Math 5040: Stochastic control theory
Find the function y(x) that minimizes the functional
J[y] = ∫ (0,1) (y′2+y2) dx
subject to the boundary conditions y (0) = 0 and y (1) =1.
A) y(x)=x
B) y(x)=sin(x)
C) y(x)=x2
D) y(x)=sin h (x)
Solution
The Euler-Lagrange equation for the functional is:
∂F ∂F d ∂F
= 2y, = 2y’, = 2y′′
∂y ∂ y' dx ∂ y
2y−2y′′ = 0 ⇒ y′′- y = 0
The general solution to this differential equation is:
y (x) = C1e x + C2e−¿ x ¿
Applying the boundary conditions y (0) =0 and y (1) =1:
y (0) = C1+ C2 = 0 ⇒ C2 =- C1
y (1) = C1e 1 – C1e−1= 1 ⇒ C1 (e – e-1) = 1
1 1
C1 = , C2 = -
e – e−1 e – e−1
1
Y(x) = (e x -e− x)
e – e−1
Correct answer: (D)
sinh(x )
y(x) =
sinh(1)
Find the function y(x) that minimizes the functional
J[y] = ∫ (0,1) (y′2+y2) dx
subject to the boundary conditions y (0) = 0 and y (1) =1.
A) y(x)=x
B) y(x)=sin(x)
C) y(x)=x2
D) y(x)=sin h (x)
Solution
The Euler-Lagrange equation for the functional is:
∂F ∂F d ∂F
= 2y, = 2y’, = 2y′′
∂y ∂ y' dx ∂ y
2y−2y′′ = 0 ⇒ y′′- y = 0
The general solution to this differential equation is:
y (x) = C1e x + C2e−¿ x ¿
Applying the boundary conditions y (0) =0 and y (1) =1:
y (0) = C1+ C2 = 0 ⇒ C2 =- C1
y (1) = C1e 1 – C1e−1= 1 ⇒ C1 (e – e-1) = 1
1 1
C1 = , C2 = -
e – e−1 e – e−1
1
Y(x) = (e x -e− x)
e – e−1
Correct answer: (D)
sinh(x )
y(x) =
sinh(1)
