π 2 ∫ ( ) ⎡ ⎤ ( ) ( ) ( ) 2 2 ′ + = ⎡ + ⎤ − ⎡ + ⎤ I sin x h x sin x h x cos x h x dx ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 0 π π 2 2 ∫ ( ) ∫ ( ) ( ) ( ) ′ = − + − 2 2 sin x cos x dx 2 h x sin x h x cos x dx ����� � ������ � 0 0 = 0 π 2 ∫ ⎡ ( ) ( ) ⎤ ′ 2 + 2 − ⎡ ⎤ h x ⎣ h x ⎦ dx ⎣ ⎦ 0 π 2 ∫ ⎡ ⎤ ( ) ( ) ( ) 2 ′ = + 2 − ⎡ ⎤ I sin x h x h x dx ⎣ ⎦ ⎣ ⎦ � � �� � 0 0
= y sin x So what’s going on at the critical function , depends π 2 ∫ ⎡ ⎤ ( ) ( ) ′ 2 2 − ⎡ ⎤ h x h x dx ⎣ ⎦ on whether the expression is always ⎣ ⎦ 0 positive, always negative, or can be either positive or negative for continuously differentiable functions h with h π ⎛ ⎞ = ( ) = ⎜ ⎟ 0 h 0 0 and . So let’s investigate. ⎝ ⎠ 2
We’ll use a special case of the Poincare Inequality to get our ≤ ≤ result: (Assume that .) 0 x 1 x = ∫ ( ) ( ) ′ h x h t dt 0 so 2 ⎡ ⎤ x ∫ ( ) ( ) ′ ⎢ ⎥ = 2 h x h t dt ⎢ ⎥ ⎣ ⎦ 0
We’ll need a version of the Cauchy-Schwarz inequality: [ ] a b , For f and g continuous functions on 2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ b b b ∫ ∫ ∫ ( ) ( ) ( ) ( ) ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ≤ 2 2 f x g x dx f x dx g x dx ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ a a a See if you can prove it. (See handout, page 5!)
From the Cauchy-Schwarz Inequality, we get x x ∫ ∫ ( ) ( ) 2 ′ ≤ ⎡ ⎤ 2 2 h x 1 dt h t dt ⎣ ⎦ 0 0 x ∫ ( ) ′ 2 ≤ ⎡ ⎤ x h t dt ⎣ ⎦ 0 And so 1 ∫ ( ) ( ) ′ 2 ≤ ⎡ ⎤ 2 h x ⎣ h x ⎦ dx 0
If we integrate this inequality, we get 1 1 ∫ ∫ ( ) ( ) ′ 2 ≤ ⎡ ⎤ 2 h x dx h x dx ⎣ ⎦ 0 0 π ≤ ≤ 1 We can do the same thing for . x 2 π 2 ∫ ( ) ( ) ′ = − h x h t dt x
so 2 ⎡ ⎤ π 2 ∫ ⎢ ⎥ ( ) ( ) ′ = − 2 h x h t dt ⎢ ⎥ ⎣ ⎦ x From the Cauchy-Schwarz Inequality, we get π π 2 2 ∫ ∫ ( ) ( ) ′ 2 ≤ ⎡ ⎤ 2 2 h x 1 dt ⎣ h t ⎦ dt x x π π 2 ⎛ ⎞ ∫ ( ) 2 ′ ≤ − ⎡ ⎤ ⎜ ⎟ x h t dt ⎣ ⎦ ⎝ ⎠ 2 x
