[PDF] Some Elementary Methods for Solving Functional Differential

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Some Elementary Methods for Solving Functional

Clement E. Falbo, Sonoma State University

Abstract

methods that use only the basic techniques taught in a ...rst course of Ordinary DDEs and FDEs are often used as modeling tools in several areas of ap- plied mathematics, including the study of epidemics, age-structured population growth, automation, tra¢ c ‡ow and problems related to the engineering of high-rise buildings for earthquake protection. We discuss the solution of constant coe¢ cient-DDEs by the "Method of Characteristics," and we show how to solve more general DDEs using Myshkis" "Method of Steps." This method is one of those "natural" procedures that are often repeatedly discovered by workers in the ...eld; it is easy to understand and "reverse time" FDEs that have interesting applications in biology. DDEs are well known to be useful in various ...elds such as age-structured popu- lation growth, control theory, and any model involving responses with non-zero delays; these include models of conveyor belts, urban tra¢ c, heat exchangers, robotics, and chatter. Chatter is described by Asi and Ulsoy [1] as "...a self- excited vibration, which is the result of an interaction between the tool structure and the cutting process dynamics." Driveret al.[5], discuss mixture problems that do not assume instantaneously perfectly mixed solutions. Applications to delays in transportation, signal transmission, genetic repression, control systems for nuclear reactors with time delays, and other diverse topics may be found in

Hale and Lunel [9], and in Wu [17].

The general ...rst order DDE has the form:y0(t) =f(t;y(t);y(td));for some givend >0:Here,y0(t)depends on the value ofyat some timetdin the past, as well as depending upon the current value ofy, and other functions oftas de- termined byf:Applications are discussed in the following additional references: Bender and Neuwirth[3], Insperger and Stepan[11], Nesbit[14], Asi[1].

1.1 Linear DDEs

y

0(t) =a1(t)y(t) +a2(t)y(td);fort0:(1)

1 Equation (1) is usually accompanied by an auxiliary condition, stated in terms of a function revealing the state of the system for a period prior to the initial time,t= 0. The auxilliary function is sometimes called the "history" function for the system, but as we shall see later, it is really a "remote control" function, describing the behavior ofyon a remote time-interval other than the of the ...rst order delay system, letd >0, anda1;a2be classC1functions on[0;d] and letp(t)be a classC1function on[d;0];then there is a unique function y(t)satisfying the system: y

0(t) =a1(t)y(t) +a2(t)y(td);fort2[0;d](2)

y(t) =p(t);fort2[d;0](3) It is quite easy to prove that the system (2),(3) cannot have more than one solution. Simply assume two solutionsu(t)andy(t), and letg(t)be their a

2(t)g(td);on[0;d]. But sinceuandyboth satisfy the same auxilliary

equation, (3) becomesg(t)0fort2[d;0];makingg(td)0fort2[0;d]. The resulting ...rst order ODE systemg0(t) =a1(t)g(t)on[0;d], withg[0] = 0 provesg(t)0on[0;d];as well as on[d;0], thusu(t)y(t)on[d;d]: The system (2),(3) has a solution, which we show by actually getting one. Of the several methods used for solving this system, we will discuss two -both a course in numerical analysis.

The Method of Characteristics (MOC)

The Myshkys Method of Steps (STEPS)

Turning ...rst to the MOC, we ...nd it is suitable for solving the simplest case of equation (2), namely the constant-coe¢ cient equation y

0(t) =a1y(t) +a2y(td)(4)

with botha1anda2constant. One of the bene...cial consequence of using the MOC is that we are introduced to an interesing function called the Lambert w-function,W(z);namely the inverse of the equation z(w) =wew(5)

1.1.1 Method of Characteristics

We will assume that the solution to (4) has the form ofy(t) =emt, for some constantm;(real or complex). So, we sety(t) =Cemt;withCarbitrary, getting Cme mt=a1Cemt+a2Cemtmd(6) Division byCemtreduces (6) to the characteristic equation 2 (ma1)emda2= 0(7) Notice that whena2= 0; m=a1;giving usy(t) =Cea1t, the solution to the ODEy0(t) =a1y(t), which is Equation (4), witha2= 0:On the other hand, in the pure delay equation, wherea1= 0, buta26= 0;the characteristic equation becomes me md=a2 Multiplying bydconverts this to the inverse of the Lambert function mde md=a2d(8)

