Capítulo 6: Sistemas Lineares de Equações Diferenciais

Chapter 6 Linear Systems of Differential Equations Do not worry too much about your difficulties in mathematics I can assure you that mine are still greater Albert Einstein 18791955 61 Linear Systems 611 Coupled Oscillators In Section 35 we saw that the numerical solution of second order equa tions or higher can be cast into systems of first order equations Such sys tems are typically coupled in the sense that the solution of at least one of the equations in the system depends on knowing one of the other solutions in the system In many physical systems this coupling takes place naturally We will introduce a simple model in this section to illustrate the coupling of simple oscillators x k m Figure 61 SpringMass system There are many problems in physics that result in systems of equations This is because the most basic law of physics is given by Newtons Second Law which states that if a body experiences a net force it will accelerate Thus F ma Since a x we have a system of second order differential equations in general for three dimensional problems or one second order differential equation for one dimensional problems for a single mass We have already seen the simple problem of a mass on a spring as shown in Figure 21 Recall that the net force in this case is the restoring force of the spring given by Hookes Law Fs kx where k 0 is the spring constant and x is the elongation of the spring When the spring constant is positive the spring force is negative and when the spring constant is negative the spring force is positive The equation for simple harmonic motion for the massspring system was found to be given 212 differential equations by m x kx 0 This second order equation can be written as a system of two first order equations in terms of the unknown position and velocity We first set y x Noting that x y we rewrite the second order equation in terms of x and y Thus we have x y y k m x 61 One can look at more complicated springmass systems Consider two blocks attached with two springs as in Figure 62 In this case we apply Newtons second law for each block We will designate the elongations of each spring from equilibrium as x1 and x2 These are shown in Figure 62 For mass m1 the forces acting on it are due to each spring The first spring with spring constant k1 provides a force on m1 of k1x1 The second spring is stretched or compressed based upon the relative locations of the two masses So the second spring will exert a force on m1 of k2x2 x1 Figure 62 System of two masses and two springs x k m x m k 1 1 1 2 2 2 Similarly the only force acting directly on mass m2 is provided by the restoring force from spring 2 So that force is given by k2x2 x1 The reader should think about the signs in each case Putting this all together we apply Newtons Second Law to both masses We obtain the two equations m1 x1 k1x1 k2x2 x1 m2 x2 k2x2 x1 62 Thus we see that we have a coupled system of two second order differential equations Each equation depends on the unknowns x1 and x2 One can rewrite this system of two second order equations as a system of four first order equations by letting x3 x1 and x4 x2 This leads to the system x1 x3 linear systems of differential equations 213 x2 x4 x3 k1 m1 x1 k2 m1 x2 x1 x4 k2 m2 x2 x1 63 As we will see in the next chapter this system can be written more com pactly in matrix form d dt x1 x2 x3 x4 0 0 1 0 0 0 0 1 k1k2 m1 k2 m1 0 0 k2 m2 k2 m2 0 0 x1 x2 x3 x4 64 We can solve this system of first order equations using matrix methods However we will first need to recall a few things from linear algebra This will be done in the next chapter For now we will return to simpler systems and explore the behavior of typical solutions in planar systems 612 Planar Systems We now consider examples of solving a coupled system of first order differential equations in the plane We will focus on the theory of linear sys tems with constant coefficients Understanding these simple systems will help in the study of nonlinear systems which contain much more interest ing behaviors such as the onset of chaos In the next chapter we will return to these systems and describe a matrix approach to obtaining the solutions A general form for first order systems in the plane is given by a system of two equations for unknowns xt and yt xt Px y t yt Qx y t 65 An autonomous system is one in which there is no explicit time dependence Autonomous systems xt Px y yt Qx y 66 Otherwise the system is called nonautonomous A linear system takes the form x atx bty et y ctx dty f t 67 A homogeneous linear system results when et 0 and f t 0 A linear constant coefficient system of first order differential equations is given by x ax by e y cx dy f 68 214 DIFFERENTIAL