And so π π 2 ⎛ ⎞ ⎠ ∫ ( ) ( ) ′ 2 ≤ − ⎡ ⎤ 2 ⎜ ⎟ h x 1 ⎣ h x ⎦ dx ⎝ 2 1 If we integrate this inequality, we get π π π 2 2 2 ⎛ ⎞ ∫ ∫ ( ) ( ) 2 ′ ≤ − ⎡ ⎤ 2 ⎜ ⎟ h x dx 1 h x dx ⎣ ⎦ ⎝ ⎠ 2 1 1 π 2 ∫ ( ) ′ 2 ≤ ⎡ ⎤ ⎣ h x ⎦ dx 1
So we have 1 1 1 ∫ ∫ ∫ ⎡ ⎤ ( ) ( ) ( ) ( ) ′ 2 ′ 2 ≤ ⎡ ⎤ ⇒ − ⎡ ⎤ ≤ 2 2 h x dx h x dx h x h x dx 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 0 0 0 and π π π 2 2 2 ∫ ∫ ∫ ⎡ ⎤ ( ) ( ) ( ) ( ) ′ 2 ′ 2 ≤ ⇒ − ≤ ⎡ ⎤ ⎡ ⎤ 2 2 h x dx h x dx h x h x dx 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 1 1
If we add them together, we get π 2 ∫ ⎡ ⎤ ( ) ( ) ′ 2 − ⎡ ⎤ ≤ 2 h x h x dx 0 ⎣ ⎦ ⎣ ⎦ 0 So π 2 ∫ ( ) ⎡ ⎤ ( ) ( ) ( ) ( ) 2 ′ + = + − ⎡ ⎤ 2 I sin x h x I sin x h x h x dx ⎣ ⎦ ⎣ ⎦ � � �� � ���� � ����� � 0 0 ≤ 0 = y sin x Which means that is a maximum.
Let’s just change the previous example a little bit. π 3 2 ∫ ( ) ( ) ⎡ ⎤ ′ = − 2 2 I y y y dx ⎣ ⎦ 0 The Euler-Lagrange equation is still ′′ + = 0 y y
The optimal solution candidate of π 3 2 ∫ ( ) ( ) ⎡ ⎤ ′ = − 2 2 I y y y dx ⎣ ⎦ 0 Subject to π ⎛ ⎞ = − 3 ( ) = ⎜ ⎟ y 1 and y 0 0 ⎝ ⎠ 2 is = y sin x
π 3 2 ∫ ( ) ⎡ ⎤ ( ) ( ) ( ) 2 ′ 2 + = ⎡ + ⎤ − ⎡ + ⎤ I sin x h x sin x h x cos x h x dx ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 0 π π 3 3 2 2 ( ) ∫ ∫ ( ) ( ) ( ) ′ = − + − 2 2 sin x cos x dx 2 h x sin x h x cos x dx ����� � ������ � 0 0 = 0 π 3 2 ∫ ⎡ ⎤ ( ) ( ) 2 ′ + 2 − ⎡ ⎤ h x ⎣ h x ⎦ dx ⎣ ⎦ 0 π 3 2 ∫ ⎡ ⎤ ( ) ( ) ( ) ′ 2 = + − ⎡ 2 ⎤ I sin x h x ⎣ h x ⎦ dx ⎣ ⎦ � � �� � 0 0
π ⎛ ⎞ 3 ( ) ( ) = − = ⎜ ⎟ sin2 h x x x h x x Consider and 1 2 ⎝ ⎠ 2 π π ⎛ ⎞ ⎛ ⎞ 3 3 ( ) ( ) = = = = ⎜ ⎟ ⎜ ⎟ 0 0 0 h h h h 1 2 1 ⎝ ⎠ 2 ⎝ ⎠ 2 2 π 3 2 ⎡ ⎤ ∫ 2 ⎡ ⎤ ( ) ( ) ′ 2 − h x h x dx Compute and ⎢ ⎥ ⎣ ⎦ 1 1 ⎣ ⎦ 0 π 3 2 ⎡ ⎤ ∫ 2 ⎡ ⎤ ( ) ( ) ′ − 2 h x h x dx ⎢ ⎥ ⎣ ⎦ ⎣ 2 2 ⎦ (See handout, 0 page 6!)
