CS计算机代考程序代写 Isoperimetric Theorem

Isoperimetric Theorem
Assume that the function x* (t) is a first-variational curve which results in the minimum of
J (x)tb f(x,x,t)h f (x,x,t)dt at11
a i.e.
t
J (x)J (x*) b f(x*,x*,t)hf(x*,x*,t) dt
 with h independent of x,t.
 aata 11
1
Further assume that the constraint is satisfied by x* (t).
Then, x*(t) is the minimal solution to the problem.
4/13/2020 @2020 New York University Tandon 345 School of Engineering

4/13/2020
@2020 New York University Tandon School of Engineering
346
Sufficient Conditions
Sufficient conditions for weak relative minimum of J(x) at x  x*(t) are:
 x* (t) is a first-variational curve, i.e. J  (strengthened) Legendre condition:
xx*(t)
 0.
*(t)f |xx >0, t[ta,tb] xx
(t)1f d f | 0, t[t,t].
* 2xx dtxxxx*(t) 
a b

Sketch of Proof
It suffices to note that, for any first-variational curve x(t), J J(xx)J(x)
1 2t 2  2J  b f xx,xx,t
2 2 2  ta 0
1tb 21 21d
 [x fx ff]dt

2 2xx 2xx dtxx
ta   
where integration by parts was used in establishing the last equality, along with the fixed-end conditions.
4/13/2020 @2020 New York University Tandon 347 School of Engineering


Solution to Bernoulli’s Brachistocrone Problem
min T x 1Y(x)2
J(Y) dt 1 dx
2gy0 Y(x) :yY(x), xxx
0 x0 subject to
4/13/2020
@2020 New York University Tandon 348 School of Engineering

with Y(x)y,Y(x)y. 0011
01

Solution (cont’d)
Comparing with J(x)  tb f x,x,tdt, (*)
x 1Y(x)2 fromJ(Y) 1 dx,
x0 it follows that
2gy0 Y(x) 1Y2
f (Y,Y) 
2gy0 Y
t a
which is independent of x [or, t using the notation in eq. (*)].
4/13/2020 @2020 New York University Tandon 349 School of Engineering

Solution (cont’d)
To obtain an extremal solution, apply
ELequation: dfxfx0 dt
 1Y2
to f(Y,Y) 2gy0Y.
It holds:
fY (Y(x),Y(x)) d fY(Y(x),Y(x))  0.
dx
4/13/2020 @2020 New York University Tandon 350 School of Engineering

Solution (cont’d)
 1Y2
Sincef(Y,Y) 2gy0Yisindependentofx, from fY (Y(x),Y(x)) d fY(Y(x),Y(x))  0,
we can derive successively
d f (Y(x),Y(x))Y(x) fY(Y(x),Y(x))
dx
Y(x)f (Y(x),Y(x)) d f (Y(x),Y(x))
0
dx
 Y dx Y  
4/13/2020 @2020 New York University Tandon 351 School of Engineering

Solution (cont’d)
f (Y(x),Y(x))Y(x) fY(Y(x),Y(x))  C, (**)
x0 xx1.
Usingf(Y,Y) 2gy0Y,
fY(Y(x),Y(x))  Thus, from (**),
Y(x)
2g y Y(x) 1Y (x)2
1Y(x)2 
Y(x)
 2gC. 352
 1Y2
yY(x)    0 y Y(x) 1Y(x)2
4/13/2020 @2020 New York University Tandon School of Engineering
0
 
0

Solution (cont’d)
After further simplication, letting A  1 / 2 gC 2 ,
yY(x)1Y(x)2 A.  
0
Solving this equation for Y (x) yields:
Y(x) Ay0 Y(x), (***) y0 Y(x)
(A graphic argument tells the slope should be negative.) In short, any extremum function Y (x) necessarily satisfies (***).
4/13/2020 @2020 New York University Tandon 353 School of Engineering

Solution (cont’d)
In order to obtain the solution Y (x) of eq. (***),
4/13/2020
@2020 New York University Tandon 354 School of Engineering
introduce a new function (x) via  (x)2
y0Y(x)Asin 2 . 
Upon substitution into (***) gives Asin(x)cos(x)(x) cos(x)/2
2 2 sin (x) / 2  (x)2
d(x) Asin 2 (x)1, with(x) dx .


