Minimization of a Functional with integral constraints
Minimization with Integral Constraints
We consider the minimization of a functional \mathcal{A}(x) defined as
\mathcal{A}(x) = \int_0^1 (x^\prime)^2 \, \mathrm{d}t,
subject to the following constraints:
- Integral Constraint: \int_0^1 x \, \mathrm{d}t = 1,
- Boundary Conditions: x(0) = 2 and x(1) = 1.
Thus, the minimization problem can be stated as
\begin{aligned} \text{minimize} \quad & \mathcal{A}(x) = \int_0^1 (x^\prime)^2 \, \mathrm{d}t, \\ \text{subject to} \quad & \int_0^1 x \, \mathrm{d}t = 1, \\ & x(0) = 2, \quad x(1) = 1. \end{aligned}
To address this problem, we define a feasible space \mathbb{W} for x, which includes functions satisfying the boundary conditions:
\mathbb{W} = \left\{ x \in C^1([0,1]) \mid x(0) = 2, \, x(1) = 1 \right\}.
The problem can thus be reformulated as finding x \in \mathbb{W} that minimizes \mathcal{A}(x).
Feasible Directions and Conditions
To analyze feasible directions for variations of x, we introduce a perturbation x + \alpha \delta x with \delta x satisfying
\begin{aligned} x(0) + \alpha \delta x(0) &= 2, \\ x(1) + \alpha \delta x(1) &= 1, \\ \int_0^1 (x(t) + \alpha \delta x(t)) \, \mathrm{d}t &= 1. \end{aligned}
Expanding these conditions yields:
- \delta x(0) = 0,
- \delta x(1) = 0,
- \int_0^1 \delta x(t) \, \mathrm{d}t = 0.
This means the set of feasible variations \delta x lies in the space
\mathbb{V} = \left\{ \delta x \in C^\infty([0,1]) \mid \delta x(0) = 0, \, \delta x(1) = 0, \, \int_0^1 \delta x \, \mathrm{d}t = 0 \right\}.
Stationarity Condition
A function x \in \mathbb{W} is stationary for this problem if
\frac{\mathrm{d}}{\mathrm{d} \alpha} \bigg|_{\alpha=0} \mathcal{A}(x + \alpha \delta x) = 0, \quad \forall \delta x \in \mathbb{V}.
Calculating this derivative, we get
\frac{\mathrm{d}}{\mathrm{d} \alpha} \bigg|_{\alpha=0} \int_0^1 (x^\prime + \alpha \delta x^\prime)^2 \, \mathrm{d}t = \int_0^1 2 x^\prime x^{\prime\prime} \, \mathrm{d}t.
Using integration by parts on \int_0^1 2 x^\prime x^{\prime\prime} \, \mathrm{d}t, we find
\int_0^1 2 x^\prime x^{\prime\prime} \, \mathrm{d}t = 2 x^\prime \delta x \Big|_{0}^{1} - 2 \int_0^1 x^{\prime\prime} \delta x \, \mathrm{d}t.
Since \delta x(0) = \delta x(1) = 0, the boundary term vanishes, leaving
0 = \int_0^1 x^{\prime\prime} \delta x \, \mathrm{d}t \quad \forall \delta x \in \mathbb{V}.
At this point, we cannot directly conclude that x^{\prime\prime} = 0 because \mathbb{V} is not rich enough for the fundamental lemma of the calculus of variations to apply.
Constructing an Appropriate Variation
To circumvent this limitation, consider an auxiliary function \delta y \in C^\infty([0,1]) with \delta y(0) = \delta y(1) = 0. Define \delta x as
\delta x(t) = \delta y(t) - \lambda t(1 - t) \int_0^1 \delta y(\tau) \, \mathrm{d}\tau.
Choosing \lambda = 6 ensures that \delta x \in \mathbb{V}.
Concluding the Solution
Using this construction in our stationarity condition, we obtain
0 = \int_0^1 \left( x^{\prime\prime}(t) - 6 \left( x(1) + x(0) - 2 \int_0^1 x(\tau) \, \mathrm{d}\tau \right) \right) \delta y(t) \, \mathrm{d}t.
Now, since \delta y satisfies the conditions for the fundamental lemma of calculus of variations, we conclude that
x^{\prime\prime}(t) = 6 \left( x(1) + x(0) - 2 \int_0^1 x(\tau) \, \mathrm{d}\tau \right).
Applying the Boundary Conditions
Substituting the boundary conditions and integral constraint, we get
x^{\prime\prime}(t) = 6, \quad x(0) = 2, \quad x(1) = 1.
Solving this system, we find that
x(t) = 3t^2-4t + 2,
which satisfies all the boundary conditions and integral constraint.
