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Ok, may be this is a silly question but consider the following.

Let $\dot x=f(x)$ be an autonomous differential equation with $f$ having enough smoothness (Say $C^2$). Let $\xi:\mathbb R\to[0,\infty)$ with compact support $[-r,r]$ for some $r>0$. Let $\phi_t(x)$ denote the flow of the given differential equation. I wan to find the following derivative.

$$ \frac{d}{dx}(\xi(\phi_t(x)) \tag{A}$$

Here is what I did. By chain rule we get \begin{align} \frac{d}{dx}(\xi(\phi_t(x)) &=\xi'(\phi_t(x)\frac{\partial}{\partial x}(\phi_t(x))) \quad \text{by chain rule}\\ &=\xi'(\phi_t(x)\frac{d}{dt}\left(\phi_t(x) \right)\cdot \frac{dt}{dx} \tag{*}\\ &=\xi'(\phi_t(x)(f(\phi_t(x)) \frac{1}{\dot x} \tag{**} \end{align}

Question? I am not convinced about the lines denoted as (* ) and (**) above. Can anybody please show me how this derivative can be found. May be I am missing some property of the flow of a differential equation. (I know that for a flow $\phi_t(x)$ of a differential equation that $\dot \phi_t =f(\phi_t)$)

Some context I eventually want to be able to compute the derivative of the following integral. $$ \eta (x) =\int_0^\infty e^{at} \xi(\phi_t(x)) dt$$ for a given $f$. Since differentiation under the integral sign can be easily justified this shouldn't be a problem if I can somehow find the derivative in $(A)$

Addendum What if $f(x)=-ax+\xi(x)?$. Does that make any difference as far as $\partial \phi_t/\partial x $ not being able to be simplified?. If $f(x)$ is the given function then I know some properties of $\phi_t$. For example the origin is asymptotically stable and so $\phi_t(x) \to 0$ as $t\to \infty$. I can also assume $\xi(0)=\xi'(0)=0$ as well. By continuous dependance theorem I can also safely say that $\phi_t$ is sufficiently smooth.

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  • $\begingroup$ Note that $x$ is the initial condition of the system, and you want to compute the derivative of the solution at time $t$ with respect to this initial condition. Th computation of the derivative is a little more involved than the above. $\endgroup$ – copper.hat Aug 26 '14 at 4:20
  • $\begingroup$ Yes, this is something that has always bothered me. Is $x$ fixed or not?. Why do they use the same letter $x$ in the differential equation and for the initial condition?. I am sure there is a good reason, but I am completely confused now. $\endgroup$ – minibuffer Aug 26 '14 at 4:34
  • $\begingroup$ (*) is incorrect. Think of $\phi_t(x)=\phi(t,x)$ as a function of two variables. You want the partial derivative with respect to $x$. $\endgroup$ – user138530 Aug 26 '14 at 4:35
  • $\begingroup$ @JackDawkins: It is a little confusing. I prefer to use a different letter to distinguish the initial condition from the solution at some time $t$ (clearly they are the same for $t=0$). $\endgroup$ – copper.hat Aug 26 '14 at 4:38
  • $\begingroup$ $\partial \phi_t/\partial x$ can't really be simplified. You can characterize it as the solution of another ODE, which you obtain by differentiating the original equation. $\endgroup$ – user138530 Aug 26 '14 at 4:39
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Remember that $\phi_t(x)$ is the solution at time $t$ to the system $\dot{y} = f(y)$ with initial condition $y(0) = x$.

The following skips a lot of details, but hopefully gives some idea of one approach to 'computing' the derivative you required.

To compute ${\partial \phi_t(x) \over \partial x}$, consider the system $L(z,x) = 0$, where $L(z,x)(t) = x+ \int_0^t f(z(\tau)) d \tau -z(t)$ and $L : C^n[0,t] \times \mathbb{R}^n \to C^n[0,t]$.

Some work shows that $({ \partial L(z,x) \over \partial z}h) (s) = \int_0^s { \partial f(z(\tau)) \over \partial y} h(\tau) d \tau-h(s)$ (with $ s \in [0,1]$), some more work shows that the map $D h = { \partial L(z,x) \over \partial z}h$ is continuous, and that the equation $D h = b$ is invertible (with $b \in C^n[0,t]$), and so the operator ${ \partial L(z,x) \over \partial z}$ is invertible.

It is straightforward to see that $({ \partial L(z,x) \over \partial x}\delta)(s) = \delta$ (with $ s \in [0,1]$), and applying the implicit function theorem (see Kantorovich & Akilov, "Functional analysis", for example) to $L(z,x) = 0$ gives the existence of some $\zeta$ such that $L(\zeta(a), a) = 0$ for $a$ in some neighbourhood of $x$. Furthermore, ${ \partial L(z,x) \over \partial z} { \partial \zeta(x) \over \partial x} = -{ \partial L(z,x) \over \partial x}$, and so, for any $\delta \in \mathbb{R}^n$, we have $({ \partial L(z,x) \over \partial z} ({ \partial \zeta(x) \over \partial x} \delta))(s) = -\delta$.

Unwinding this, and letting $\eta = { \partial \zeta(x) \over \partial x} \delta $, gives $\eta$ as the solution to the equation $\int_0^s { \partial f(z(\tau)) \over \partial y} \eta(\tau) d \tau-\eta(s) = -\delta$, or in other words, $\eta$ solves $\dot{\eta}(s) = { \partial f(z(s)) \over \partial y} \eta(s)$, with $\eta(0) = \delta$.

Suppose $\Phi(t_1,t_2)$ is the state transition matrix for $\dot{\eta}(s) = { \partial f(z(s)) \over \partial y} \eta(s)$, then we have ${ \partial \phi_t(x) \over \partial x} \delta = \eta(t) = \Phi(t,0) \delta$, the solution of the 'linearised' differential equation at time $t$.

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  • $\begingroup$ Thanks for the hints. What I wanted to show was $\eta(x)$ was $C^1$. May be there is an indirect way to show this without actually 'computing' the derivative under the integral sign. $\endgroup$ – minibuffer Aug 27 '14 at 0:51
  • $\begingroup$ Well, $C^1$ will follow from the continuity of the partials of $L$. $\endgroup$ – copper.hat Aug 27 '14 at 4:11
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Your confusion comes from the abuse of notation. The derivative is taken with respect to the initial value $x(0)$, rather than position $x(t)$. The integral is ought to be $$\eta(x(0))=\int_0^\infty e^{at}\xi(\phi_t(x(0)))dx$$ But it looks rather tedious and hence ignore the parenthesis. To make a clear statement, avoiding $x$ explicitly, let's assume $x(t)=\phi_t(p)$, with $x(0)=\phi_0(p)=p$, then we can write $$\frac{d}{dp}\xi(\phi_t(p))=\frac{d\xi(\phi_t)}{d\phi_t}\frac{d\phi_t}{dp}$$

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