That is,md=W(a2d);thusm=1d

W(a2d): W(a2d)has either no, one or

two real roots depending upon whethera2is negative and<;=, or>1de and it has only one real root ifa2>0:Under any of these conditionsW(a2d)has in...nitely many complex roots,rk+isk. Thus, the MOC produces the following solution to Equation (4) witha1= 0: y(t) =c1em1t+c2em2t+1X k=1e rktc(1;k)cos(skt) +c(2;k)sin(skt)(9) withc1orc2zero or not zero according to the above restrictions ona2and1de

For a derivation of this, see my paper, Falbo[6].

In order to determine the coe¢ cientsc1;c2;c(1;k);c(2;k);we could use the history functionp(t)and its derivatives, or we could usep(t)to create a least To solve Equation (4) where botha1anda2are not zero,we simply notice that a change in variables in Equation (7) can be used to reduce the equation to the form of Equation (8).Viz, letma1=n, then from Equation (7), we get ne nd+da1a2= 0 So, ne nd=a2eda1

Multiply through byd

nde nd=a2dea1d(10) Usingnd=W(a2dea1d);we get the real and complex roots of Equation (10). Divide bydto getn, then adda1to get the real and complex valuesm satisfying the characteristic equaton (7). It should be noted that the MOC can be applied to a higher order constant coe¢ cient linear DDE. This is done in the usual way of writing a linear nth 3

1.1.2 Method of Steps

The method of steps is much more intuitive and can be used to solve DDEs with variable coe¢ cients. Although this method may have been discovered many times by several workers, we will cite as our reference, Myshkis [13] from the

Soviet Encyclopeadia of Mathematics.

This method converts the DDE on a given interval to an ODE over that interval, by using the known history function for that interval. The resulting equation is solved, and the process is repeated in the next inerval with the newly found solution serving as the history function for the next interval. We will show how to apply it to the system of equations (2) and (3). Step 1On the interval[d;0], the functiony(t)is the given functionp(t), soy(t)is known there. Thus, we say the equation is "solved" for the interval [d;0], call this solutiony0(t). Note 1:Whent2[0;d],td2[d;0], soy(td)becomesy0(td)on [0;d]: Step 2In the interval[0;d], the system (2),(3) becomes y

0(t) =a1(t)y(t) +a2(t)y0(td), on[0;d](11)

y(0) =p(0) Equation (11) is an ODE andnota delay equation becausey0(td)isknown; it is simplyp(td). Thus, we solve this ODE on[0;d], usingy(0) =p(0)as our initial condition. Denote byy1(t)this solution on the interval[0;d]: Note 2:Solving Equation (11) may be accomplished by treating it as the nonhomogeneous equation, y

0(t)a1(t)y(t) =a2(t)p(td);on[0;d]

y(0) =p(0) as an Integrating FactorIF=eRa1(t)dt;for a closed form solution, or by numerical methods, for an approximate solution.

Step 3On the interval[d;2d], the system becomes

y

0(t) =a1(t)y(t) +a2(t)y1(td), on[d;2d](12)

y(d) =y1(d) which is again an ODE. We solve this, using the initial condition atdand get a solutiony2(t)for our system on[d;2d]. These steps may be continued for subsequent intervals. 4

Figure 1:Solution to (13)

EXAMPLE 1

Problem. Find one step of the solution to the system (2)-(3) for the following data d= 5 p(t) =dt(t+d) a 1=1 a

2= 0:5:

Solution. Assigning these values, we get

y

0(t) =y(t) + 0:5y(td), on[0;5](13)

y(t) =dt(t+d)on[5;0] Now replacey(td)byp(td);and use the history function to get the initial conditiony(0) =p(0). On the ...rst interval the solutiony1(t)will be the function satisfying y

0(t) =y(t) + 0:5p(td)

y(0) =p(0) Whose solution we can obtain directly or numerically; its graph is shown in

Figure 1.

