Problem understanding "formal proof" of Duhamel principle

Question

I am now studying PDEs. My teacher referred to these notes. In particular, I'm having trouble understanding the proof on pages 20-21 of this part. This is a non-rigorous, formal proof of Duhamel's principle. It starts by defining the solution operator $S(t)$ such that $S(t)\Phi$ is the solution of the problem $U_t=AU$, where $A$ is a matrix (or matrix operator) and $U$ is a vector function which is the unknown in this ODE/PDE. Then it sets:

The aim is to prove that this solves $U_t=AU+F$, where $A$ is as before, $U$ is unknown and $F$ is an inhomogeneous term. The proof consists of basically just the following equality chain:

What I do not understand is where the term $S(0)F(\vec x,t)$ is coming from. The first term is simply the derivative of the first term in $U$. Then we should derive the second term. I understand that the third term comes from deriving the integrand, but I do not see how the second term comes from "deriving the extreme". I tried some calculations and I find $S(t)F(\vec x,0)$, which of course must be wrong. Could anyone explicitly do all the passages that lead to that term?

Answer 1

The differentiation of the second term is an application of Leibniz's Rule: $$\frac{\mathrm{d}}{\mathrm{d}\theta} \left (\int_{a(\theta)}^{b(\theta)} f(x,\theta)\,\mathrm{d}x \right )= f\big(b(\theta),\theta\big)\cdot b'(\theta) \,-\, f\big(a(\theta),\theta \big)\cdot a'(\theta)\,+\, \int_{a(\theta)}^{b(\theta)}\partial_{\theta} f(x,\theta)\,\mathrm{d}x $$ Note that the first two terms come from the fundamental theorem of calculus, whereas the the last one must be proven additionally. It is the last one that you accepted above, and the first two that are unclear:

\begin{align*} U_t ={}& (S(t)\Phi(x))_t + \frac{d}{dt}\int_0^t S(t-s)F(x, s) ds \\ {}={}& (S(t)\Phi(x))_t + \left(S(t-s)F(x, s)\right)\big|_{s = t} + \int_0^t \frac{d}{dt}S(t-s)F(x, s) ds \\ {}={}& (S(t)\Phi(x))_t + S(0)F(x, t) + \int_0^t AS(t-s)F(x, s) ds \\ \end{align*}

Edit: the rule proved

We are in the case where we have: $$\phi(\alpha)=\hspace{-4mu}\int\limits_{a(\alpha)}^{b(\alpha)}\hspace{-8mu}f(x,\alpha)\,dx,$$ and need to compute $\phi'(\alpha_0)$, the derivative of $\phi$ with respect to $\alpha$ in the point $\alpha_0$. We first write out the "variation" of $\phi$, that is: $$\Delta\phi=\phi(\alpha+\Delta\alpha)-\phi(\alpha)=\hspace{-16mu}\int\limits_{a(\alpha+\Delta\alpha)}^{b(\alpha+\Delta\alpha)}\hspace{-20mu}f(x,\alpha+\Delta\alpha)\,dx-\hspace{-4mu}\int\limits_{a(\alpha)}^{b(\alpha)}\hspace{-8mu}f(x,\alpha)\,dx.$$ We split the first integral into three parts:

1. The integral from $a(\alpha)$ to $a(\alpha+\Delta\alpha)$, which will have a minus sign since the limits must be swapped to get to it;
2. The integral from $a(\alpha)$ to $b(\alpha)$, which we then put together with the $-\phi(\alpha)$;
3. The integral from $b(\alpha)$ to $b(\alpha+\Delta\alpha)$.

With that, we get that: $$\Delta\phi=-\hspace{-16mu}\int\limits_{a(\alpha)}^{a(\alpha+\Delta\alpha)}\hspace{-20mu}f(x,\alpha+\Delta\alpha)\,dx+\hspace{-4mu}\int\limits_{a(\alpha)}^{b(\alpha)}\hspace{-8mu}[\phi(\alpha+\Delta\alpha)-\phi(\alpha)]\,dx+\hspace{-16mu}\int\limits_{b(\alpha)}^{b(\alpha+\Delta\alpha)}\hspace{-20mu}f(x,\alpha+\Delta\alpha)\,dx.$$ By the mean value theorem, the first integral is $f(\xi_1,\alpha+\Delta\alpha)\cdot(a(\alpha+\Delta\alpha)-a(\alpha))$, where $\xi_1\in[a(\alpha),a(\alpha+\Delta\alpha)]$ (or with swapped extremes if $a$ is decreasing in $[\alpha,\alpha+\Delta\alpha]$), and the third integral can be similarly manipulated. We then place the middle term at the end and get: \begin{align*} \Delta\phi={}&-(a(\alpha+\Delta\alpha)-a(\alpha))\cdot f(\xi_1,\alpha+\Delta\alpha)+(b(\alpha+\Delta\alpha)-b(\alpha))\cdot f(\xi_2,\alpha+\Delta\alpha)+{} \\ &{}+\hspace{-4mu}\int\limits_{a(\alpha)}^{b(\alpha)}\hspace{-8mu}[\phi(\alpha+\Delta\alpha)-\phi(\alpha)]\,dx. \end{align*} What we now want is to compute the limit of $\frac{\Delta\phi}{\Delta\alpha}$ for $\Delta\alpha\to0$. Notice how:

1. $\xi_1\to a(\alpha)$ and $\xi_2\to b(\alpha)$, as can be deduced by the constraints of the mean value theorem;
2. Supposing $a,b$ are differentiable, the coefficients of the evalued $f$, divided by $\Delta\alpha$, tend to $a'(\alpha),b'(\alpha)$;
3. Supposing $f$ is continuous, the $\Delta\alpha$ vanishes when it tends to zero, yielding $f(a(\alpha),\alpha)$ and $f(b(\alpha),\alpha)$;
4. Provided the limit can harmlessly pass under the integral, we get the integral of the derivative of $f$ as a last term.

In summary, if nothing goes wrong in the above, we get: $$\phi'(\alpha)=b'(\alpha)\cdot f(b(\alpha),\alpha)-a'(\alpha)\cdot f(a(\alpha),\alpha)+\hspace{-4mu}\int\limits_{a(\alpha)}^{b(\alpha)}\hspace{-8mu}f_\alpha(x,\alpha)\,dx,$$ as we wanted to prove. $\square$

Proof adapted from Wikipedia.