The LQ Problem

We now meet the one optimal-control problem that yields completely to analysis — the Linear-Quadratic (LQ) problem. Its name is its recipe: linear dynamics and a quadratic cost. Everything earlier — the cost functional, the maximum principle, the Pontryagin–dynamic-programming comparison — was the general theory. Here that theory crystallises into formulas you can compute.

Linear dynamics, quadratic cost

The state evolves under a linear law, driven by the control through a constant matrix:

\dot{x} = A\,x + B\,u, \qquad x(0) = x_0.

Here x \in \mathbb{R}^n is the state, u \in \mathbb{R}^m the control, A the n\times n system matrix and B the n\times m input matrix. This is the linear system whose free response is the matrix exponential e^{At}.

The cost is a quadratic form in the state and the control, integrated over the horizon and capped by a terminal penalty:

J = \tfrac12 \int_0^T \Big( x^{\mathsf{T}} Q\,x + u^{\mathsf{T}} R\,u \Big)\,dt \;+\; \tfrac12\, x(T)^{\mathsf{T}} S\, x(T).

The factor \tfrac12 is a convention that keeps later derivatives tidy. Each term is a quadratic form: x^{\mathsf{T}} Q x charges the state for drifting from zero, u^{\mathsf{T}} R u charges control effort, and x(T)^{\mathsf{T}} S x(T) charges the final miss.

The conditions on the weights

The weight matrices are symmetric, and their definiteness is exactly what makes the problem well-posed.

Q \succeq 0 and S \succeq 0 (positive-semi-definite): a state penalty is never negative, so straying from the target can only ever cost you. Semi-definite is enough — some directions in the state may simply not be penalised.
R \succ 0 (strictly positive-definite): every non-zero control incurs a genuine, strictly positive penalty.

Why R must be strictly positive-definite. If some non-zero control direction u_\star had u_\star^{\mathsf{T}} R\, u_\star = 0, then effort along u_\star would be free: the optimiser could pour unlimited, impulsive control in that direction at no charge, and the minimisation would have no finite solution. Demanding R \succ 0 — every eigenvalue of R strictly positive — closes that loophole, so the optimal control stays finite and unique. It is the positive-definite R that makes the cost a genuine bowl in u, with a single bottom to roll to. And because R is invertible, R^{-1} exists — a fact the optimal control law will lean on directly.

Dynamics are linear: \dot{x} = A x + B u, x(0) = x_0.
Cost is quadratic: J = \tfrac12\int_0^T (x^{\mathsf{T}} Q x + u^{\mathsf{T}} R u)\,dt + \tfrac12 x(T)^{\mathsf{T}} S x(T).
Weights are symmetric with Q \succeq 0, S \succeq 0 and R \succ 0; the strict positive-definiteness of R guarantees a finite, unique optimum and an invertible R.

The harmonic oscillator of control theory

Why does this one problem deserve a whole stage? Because it plays the role in control that the simple harmonic oscillator plays in physics: the one nontrivial case solvable in closed form, and the local model of everything else.

It is exactly solvable. A quadratic cost minimised over linear dynamics has a value function that is itself quadratic, and the optimal control turns out to be a simple linear feedback. No other interesting class collapses so cleanly — the rest of this stage derives that collapse, term by term.
It is the universal local approximation. Take any smooth nonlinear problem and sit at an operating point. Linearise the dynamics there (the Jacobian gives A and B) and quadraticise the cost (the Hessian gives Q and R), and the LQ problem is what you are left with. So the LQ solution is the leading-order controller for almost any system you will ever meet — which is why we invest in solving it exactly.

Steepening the bowl

The quadratic cost is, in each variable, a bowl. Take the scalar slice \tfrac12 q\,x^2 for the state and \tfrac12 r\,u^2 for the control. Slide the weights q and r and watch each parabola steepen: a larger weight makes that bowl narrower, so the optimiser is punished harder for the slightest deviation. With q, r > 0 both are genuine bowls curving up from the origin — the geometric picture behind “positive-definite”. That balance between the two steepnesses is the whole design knob of quadratic control.

The combination is not an accident of convenience — it is the richest pairing that still keeps the door to closed-form solution open. Linear dynamics are the first term of any Taylor expansion of the true dynamics; a quadratic cost is the first nontrivial term of any smooth penalty (the constant and linear terms vanish at a well-chosen operating point). So “linear plus quadratic” is precisely the second-order model of reality, and the next page shows that solving it reduces the entire problem to a single matrix differential equation.