Adjoint state method

Remarks

The formula follows from the chain rule. Differentiating $F(u(\theta), \theta) = 0$ gives $F_u \, u'(\theta) = -F_\theta$, so $\nabla J^T = j_u u'(\theta) + j_\theta = -j_u F_u^{-1} F_\theta + j_\theta = -\lambda^T F_\theta + j_\theta$. By the implicit function theorem, invertibility of $F_u$ at a solution gives the smooth map $u(\theta)$ locally. The forward method needs $p$ linear solves with $F_u$, one for each column of $u'(\theta)$; the adjoint method needs one solve with $F_u^T$. For a time-dependent state equation the adjoint equation runs backward in time; in optimal control it is the costate equation.

Example: $F(u, \theta) = A u - \theta$ with $A$ an invertible $n \times n$ matrix and $p = n$, and $j(u, \theta) = \frac{1}{2} \abs{u - d}^2$ for a fixed $d \in \R^n$. Then $F_u = A$, $F_\theta = -I$ and $\nabla_\theta j = 0$, so $A^T \lambda = u - d$ and $\nabla J(\theta) = \lambda$. Directly, $J(\theta) = \frac{1}{2} \abs{A^{-1} \theta - d}^2$ has gradient $A^{-T} (A^{-1} \theta - d) = \lambda$, as it should.

Non-example: let $n = p = 1$ and $F(u, \theta) = u^2 - \theta$, at the point $u = \theta = 0$, where $F_u = 0$ is singular. The adjoint equation $0 \cdot \lambda = j_u$ need not have a solution, and the state is not a differentiable function of $\theta$ there.