π 3 π 2 ⎡ ⎤ ∫ 2 ⎡ ⎤ 9 ( ) ( ) ′ 2 − = − < h x h x dx 0 ⎢ ⎥ ⎣ ⎦ 1 1 ⎣ ⎦ 4 0 π 3 π ⎛ ⎞ π 2 ⎡ ⎤ 2 ∫ 2 9 9 ⎡ ⎤ ( ) ( ) ′ 2 − = − > 3 ⎜ ⎟ h x h x dx 1 0 ⎢ ⎥ ⎣ ⎦ 2 2 ⎣ ⎦ ⎝ ⎠ 8 40 0
= y sin x So in this case, is neither a maximum or minimum. π 3 ⎡ ⎤ 2 2 ∫ ⎡ ⎤ ′ ( ) ( ) ( ) + ε = + ε − 2 2 ⎢ ⎥ I sin x sin2 x I sin x sin 2 x sin2 x dx ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 0 ( ) < I sin x ⎡ ⎤ 2 π ⎡ ⎤ ′ 3 ⎛ π ⎞ π 2 ⎛ π ⎞ 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ∫ ⎢ ⎥ 3 3 3 ( ) ⎢ ⎥ + ε − = + ε − − − 2 2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ I ⎜ sin x x x ⎟ I sin x x x ⎜ x x ⎟ dx ⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 2 2 ⎢ ⎣ ⎦ ⎥ ⎣ ⎦ 0 ( ) > I sin x
See if you can find the Euler-Lagrange equation for ln 2 ∫ ( ) ( ) ⎡ ⎤ ′ = + 2 2 I y y y dx ⎣ ⎦ 0 (See handout, page 7!)
Answer: ′′ − = y y 0
Find all optimal solution candidates of ln 2 ∫ ( ) ( ) ⎡ ⎤ ′ = + 2 2 I y y y dx ⎣ ⎦ 0 Subject to 3 ( ) ( ) = = ln2 and y y 0 0 4 (See handout, page 8!)
Answer: = y sinh x See if you can determine if it’s a maximum or minimum. (See handout!, page 9)
ln 2 ∫ ( ) ⎡ ⎤ ( ) ( ) ( ) 2 ′ 2 + = ⎡ + ⎤ + ⎡ + ⎤ I sinh x h x ⎣ sinh x h x ⎦ ⎣ cosh x h x ⎦ dx ⎣ ⎦ 0 ln 2 ln 2 ( ) ∫ ∫ ( ) ( ) ( ) ′ = + + + 2 2 sinh cosh 2 sinh cosh x x dx h x x h x x dx ������ � ������� � 0 0 = 0 ln 2 ∫ ⎡ ⎤ ( ) ( ) ′ 2 + + ⎡ ⎤ 2 h x h x dx ⎣ ⎦ ⎣ ⎦ 0 ln 2 ∫ ⎡ ⎤ ( ) ( ) ( ) ′ 2 = + + ⎡ ⎤ 2 I sinh x h x ⎣ h x ⎦ dx ⎣ ⎦ � � �� � ���� � ����� � 0 1 ≥ 0 So, it’s a minimum!
See if you can find the Euler-Lagrange equation for b ∫ ( ) ( ) ′ = + 2 1 I y y dx a (See handout, page 10!)
Answer: ⎛ ⎞ ′ d y ⎜ ⎟ = 0 ⎜ ⎟ ( ) ′ dx + 2 1 y ⎝ ⎠ ′ y = C ( ) ′ + 2 1 y ± C ′ = ( ) ( ) = ⎡ ⎤ 2 1 ′ ′ 2 = + 2 y K y C y ⎣ ⎦ + 2 1 C
Find all optimal solution candidates of b ∫ ( ) ( ) ′ = + 2 I y 1 y dx �� � ��� � a length of y Subject to ( ) ( ) = = and y b d y a c (See handout, page 11!)
Answer: − ⎛ ⎞ d c ( ) = − + ⎜ ⎟ y x a c − ⎝ ⎠ b a Do you think that it’s a maximum or minimum? Shortest path between two points in the plane?