The error graph for this numerical solution is given in Figure 2.

1.2 The General First Order DDE

written y

0(t) =f(t;y(t);y(td))fort2[0;d](14)

y(t) =p(t);fort2[d;0] with appropriate conditons forfandp:Just as in the linear cases described in he preceding section, STEPS works here as well. 5

Figure 2:Error for Figure 1

Step 1.On the interval[d;0]the functiony(t)is the given functionp(t); this isy0(t). Step 2.In the interval[0;d];the system (14) becomes y

0(t) =f(t;y(t);y0(td))on[0;d](15)

y(0) =p(0) Equation 15 is an ODE andnota delay equation becausey0(td)is known; it is simplyp(td)fort2[0;d]. Solving (15), we obtain a solution on[0;d], call ity1(t). On the next interval,[d;2d], we solvey0(t) =f(t;y(t);y1(td)), withy(d) =y1(d), etc. Here is why STEPS works in these nonlinear DDEs. The equation is de...ned on some given domain,[0;d]but the expressiony(td)is reading its values from another domain,[d;0]And on[0;d],y(td)is known because for allt in[d;0],y(t) =p(t):. Therefore in[0;d], we can replacey(td)withp(td). The process then can be continued for the next step. This is because the solution on[0;d]becomes the known function that replacesy(td)in the next interval. a term in which the "unknown" functionyis given for values in some other domain. In such a case, the expression foryon the remote domain becomes a known function. Such a remote domain need not be the interval immediately preceding the current one. In which case it may not be possible to compute a second step. Nevertheless we can apply STEPS, at least once, to many FDE that are not just DDEs. We will discuss some of these in the next section.

2 Remote Control Dynamical Systems

If you send a signal to a robot telling it to turn, stop, or perform some other task, there will be some lag between the time you initiate the signal and the 6 time the robot responds. It takes another delay for you to see what the robot did and then to make use of this feedback to in‡uence your next decision about what new signal to send. For another example, if you are trying to row a boat you may push your oar through the water and then wait to see the heading of the boat before dipping the oar again. However, if you are heading for a dangerous obstacle, you may not wait after each stroke, but simply decide to execute a series of pre-planned back strokes before getting feedback. Typically, controls are not sent as indiviudal signals, one at a time, but rather as a pre-set pattern, a template. Almost any "automated" process works from a template. For instance, a pre-determined design can be programmed into a weaving machine to produce a desired pattern in a rug. The methods we have already discussed may be used to solve equations employed in modelling a dynamic process with a given pre-set template, or control function The control function need not be simply apastaction, but can express a desired future goal or target. It may not even be related to the remote action in the usual sense of "time." Speci...cally, we want to solve a Remote Control Dynamical System (RCDS) which is de...ned as one whose dynamical equation is the FDE y

0(t) =f(t;y(t);y(h(t))); t2 I(16)

whereI, is an open interval called theoperational interval. Thedeviating argument,h(t)2 C1[I];is one whose range,h[I], (called theremote domain orremote interval) is disjoint fromI. Initially, the system is assumed to be subject to acontrol function,p(t)2 C1[h[I]]de...ned on the remote domain. Thus, theoutput function,y(t);is the solution of Equation (16) onI, andy(t) =p(t)onh[I]. Whend >0and the deviating argumenthis de...ned byh(t) =tdwe get the DDE. The functiontdis called theretarded argument.Ifh(t) =t+d the equation is called anadvanced argument. If, the derivative is also of the formy0(td);the equation is calledneutral. Otherwise, some possible deviating arguments are: accelerated delaysh(t) =mtd;wherem >0;or nonlinear delays,h(t) =g(t)d;ifg(t)2 C1[I]andg(t)d =2 I. Several applications that involve ...rst and second order dynamical systems using delay equations with feedback, can be found in Olgaket al.[15].An interesting version of this type of feedback, balancing an inverted pendulum, is discussed in Atay [2]. Another application using acceleration feedback in- cludes connecting two buildings together by a shock absorber in order to reduce earthquake damage, Christenson and Spencer [4].

