Math - Lagrange Multiplier

Lagrange Multiplier

Multivating example: maximize $f(x,y)$, where $xy+1$, given constraint $g(x,y) = x^2+y^2-1 = 0$

Geometric Intuition: the value of f(x,y) and the constraint it must stay on are tangent to each other. That is, a small perturbation along the constraint curve will not cause change in the value function, hence a potential extrema is achieved.

So this is equivalent to: $L=f(x)-\lambda g(x)$, and get $[\frac{\partial{L}}{\partial{x}}, \frac{\partial{L}}{\partial{\lambda}}] = 0$

$\lambda$ is the lagrange multiplier.

To solve:

1. Define Lagrangian $L = f(x) + \sum_i \lambda_i g_i(x,y)$. In this case, it’s simply:

\[\begin{gather*} \begin{aligned} & L = f(x) + g(x,y) = xy + 1 + \lambda(x^2+y^2-1) \end{aligned} \end{gather*}\]

1. Now, get the first order derivatives:

\[\begin{gather*} \begin{aligned} & \nabla L = [\frac{\partial L}{\partial x}, \frac{\partial L}{\partial y}, \frac{\partial L}{\partial \lambda}] \\ & = [y - 2 \lambda x, x - 2\lambda y, x^2 + y^2 - 1] \end{aligned} \end{gather*}\]

1. We now let the derivates to be 0 (note we are explicitly adding constraint here):

\[\begin{gather*} \begin{aligned} & y = 2 \lambda x \\ & x = 2 \lambda y \\ & x^2 + y^2 - 1 = 0 \end{aligned} \end{gather*}\]

We can get the solution: $\lambda = \frac{1}{2}, x = y = \sqrt{\frac{1}{2}}$

Why does Lagrange Multiplier work?

Since the optimal point is on multiple constraint surfaces, the pertabtion on each surface must be perpendicular to its surface normal: $\nabla g_k(x, y)$:

\[\begin{gather*} \begin{aligned} & \nabla g_k(x, y) = [\frac{\partial g(x, y)}{\partial x}, \frac{\partial g(x, y)}{\partial y}] \\ & d^T \nabla g_k(x, y) = 0 \end{aligned} \end{gather*}\]

At the optimal point, any disturbance d would also satisfy:

\[\begin{gather*} \begin{aligned} & d^T \nabla f(x,y) = 0 \end{aligned} \end{gather*}\]

So the surface normals and the surface normal of the surface under optimization are parallel:

\[\begin{gather*} \begin{aligned} & \nabla f(x,y) = \lambda_1 \nabla g_1(x, y) \\ & ... \\ & \nabla f(x,y) = \lambda_k \nabla g_k(x, y) \\ & \Rightarrow \nabla f(x,y) + \sum_k \lambda_k g_k(x, y) = 0 \Rightarrow \nabla L = 0 \end{aligned} \end{gather*}\]

Pontryagin’s Minimum Principle

If f(x,y, t) is a function of time, we will have “costates” $\lambda_k(t)$. So:

If we define our state transition to be: \(\begin{gather*} \begin{aligned} & x'(t) = f(x(t), u(t)), x(0) = u(0) \end{aligned} \end{gather*}\)

And we want to minimize totla control effort u:

\[\begin{gather*} \begin{aligned} & J = \int_0^T L(x(t), u(t)) dt + h(x(T)) \end{aligned} \end{gather*}\]

Pontryagin’s Minimum Principle says any optimal pair x*(.), u*(.) must have a costate $\lambda(t)$ satisfying the hamiltonian:

\[\begin{gather*} \begin{aligned} & H(x, u, \lambda) = L(x,u) + \lambda^T f(x,u) \end{aligned} \end{gather*}\]

Then State Dynamics:

\[\begin{gather*} \begin{aligned} & \dot x^*(t) = \frac{\partial H}{\partial \lambda} (x^*(t), u^*(t), \lambda(t)), x^*(0) = x_0 \end{aligned} \end{gather*}\]

$\lambda(t)$:

\[\begin{gather*} \begin{aligned} & \lambda(t) = -\frac{\partial H}{\partial x} (x^*(t), u^*(t), \lambda(t)) \end{aligned} \end{gather*}\]

With the boundary condition:

\[\begin{gather*} \begin{aligned} & \lambda(T) = \frac{\partial H}{\partial x}(x^*(T)) \end{aligned} \end{gather*}\]

Optimal Control Output is:

\[\begin{gather*} \begin{aligned} & u^*(t) = argmin H(x^*, u, \lambda) \end{aligned} \end{gather*}\]

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