Is there a 2 nd Derivative Test? Yes. If 1. y is a stationary function F ′ ′ > 0 2. along y y y ] ( a b , 3. There is no number z in such that ⎛ ⎞ ( ) d d ′ − + − = ⎜ ⎟ F h F F h 0 ′ ′ ′ y y ⎝ yy yy ⎠ dx dx ( ) ( ) = = h a h z 0 has a nontrivial solution along y. Then y is a local minimum.
b ∫ ( ) ( ) ( ) ′ y ′ = + 2 = + 2 For , I y 1 y dx 1 . F a ′ y 1 = = = = F 0 , F 0 , F , F ′ ′ ′ ′ y yy y ( ) y y 3 ( ) ′ ⎡ ⎤ + 2 ′ + 2 2 1 y 1 y ⎣ ⎦ − ⎛ ⎞ d c ( ) = − + ⎜ ⎟ y x a c Along the stationary curve , − ⎝ ⎠ b a 1 ′ ′ = > F 0 y y 3 ⎡ ⎤ − 2 ⎛ ⎞ 2 d c + ⎢ ⎥ ⎜ ⎟ 1 − ⎝ ⎠ ⎢ ⎥ b a ⎣ ⎦
Along y, the boundary value problem is ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ d 1 d ′ − + − = ⎜ ⎟ ⎜ ⎟ h 0 0 h 0 3 ⎝ ⎠ ⎜ ⎡ ⎤ ⎟ dx dx − 2 ⎛ ⎞ 2 d c + ⎜ ⎢ ⎥ ⎟ ⎜ ⎟ 1 ⎜ ⎟ − ⎝ ⎠ ⎢ ⎥ b a ⎣ ⎦ ⎝ ⎠ ( ) ( ) = = h a h z 0 Or simply ( ) ′′ = h x 0 ( ) ( ) = = h a h z 0
The boundary value problem has only the trivial solution for − ⎛ ⎞ d c [ ] ( ) = − + ⎜ ⎟ y x a c a b , all choices of z in , and so is − ⎝ ⎠ b a at least a local minimum. See if you can show that the boundary value problem has only the trivial solution. (See handout, page 12!)
See if you can find the Euler-Lagrange equation for 1 ∫ ( ) ( ) ⎡ ⎤ ′ = 2 + 12 I y y xy dx ⎣ ⎦ 0 (See handout, page 13!)
x 6 ′′ = y Answer:
Find all optimal solution candidates of 1 ∫ ( ) ( ) ⎡ ⎤ ′ = 2 + 12 I y y xy dx ⎣ ⎦ 0 Subject to ( ) ( ) = = and y 1 1 y 0 0 (See handout, page 14!)
Answer: = 3 y x See if you can determine if it’s a maximum or minimum? (See handout, page 15!)
1 ( ) ∫ ⎡ ⎤ ( ) ( ) ( ) 2 ⎡ ′ ⎤ ⎡ ⎤ + = + + + 3 2 3 I x h x 3 x h x 12 x x h x dx ⎣ ⎦ ⎣ ⎦ ⎢ ⎥ ⎣ ⎦ 0 1 1 1 ∫ ∫ ∫ ( ) ( ) ( ) ′ 2 ′ ⎡ ⎤ = + ⎡ ⎤ + + 4 2 21 x dx h x dx 12 xh x 6 x h x dx ⎣ ⎦ ⎣ ⎦ ����� ����� � 0 0 0 = 0 1 ( ) ∫ ( ) ′ 2 = + ⎡ ⎤ 3 I x h x dx ⎣ ⎦ � �� � ��� � 0 21 ≥ 5 0 So, it’s a minimum!
Find all optimal solution candidates of 1 = ∫ ( ) ′ I y xyy dx 0 Subject to ( ) ( ) = = and y 1 1 y 0 0 (See handout, page 16!)
A Minimal Surface Area of Revolution Problem
a b a −
a b a −
( ) ( ) ( ) − = = We want to find a function, , with y x y a y a b which makes the surface area a ∫ ( ) ( ) ′ = π + 2 A y 2 y 1 y dx − a as small as possible.