EQUATIONS We will focus on linear homogeneous systems of constant coefficient first A linear homogeneous system of con order differential equations stant coefficient first order differential equations in the plane x axby yo cxdy 69 As we will see later such systems can result by a simple translation of the unknown functions These equations are said to be coupled if either b 0 orc 0 We begin by noting that the system 69 can be rewritten as a second or der constant coefficient linear differential equation which we already know how to solve We differentiate the first equation in system 69 and system atically replace occurrences of y and y since we also know from the first equation that y x ax Thus we have x ax by axbcxdy ax bexdx ax 610 Rewriting the last line we have x adx ad bcx 0 611 This is a linear homogeneous constant coefficient ordinary differential equation We know that we can solve this by first looking at the roots of the characteristic equation r adradbe 0 612 and writing down the appropriate general solution for xt Then we can find yt using Equation 69 1 Y5 x ax We now demonstrate this for a specific example Example 61 Consider the system of differential equations x x6y y x2y 613 Carrying out the above outlined steps we have that x 3x 4x 0 This can be shown as follows x x46 y x6x2y x6x12 6 3x4x 614 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 215 The resulting differential equation has a characteristic equation of r2 3r4 0 The roots of this equation are r 14 Therefore xt cye coe But we still need yt From the first equation of the system we have ly 1 t 4t yt 6 x g 201 3cpe Thus the solution to the system is xt cyebce yt 34cye tee 615 Sometimes one needs initial conditions For these systems we would specify conditions like x0 xo and y0 yo These would allow the determination of the arbitrary constants as before Solving systems with initial conditions Example 62 Solve x x6y y x2y 616 given x0 2 y0 0 We already have the general solution of this system in 615 In serting the initial conditions we have 2 c 0 5C1 502 617 Solving for cj and cp gives cy 65 and cp 45 Therefore the solution of the initial value problem is xt 2 3e2e yt 2ee 618 613 Equilibrium Solutions and Nearby Behaviors IN STUDYING SYSTEMS OF DIFFERENTIAL EQUATIONS it is often useful to study the behavior of solutions without obtaining an algebraic form for the solution This is done by exploring equilibrium solutions and solutions nearby equilibrium solutions Such techniques will be seen to be useful later in studying nonlinear systems We begin this section by studying equilibrium solutions of system 68 For equilibrium solutions the system does not change in time Therefore equilibrium solutions satisfy the equations x 0 and y 0 Of course this can only happen for constant solutions Let xp and yo be the constant equilibrium solutions Then xp and yo must satisfy the system Equilibrium solutions 0 axotbyoe 0 cxodyo f 619 216 DIFFERENTIAL EQUATIONS This is a linear system of nonhomogeneous algebraic equations One only has a unique solution when the determinant of the system is not zero ie ad bc 0 Using Cramers determinant Rule for solving such systems we have e b ae mt 4 y SI 620 a b a b c d c d If the system is homogeneous e f 0 then we have that the origin is the equilibrium solution ie x9 Yo 00 Often we will have this case since one can always make a change of coordinates from xy to uv by uxxg and vy yo Then ug vp 0 Next we are interested in the behavior of solutions near the equilibrium solutions Later this behavior will be useful in analyzing more complicated nonlinear systems We will look at some simple systems that are readily solved Example 63 Stable Node sink Consider the system x 2x y y 621 This is a simple uncoupled system Each equation is simply solved to give xt cye 7 and yt coe In this case we see that all solutions tend towards the equilibrium point 00 This will be called a stable node or a sink Before looking at other types of solutions we will explore the stable node in the above example There are several methods of looking at the behavior of solutions We can look at solution plots of the dependent versus the independent variables or we can look in the xyplane at the parametric 44 curves xt yf 34 Solution Plots One can plot each solution as a function of t given a set 2 oa of initial conditions Examples are shown in Figure 63 for several initial a conditions Note that the solutions decay for large t Special cases result for 0 2S various initial conditions Note that for t 0 x0 cy and y0 cp Of 1 Ah course one can provide initial conditions at any fo It is generally easier 5 to pick t 0 in our general explanations If we pick an initial condition 5 with co0 then xt 0 for all t One obtains similar results