Min/Max Variation Problem 1
Problem 1
Among all smooth (sufficiently differentiable) functions f(x) defined on segment [a,b] and taking values at endpoints f(a)=A and f(b)=B find the one with the shortest graph between points (a,A) and (b,B).
Solution
First of all the length of a curve representing a graph of a function is a functional with that function as an argument. Let's determine its explicit formula in our case.
The length ds of an infinitesimal segment of a curve that represents a graph of function y=f(x) is
ds = [(dx)² + (dy)²]½ =
= [(dx)² + (df(x))²]½ =
= [1 + (df(x)/dx)²]½·(dx) =
= [1 + f '(x)²]½·(dx)
The length of an entire curve would then be represented by the following functional of function f(x):
Φ[f(x)] = ∫[a,b][1 + f '(x)²]½dx
We have to minimize this functional within a family of smooth functions defined on segment [a,b] and satisfying initial conditions
f(a)=A and f(b)=B
As explained in the previous lecture, if functional Φ[f(x)] has local minimum at function-argument f0(x), the variation (directional derivative)
d/dt Φ[f0(x)+t(f1(x)−f0(x))]
at function-argument f0(x) (that is, for t=0) in the direction from f0(x) to f1(x) should be equal to zero regardless of location of f1(x) in the neighborhood of f0(x).
Let f1(x) be another function from the family of functions defined on segment [a,b] and satisfying initial conditions
f(a)=A and f(b)=B
Let Δ(x) = f1(x) − f0(x).
It is also defined on segment [a,b] and, according to its definition, satisfies the initial conditions Δ(a)=0 and Δ(b)=0.
Using an assumed point (function-argument) of minimum f0(x) of our functional Φ[f(x)], another point f1(x) that defines the direction of an increment of a function-argument and real parameter t, we can describe a subset of points (function-arguments) linearly dependent on f0(x) and f1(x) as
f0(x)+t·(f1(x)−f0(x)) =
= f0(x) + t·Δ(x)
Let's calculate the variation (we will use symbol δ for variation) of functional Φ[f(x)] at any point (function-argument) defined above by function-argument f0(x) minimizing our functional, directional point f1(x) and real parameter t :
δ[f0,f1,t] Φ[f(x)] =
= d/dt Φ[f0(x)+t(f1(x)−f0(x))] =
= d/dt Φ[f0(x)+t·Δ(x)] =
(use the formula for a length of a curve)
= d/dt ∫[a,b][1+((f0+t·Δ)')²]½dx
In the above expression we dropped (x) to shorten it.
The derivative indicated by an apostrophe is by argument x of functions f(x) and Δ(x).
Under very broad conditions, when smooth functions are involved, a derivative d/dt and integral by dx are interchangeable.
So, let's take a derivative d/dt from an expression under an integral first, and then we will do the integration by dx.
d/dt [1+((f0(x)+t·Δ(x))')²]½ =
= d/dt [1+(f0'(x)+t·Δ'(x))²]½ =
| = |
|
= |
| = |
|
Let's use the integrating by parts using a known formula for two functions u(x) and v(x)
∫[a,b]u·dv = u·v|[a,b] − ∫[a,b]v·du
Use it for
| u(x) = |
|
and, therefore,
dv(x) = dΔ(x) = Δ'(x)·dx
with all participating functions assumed to be sufficiently smooth (differentiable to, at least, second derivative).
Since
v(a) = Δ(a) = 0 and
v(b) = Δ(b) = 0,
the first component of integration is zero
u·v|[a,b]=u(b)·v(b)−u(a)·v(a)=0
Now the variation of our functional is
δ[f0,f1,t] Φ[f(x)] =
= −∫[a,b]v(x)·du(x,t)
where
| u(x,t) = |
|
As we know, the necessary condition for a local minimum of functional Φ[f(x)] at function-argument f0(x) is equality to zero of all its directional derivatives at point f0(x) (that is at t=0).
It means that for any direction defined by function f1(x) or, equivalently, defined by any Δ(x)=f1(x)−f0(x), the derivative by t of functional Φ[f0(x)+t·Δ(x)] should be zero at t=0.
So, in our case of minimizing the length of a curve between two points in space, the proper order of steps would be
(1) calculate the integral above getting variation of a functional δ[f0,f1,t]Φ[f0(x)+t·Δ(x)] which is a functional of three variables:
- real parameter t,
- function f0(x) that is an argument to an assumed minimum of functional Φ[f(x)],
- function Δ(x) that signifies an increment of function f0(x) in the direction of function f1(x);
(2) set t=0 obtaining a directional derivative of functional Φ[f(x)] at assumed minimum function-argument f0(x) and increment Δ(x):
δ[f0,f1,t=0]Φ[f0(x)];
(3) equate this functional to zero and find f0(x), that solves this equation regardless of argument shift to f1(x).
Integration in step (1) above is by x, while step (2) sets the value of t.
Since x and t are independent variables, we can exchange the order and, first, set t=0 and then do the integration.
This simplifies the integration to the following
δ[f0,f1,t=0] Φ[f(x)] =
= −∫[a,b]Δ(x)·du(x,0)
where
| u(x,0)= |
|
| = |
|
δ[f0,f1,t=0] Φ[f(x)] =
| = −∫[a,b]Δ(x)·d |
|
| −∫[a,b]Δ(x)·[ |
|
]'dx |
For δ[f0,f1,t=0] Φ[f(x)] to be equal to zero for t=0 regardless of Δ(x), or, in other words, for the integral above to be equal to zero at t=0 regardless of function Δ(x), function u'(x,0) (the one in [...]') must be identically equal to zero for all x∈[a,b].
If u'(x,0) is not zero at some point x (and in some neighborhood of this point since we deal with smooth functions), there can always be constructed such function Δ(x) that makes the integral above not-zero.
From this follows that u(x,0)=const.
Therefore,
|
= u(x,0) = const |
All that remains is to find a linear function f0(x) that satisfies initial conditions f0(a)=A and f0(b)=B.
Obviously, it's one and only function
f0(x) = (B−A)·(x−a)/(b−a) + A
whose graph in (X,Y) Cartesian coordinates is a straight line from (a,A) to (b,B).
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