Discrete (Difierence) Equations - School of …

1 Discrete (Difference) EquationsA Discrete , or difference, equation expresses a relationship between the elements of a sequence,{yn},wheren N0 {0,1,2, ..}. For example, the trivial difference equationyn+1=yn,(1)has the solutionyn=y0, which means that the sequence{yn}may be any constant mathematical models are posed in the form of Discrete Equations . The (in)famous logisticmapyn+1=ryn(1 yn),(2)was introduced as a model for the growth of a population. Here,yn [0,1] represents the (scaled)population in then-th year,y0is the initial population; andr >0 is a combined birth and Equations also arise when solving continuous models using numerical methods, a nec-essary task for all but the most simple models.

2 Computers can only work with Discrete data, socontinuous Equations must be discretised1before they can be solved continuous differential equationdydx= 0;(3)has the simple solutiony=y(0), a constant function. A simple discretisation method is to usefinite differences and approximate the derivative by the expressiondydx(x) y(x+h) y(x)h,wherehis , we define the sequence{yn}, such thaty0=y(0),y1=y(h),y2=y(h+h) =y(2h), ..yn=y(nh). Thus,dydx(nh) y(nh+h) y(nh)h=yn+1 Discrete approximation to the differential equation (3) is, therefore,yn+1 ynh= 0 yn+1 yn= 0 yn+1=yn,our original example (1).

3 In this simple example, the solutions to the Discrete equation and contin-uous equation are exactly the same and independent ofh, why?In fact, there is a strong connection between difference Equations and differential a differential equation can always be approximated by a difference equation, can you seehow?1 Discretisation is the general term used to describe the procedure of converting a continuous system into a set ofdiscrete linear difference Equations with constant coefficientsA generalm-th order difference equation with constant coefficients can be written in the formamyn+m+am 1yn+m 1+ +a1yn+1+a0yn=rn,(4)where{ak},k 0.

4 , mare constants, andrn=r(n) is an arbitrary function. Note that the axishas been re-scaled so that the distance between points in the sequence{yn}ish= 1, the standardconvention when working directly with difference solution of equation (4) can be split into two separate procedures: Solution of the homogeneous equation (rn= 0) Calculation of a particular solution(any function that satisfies the entire inhomogeneous equation)The solution of the equation is thesumof the particular solution and the solution of thehomogeneous equation, known as the complementary function.

5 Note that this linear superposition(adding the two solutions together) is only possible because the difference equation is equations1st orderA homogeneous difference equation is one in which the right-hand side is zero,i.. 0, inequation (4). Initially, the simplest case to consider is a first-order, linear , homogeneous differenceequation with constant coefficients,yn+1+pyn= 0,(5)wherepis a find the solution by an inductive procedure; rearranging equation (5) givesyn= pyn 1,(6)and because the equation (6) must be true for anyn(>1), thenyn 1= pyn 2,provided thatn >2.

6 (7)Combining Equations (6) and (7) givesyn= ( p)2yn 2= = ( p)ny0;we have continued the inductive procedure until we reachedy0, which must be given as an initialcondition Thus, the fundamental solution of the equation (5) isy0Cn, whereCis a constant thatis the solution of the characteristic (or auxiliary) equationC+p= orderOur result may be generalised to homogeneous difference Equations of any order:amyn+m+am 1yn+m 1+ +a1yn+1+a0yn= 0.(8)The equation (8) can only be satisfied ifyn+1is equal toynmultiplied by a constant,yn+1=Cyn;in which case, each term in the equation is equal to any other term multiplied by a constant,e.

7 +2=Cyn+1=C2yn, etc. Note that the constant will not necessarily be the same for each pair ofterms. The condition is merely equation(5) in another guise, however, and we have already shownthat the solution has the formyn=Cn. On substitution of this result into equation (8), we obtainamCn+m+am 1Cn+m 1+ +a1Cn+1+a0Cn= though byCn,2we obtain the characteristic equationamCm+am 1Cm 1+ +a1C+a0= characteristic equation is anm-th order polynomial and will havemroots, each of whichrepresents a different solution of the difference equation.

8 The general solution may by found by alinear combination of thesemsolutions:yn=A1Cn1+A2Cn2+ +AmCnm,where{Ak}are constants and{Ck}are the roots of the characteristic equation. Note that calcula-tion of the constants{Ak}requiresmboundary will see in MT1222/MT1232 that the same general philosophy can be used to deduce thegeneral solution of a linear homogeneous differential equation with constant +am 1dm 1ydxm 1+ +a1dydx+a0y= 0.(9)In this case the condition to be satisfied is thatdydx=Cy, which implies thaty= eCx. Substitutionof this form into equation (9) gives the resultamCmeCx+am 1Cm 1eCx+ +a1 CeCx+a0eCx= now divide through by eCxto obtain the same characteristic equation as in the case of thedifference equation:amCm+am 1Cm 1+ +a1C+a0= 0,but the general solution of the differential equation isy(x) =A1eC1x+A2eC2x+ + note that the trivial sequenceyn= 0 is always a solution of a homogeneous equation, and we assume thatC6= 0, in order to find a non-trivial Express the series{5,10,20,40.}

9 }as a difference equation and solve it to find an explicitexpression difference equation isyn+1= 2yn yn+1 2yn= characteristic equation is, therefore,C 2 = 0, and so the general solution isyn=A12nThe initial conditiony0= 5, can be used to findA1= 5 and soyn= 5 solutionsIn order to calculate a particular solution, we use the method of substitution , which is fancy wayof saying that we determine the solution by working from a general (guessed) Find a particular solution of the difference equationyn+1+ 2yn=n.(10)If the right-hand side is a polynomial inn, then we should try a general polynomial solutionof the same order.

10 In this case, we try a solution of the formyn=An+B. On substitutioninto the equation we obtainA(n+ 1) +B+ 2An+ 2B= find the unknown valuesAandBby equating coefficients of powers ofn:n1: 3A= 1,n0:A+ 3B= 0,Solving these two simultaneous Equations we find the solutionA= 1/3, B= 1/9 yn=3n complementary function of the equation (10) isA1( 2)n, because the characteristic equationisC+ 2 = 0. The general solution of the difference equation (10) is, thereforeyn=3n 19+A1( 2)n,and the constantA1will depend on the initial Non- linear difference equationsIn a linear difference equation, every term of the equation contains at most one of the elementsof the sequence{yn}and the elements occur only as themselves are not raised to anypower (other than one) and are not an argument to a function.