Discrete continuous and pde adjoint sensitivities — Examples

Direct substitution gives $x_1=2\theta$ and $x_2=6\theta$, so $J(\theta )=\frac{1}{2}(1+4+36){\theta }^2=\frac{41}{2}{\theta }^2$ and $J^{\mathrm{\prime }}(\theta )=41\theta$.

The terminal adjoint seed is $p_2=x_2=6\theta$. The reverse recursion [F1] gives $p_1=x_1+3p_2=20\theta$ and $p_0=x_0+2p_1=41\theta$. Since $D{x}_0\mathrm{/}D\theta =1$, the adjoint gradient is $p_0=41\theta$, agreeing with the direct derivative.

To apply [F1] with the prescribed center $u_0$, use the larger closed ball $\mathcal{B}=[0,2u_0]$ of radius $R=u_0$. For $F(Y)=-kY$, take $M=2k{u}_0$ and $L=k$; then $hM=2hk{u}_0<R$ and $q=hk<1\mathrm{/}2$. The field is $C^1$ on $\mathbb{R}$ with derivative bounded by $L$, so the nonsingularity hypotheses inherited by [F2] hold on this ball. Solving the scalar residual gives $(1+hk)Y=u_0$, hence $Y=u_0\mathrm{/}(1+hk)$. Since $1+hk>0$, this is the unique root in the stated interval and agrees with the branch in [F1].

Differentiate the residual $(1+hk)Y-u_0=0$: $(1+hk)Y_k+hY=0$, so [F2] gives $Y_k=-h{u}_0\mathrm{/}(1+hk{)}^2$. Therefore $J_k=Y{Y}_k=-h{u}_0^2\mathrm{/}(1+hk{)}^3$.

Put $w(x)=x(1-x)\mathrm{/}2$. Direct differentiation gives $w^{\mathrm{\prime }\mathrm{\prime }}=-1$, $w(0)=w(1)=0$, and [F3] gives $w\in V$. By [F9], $V$ is a real Hilbert space. Holder's inequality [F8] gives $\mathrm{\mid }\mathrm{\ell }(v)\mathrm{\mid }=\mathrm{\mid }{\int }_0^{1}v\mathrm{\mid }\le \mathrm{\parallel }1{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$, so $\mathrm{\ell }\in V^{\ast }$; the objective $J(u,\theta )=\mathrm{\ell }(u)$ is bounded linear in $u$ and independent of $\theta$, hence $C^1$ with $D_{u}J[v]=\mathrm{\ell }(v)$ and $D_{\theta }J=0$. Also $\mathrm{\mid }{a}_{\theta }(u,v)\mathrm{\mid }\le (1+\theta )\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$ for $\theta >-1$, and $A_{\theta }=(1+\theta )A_0$ is operator-norm $C^1$, while ${\mathrm{\ell }}_{\theta }=\mathrm{\ell }$ is constant. For each $v\in V$ and $\epsilon >0$, [F5] supplies $v_{\epsilon }\in C_c^{\mathrm{\infty }}(0,1)$ with $\mathrm{\parallel }v-v_{\epsilon }{\mathrm{\parallel }}_{H^1}<\epsilon$. By the product rule [F7], $(w^{\mathrm{\prime }}{v}_{\epsilon }{)}^{\mathrm{\prime }}=w^{\mathrm{\prime }\mathrm{\prime }}{v}_{\epsilon }+w^{\mathrm{\prime }}{v}_{\epsilon }^{\mathrm{\prime }}$; since $v_{\epsilon }$ vanishes at both endpoints, [F6] gives ${\int }_0^1{w}^{\mathrm{\prime }}{v}_{\epsilon }^{\mathrm{\prime }}={\int }_0^1{v}_{\epsilon }$. By [F8], the difference between the two sides with $v$ in place of $v_{\epsilon }$ is at most $(\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}+1)\epsilon$, so the weak identity ${\int }_0^1{w}^{\mathrm{\prime }}{v}^{\mathrm{\prime }}={\int }_0^{1}v$ follows as $\epsilon \downarrow 0$. Also, [F4] gives $a_{\theta }(v,v)\ge (1+\theta )\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}^2\mathrm{/}2$; on a neighborhood of each $\theta >-1$ this is a uniform coercivity bound. Hence $u_{\theta }=w\mathrm{/}(1+\theta )$ solves the state equation and is unique. Finally ${\int }_0^{1}w=1\mathrm{/}12$, so $j(\theta )=1\mathrm{/}(12(1+\theta ))$.