Using 2 sin(x)/2 2 1cos, the solution is 
xx0 Asin. 2
Finally, we obtain the extremum curve  : x  Y (x) using the new parameter :
xx Asin,  0 2
:
 y  y 0  2 1  c o s 
 A 
for 0    1, with 0 ,1  corresponding to
the two fixed end points.
4/13/2020 @2020 New York University Tandon 355 School of Engineering

4/13/2020
@2020 New York University Tandon
356
y
A = radius of the circle
Optimal Curve = Cycloid
y0
P0
0
x0
School of Engineering
x
P
P1

4/13/2020 @2020 New York University Tandon 357 School of Engineering

Minimum Time from P0 to P1
Clearly,at0 0,xx0,yy0.
At the other end point P  x , y , we can determine
A and  0,2 from the equations: 1
A sin 2x x , 1 1 10
 A 1  c o s  1   2 y 1  y 0 . 
111
4/13/2020 @2020 New York University Tandon 358 School of Engineering

4/13/2020
@2020 New York University Tandon 359 School of Engineering
Minimum Time from P0 to P1
Therefore, it follows
x 1Y(x)2 T1 dx
min
x0
2gy0 Y(x) 22

dx/d dy/d d 2gy y
0
01
1
A A 21cos d  .
1
20 gA1cos 2g
1

4/13/2020
@2020 New York University Tandon 360 School of Engineering
g
B
C
A Related Question
Slide three identical beads at different points, say A, B, C, along a cycloid.
Which bead will arrive at D first ? A
D

Answer by
Christian Huygens (1629‐1695)
They will arrive at the same time!
This property of the cycloid is the foundation
of Huygens’s tautochronic pendulum clocks.
4/13/2020 @2020 New York University Tandon 361 School of Engineering

4/13/2020
@2020 New York University Tandon 362 School of Engineering
Problem in Investment Planning
Goal: find, if possible, an optimal consumption policy with terminal savings constraints during a period of inflation
Annual income I Annual return R
Savings S(t) at time t
Annual consumption C

Example (cont’d)
Mathematical model:

SIRC, S(0)S0 0. Assumption 1:
R   S , 0    1.
4/13/2020 @2020 New York University Tandon 363 School of Engineering

4/13/2020
@2020 New York University Tandon 364 School of Engineering
Example (cont’d)
Then, the mathematical model becomes:

SSIC, S(0)S0 0 leading to
t tt 
S(t)e S0e 0e [I()C()]d.

4/13/2020
@2020 New York University Tandon 365 School of Engineering
Problem of Optimization
Maximize a metric of satisfaction over [0, T ]: T F(t,C(t))dt
0 subject to
S (T )  ST (desired constant) or equivalently,
K(C)  T etC(t)dt  k0 (constant) 0
with
k0 S0 eTST T etI(t)dt.
0

Problem of Optimization
Notice that any realistic candidate for the satisfaction measure, say F(t,C), must satisfy the property:
F(t,C) is an increasing function in C!! Let’s take for example
F(t,C)et log(1C)
where   0 accounts for inflation.
4/13/2020 @2020 New York University Tandon 366 School of Engineering

4/13/2020
@2020 New York University Tandon 367 School of Engineering
Problem of Optimization
So, the optimization problem reduces to maxJ(C)Tet log(1C(t))dt

0
C(t)D subject to
K(C)  T etC(t)dt  k0 0
where
D C: 0, T  iscontinuous .
 

Comments
• This problem may not have an optimal solution if the person sets too high as a saving task.
• When an optimal solution exists, the Euler‐ Lagrange multiplier method can be used to find such an extremum function.
4/13/2020 @2020 New York University Tandon 368 School of Engineering

4/13/2020
@2020 New York University Tandon 369 School of Engineering
First Variations
ForanyCC0[0,T], anyCD, KC;ClimKCCK(C) KC
0  T etC(t)dt
0
and
JC;CT  F(t,C(t))C(t)dt
0 C
 T et C(t)dt.
0 1C(t)

As a consequence of the Euler-Lagrange multiplier theorem,  a Lagrange multiplier  so that
J(C*;C)K(C*;C), C if C* is a local extremum function.
Therefore,
T et etC(t)dt0,C
1C*(t)  0
T  et t 2 et t  1C*(t)e  dt0, ifC[1C*(t)e ].
0
4/13/2020 @2020 New York University Tandon 370 School of Engineering

T  et t 2 1C*(t)e  dt0
0
 et et 0
1 C* (t)
 C*(t)11et, 0tT.