when setting 4 y0 0 Phase Portrait There are other types of plots which can provide addi tional information about the solutions even if we cannot find the exact so Figure 63 Plots of solutions of Example lutions as we can for these simple examples In particular one can consider 63 for several initial conditions LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 217 the solutions xt and yt as the coordinates along a parameterized path or curve in the plane r xt yt Such curves are called trajectories or orbits The xyplane is called the phase plane and a collection of such orbits gives a phase portrait for the family of solutions of the given system One method for determining the equations of the orbits in the phase plane is to eliminate the parameter t between the known solutions to get a relationship between x and y Since the solutions are known for the last example we can do this since the solutions are known In particular we have 2 x cye C Ay C2 Another way to obtain information about the orbits comes from noting that the slopes of the orbits in the xyplane are given by dydx For au tonomous systems we can write this slope just in terms of x and y This leads to a first order differential equation which possibly could be solved analytically or numerically First we will obtain the orbits for Example 63 by solving the correspond ing slope equation Recall that for trajectories defined parametrically by x xt and y yt we have from the Chain Rule for y yxf that dy dydx dt dx dt Therefore The Slope of a parametric curve 3 dx x 622 For the system in 621 we use Equation 622 to obtain the equation for y 0 the slope at a point on the orbit 4 dy ye 2 dx 2x The general solution of this first order differential equation is found using 21 0 2 3 separation of variables as x Ay for A an arbitrary constant Plots of these i Figure 64 Orbits for Example 63 solutions in the phase plane are given in Figure 64 Note that this is the same form for the orbits that we had obtained above by eliminating from the solution of the system Once one has solutions to differential equations we often are interested in the long time behavior of the solutions Given a particular initial condition xo Yo how does the solution behave as time increases For orbits near an equilibrium solution do the solutions tend towards or away from the equilibrium point The answer is obvious when one has the exact solutions xt and yt However this is not always the case Lets consider the above example for initial conditions in the first quad rant of the phase plane For a point in the first quadrant we have that dxdt 2x 0 218 differential equations meaning that as t xt get more negative Similarly dydt y 0 indicating that yt is also getting smaller for this problem Thus these orbits tend towards the origin as t This qualitative information was obtained without relying on the known solutions to the problem x y 1 2 1 2 1 2 1 2 1 1 1 1 1 1 1 1 Figure 65 Sketch of tangent vectors us ing Example 63 Direction Fields Another way to determine the behavior of the solutions of the system of differential equations is to draw the direction field A direction field is a vector field in which one plots arrows in the direction of tangents to the orbits at selected points in the plane This is done because the slopes of the tangent lines are given by dydx For the general system 69 the slope is dy dx cx dy ax by This is a first order differential equation which can be solved as we show in the following examples Example 64 Draw the direction field for Example 63 Figure 66 Direction field for Example 63 We can use software to draw direction fields However one can sketch these fields by hand We have that the slope of the tangent at this point is given by dy dx y 2x y 2x For each point in the plane one draws a piece of tangent line with this slope In Figure 65 we show a few of these For x y 1 1 the slope is dydx 12 So we draw an arrow with slope 12 at this point From system 621 we have that x and y are both negative at this point Therefore the vector points down and to the left We can do this for several points as shown in Figure 65 Sometimes one can quickly sketch vectors with the same slope For this example when y 0 the slope is zero and when x 0 the slope is infinite So several vectors can be provided Such vectors are tangent to curves known as isoclines in which dy dx constant Figure 67 Phase portrait for Example 63 This is a stable node or sink It is often difficult to provide an accurate sketch of a direction field Com puter software can be used to provide a better rendition For Example 63 the direction field is shown in Figure 66 Looking at this direction field one can begin to see the orbits by following the tangent vectors Of course one can superimpose the