EXAMPLE 2

Problem: Using the following data

d= 5 a 1=1 a

2= 0:5

h(t) =t+d 7

Figure 3:Solution to (17)

Figure 4:Error graph for the solution to (17)

p(t) =d(td)(t2d) Solve on the interval[0;d];the following remote control problem whch re- quires that the solutionyon[0;d], take on the values ofp(t)in the future interval[d;2d]: y

0(t) =a1y(t) +a2y(h(t));on[0;d]

y(t) =p(t);on[d;2d](17) Solution: We use the ...rst step of the method of STEPS, by replacingy(h(t)) on[0;d]by its known valuep(h(t))on the intervalh[0;d]which is[d;2d], and we get y

0(t) =a1y(t) +a2p(t+d)

y(d) =p(d)(18)

The graphical solution is shown in Figure 3.

A plot of the expressiony0(t)a1y(t)a2y(h(t))is shown in Figure 4. This represents the error in solving (17). The following example shows that we can have an in...nite remote domain. 8

Figure 5:Solution to the system (19)

EXAMPLE 3

y

0(t) = 1:5y(t2)for allt2[0;1)

y(t) =1t1for allt2(1;0](19) In Figure 5 the graphical solution to the system (19) is shown on[30;30], not(1;1), but close enough!

2.0.1 A Time reversal Problem

In an experiment measuring the population growth of a species of water ‡eas, Nesbit [14], used a DDE model in his study. In simpli...ed form his population equation wasN0(t) =a1N(td) +a2N(t). He ran into di¢ culty with this model because he did not have a reasonable history function to carry out the solution of this equation. To overcome this roadblock he proposed to solve a "time reversal" problem in which he sought the solution to an FDE that is neither a DDE, nor a RCDS. He used a "time reversal" equation to get the juvenile population prior to the beginning timet= 0. We express this as p

0(t) =ap(t) +bp(t)

p(0) =c(20) Wherecis the juvenile population at timet= 0. The equation also tells us thatp0(0) = (a+b)c: At ...rst glance, it may seem that it is madness to try to solve such an equation. It turns out, however, that this equation can be solved and it can be shown to have a unique solution. The fundamental form of the solution is p(t) =Aert+Bert(21)

Wherer=pb

2a2;and is real or imaginary according to whetherb2a2or

9 The time reversal problem in the preceding section is a special case of a type of equations of the form y

0(t) =f(t;y(t);y(u(t))

y(t0) =y0(22) Whereu(t)is idempotent, that isu(u(t)) =t, andt0is a ...xed point ofu. For details see my paper, Falbo[7]. In Example 4, below, we consider the simplest IDE, one in which the deviat- ing argument isu(t) =dt:This function is idempotent sinceu(u(t)) =u(dt), which isd(dt) =t:Notedtis not the "delay" functiontd:

EXAMPLE 4

Problem:

Ifd >0, andaandx0are any real numbers, ...nd a functionx(t)2 C1on some open intervalIcontainingd=2that satis...es the system x

0(t) =ax(dt)

x(d=2) =x0(23) Although Equation(23) is ...rst order and linear, it has the following periodic function as its only solution. x(t) =x0cos(atad2 ) +x0sin(atad2 )(24) It is easy to show by substitution that (24)isa solution to the system (23), but how do we know it is theonlysolution? We will take the time to show that it has only one solution because in proving uniqueness, we will also introduce a method for getting the solution. Suppose that there is a "second" solutiony(t)2 C1that satis...es (23) and its initial condition, then y

0(t) =ay(dt)

y(d=2) =x0(25) y

00(t) =ay0(dt)(26)

Now, for some positive number;ift2[d2

;d2 +], thendtis in this same interval. Thus,y0(dt)exists and by equation (25), it is equal toay(t). Substituting this into Equation (26) and noting thaty0(d=2) =ay(dd=2) = ax

0,we get

y

00(t) =a2y(t)(27)

y(d=2) =x0 y

0(d=2) =ax0

10 solution, namely the one de...ned in Equation (24). Thus, this supposedlyquotesdbs_dbs47.pdfusesText_47

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