Single-step autonomous delay differential equation: Difference between revisions

Revision as of 16:06, 8 July 2012

Definition

Form of the differential equation

This is a particular type of first-order first-degree autonomous delay differential equation, given explicitly as:

$\frac{d (x (t))}{d t} = f (x (t), x (t - τ))$

where $f$ is a known function and $τ > 0$ is also known.

Nature of initial value specification

The initial value specification for this type of delay differential equation is a description of $x$ as a function of $t$ on an interval of length $τ$ , typically the left-most such interval in our domain.

Solution method: moving forward

The solution method is called the method of steps. The idea is that, if the function is known on an interval of the form $[a - τ, a]$ , we can figure out what it is on $[a, a + τ]$ , and then repeat the process to determine what the function is on $[a + τ, a + 2 τ]$ , and continue to proceed in this way to determine the function everywhere on $[a, \infty)$ .

Let us say that we know that $x (t) = φ (t)$ on the interval $[a - τ, a]$ . Then, $x (t)$ is the solution $ψ (t)$ to the following equation on $[a, a + τ]$ subject to the condition $ψ (a) = φ (a)$ :

$\frac{d}{d t} (ψ (t)) = f (ψ (t), φ (t - τ))$

This is an ordinary first-order first-degree differential equation in $ψ$ with an initial-value specification, so we expect it to have a unique solution.

Solution method: moving backward

We can also do a similar process to move backward. Explicitly, suppose $x (t) = ψ (t)$ on an interval of the form $[a, a + τ]$ . We want to find out what it looks like on $[a - τ, a]$ . We set $x (t) = φ (t)$ on this interval, and we want to solve the following for $t \in [a - τ, a]$ subject to the initial value condition $φ (a) = ψ (a)$ :

$ψ^{'} (t + τ) = f (ψ (t + τ), φ (t))$

Note that this is just an equation in $φ$ without derivatives, i.e., it is an ordinary equation (a zeroth-order differential equation). However, depending on the nature of $f$ , $ψ$ , and $ψ^{'}$ , we may have difficulty getting an explicit functional form for $φ$ , and it may be far from unique. Thus, unlike forward motion, which we expect to be uniquely determined by the initial value specification, backward motion may not be uniquely determined.

@@ Line 13: / Line 13: @@
 The ''initial value'' specification for this type of delay differential equation is a description of <math>x</math> as a function of <math>t</math> on an interval of length <math>\tau</math>, typically the left-most such interval in our domain.
-===Solution method===
+===Solution method: moving forward===
 The solution method is called the ''method of steps''. The idea is that, if the function is known on an interval of the form <math>[a - \tau,a]</math>, we can figure out what it is on <math>[a,a+\tau]</math>, and then repeat the process to determine what the function is on <math>[a+\tau,a+2\tau]</math>, and continue to proceed in this way to determine the function everywhere on <math>[a,\infty)</math>.
@@ Line 21: / Line 21: @@
 <math>\frac{d}{dt}(\psi(t)) = f(\psi(t),\varphi(t - \tau))</math>
-This is an ''ordinary'' [[first-order first-degree differential equation]] in <math>\varphi</math> with an initial-value specification, so we expect it to have a unique solution.
+This is an ''ordinary'' [[first-order first-degree differential equation]] in <math>\psi</math> with an initial-value specification, so we expect it to have a unique solution.
+===Solution method: moving backward===
+We can also do a similar process to move ''backward''. Explicitly, suppose <math>x(t) = \psi(t)</math> on an interval of the form <math>[a,a+\tau]</math>. We want to find out what it looks like on <math>[a-\tau,a]</math>. We set <math>x(t) = \varphi(t)</math> on this interval, and we want to solve the following for <math>t \in [a - \tau,a]</math> subject to the initial value condition <math>\varphi(a) = \psi(a)</math>:
+<math>\psi'(t + \tau) = f(\psi(t + \tau),\varphi(t))</math>
+Note that this is just an equation in <math>\varphi</math> without derivatives, i.e., it is an ordinary equation (a zeroth-order differential equation). However, depending on the nature of <math>f</math>, <math>\psi</math>, and <math>\psi'</math>, we may have difficulty getting an explicit functional form for <math>\varphi</math>, and it may be far from unique. Thus, unlike forward motion, which we expect to be uniquely determined by the initial value specification, backward motion may not be uniquely determined.