Final Remark
While this approach correctly finds the solution, it would be advantageous to simplify the derivation by seeking a direct application of the fundamental lemma of calculus of variations. This simplification would streamline the solution process and avoid the need for constructing additional variations.
Derivation Using Finite Differences
To approximate the variational problem
\begin{aligned} \text{minimize} \quad & \mathcal{A}(x) = \int_0^1 (x^\prime)^2 \, \mathrm{d}t, \\ \text{subject to} \quad & \int_0^1 x \, \mathrm{d}t = 1, \\ & x(0) = 2, \quad x(1) = 1, \end{aligned}
we apply the finite difference method. This approach transforms the continuous problem into a discrete one, allowing us to approximate the solution numerically.
Discretization Setup
Divide the Interval: Let n be the number of intervals, and define h = \frac{1}{n} as the interval width. We set sample points t_k = k h for k = 0, 1, \dots, n, where t_0 = 0 and t_n = 1.
Approximate the Solution: Define y_k \approx x(t_k) as the approximate value of the solution at each sample point t_k.
Quadrature Formula for the Integral: To approximate the integral of (x^\prime)^2, we use the midpoint rule:
\int_0^1 (x^\prime)^2 \, \mathrm{d}t \approx h \sum_{k=1}^n \left(x^\prime(t_{k - \frac{1}{2}})\right)^2 + \mathcal{O}(h^2),
where x^\prime(t_{k - \frac{1}{2}}) represents the approximate derivative at the midpoint between t_{k-1} and t_k.
Finite Difference Approximation for the Derivative: The first derivative at t_{k - \frac{1}{2}} can be approximated by
x^\prime(t_{k - \frac{1}{2}}) = \frac{x(t_k) - x(t_{k-1})}{h} + \mathcal{O}(h^2).
Quadrature Formula for the Integral constraint: To approximate the integral of x, we use the trapezoidal rule:
\int_0^1 (x^\prime)^2 \, \mathrm{d}t = \frac{h}{2}(x(0)+x(1))+h \sum_{k=1}^{n-1}x(t_{k}) + \mathcal{O}(h^2),
Using this setup, we can now write the discretized form of the problem.
Discretized Problem Formulation
Substituting the finite difference approximation into the functional, the discretized problem becomes
\begin{aligned} \text{minimize} \quad & \mathcal{A}_h(\bm{x}) = h \sum_{k=1}^n \left(\frac{x_k - x_{k-1}}{h}\right)^2, \\ \text{subject to} \quad & h_1(\bm{x}) = \frac{h}{2}(x_0+x_n)+ h \sum_{k=0}^n x_k - 1 = 0, \\ & h_2(\bm{x}) = x_0-2 = 0, \\ & h_3(\bm{x}) = x_n-1 = 0. \end{aligned}
This formulation transforms our continuous problem into a finite-dimensional optimization problem, where we seek values of \bm{x} that minimize \mathcal{A}_h(\bm{x}) while satisfying the discretized constraints.
Lagrange Function Formulation
Let \lambda_1, \lambda_2, and \lambda_3 be the Lagrange multipliers corresponding to constraints h_1(\bm{x}), h_2(\bm{x}), and h_3(\bm{x}), respectively. We define the Lagrange function as
\begin{aligned} \mathcal{L}(\bm{x}, \lambda_1, \lambda_2, \lambda_3) & = \frac{1}{h} \sum_{k=1}^n (x_k - x_{k-1})^2 \\ & - \lambda_1 \left(\frac{h}{2}(x_0 + x_n) + h \sum_{k=1}^{n-1} x_k - 1\right) \\ & - \lambda_2 \cdot (x_0-2) + \lambda_3 \cdot (x_n - 1). \end{aligned}
First-Order Conditions
To find the stationary points of \mathcal{L}, we compute the partial derivatives with respect to each x_k and set them equal to zero:
For k = 1, 2, \ldots, n-1: \begin{aligned} \frac{\partial \mathcal{L}}{\partial x_k} &= \frac{2}{h} (x_k - x_{k-1}) - \frac{2}{h} (x_{k+1} - x_k) - \lambda_1 \cdot h\\ &= h\left(\frac{2}{h^2} (2x_k - x_{k-1}- x_{k+1}) - \lambda_1\right) = 0. \end{aligned}
For k = 0 (boundary point): \frac{\partial \mathcal{L}}{\partial x_0} = \frac{2}{h} (x_1 - x_0) - \lambda_1 \cdot \frac{h}{2} - \lambda_2 = 0.
