When you directly minimize the cost functional (for instance using a discretization in time so that the integral becomes a summation), you are solving the optimal control problem for a single initial point $x_0$. Moreover, you will find an open-loop control, i.e. a function $t \mapsto u^*(t)$, which is not robust to perturbations in the dynamics (see Open-loop vs closed-loop).
When you solve the HJB equation, instead, you are simultaneously solving the problem for all values of $x_0$, and you can even provide an optimal control in feedback form.
Regarding the question on the double minimization problem, you are right. There is a double minimization problem, but
- The dimension of the minimization problem is the dimension of the control space, which is typically very small, so it not very expensive.
- Simulations have shown that the first minimization, the one while solving the HJB in step 1, does not need to be solved accurately, so one can discretize the control space with few controls and minimize over a finite set of possible control values. The second minimization problem, however, must be solved more accurately.
I recommend this book for more details.
For reference, I am re-posting the answer from MO here:
On the role of verification theorem: it is an issue related to the existence-uniqueness of solutions in the classical sense for the HJB PDE. In applying the verification theorem, we ignore such issues, guess the structure of a smooth value function, formally verify (by substitution) that the guessed structural form satisfies the HJB PDE under consideration, and then use the Bellman's principle of optimality to compute the optimal control. Whether such verification is valid remains contingent on the existence-uniqueness of smooth enough classical solution (at least $C^1$ in the deterministic case and $C^2$ in the stochastic case) for the HJB PDE.
On deterministic versus stochastic: The above verification/HJB classical solution issue is for both the deterministic and the stochastic case. For example, see Ch. 4, Sec. 2 of 1, which specifically talks about verification theorems for the first order HJB PDEs in deterministic optimal control. Example 2.3 there is about an 1D deterministic optimal control problem whose HJB PDE does not admit any $C^{1}([0,T],\mathbb{R})$ solution. Ch. 4 and Ch. 5 of that book discusses details on the verification theorems for both the deterministic and stochastic optimal control problems, and also the viscosity solutions.
1 J. Yong and X.Y. Zhou, Stochastic Controls: Hamiltonian Systems and HJB Equations, vol 43. Springer, New York, 1999.
Best Answer
My simple answer would be that they are quite the same thing.
Explanation
Riccati Equations can be derived from Hamilton-Jacobi-Bellman equations in the particular case of LQR problem, an optimal control problem where the dynamics is linear and the cost is quadratic.
Consider the finite horizon LQR problem. In this particular case, the Hamilton-Jacobi-Bellman equation has the expression (for semplicity I put $N=0$)
$$ \partial_t V(x,t) + \min_u \left\{ \partial_x V(x,t) \cdot (Ax+Bu) + x^T Q x + u^T Ru \right\} = 0 $$ with the terminal condition $$ V(x,T) = x^T Q_f x . $$
Now we look for solutions of the form $V(x,t) = x^T P(t) x$, where $P(t)$ is a symmetric matrix for each $t \in [0,T]$. If we substitute this expression in the (HJB) equation, we get
$$ x^T P'(t) x + \min_u \left\{ 2 P(t)x \cdot (Ax+Bu) + x^T Q x + u^T Ru \right\} = 0 . $$
We can explicitly find the minimum of the expression inside the curly brackets. For a given couple $(t,x)$, let us define $\Phi$ as $$ \Phi(u) = 2 P(t)x \cdot (Ax+Bu) + x^T Q x + u^T Ru .$$
The minimum is obtained when $∇\Phi(u) = 0$, that is when $$ 2B^T P(t) x + 2 Ru = 0,$$ so the optimal control is $$ u^*(t,x) = -R^{-1} B^T P(t)x ,$$ with $$ \Phi(u^*(t,x)) = 2 x^T P(t)^T \left(Ax-BR^{-1} B^T P(t)x \right) + x^T Q x + x^T P(t)^T B R^{-1} B^T P(t)x $$
So we can rewrite again the (HJB) equation without the minimization term: $$ x^T P'(t) x + 2 x^T P(t)^T \left(Ax-BR^{-1} B^T P(t)x \right) + x^T Q x + x^T P(t)^T B R^{-1} B^T P(t)x = 0 , $$ and by grouping the $x^T$ and the $x$ term and doing some simple algebraic steps ($P(t)$ is symmetric) we get $$ x^T \left( P'(t) + 2 P(t) A - P(t) BR^{-1} B^T P(t) + Q \right) x = 0 , $$
Since the above equation must hold for each $x$, it is equivalent to the matrix differential equation
$$ P'(t) + 2 P(t) A - P(t) BR^{-1} B^T P(t) + Q = 0 .$$
Finally, in order to satisfy the final condition, it must be $$ P(T) = Q_f .$$
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