orbits on the direction field This is shown in Figure 67 Are these the patterns you saw in Figure 66 In this example we see all orbits flow towards the origin or equilibrium point Again this is an example of what is called a stable node or a sink Imagine what happens to the water in a sink when the drain is unplugged This is another uncoupled system The solutions are again simply gotten by integration We have that xt c1et and yt c2et Here we have that x decays as t gets large and y increases as t gets large In particular if one picks initial conditions with c2 0 then orbits follow the xaxis towards linear systems of differential equations 219 the origin For initial points with c1 0 orbits originating on the yaxis will flow away from the origin Of course in these cases the origin is an equilibrium point and once at equilibrium one remains there In fact there is only one line on which to pick initial conditions such that the orbit leads towards the equilibrium point No matter how small c2 is sooner or later the exponential growth term will dominate the solution One can see this behavior in Figure 68 Figure 68 Plots of solutions of Example 65 for several initial conditions Example 65 Saddle Consider the system x x y y 623 Similar to the first example we can look at plots of solutions orbits in the phase plane These are given by Figures 6869 The orbits can be obtained from the system as dy dx dydt dxdt y x The solution is y A x For different values of A 0 we obtain a family of hyperbolae These are the same curves one might obtain for the level curves of a surface known as a saddle surface z xy Thus this type of equilibrium point is classified as a saddle point From the phase portrait we can verify that there are many orbits that lead away from the origin equilibrium point but there is one line of initial conditions that leads to the origin and that is the xaxis In this case the line of initial conditions is given by the xaxis Figure 69 Phase portrait for Example 65 This is a saddle Example 66 Unstable Node source x 2x y y 624 This example is similar to Example 63 The solutions are obtained by replacing t with t The solutions orbits and direction fields can be seen in Figures 610611 This is once again a node but all orbits lead away from the equilibrium point It is called an unstable node or a source Figure 610 Plots of solutions of Exam ple 66 for several initial conditions Example 67 Center x y y x 625 This system is a simple coupled system Neither equation can be solved without some information about the other unknown function However we can differentiate the first equation and use the second equation to obtain x x 0 220 differential equations We recognize this equation as one that appears in the study of simple harmonic motion The solutions are pure sinusoidal oscillations xt c1 cos t c2 sin t yt c1 sin t c2 cos t In the phase plane the trajectories can be determined either by look ing at the direction field or solving the first order equation dy dx x y Performing a separation of variables and integrating we find that x2 y2 C Thus we have a family of circles for C 0 Can you prove this using the general solution Looking at the results graphically in Figures 612613 confirms this result This type of point is called a center Figure 611 Phase portrait for Example 66 an unstable node or source Figure 612 Plots of solutions of Exam ple 67 for several initial conditions Example 68 Focus spiral x αx y y x 626 In this example we will see an additional set of behaviors of equi librium points in planar systems We have added one term αx to the system in Example 67 We will consider the effects for two spe cific values of the parameter α 01 02 The resulting behaviors are shown in the Figures 615618 We see orbits that look like spi rals These orbits are stable and unstable spirals or foci the plural of focus We can understand these behaviors by once again relating the sys tem of first order differential equations to a second order differential equation Using the usual method for obtaining a second order equa tion form a system we find that xt satisfies the differential equation x αx x 0 We recall from our first course that this is a form of damped simple harmonic motion The characteristic equation is r2 αr 1 0 The solution of this quadratic equation is r α α2 4 2 Figure 613 Phase portrait for Example 67 a center There are five special cases to consider as shown in the below clas sification linear systems of differential equations 221 Classification of Solutions of x αx x 0 1 α 2 There is one real solution This case is called critical damping since the solution r 1 leads to exponential decay The solution is xt c1 c2tet 2 α 2 There are two real negative solutions r µ ν µ ν 0 The solution is xt c1eµt c2eνt