For k = n (boundary point): \frac{\partial \mathcal{L}}{\partial x_n} = -\frac{2}{h} (x_n - x_{n-1}) - \lambda_1 \cdot \frac{h}{2} - \lambda_3 = 0.
These equations, combined with the constraints, form a system of equations that can be solved for \bm{x}, \lambda_1, \lambda_2, and \lambda_3.
\left\{ \begin{aligned} 2\dfrac{x_{k+1} - 2x_{k} + x_{k-1}}{h^2} + \lambda_1 &= 0. \\ x_0 & = 2 \\ x_n & = 1 \\ \frac{2}{h} (x_1 - x_0) - \lambda_1 \cdot \frac{h}{2} & = \lambda_2 \\ -\frac{2}{h} (x_n - x_{n-1}) - \lambda_1 \cdot \frac{h}{2} & = \lambda_3 \end{aligned} \right.
notice that \lambda_2 and \lambda_3 are trivially solved and can be removed to the nonlinear system to obtain
\left\{ \begin{aligned} 2\dfrac{x_{k+1} - 2x_{k} + x_{k-1}}{h^2} + \lambda_1 &= 0. \\ x_0 & = 2 \\ x_n & = 1 \\ \end{aligned} \right.
which can be interpreted as the discrete approximation of the Boundary Value Problem (BVP)
\left\{ \begin{aligned} 2x^{\prime\prime} + \lambda_1 &= 0. \\ x(0) & = 2 \\ x(1) & = 1 \\ \end{aligned} \right.
solving this last BVP
x(t) = -\dfrac{\lambda_1}{4} t^2 + C_1 t + C_2
using BC \begin{aligned} x(0) & = C_2 = 2,\quad & C_2 & = 2, \\ x(1) & = -\dfrac{\lambda_1}{4} + C_1 + C_2 = 1, \quad & C_1 & = \dfrac{\lambda_1}{4} - 1 \end{aligned}
Using integral constraint
\begin{aligned} 0 & = \int_0^1 x(t)\,\mathrm{d}t-1 \\ & = \int_0^1 -\dfrac{\lambda_1}{4} t^2 + \left(\dfrac{\lambda_1}{4}-1\right)t + 2\,\mathrm{d}t-1 \\ &= -\frac{\lambda_1}{12} + \frac{\lambda_1-4}{8} + 2-1 \\ &= \frac{\lambda_1}{24} + \frac{1}{2} \end{aligned}
so that \lambda_1 = -12 and
x(t) = -\dfrac{\lambda_1}{4} t^2 + C_1 t + C_2 = 3 t^2 -4t+2
the same solution.
Solving via Analogy
The finite difference approach motivates solving the problem
\begin{aligned} \text{minimize} \quad & \mathcal{A}(x) = \int_0^1 (x')^2 \, \mathrm{d}t, \\ \text{subject to} \quad & \int_0^1 x \, \mathrm{d}t = 1, \\ & x(0) = 0, \quad x(1) = 2, \end{aligned}
by introducing a Lagrangian functional that incorporates the integral constraint. We define:
\begin{aligned} \mathcal{L}(x, \lambda) &= \mathcal{A}(x) - \lambda \left(\int_0^1 x \, \mathrm{d}t - 1\right) \\ &= \int_0^1 (x')^2 \, \mathrm{d}t - \lambda \left(\int_0^1 x \, \mathrm{d}t - 1\right). \end{aligned}
The stationary points of this Lagrangian, \mathcal{L}(x, \lambda), are candidates for the minimizers. To find these stationary points, we perform the first variation of \mathcal{L}.
First Variation of \mathcal{L}
The variation of \mathcal{L}(x, \lambda) with respect to both x and \lambda is:
\begin{aligned} \delta \mathcal{L}(x, \lambda; \delta x, \delta \lambda) &= \delta \int_0^1 (x')^2 \, \mathrm{d}t - \delta \lambda \left(\int_0^1 x \, \mathrm{d}t - 1\right) - \lambda \delta \int_0^1 x \, \mathrm{d}t \\ &= \int_0^1 2 x' \delta x' \, \mathrm{d}t - \delta \lambda \left(\int_0^1 x \, \mathrm{d}t - 1\right) - \lambda \int_0^1 \delta x \, \mathrm{d}t. \end{aligned}
Using integration by parts on the term involving x' and using \delta x(0)=\delta x(1)=0, we get:
\begin{aligned} \delta \mathcal{L}(x, \lambda; \delta x, \delta \lambda) &= \int_0^1 2 x' \delta x' - \lambda \delta x \, \mathrm{d}t - \delta \lambda \left(\int_0^1 x \, \mathrm{d}t - 1\right) \\ &= -\int_0^1 (2 x^{\prime \prime} + \lambda) \delta x \, \mathrm{d}t - \cancel{2 x' \delta x \Big|_{t=0}^{t=1}} - \delta \lambda \left(\int_0^1 x \, \mathrm{d}t - 1\right). \end{aligned}
Applying the fundamental lemma of calculus of variations yields the following conditions:
\left\{ \begin{aligned} 2 x^{\prime \prime} + \lambda &= 0, \\ x(0) &= 0, \\ x(1) &= 2, \\ \int_0^1 x(t) \, \mathrm{d}t &= 1. \end{aligned} \right.
