A two-interval shooting residual has a block-bidiagonal sensitivity Jacobian

The function $u(t)=e^{\theta t}s$ satisfies $u^{\mathrm{\prime }}=\theta u$ and $u(0)=s$. Conversely, for any solution $v$ with this initial value, the product and chain rules give $(e^{-\theta t}v(t){)}^{\mathrm{\prime }}=0$. By [F4] this product is constantly $s$ on its solution interval, and positivity and the reciprocal identity in [F3] imply $v(t)=e^{\theta t}s$. Thus the displayed solution exists globally and is unique.

Verify the actual common-tube hypotheses locally at any $(s_{\ast },{\theta }_{\ast })$. Fix $\epsilon ,\delta ,\eta >0$ and restrict the parameters to $\mathrm{\mid }s-s_{\ast }\mathrm{\mid }<\epsilon$, $\mathrm{\mid }\theta -{\theta }_{\ast }\mathrm{\mid }<\delta$. For $-\eta \le t\le \tau +\eta$, $\mathrm{\mid }{e}^{\theta t}s\mathrm{\mid }\le B:=(\mathrm{\mid }{s}_{\ast }\mathrm{\mid }+\epsilon )\exp ((\mathrm{\mid }{\theta }_{\ast }\mathrm{\mid }+\delta )(\tau +\eta ))\mathrm{.}$ Treat $(s,\theta )$ as the flow parameter, with initial map $(s,\theta )\mapsto s$ and field $(t,u,s,\theta )\mapsto \theta u$ on ${\mathbb{R}}^4$. The bounded convex open set $(-\eta ,\tau +\eta )\times (-B-1,B+1)\times (s_{\ast }-2\epsilon ,s_{\ast }+2\epsilon )\times ({\theta }_{\ast }-2\delta ,{\theta }_{\ast }+2\delta )$ has compact closure inside that field's domain and contains all these trajectories for $0\le t\le \tau$ and the restricted parameters. The field and initial map are $C^1$. Step 1.1 supplies existence and uniqueness, so [F1] applies. Each shooting interval uses this same argument with its own initial node; since there are two intervals, a common finite bound may also be taken.

The exact interval map is therefore ${\mathrm{\Phi }}_{\tau }(s,\theta )=e^{\tau \theta }s$. Differentiate the residuals with respect to $(z_0,z_1,z_2)$ and $\theta$, writing $E=e^{\tau \theta }$: $D_{z}R=\left(\begin{array}{ccc}1& 0& 0\\ -E& 1& 0\\ 0& -E& 1\end{array}\right),D_{\theta }R=\left(\begin{array}{c}0\\ -\tau E{z}_0\\ -\tau E{z}_1\end{array}\right)\mathrm{.}$ The state Jacobian is lower bidiagonal, and the parameter column contains the two local flow sensitivities.

Solving $R=0$ gives $z_0=a$, $z_1=Ea$, and $z_2=E^{2}a$. Direct differentiation yields $D_{\theta }{z}_2=2\tau E^{2}a$. The discrete adjoint for the two flow steps seeds $p_2=1$, then gives $p_1=E$ and $p_0=E^2$; its parameter sum is $\tau E{z}_0{p}_1+\tau E{z}_1{p}_2=2\tau E^{2}a$. Thus the reverse gradient agrees with the differentiated shooting solution.