In this case we have what is called overdamped motion There are no oscillations 3 2 α 0 There are two complex conjugate solutions r α2 iβ with real part less than zero and β 4α2 2 The solution is xt c1 cos βt c2 sin βteαt2 Since α 0 this consists of a decaying expo nential times oscillations This is often called an underdamped oscillation 4 α 0 This leads to simple harmonic motion 5 0 α 2 This is similar to the underdamped case except α 0 The solutions are growing oscillations 6 α 2 There is one real solution The solution is xt c1 c2tet It leads to unbounded growth in time 7 For α 2 There are two real positive solutions r µ ν 0 The solution is xt c1eµt c2eνt which grows in time Figure 614 Plots of solutions of Ex ample 68 for several initial conditions α 02 Figure 615 Plots of solutions of Ex ample 68 for several initial conditions α 01 For α 0 the solutions are losing energy so the solutions can oscil late with a diminishing amplitude See Figure 614 For α 0 there is a growth in the amplitude which is not typical See Figure 615 Of course there can be overdamped motion if the magnitude of α is too large Figure 616 Phase portrait for 69 This is a degenerate node Example 69 Degenerate Node For this example we will write out the solutions It is a coupled system for which only the second equa tion is coupled x x y 2x y 627 There are two possible approaches a We could solve the first equation to find xt c1et Inserting this solution into the second equation we have y y 2c1et This is a relatively simple linear first order equation for y yt The integrating factor is µ et The solution is found as yt c2 2c1tet b Another method would be to proceed to rewrite this as a second order equation Computing x does not get us very far So we look at y 2x y 222 differential equations 2x y 2y y 628 Therefore y satisfies y 2y y 0 The characteristic equation has one real root r 1 So we write yt k1 k2tet This is a stable degenerate node Combining this with the solution xt c1et we can show that yt c2 2c1tet as before Figure 617 Phase portrait for Example 68 with α 02 This is a stable focus or spiral In Figure 616 we see several orbits in this system It differs from the stable node show in Figure 64 in that there is only one direction along which the orbits approach the origin instead of two If one picks c1 0 then xt 0 and yt c2et This leads to orbits running along the yaxis as seen in the figure Figure 618 Phase portrait for Example 69 This is a degenerate node Example 610 A Line of Equilibria Zero Root x 2x y y 2x y 629 Figure 619 Plots of direction field of Ex ample 610 In this last example we have a coupled set of equations We rewrite it as a second order differential equation x 2x y 2x 2x y 2x 2x x 2x 3x 630 So the second order equation is x 3x 0 and the characteristic equation is 0 rr 3 This gives the general solution as xt c1 c2e3t and thus y 2x x 2c1 c2e3t 3c2e3t 2c1 c2e3t In Figure 619 we show the direction field The constant slope field seen in this example is confirmed by a simple computation dy dx 2x y 2x y 1 Furthermore looking at initial conditions with y 2x we have at t 0 2c1 c2 2c1 c2 c2 0 Therefore points on this line remain on this line forever x y c1 2c1 This line of fixed points is called a line of equilibria linear systems of differential equations 223 614 Polar Representation of Spirals In the examples with a center or a spiral one might be able to write the solutions in polar coordinates Recall that a point in the plane can be described by either Cartesian x y or polar r θ coordinates Given the polar form one can find the Cartesian components using x r cos θ and y r sin θ Given the Cartesian coordinates one can find the polar coordinates using r2 x2 y2 and tan θ y x 631 Since x and y are functions of t then naturally we can think of r and θ as functions of t Converting a system of equations in the plane for x and y to polar form requires knowing r and θ So we first find expressions for r and θ in terms of x and y Differentiating the first equation in 631 gives rr xx yy Inserting the expressions for x and y from system 69 we have rr xax by ycx dy In some cases this may be written entirely in terms of rs Similarly we have that θ xy yx r2 which the reader can prove for homework In summary when converting first order equations from rectangular to polar form one needs the relations below Derivatives of Polar Variables r xx yy r θ xy yx r2 632 Example 611 Rewrite the following system in polar form and solve the resulting system x ax by y bx ay 633 We first compute r and θ rr xx yy xax by ybx ay ar2 224 differential equations r2θ xy yx xbx ay yax by br2 This leads to simpler system r ar θ b 634 This system is uncoupled The second equation in this system in dicates that we traverse the orbit at a