A Service of Leibniz-Informationszentrum econstor Wirtschaft Leibniz Information Centre Make Your Publications Visible. zbw for Economics Todorova, Tamara Book Part — Published Version Advanced Differential and Difference Equations Suggested Citation: Todorova, Tamara (2010) : Advanced Differential and Difference Equations, In: Tamara Todorova, Problems Book to Accompany Mathematics for Economists, ISBN 978-0-470-59181-9, Wiley, Hoboken, pp. 615-701, http://eu.wiley.com/WileyCDA/WileyTitle/productCd-EHEP001511.html This Version is available at: http://hdl.handle.net/10419/148409 Standard-Nutzungsbedingungen: Terms of use: Die Dokumente auf EconStor dürfen zu eigenen wissenschaftlichen Documents in EconStor may be saved and copied for your Zwecken und zum Privatgebrauch gespeichert und kopiert werden. personal and scholarly purposes. Sie dürfen die Dokumente nicht für öffentliche oder kommerzielle You are not to copy documents for public or commercial Zwecke vervielfältigen, öffentlich ausstellen, öffentlich zugänglich purposes, to exhibit the documents publicly, to make them machen, vertreiben oder anderweitig nutzen. publicly available on the internet, or to distribute or otherwise use the documents in public. Sofern die Verfasser die Dokumente unter Open-Content-Lizenzen (insbesondere CC-Lizenzen) zur Verfügung gestellt haben sollten, If the documents have been made available under an Open gelten abweichend von diesen Nutzungsbedingungen die in der dort Content Licence (especially Creative Commons Licences), you genannten Lizenz gewährten Nutzungsrechte. may exercise further usage rights as specified in the indicated licence. www.econstor.eu Chapter 11. Advanced Differential and Difference Equations In this chapter we deal with harder differential and difference equations. We already discussed first- order equations in which a first-order derivative or difference is involved. Some more sophisticated cases are second-, third-, or higher-order differential or difference equations. The chapter is split in two: first we cover more advanced differential equations, and then we turn onto their discrete-time counterpart, higher-order difference equations. Second-order Differential Equations Consider the linear differential equation ()nn ( 1) ytutyt()11 () () ... utytutytvtnn () () () () () Since it contains the nth derivative ytn () of the function yt(), it is an n -th order differential equation with variable coefficients. It is easy to notice that when only a first-derivative yt() is involved, the equation becomes the special case of a first-order differential equation dy ut() y vt () dt which we are already familiar with. By analogy with the constant coefficient case, we have the general linear nth order equation ()nn ( 1) ytayt()11 () ... aytaybnn () th ()n where again the n derivative yt() is involved; but this time, the functions uti () and vt() ( in1,2.., ) correspond to the constants ai and b , respectively. Similar to the first-order equation case when only the first derivative yt() is involved, we have the familiar equation dy ay b dt We found the general solution to this simple first-order differential equation to be b yt() y y Aeat cp a b where y is the particular integral giving the intertemporal equilibrium. This implies that we p a have the simplest possible type of solution for yt(); that is, yt() c where the function y is constant dy in time and the derivative is zero. Consider now the case dt yt() ayt12 () ayb where the highest derivative is the second-order derivative yt(). If we again assume the simplest possible type, that is, y being a constant, we should have yt() yt () 0 615 616 Problems Book to Accompany Mathematics for Economists and the particular integral is b y p a2 0 a2 Example: Find the particular integral of the equation yt() yt () 2 y 6. Since a2 2 and 6 b 6 , substituting in the expression for the particular integral yields y 3 . p 2 What if a2 0 so the expression for the particular integral is undefined? Then it must be that y is no longer constant. A simple case to consider is yct where again cconst . Then the differential equation becomes yt() ayt1 () b a2 0 Since yct , it follows that yt() c and yt() 0, which reduces the equation to b yt() a1 We find the particular integral by integrating yt() with respect to t , which gives b ytp a2 0 a1 0 a1 Given that this time y p is a nonconstant function of time, it constitutes a moving equilibrium. In the case when aa120 , the second-order differential equation becomes yt() b Integrating yt() twice with respect to t gives bt 2 y aa0 p 2 12 The Complementary Function