The adjoint equation is $a_{\theta }(v,p)=D_{u}J[v]={\int }_0^{1}v$ for all $v\in V$, so symmetry of $a_{\theta }$ and uniqueness give $p=u_{\theta }$. The direct parameter derivatives of $J$ and $\mathrm{\ell }$ vanish. For the scalar parameter direction $1$, $D_{\theta }a[1](u,p)={\int }_0^1{u}^{\mathrm{\prime }}{p}^{\mathrm{\prime }}dx=(1+\theta {)}^{-2}{\int }_0^1(w^{\mathrm{\prime }}{)}^{2}dx$. Direct integration gives ${\int }_0^1(w^{\mathrm{\prime }}{)}^{2}dx=\frac{1}{4}{\int }_0^1(1-2x{)}^{2}dx=1\mathrm{/}12$, hence $D_{\theta }a[1](u,p)=1\mathrm{/}(12(1+\theta {)}^2)$.

Formula [F1] therefore gives $j^{\mathrm{\prime }}(\theta )=-1\mathrm{/}(12(1+\theta {)}^2)$, equal to the derivative of the explicit reduced objective in step 1.1.

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.

The product is $P_2=A_2{A}_1=\mathrm{diag}(2\mathrm{/}3,4)$, so $\log \mathrm{\mid }\det P_2\mathrm{\mid }=\log (8\mathrm{/}3)$. The positive-diagonal QR factors have diagonal entries $(2,1)$ and $(1\mathrm{/}3,4)$.

Their accumulated log sum is $\log 2+\log 1+\log (1\mathrm{/}3)+\log 4=\log (8\mathrm{/}3)$, which matches [F1]. The calculation is finite-time and asserts no limit.

At ${\theta }_{\ast }=1\mathrm{/}3$, the orbit is $x_0=1\mathrm{/}3$ and $x_n=2\mathrm{/}3$ for every $n\ge 1$. If $\mathrm{\mid }\theta -1\mathrm{/}3\mathrm{\mid }<1\mathrm{/}(6\cdot 2^N)$, then the first state remains below $1\mathrm{/}2$ and each later state remains within $1\mathrm{/}6$ of $2\mathrm{/}3$ through time $N$; hence this whole open neighborhood has the same strict branch itinerary.

On that neighborhood the first derivative is $T^{\mathrm{\prime }}(x_0)=2$, and each of the next $N-1$ derivatives is $-2$. The chain rule gives $J^{\mathrm{\prime }}(1\mathrm{/}3)=2(-2{)}^{N-1}$, whose magnitude is $2^N$. By [F1], reverse accumulation returns this same finite-horizon derivative. No infinite-time exponent or orbit-shadowing claim is involved.

Directly, $J(\theta )=\frac{1}{2}{\theta }^2$, so $J^{\mathrm{\prime }}(0)=0$. The correct seed is $p_1=x_1=0$ at $\theta =0$ and the transition derivative is $D_{\theta }{x}_1=1$, so the exact adjoint gradient is also zero.

If the terminal seed is incorrectly replaced by $p_1=1$, reverse accumulation returns $p_1{D}_{\theta }{x}_1=1$ at $\theta =0$, which differs from $J^{\mathrm{\prime }}(0)$. Thus the terminal condition is load-bearing for the discrete gradient.

For every $\theta$ and $h$, $\mathrm{\mid }{u}_h(\theta )-u(\theta )\mathrm{\mid }=h^2\mathrm{\mid }\sin (\theta \mathrm{/}{h}^3)\mathrm{\mid }\le h^2$. Thus the state error tends uniformly to zero as $h\downarrow 0$.

Differentiating gives $u_h^{\mathrm{\prime }}(\theta )=h^{-1}\cos (\theta \mathrm{/}{h}^3)$ and $u^{\mathrm{\prime }}(\theta )=0$. At $\theta =0$, the gradient error is $\mathrm{\mid }{u}_h^{\mathrm{\prime }}(0)-u^{\mathrm{\prime }}(0)\mathrm{\mid }=h^{-1}$, which is unbounded as $h\downarrow 0$. This refutes the stated implication; the example lacks a uniform derivative-data bound of the type required by [F1].