Single-step autonomous delay differential equation - Revision history

Vipul: /* Time translation invariance of solutions */

2012-07-16T01:40:19Z

Time translation invariance of solutions

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	===Time translation invariance of solutions===		===Time translation invariance of solutions===

	If <math>x(t)</math> is a solution function on an interval, then <math>t \mapsto x(t - \alpha)</math> is a solution function on the right translate of the interval by <math>\alpha</math>.		If <math>x(t)</math> is a solution function on an interval, then <math>t \mapsto x(t - \alpha)</math> is a solution function on the right translate of the interval by <math>\alpha</math>, for any fixed <math>\alpha \in \R</math>. This has specifically to do with the differential equation being autonomous.

	==Examples==		==Examples==

Vipul: /* Facts */

2012-07-16T01:39:37Z

Facts

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	Even if the initial value specification is an infinitely differentiable function, it is likely that when we extend it using the method of steps, the solution will have nice differentiability properties within each interval of length <math>\tau</math>, but not ''at'' the endpoints shared by the interval.		Even if the initial value specification is an infinitely differentiable function, it is likely that when we extend it using the method of steps, the solution will have nice differentiability properties within each interval of length <math>\tau</math>, but not ''at'' the endpoints shared by the interval.

			===Time translation invariance of solutions===

			If <math>x(t)</math> is a solution function on an interval, then <math>t \mapsto x(t - \alpha)</math> is a solution function on the right translate of the interval by <math>\alpha</math>.

	==Examples==		==Examples==

Vipul: /* Solution method: moving forward */

2012-07-16T01:37:29Z

Solution method: moving forward

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	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.		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.

			'''NOTE''': The method of steps also works for single-step ''non''-autonomous first-order first-degree delay differential equations, where <math>f</math> is replaced by a function that also depends on <math>t</math>. The difference now is that the solutions are no longer invariant under time translations. For a number of reasons, the autonomous case is encountered much more frequently than the non-autonomous case.

	==Facts==		==Facts==

Vipul: /* Solution method: moving backward */

2012-07-09T01:47:37Z

Solution method: moving backward

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	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.		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.

	==Facts==		==Facts==

Vipul: /* Definition */

2012-07-09T00:38:04Z

Definition

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	where <math>f</math> is a known function and <math>\tau > 0</math> is also known.		where <math>f</math> is a known function and <math>\tau > 0</math> is also known.

			===Solution concept===

			An explicit functional solution is a function <math>x(t)</math> that satisfies the following condition:

			* If the domain is an interval bounded from below: The delay differential equation's interval of validity starts at a distance <math>\tau</math> after the beginning of the interval of definition of the function. In other words, if the function is defined on an interval <math>[r,s]</math>, the delay differential equation needs to be valid only on <math>[r + \tau,s]</math> (with one-sided derivative used at the upper endpoint).
			* If the domain is an interval not bounded from below: In this case, the delay differential equation has to be valid everywhere on the domain of the function.


	===Nature of initial value specification===		===Nature of initial value specification===

Vipul: /* Nature of initial value specification */

2012-07-09T00:34:19Z

Nature of initial value specification

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	===Nature of initial value specification===		===Nature of initial value specification===

	~~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.		If we are looking for solutions that are once differentiable, and are not insisting on higher order differentiability, the initial value specification is as follows: it is a description of <math>x</math> as a continuous function of <math>t</math> on an interval of length <math>\tau</math>, chosen as the left-most interval of the domain where we want the function to be defined. Note that if the left-most interval is <math>[a - \tau,a]</math>, then the delay differential equation becomes active only on <math>[a,\infty)</math>.

			Further, the initial value specification must satisfy the additional condition that the [[left-hand derivative]] of <math>x</math> at <math>a</math> is <math>f(x(a),x(a - \tau))</math>.