In the case of the first-order linear differential equation, its complementary function was the general solution of the homogeneous (reduced) equation yt() ayt () 0, i.e., yt() Aeat . Generally, an expression of the form Aert fits well into complementary functions. One reason why we can apply this exponential term to a second-order differential equation is that the latter is a second-order generalization of the first-order homogeneous equation. If we assume the solution for the function yt() to be of an exponential type yt() Aert , then we have yt() rAert and yt() rAe2 rt Substituting these values of the derivatives and the parental function in the homogeneous second-order differential equation yields 2 rt rt rt rAe arAe12 aAe 0 2 which gives rise to the characteristic equation rara 120 and the two characteristic roots aaa2 4 r 112 1,2 2 Chapter 11. Advanced Differential and Difference Equations 617 1 where by Viete’s formula rr12 a 1 and rr12 a 2. These two roots result in two solutions for yt() Aert , respectively rt1 rt2 yAe11 and yAe22 where A1 and A2 are two arbitrary constants and the complementary function of the nonhomogeneous (complete) equation is yyyc 12. Three possible situations exist in relation to the characteristic roots r1 and r2 . Case 1. Distinct real roots 2 If aa12 4 , then both roots r1 and r2 are distinct real numbers and we can write rt12 rt yyyAeAec 12 1 2 rr12 For particular values of the two constants A1 and A2 implied by some initial conditions of yt() and its derivatives, we can find the general solution to the complete equation as the sum of the complementary function and the particular integral rt12 rt yt() ycp y y12 y y p Ae 1 Ae 2 y p Example: Solve the differential equation yt() yt () 2 y 6. We already found the particular integral of this nonhomogeneous equation to be y p 3. How to find the complementary function? 2 We see that the equation fits this first case since a1 1, a2 2 and aa12 4 because 18 . Furthermore, the characteristic roots are 11813 r 1,2 22 r1 1 r2 2 rt12 r t tt2 yt() Ae12 Ae yp Ae 12 Ae 3 To find the particular values of the constants A1 and A2 , we need two initial conditions. Suppose y(0) 10 and y(0) 2 where the initial moment is t 0 . Substituting for t 0 we obtain 02(0) yAeAeAA(0)12 3 12 3 10 Differentiating yt() with respect to t , we get tt2 yt() Ae12 2 Ae Then at t 0, 02(0) yAeAeAA(0)12 2 12 2 2 which leads to the system of equations AA127 AA1222 with solutions A1 4 and A2 3 . Substituting to obtain the definite solution of the second-order differential equation, 1 Named after the French mathematician François Viete (1540-1603). 618 Problems Book to Accompany Mathematics for Economists yt() 4 ett 3 e2 3 Case 2. Single real root a If aa2 4 , there is only one root (also called a coincident or repeated real root) r 1 . Then the 12 2 complementary function is rt yAeAhtc 12() where ht() is a function that cannot be a constant multiple of ert . Therefore, we set ht() tert , and the general solution to the second-order differential equation is rt rt yt() ycp y Ae12 Ate y p 2 Example: Solve the differential equation yt() 2 yt () y 5. We can easily notice that aa12 4 a 2 since a 2 , a 1 and 24(1)2 . Thus, the example is one of a sinlge real root r 1 1 1 2 22 tt and y p 5. Therefore, yt() ycp y Ae12 Ate 5. Case 3. Complex roots 2 What if aa12 4 ? Then the roots r1 and r2 contain the square root of a negative number i 1 called an imaginary number. The very roots are called complex numbers as they contain a real part and an imaginary part, for instance, (5 i ), where we already defined i . Complex numbers cannot be ordered along the real line and, therefore, do not belong to the real-number system. They can generally be represented in the form ()mni where m and n are two real numbers. A complex number can be represented graphically in the xy -plane where x is the real-number axis and y is the imaginary-number axis. In this two-dimensional diagram known as the Argand diagram (shown by Figure 1) m is plotted on the horizontal axis and n on the vertical. Thus when n 0 , the complex number does not have an imaginary part and reduces to a real one. When m 0 , it is solely an imaginary number. By Pythagoras theorem the length of the ON line is found as the radius vector R mn22. Imaginary axis N(m,n) R m 2 n 2 n 0 m M Real axis Figure 1 Argand diagram 2 When aa12 4 , the two roots of the characteristic equation are a pair of conjugate complex numbers: a 4aa 2 rmni where m 1 n 21 and i 1 1,2 2 2 In the complex-root case the complementary function of the differential equation becomes Chapter 11.