h), where u is another function entirely different from f.
-The evaluated Kf is a function mapping R2->R1. That is, g, and h are
ordinary numbers (R1 elements). (The symbols g and h are used here to
represent numbers rather than in the more traditional appointment as
symbols for functions.)
u(p,q) =def 1 for all p, q. That is, u is the unit function.
Note that (Ku)(p,q) is actually just [(x't)^2+q^2]^(1/2). Thus, the
derivative of (Ku)(p,q) w/r to p is zero.
L(y) =def Int[(Kf)(yt,y't) dt] / Int[(Ku)(yt,y't) dt]
L is a functional mapping (R1->R1) into R1.
Q =def Kf - L(y)*Ku
-Note that Q is a mapping of the same type as Kf; ie, an R2->R1 map.
-Q has functional dependency on y (through L) which for brevity is not
shown explicitly.
P1Q represents the partial derivative of Q w/r to the first argument, ie
P1Q(g,h) = (d/dg)(Q(g,h)). (Again, g, h are used here to represent
ordinary numbers.) P2Q is the partial derivative of Q w/r to the second
argument.
We can now give the NECESSARY CONDITION. NC(y) iff(def)
a) (P1Q)(yt,y't) = (d/dt)[(P2Q)(yt,y't)] for all 0<=t<=1
b) (P2Q)(y0,y'0) = 0
c) (P2Q)(y1,y'1) = 0
It is worth noting that P1Q, P2Q have the concealed functional dependency
on y, so this system is more complicated then it appears on the surface.
Note that in "a" the differentiation w/r to t sees t dependency in both the
arguments yt, y't and also within the expression P2Q itself. Writing this
out, we have in part:
(P2Q)(yt,y't) = (P2(Kf))(yt,y't) - etc
= (d/dh){(Kf)(g,h)}(g=yt,h=y't) - etc
= (d/dh){f(xt,g)*[(x't)^2+h^2]^(1/2)}(g=yt,h=y't) - etc
= f(xt,g)*[(x't)^2+h^2]^(-1/2)*h}(g=yt,h=y't) - etc
= f(xt,yt)*[(x't)^2+(y't)^2]^(-1/2)*y't - etc
CONCLUSION
We have the following result:
If y relatively maximizes E in the unconstrained endpoint sense,
then NC(y) holds true.
The problem of finding the maximizing element however is only begun. We
first have to solve the complicated integro-differential system given by
NC(y). In most cases, this will be quite difficult, and series methods
will have to employed. A further difficulty is that such solutions may or
may not be relative maxima. It is a characteristic of the stationary value
strategy that such values may be (relative) maxima, minima, or "other", and
the literature of the calculus of variations includes various tests to
distinguish between the three situations. This seems to be a complicated
subject and I can't really offer any firm information -- my impression is
that inflection points and saddle points will occur as in vector
transformations but do not know if there are other situations as well.
Let us finish by thinking about how the system NC(y) might be solved. I
would suggest the following approach:
1) Expand y in a (generalized) Fourier series. In other words, set y equal
to the inverse Fourier transform of some infinite-vector c.
2) Left-operate on "a" by the (forward) Fourier series transform.
3) The result is that "a" is transformed into an (infinite-vector into
infinite-vector) transformation.
4) "b" and "c" give two additional relations involving the infinite-vector
c; Fourier series expansion of y is assumed as before.
5) Choose the order of accuracy desired and truncate the Fourier series
expansion to the first n terms.
6) Retain the first n-2 equations from the infinity produced by "2".
7) Append the two additional equations supplied by "4".
8) The result is a system of n equations in n unknowns (truncated c
vector), and can be solved in various ways, e.g., Newton-Raphson iteration.
9) With the values of truncated c vector known explicitly, evaluate
(inverse Fourier transform onto truncated c) and obtain the approximate y.
Sufficiency tests are discussed in:
-Korn and Korn, "Mathematical Handbook for Scientists and Engineers", 2nd
Edition, McGraw-Hill, 1967, ISBN 0070353700;
-Bronshtein and Semendyayev, "Handbook of Mathematics", Van Nostrand
Reinhold, New York, 1985, ISBN 0-442-21171-6.
David Ziskind
zis...@ntplx.net
David Ziskind
which had identifiers: David Ziskind <zis...@ntplx.net>, and
<01be7ccf$caea9bc0$LocalHost@default>.
In the problem discussed in that post, there is a difficulty in the problem
formulation that was overlooked -- as stated, the problem has a trivial
least upper bound and typically there is no maximum.