Substituting this into the constraint leads to 1k1eT  .
 0  1eT 
4/13/2020 @2020 New York University Tandon 371 School of Engineering

Comments
(1) The extremum function C* (t) may be infeasible, if it takes negative values.
(2) Even when feasible, we need to check if or not
4/13/2020
@2020 New York University Tandon 372 School of Engineering
C* (t) is maximizing the satisfaction functional J (C). It suffices to verify
JC* CJC*0, admissibleC. (left as an exercise)

Comments
(3) The physical meaning of C* (t)  1 1 e(  )t 
is as follows:
when    , better to invest more and consume less
in early years, and then consume more in later years; when    , better to invest less and consume more
in early years, and then consume less and invest more
in later years;
when    , consume at a constant rate during [0, T ].
4/13/2020 @2020 New York University Tandon 373 School of Engineering

Exercise 1
Let J be a typical real-valued function of arguments x,x ,,x .
12n
What is the first variation of J at x ?
4/13/2020 @2020 New York University Tandon 374 School of Engineering

Exercise 2
Find the maximum value of the functional
J x 1 xtdt 0
over all x t C0 0,1 , subject to 
12 11 0xt tx(t)dt312.
4/13/2020 @2020 New York University Tandon 375 School of Engineering

Extension:
Variable end‐point conditions
Some applications involve the case when one of the end points, or both, are variable (not fixed).
4/13/2020 @2020 New York University Tandon 376 School of Engineering

Classical Example
Context: A boatman wants to find the optimal path leading to the shortest time when crossing a river
from a fixed initial point on one bank to an unspecified terminal point on the other bank.
4/13/2020 @2020 New York University Tandon 377 School of Engineering

4/13/2020
@2020 New York University Tandon School of Engineering
378
Left bank
y

Right bank
0l
x
Water current

4/13/2020
@2020 New York University Tandon 379 School of Engineering
Standing Assumptions
H1: There is no cross current so that the current velocity is everywhere directed downstream along the y-direction.
H2: The downstream current speed w depends only on x :
ww(x), 0xl.
H3: The boat travels at a constant natural speed v0 .


 x   ( t )  v 0 c o s  ( t ) ,
Modeling
If the path of the boat is represented by a curve  :  x    t 
yt for0tT
for suitable functions  t ,  t , then the absolute
velocity of the boat satisfies
  
where (t) is the steering angle of the boat from the x-axis.

 y   ( t )  v 0 s i n  ( t )  w  t
4/13/2020 @2020 New York University Tandon 380 School of Engineering

0 0dx dt Then,
T  T dx
0 v0 cos
Time of Transit
The time of transit T of the boat can be calculated as TTdtT dtdx, notingdx(t).
. note: cos  0
Now, we use x as the parameter(why possible?) for the curve  :
: yY(x), 0xl. Indeed,Y(x)  (x) .
1 
4/13/2020 @2020 New York University Tandon 381 School of Engineering

Note that
Y(x)  dy  (t)  sin  e(x) ,
dx (t) cos where e(x)w(x).
v0 Then,
Ycosesine 1cos2 e(x)Y(x) 1e(x)2 Y(x)2
 cos 1Y(x)2
.
4/13/2020 @2020 New York University Tandon School of Engineering
382

Problem Formulation
Summarizing the above, the minimum transit time problem becomes:
1 l 1Y(x)2 minTT(Y)v0 0 e(x)Y(x) 1e(x)2 Y(x)2 dx
where
Y C1[0,l] the class of continuously differentiable functions
over 0  x  l.
4/13/2020 @2020 New York University Tandon 383 School of Engineering

Comment
Clearly, the above min problem is subject to one constraint at the starting point:
Y(x0)y0, inthepresentcase, x0,y0(0,0)
along with the terminal constraint C:(x,y)0, inthiscase,xl0.
No constraint on y ! Such a modified problem is often called:
James Bernoulli’s Brachistochrone Problem (1697).
4/13/2020 @2020 New York University Tandon 384 School of Engineering

Variable End‐point Conditions
It corresponds to the following situation: for the optimization problem
minJtb fx,x,tdt, t
a
thevaluesofta, tb, xta, xtbarenot
necessarily fixed.
Question: what are the end-point conditions
that the optimal variables must satisfy?
4/13/2020 @2020 New York University Tandon 385 School of Engineering

The Transversality Condition
(Necessary Condition)
4/13/2020
@2020 New York University Tandon 386 School of Engineering
Optimal values of ta , tb , xta , xtb  must satisfy:
t
f xf tb f x(t)b 0.
t  x ta x ta