Introduction
This question involves two problems related to the calculus of variations, which is a branch of mathematics concerned with finding functions that optimize functionals (integrals depending on functions and their derivatives). We are asked to find the extremals — functions y(t) or x(t) — that make the given functionals take on a maximum or minimum value, under certain boundary conditions.
Part a) Finding the Extremal for:
J(y) = ∫01 (y′² − y² + 2ty) dt with boundary conditions: y(0) = 0 and y(1) = 1
Step 1: Euler–Lagrange Equation
The general form of the Euler–Lagrange equation is:
∂F/∂y − d/dt (∂F/∂y′) = 0
For the given functional, the integrand is:
F(t, y, y′) = y′² − y² + 2ty
Step 2: Compute Partial Derivatives
∂F/∂y = −2y + 2t∂F/∂y′ = 2y′d/dt(∂F/∂y′) = 2y″
Step 3: Euler–Lagrange Equation
−2y + 2t − 2y″ = 0 → y″ − y + t = 0
Step 4: Solve the Differential Equation
This is a second-order non-homogeneous differential equation.
Homogeneous part: y″ − y = 0
Solution: yh(t) = C₁et + C₂e−t
Particular solution: Try yp = At + B
Then y″ = 0, so substitute into the equation:
0 − (At + B) + t = 0 → −At − B + t = 0 → A = 1, B = 0
So yp = t
General solution:
y(t) = C₁et + C₂e−t + t
Step 5: Apply Boundary Conditions
y(0) = C₁ + C₂ + 0 = 0 → C₁ + C₂ = 0y(1) = C₁e + C₂e−1 + 1 = 1
Substitute C₂ = −C₁ into second equation:
C₁(e − e−1) + 1 = 1 → C₁(e − e−1) = 0 → C₁ = 0
Thus, C₁ = 0, C₂ = 0
Final extremal: y(t) = t
Part b) Find extremals for:
J(x) = ∫01 [x′² + 10xt] dt with boundary conditions x(0) = 2, x(1) = 3
Step 1: Define the integrand
F(t, x, x′) = x′² + 10xt
∂F/∂x = 10t∂F/∂x′ = 2x′d/dt(∂F/∂x′) = 2x″
Euler–Lagrange Equation
10t − 2x″ = 0 → x″ = 5t
Step 2: Integrate twice
- First integration:
x′ = (5/2)t² + C₁ - Second integration:
x(t) = (5/6)t³ + C₁t + C₂
Step 3: Apply boundary conditions
x(0) = C₂ = 2x(1) = (5/6) + C₁ + 2 = 3→C₁ = 3 − 2 − (5/6) = 1/6
Final extremal: x(t) = (5/6)t³ + (1/6)t + 2
Conclusion
Using calculus of variations, we solved two functional optimization problems. In the first, the extremal function was found to be y(t) = t, and in the second, we derived a cubic polynomial satisfying the required boundary conditions. These problems are typical in physics, engineering, and economics where optimal paths or strategies are required over time.