To determine a lub, one would simply determine the (xt,yt) giving greatest
f(xt,yt) (an interior maximum is assumed) and spend as much time there is
possible. Thus, we would construct a path with a very high frequency
sinusoid which repeatedly bracketed the highest mountain top with many
small North-South excursions (and minimal West-to-East travel if x't=1).
Such an extremely long path within a very small region would dominate the
average to as great a degree as desired.
This trivial result is not what we are after. The deduction and suggested
series calculation in the prior post leads to a result which is strongly
controlled by the discarding of the high-frequency terms in the finite
calculation. Possibly, we would obtain results which coincide with
intuitive notions of following the highest ridge but it seems difficult to
put this on a rigorous footing.
The original problem seems to express a fairly simple idea -- travel from
start line to finish line, keeping as high as possible, and do not spend a
lot of time wandering back-and-forth across the ridge line. Accordingly, I
think the problem statement can be revived by redefining E in the following
way:
E(z) =def Int[f(xt,zt)*((x't)^2+(z't)^2)^(1/2) dt]
/ Int[{1+b*k(x't,x''t,z't,z''t)}*((x't)^2+(z't)^2)^(1/2) dt]
This is similar to before, with the new factor of 1+b*k(...) in the
denominator which introduces a penalty for non-zero path curvature. k(...)
here is the absolute measure of curvature for a path in the plane and is
given by the traditional formula
|(x't*z''t-z't*x''t)/((x't)^2+(z't)^2)^(3/2)|. The greater the curvature,
the greater is the penalty. For nearly straight segments, the penalty is
nearly zero and the weight is nearly one. b is a parameter determining how
strongly the penalty is felt.
The analysis, calculation, and results for this new E are very similar to
those in the prior post. The main difference is that the Euler-Lagrange
system now incorporates third order derivatives. Referring to the prior
post, we make the following changes.
Two new quantities k and w are introduced. (There is no relationship
between k and the forthcoming K, they are two entirely different objects).
k(p,q,r,s) =def |(p*s-r*q)/(p^2+r^2)^(3/2)|
w(p,q,r,s) =def 1+b*k(p,q,r,s)
We now also have two unit functions:
u(p,q) =def 1 for all p,q
v(p,q,r,s) =def 1 for all p,q,r,s
The remaining intermediates parallel those in the prior post:
(Kfw)(g,h,i) =def f(xt,g)*w(x't,x''t,h,i)*[(x't)^2+h^2]^(1/2)
g,h,i are ordinary numbers (reals). f, w are used in this line as dummy
symbols and are substitutable -- thus, e.g., w is not restricted to 1+b*k.
L(y) =def Int[(Kfv)(yt,y't,y''t) dt] / Int[(Kuw)(yt,y't,y''t) dt]
Q =def Kfv - L(y)*Kuw
To express the Euler-Lagrange system compactly, it is convenient to define
a new quantity J:
J(t) =def (P2Q)(yt,y't,y''t) - (d/dt)[(P3Q)(yt,y't,y''t)]
Note that J(t) has t dependency in two ways: a) through the arguments yt,
y't, and y''t; and b) through Q which contains concealed dependency on xt,
x't, and x''t.
The necessary condition is now given by: NC(y) iff(def)
a) (P1Q)(yt,y't,y''t) = (d/dt)(J(t)) for all 0<=t<=1
b) J(0) = 0
c) J(1) = 0
The condition NC(y) is obtained when the maximum is computed over an
appropriate set U, say, a set of (Y,g,h) triples. (The preceding u is
entirely different from U.) A suggested definition for
"(Y,g,h) is in U" is:
Y is in R3->R1
and g,h are in R1
and (there exist c,d such that)
(for all a,b)
( Y(a,b,0)=a
and Y(a,b,1)=b
and (P3Y)(a,b,0)=c
and (P3Y)(a,b,1)=d) .
The problem is now a mix of unconstrained and constrained end-point
conditions. The y in NC(y) is given by y(t)=Y(a*,b*,t) with a*, b*
indicating the maximizing end-points.
In the prior post, the exemplary expansion of P2Q is specific to the
earlier form and does not hold for the new definition.
The nine step suggested program for solving NC(y) carries over unchanged.
ADDITIONAL ITEMS: (1) In the prior post, the "result" given following
"CONCLUSION" is subject to the hypothesis that the three mentioned
derivatives of the functional "e" do in fact exist. This qualification
continues in the new formulation. (2) In paragraph five of the prior post,
"is in" should be "as in".
David Ziskind
zis...@ntplx.net