Here’s the Euler-Lagrange equation(we essentially found it earlier): ( ) ( ) ′ ′′ ′ 4 + + 2 y yy y ( ) ′ π + − π = 2 2 1 y 2 0 3 ( ) ⎡ ⎤ ′ + 2 2 1 y ⎣ ⎦ Or simply ( ) ′′ ′ − 2 − = yy y 1 0
So we have to solve ( ) ′ ′ − − = 2 yy y 1 0 Subject to ( ) ( ) − = = y a y a b dp dy dp dp dy ′′ = = = ⋅ = ⋅ p y p If we let , then , and the dx dx dx dy dy differential equation becomes dp dy − − = 2 yp p 1 0
Separation of variables leads to p 1 = dp dy + 2 1 p y ∫ ∫ p 1 = dp dy + 2 1 p y ( ) + = + 2 2 ln p 1 ln y C 1
+ = 2 2 p 1 C y 1 = − 2 2 1 p C y 1 ′ = ± − 2 1 y C y 1 Another separation of variables leads to
∫ ∫ dy = ± dx − 2 1 C y 1 ∫ dy = ± + x C 2 − 2 C y 1 1 ( ) 1 cosh − = ± + 1 C y x C 1 2 C 1
( ) ( ) − = ± + 1 cosh C y C x C 1 1 2 ( ) ⎡ ⎤ ± + cosh C x C ⎣ ⎦ 1 2 = y C 1 ( ) ± + cosh C x C = 1 2 y C 1 ( ) ( ) − = = y a y a b
From the symmetry and boundary conditions, we get: ( ) = cosh C a bC 1 1 b = r Let a ⎛ ⎞ = cosh bC bC ⋅ 1 1 ⎜ ⎟ r ⎝ ⎠ r r So if we can solve the equation ( ) = cosh x rx
r > 0 For , then we’ll have a stationary function for each ( ) = cosh x rx solution of the equation Two stationary solutions One stationary solution No stationary solution
r < 1.508879563 If then no stationary solution. r = 1.508879563 If then one stationary solution. r > 1.508879563 If then two stationary solutions.
( ) a b , In other words, if the distance from the point to the y- a axis, , is larger than , then there is no .6627434187 b a .6627434187 b stationary solution. If is equal to , then there a .6627434187 b is one stationary solution. If is less than , then there are two stationary solutions. > a .6627434187 b It can be shown that for , the two disks give the absolute minimum area, and there are no other local < < .5276973968 b a .6627434187 b minima; for , the two disks give the absolute minimum area and the two stationary < a .5276973968 b solutions are local minima; for , then one stationary solution is the absolute minimum, the other is a local minimum, and the two disks are a local minimum.
By the way, surfaces of revolution generated by catenaries(graphs of hyperbolic cosines) are called catenoids.
Let’s check the mathematical prediction with reality. The large rings have a radius of approximately 12.75 cm. The medium rings have a radius of approximately 10 cm. The small rings have a radius of approximately 5 cm.
So for the large rings: You should see a catenoid for separation distances between 0 and 13.5 cm.(The other catenoid and two discs are local minima.) You should see the two discs as the absolute minimum for separation distances between 13.5 cm and 16.9 cm.(The two catenoids are local minima.) You should have the two discs as the absolute minimum for separation distances larger than 16.9 cm.(There are no other local minima.)
So for the medium rings: You should see a catenoid for separation distances between 0 and 10.6 cm.(The other catenoid and two discs are local minima.) You should see the two discs as the absolute minimum for separation distances between 10.6 cm and 13.3 cm.(The two catenoids are local minima.) You should have the two discs as the absolute minimum for separation distances larger than 13.3 cm.(There are no other local minima.)
So for the small rings: You should see a catenoid for separation distances between 0 and 5.3 cm.(The other catenoid and two discs are local minima.) You should see the two discs as the absolute minimum for separation distances between 5.3 cm and 6.6 cm.(The two catenoids are local minima.) You should have the two discs as the absolute minimum for separation distances larger than 6.6 cm.(There are no other local minima.) (See handout, pages 17-19!)