	===Solution method: moving forward===		===Solution method: moving forward===

Vipul: /* Examples */

2012-07-08T16:29:13Z

Examples

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	<math>x(t) = \left\lbrace \begin{array}{rl} t, & t \in [0,1] \\ -t + 2e^{t-1}, & t \in [1,2] \\ t + (2t - 8)e^{t-2} + 2e^{t-1}, & t \in [2,3]\end{array}\right.</math>		<math>x(t) = \left\lbrace \begin{array}{rl} t, & t \in [0,1] \\ -t + 2e^{t-1}, & t \in [1,2] \\ t + (2t - 8)e^{t-2} + 2e^{t-1}, & t \in [2,3]\end{array}\right.</math>

			Note that the function has a piecewise definition, and at each transition point, it is <math>C^1</math> (i.e., [[continuously differentiable function\|continuously differentiable]]) but not twice differentiable.

Vipul: /* Examples */

2012-07-08T16:27:23Z

Examples

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	<math>\psi(t) = t + (2t - 8)e^{t-2} + 2e^{t-1}</math>		<math>\psi(t) = t + (2t - 8)e^{t-2} + 2e^{t-1}</math>

			So:

			<math>\psi(t) = t + (2t - 8)e^{t-2} + 2e^{t-1}, \qquad t \in [2,3]</math>

			Overall, we have:

			<math>x(t) = \left\lbrace \begin{array}{rl} t, & t \in [0,1] \\ -t + 2e^{t-1}, & t \in [1,2] \\ t + (2t - 8)e^{t-2} + 2e^{t-1}, & t \in [2,3]\end{array}\right.</math>

Vipul: /* Facts */

2012-07-08T16:24:24Z

Facts

← Older revision		Revision as of 16:24, 8 July 2012
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	Even if the initial value specification is an infinitely differentiable function, it is likely that when we extend it using the method of steps, the solution will have nice differentiability properties within each interval of length <math>\tau</math>, but not ''at'' the endpoints shared by the interval.		Even if the initial value specification is an infinitely differentiable function, it is likely that when we extend it using the method of steps, the solution will have nice differentiability properties within each interval of length <math>\tau</math>, but not ''at'' the endpoints shared by the interval.

			==Examples==

			Consider the delay differential equation:

			<math>\frac{d}{dt}(x(t)) = x(t) + x(t - 1)</math>

			Suppose that we are given that <math>x(t) = t</math> on <math>[0,1]</math>. We note that this initial value specification is consistent because the derivative <math>x'(1)</math> equals <math>x(1) + x(1 - 1)</math>.

			We use the method of steps. Our first goal is to determine <math>x</math> on <math>[1,2]</math>. Explicitly, we are trying to find a function <math>\psi</math> on <math>[1,2]</math> such that:

			<math>\frac{d}{dt}(\psi(t)) = \psi(t) + (t - 1)</math>

			The differential equation with dependent variable <math>\psi</math> and independent variable <math>t</math> is:

			<math>\psi' - \psi = t - 1</math>

			This is a [[linear differential equation]]. The general solution would be:

			<math>\psi(t) = -t + Ce^t</math>

			We must choose <math>C</math> such that <math>\psi(1) = \varphi(1) = 1</math>, so <math>C = 2/e</math>. We get:

			<math>\psi(t) = -t + 2e^{t-1}</math>.

			Thus, we have:

			<math>x(t) = -t + 2e^{t-1}, \qquad t \in [1,2]</math>

			We can now do a similar procedure to find what <math>x(t)</math> looks like in <math>[2,3]</math>. Note that we will still get a linear differential equation but with a new particular solution:

			<math>\psi' - \psi = 1 - t + 2e^{t-2}</math>

			The general solution is:

			<math>\psi(t) = t + 2te^{t-2} + Ce^t</math>

			Plugging in that <math>\psi(2) = x(2) = 2e - 2</math>, we get <math>C = (2e - 8)/e^2</math>, so we get:

			<math>\psi(t) = t + (2t - 8)e^{t-2} + 2e^{t-1}</math>

Vipul: /* Definition */

2012-07-08T16:09:20Z

Definition

← Older revision		Revision as of 16:09, 8 July 2012
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	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.		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.

			==Facts==

			===Expect piecewise definitions for solutions===

			Even if the initial value specification is an infinitely differentiable function, it is likely that when we extend it using the method of steps, the solution will have nice differentiability properties within each interval of length <math>\tau</math>, but not ''at'' the endpoints shared by the interval.