18 Isoperimetric 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: \displaystyle\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^2([0,1]) \ \middle|\ x(0) = 2, \, x(1) = 1,\, \int_0^1x(t)\,\mathrm dt=1 \right\}.
The problem can thus be reformulated as finding x \in \mathbb{W} that minimizes \mathcal{A}(x).
18.1 Feasible Variations
Feasible Directions
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
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\delta x^\prime \, \mathrm{d}t.
Using integration by parts, we find
\int_0^1 2 x^\prime\delta x^\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 a Feasible 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}.
Differential Equation
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).
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.
Because the objective is strictly convex on the affine feasible set, this stationary function is the unique global minimizer.
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.
18.2 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(t) \, \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=1}^{n-1} 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.
Discrete Lagrangian
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_0 - x_1) - \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\frac{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\frac{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) = -\frac{\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) & = -\frac{\lambda_1}{4} + C_1 + C_2 = 1, \quad & C_1 & = \frac{\lambda_1}{4} - 1 \end{aligned}
Using integral constraint
\begin{aligned} 0 & = \int_0^1 x(t)\,\mathrm{d}t-1 \\ & = \int_0^1 -\frac{\lambda_1}{4} t^2 + \left(\frac{\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) = -\frac{\lambda_1}{4} t^2 + C_1 t + C_2 = 3 t^2 -4t+2
the same solution.
18.3 Lagrange-Multiplier Method
Worked Example
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) = 2, \quad x(1) = 1, \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
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) &= 2, \\ x(1) &= 1, \\ \int_0^1 x(t) \, \mathrm{d}t &= 1. \end{aligned} \right.
Solving this boundary value problem gives the same solution as obtained before.
18.4 General Formulation
Consider a variational problem with an objective functional and one or more integral constraints. The problem can be stated as follows:
Problem
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).
Augmented 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}
Stationarity
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.
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}
Augmented Functional
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}
Stationarity Conditions
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.
Constraint Qualification
The multiplier rule requires a constraint qualification: the first variations of the active constraints must be linearly independent. Without it, an abnormal multiplier associated with the objective may be needed. Under the regularity assumption, the Euler–Lagrange equations of the augmented integrand, the original integral constraints, and the boundary conditions form the complete first-order necessary system. They identify stationary candidates, not automatically minimizers.
18.5 Normal and Abnormal Multiplier Rules
For equality-constrained variational problems it is useful to state the multiplier rule in homogeneous form. Introduce a scalar multiplier \lambda_0\geq0 for the objective and multipliers \lambda_1,\ldots,\lambda_m for the constraints, not all zero. The augmented integrand is
M=\lambda_0 L+\sum_{k=1}^m\lambda_k g_k.
If a constraint qualification holds, one may normalize \lambda_0=1; this is the normal case used in the preceding derivations. If the qualification fails, \lambda_0 may vanish and an abnormal extremal is possible. Consequently, writing the objective multiplier implicitly as one is justified only after a regularity assumption has been stated.