The Brachistochrone Problem (Curve of quickest descent)
( ) ( ) a c , , b d Consider the two points in the plane and . ( ) a c , ( ) b d ,
We want to join the two points by a curve so that a particle ( ) starting from rest at and moving along the curve a c , ( ) under the influence of gravity alone will reach the point b d , in minimum time. ( ) a c , ( ) b d ,
To formulate the problem, we’ll determine the velocity of the particle in two ways: Since the velocity, v , of the particle is determined by gravity alone, we must have dv dv dy dv = = ⋅ = ⋅ g v dt dy dt dy So we get the initial value problem: dv ( ) = = , 0 v g v a � dy < 0
The solution of the initial value problem is ( ) = − v 2 g y c But the velocity must also equal the rate of change of arclength along the curve with respect to time, so we get ds ds dx dx ( ) ′ = = ⋅ = + 2 v 1 y dt dx dt dt
Equating the two formulas for velocity leads to ( ) ′ + 2 1 y = dt dx ( ) − 2 g y c Which means that the total travel time for the particle is given by ( ) ( ) ′ ′ + 2 + 2 b b 1 y 1 y ∫ ∫ 1 = dx dx ( ) ( ) − − − c y 2 g y c 2 g a a
So the mathematical statement of the problem is to find a continuously differentiable function, y , that minimizes ( ) ′ + 2 b 1 y ∫ ( ) = I y dx − c y a ( ) ( ) = = y b d y a c Subject to and . (So in other words, it’s a fixed endpoint Calculus of Variations problem.)
Equivalent versions of the Euler-Lagrange Equation in special cases: ⎛ ⎞ d − = ⎜ ⎟ F F 0 ′ y y ⎝ ⎠ dx Case I: If F doesn’t involve y , then the Euler-Lagrange d F ′ = 0 equation reduces to , which can be y dx ′ = F C integrated into . y
Case II: If F doesn’t involve x , then the Euler-Lagrange ′ − = F y F C equation reduces to . ′ y ( ) ′ + 2 1 y ( ) ′ = F x y y , , For the Brachistochrone Problem , − c y which is independent of x .
′ y = F ′ y ( ) ( ) ⎡ ⎤ ′ − + 2 c y 1 y ⎣ ⎦ So using Case II, we get the Euler-Lagrange Equation as ( ) ( ) ′ + 2 ′ 2 1 y y − = C − ( ) ( ) ⎡ ⎤ c y ′ − + 2 c y 1 y ⎣ ⎦
( ) ( ) ⎡ ⎤ ′ − + 2 = 2 C c y 1 y 1 Simplifying algebraically, we get , ⎣ ⎦ ( ) − − 2 1 C c y ′ = ± y and solving for y’ yields . If we make ( ) − 2 C c y 1 sin ( ) − = 2 t c y the substitution , where t is a parameter, 2 2 C ( ) ( ) − 2 t t 1 sin cos ′ = ± = − 2 2 we get . The simplification y ( ) ( ) 2 t t sin sin 2 2 was chosen to reflect the fact that initially the curve will have a negative slope.
dy dy dt = ⋅ From the Chain Rule, , so dx dt dx ( ) ⎡ ⎤ 1 sin ( ) t cos 1 sin ( ) ( ) dt = − = − 2 t 2 dx dt and . t t cos ⎢ ⎥ ( ) 2 2 2 2 ⎣ ⎦ 2 C t sin C dx 2 1 ( ) = − + x t sin t k This gives us . We can solve for k if 2 2 C = t = x a 0 we insist that when . See if you can verify the formula for x and find the value of k . (See handout, page 20!)
1 = = A k a So ,and we’ll let to get the parametric 2 2 C equations ( ) = − + x A t sin t a ( ) = − + y A cos t 1 c ( ) = y b d To satisfy the condition , you’d have to solve ( ) − = − b a A t sin t ( ) − = − d c A cos t 1
( ) ( ) ( ) ( ) a c = b d = − , 0,0 , 1, 1 For example: Suppose that and . The system would be ( ) = − 1 A t sin t ( ) − = − 1 A cos t 1 If we eliminate A , we get ( ) − = − − t sin t cos t 1
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