Solving this boundary value problem gives the same solution as obtained before.
General Solution for Variational Problems with Integral Constraints
Consider a variational problem with an objective functional and one or more integral constraints. The problem can be stated as follows:
Problem Formulation
Given:
\begin{aligned} \text{minimize} \quad & \mathcal{F}(x) = \int_a^b F(x, x', x'', \ldots, x^{(n)}, t) \, \mathrm{d}t, \\ \text{subject to} \quad & \mathcal{G}(x) = \int_a^b G(x, x', x'', \ldots, x^{(n)}, t) \, \mathrm{d}t = 0. \end{aligned}
we want to find functions x(t) that minimize the objective functional \mathcal{F}(x) while satisfying the constraint \mathcal{G}(x).
Solution Using the Lagrangian Functional
To solve this constrained problem, we introduce a Lagrange multiplier \lambda and form a Lagrangian functional \mathcal{L}(x, \lambda):
\begin{aligned} \mathcal{L}(x, \lambda) &= \mathcal{F}(x) - \lambda \mathcal{G}(x) \\ &= \int_a^b \left(F(x, x', x'', \ldots, x^{(n)}, t) - \lambda G(x, x', x'', \ldots, x^{(n)}, t) \right) \, \mathrm{d}t. \end{aligned}
Stationary Points of the Lagrangian Functional
The functions x(t) that are candidates for minimizing \mathcal{F}(x) must be stationary points of the Lagrangian \mathcal{L}(x, \lambda). To find these stationary points, we compute the first variation of \mathcal{L}(x,\lambda) with respect to x(t) and \lambda, and set it to zero. This will lead to a set of Euler-Lagrange equations that must be satisfied by x(t) and \lambda.
Extension to Multiple Constraints
If the problem includes multiple integral constraints, we can generalize the formulation. Let the constraints be given by:
\begin{aligned} \text{minimize} \quad & \mathcal{F}(x) = \int_a^b F(x, x', x'', \ldots, x^{(n)}, t) \, \mathrm{d}t, \\ \text{subject to} \quad & \mathcal{G}_1(x) = \int_a^b G_1(x, x', x'', \ldots, x^{(n)}, t) \, \mathrm{d}t = 0, \\ & \mathcal{G}_2(x) = \int_a^b G_2(x, x', x'', \ldots, x^{(n)}, t) \, \mathrm{d}t = 0, \\ & \vdots \\ & \mathcal{G}_p(x) = \int_a^b G_p(x, x', x'', \ldots, x^{(n)}, t) \, \mathrm{d}t = 0. \end{aligned}
Lagrangian for Multiple Constraints
To incorporate these constraints, we introduce a Lagrange multiplier \lambda_k for each constraint \mathcal{G}_k(x), resulting in the Lagrangian functional:
\begin{aligned} \mathcal{L}(x, \lambda_1, \lambda_2, \ldots, \lambda_p) &= \mathcal{F}(x) - \sum_{k=1}^p \lambda_k \mathcal{G}_k(x) \\ &= \int_a^b \left(F(x, x', x'', \ldots, x^{(n)}, t) - \sum_{k=1}^p \lambda_k G_k(x, x', x'', \ldots, x^{(n)}, t)\right) \, \mathrm{d}t. \end{aligned}
Finding Stationary Points
The stationary points of \mathcal{L} with respect to x(t) and \lambda_k (for k = 1, 2, \ldots, p) provide candidates for the solution. Setting the first variations of \mathcal{L}(x,\lambda_1,\ldots,\lambda_p) to zero leads to a system of equations analogous to the Euler-Lagrange equations for unconstrained problems, now modified to account for the integral constraints.
Conclusion
This approach provides a general framework for handling variational problems with one or more integral constraints. By introducing appropriate Lagrange multipliers and forming the Lagrangian functional, we can convert a constrained variational problem into an unconstrained problem whose solutions must satisfy a system of differential equations determined by the first variation of the Lagrangian.
These equations, together with the boundary conditions, yield the function x(t) that minimizes \mathcal{F}(x) under the given integral constraints.