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I'm reading Steven Johnson's notes on Adjoint Methods and have made it to the Initial Value Problems section.

We have $A = e^{-Bt}$, where $B$ depends on a parameter vector $\vec{p}$ and therefore $A$ does as well through its relationship to $B$.

We now want to find $A_{\vec{p}}$, the partial derivative of $A$ w.r.t. the parameters. Johnson gives the answer as:

$$ A_{\vec{p}} = - \int\limits_0^t e^{-B\tau} B_{\vec{p}} e^{-B(t-\tau)} d\tau $$

I can't see why this is true. Can someone help me out?

Moreover, he goes on to use this equation in a different one that left-multiplies $A_{\vec{p}}$ by $-\vec{\lambda}^T$ and right-multiplies by $\vec{x}$, which produces the term:

$$ \int\limits_0^t \vec{\lambda}^T(t-\tau) \cdot B_{\vec{p}} \cdot \vec{x}(\tau) d\tau $$

which seems to come from substituting in $\vec{\lambda}(\tau) = e^{B^T \tau} g_{\vec{x}}^T$, which when transposed becomes $\vec{\lambda}^T(\tau) = g_{\vec{x}} e^{B \tau}$, which can cancel the $e^{-B\tau}$ to the left inside the integral. But then to produce $\vec{\lambda}^T(t-\tau)$, you need the $e^{-B(t-\tau)}$ to be next to $g_{\vec{x}}$, but they're separated by matrix $B_{\vec{p}}$, and it's not obvious these commute!

I'm confused. It makes me think the equation for $A_{\vec{p}}$ is somehow wrong. Hoping the derivation can help me reconcile.

Rodrigo de Azevedo
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Pavel Komarov
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    This answer mentions a Wiki article on the derivative of the exponential map, which uses some Lie group knowledge: https://math.stackexchange.com/a/2047/171839 – Sean Roberson Jan 24 '25 at 02:34
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    That Wiki article just states the formula again, doesn't prove it, unfortunately. But at least this indicates it probably doesn't have an error. – Pavel Komarov Jan 24 '25 at 02:50
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    There is an article proving it. https://en.m.wikipedia.org/wiki/Derivative_of_the_exponential_map Sadly, I don't know much about Lie groups so I can't expand on it. – Sean Roberson Jan 24 '25 at 02:53

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$$\int_0^1 e^{tA}He^{(1-t)A}dt=\sum_{i,j\geq 0}\frac{A^i}{i!}H\frac{A^j}{j!}\int_0^1t^i(1-t)^{j}dt=\sum_{i,j\geq 0}\frac{1}{(i+j+1)!}A^iHA^j$$Now $$(A+H)^{n+1}-A^{n+1}=\sum_{i=0}^nA^iHA^{n-i}+o(H)$$ $$e^{A+H}-e^A=\sum_{n=0}^{\infty}\frac{1}{(n+1)!}((A+H)^{n+1}-A^{n+1})$$$$=\sum_{i,j\geq 0}\frac{1}{(i+j+1)!}A^iHA^j+o(H).$$ This proves that the differential of the map $A\mapsto e^A$ at point $A$ is the linear map $H\mapsto \int_0^1 e^{tA}He^{(1-t)A}dt,$ a beautiful formula.

Letac Gérard
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  • Unfortunately I don't understand most of what you did there. Beauty is definitely in the eye of the beholder. I vaguely follow the sketch of the argument, but I doubt this was how the original formula was derived. How did people first figure this out? And how does the 0 to 1 integral become 0 to t in my problem? – Pavel Komarov Jan 27 '25 at 21:20
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    I apologize for having been terse. For your last question the differential of $A\mapsto e^{sA}$ is the linear map $$H\mapsto \int_0^1e^{stA}se^{sA(1-t)}dt=\int_0^se^{\tau A}He^{(s-\tau)A} d\tau$$ For understanding the differential of $p\mapsto e^{sA(p)}$ you have to use the composition of differentials, but explanations here are a bit long, and you could try to get help around to get acquainted with the differentials. – Letac Gérard Jan 28 '25 at 16:07