constant rate in the clockwise direction Solving these equations we have that rt r0eat θt θ0 bt Eliminating t between these solutions we finally find the polar equation of the orbits r r0eaθθ0tb If you graph this for a 0 you will get stable or unstable spirals Example 612 Consider the specific system x y x y x y 635 In order to convert this system into polar form we compute rr xx yy xy x yx y r2 r2θ xy yx xx y yy x r2 This leads to simpler system r r θ 1 636 Solving these equations yields rt r0et θt t θ0 Eliminating t from this solution gives the orbits in the phase plane rθ r0eθθ0 A more complicated example arises for a nonlinear system of differential equations Consider the following example Example 613 x y x1 x2 y2 y x y1 x2 y2 637 Transforming to polar coordinates one can show that in order to convert this system into polar form we compute r r1 r2 θ 1 This uncoupled system can be solved and this is left to the reader LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 225 62 Applications IN THIS SECTION WE WILL DESCRIBE SOME SIMPLE APPLICATIONS leading to systems of differential equations which can be solved using the methods in this chapter These systems are left for homework problems and the as the start of further explorations for student projects 621 MassSpring Systems THE FIRST EXAMPLES THAT WE HAD SEEN involved masses on springs Re call that for a simple mass on a spring we studied simple harmonic motion which is governed by the equation mk kx 0 This second order equation can be written as two first order equations xy k j xX 6 8 y7 638 or xy 2 y wx 639 where w x The coefficient matrix for this system is 0 1 A 2 m4 m9 Figure 620 System of two masses and Z ky ky two springs 2 YUUUU JUUJUU 2 ee 2X4 Xp We also looked at the system of two masses and two springs as shown in Figure 620 The equations governing the motion of the masses is mx kyxy ko x2 x1 myXy kyx2 x1 640 226 DIFFERENTIAL EQUATIONS We can rewrite this system as four first order equations xy x3 xX x4 ky ky xxx x3 molt my 2 x1 ke x2 x1 641 4 Mp 2 4 4 The coefficient matrix for this system is 0 0 10 A 0 0 01 ky k k i 2 my 0 0 mn 0 0 We can study this system for specific values of the constants using the meth Writing the springblock system as a sec ods covered in the last sections ond order vector system Actually one can also put the system 640 in the matrix form m 0 x4 ky ko ky x4 ny e a2 0 m Xo ko k x2 This system can then be written compactly as Mx Kx 643 where 0 ky k k ma Ka 1 kg 2 0 m kp ko This system can be solved by guessing a form for the solution We could guess x ae or x coswt 61 ay coswt d2 where 6 are phase shifts determined from initial conditions Inserting x ae into the system gives KwMa 0 This is a homogeneous system It is a generalized eigenvalue problem for eigenvalues w and eigenvectors a We solve this in a similar way to the standard matrix eigenvalue problems The eigenvalue equation is found as det K wM 0 Once the eigenvalues are found then one determines the eigenvectors and constructs the solution LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 227 Example 614 Let m m2 m and ky kp k Then we have to solve the system ww m O a 2k k ay 0 m ay k k ay The eigenvalue equation is given by 0 2k mw k 7 k k mw 2kmwk mw k mw 3kmu k 644 Solving this quadratic equation for w we have 341k 2 m For positive values of w one can show that 1 k w 5 414 v5 2 m The eigenvectors can be found for each eigenvalue by solving the homogeneous system 2k mw k a 0 k k mu ap The eigenvectors are given by v5 v51 ay an 1 1 We are now ready to construct the real solutions to the problem Similar to solving two first order systems with complex roots we take the real and imaginary parts and take a linear combination of the so lutions In this problem there are four terms giving the solution in the form xt cyaycoswyt cpasinw1t cza2coswyt cyazsinwyt where the ws are the eigenvalues and the as are the corresponding eigenvectors The constants are determined from the initial conditions x0 xp and x0 vo 622 Circuits IN THE LAST CHAPTER WE INVESTIGATED SIMPLE SERIES LRC CIRCUITS More complicated circuits are possible by looking at parallel connections or other combinations of resistors capacitors and inductors This results 228 differential equations in several equations for each loop in the circuit leading to larger systems of differential equations An example of another circuit setup is shown in Figure 621 This is not a problem that can be covered in the first year physics course There are two loops indicated in Figure 622 as traversed clockwise For each loop we need to apply Kirchoffs Loop Rule There are three oriented currents labeled Ii i 1 2 3 Corresponding to each current is a changing charge qi such that Ii dqi dt i 1 2 3 We have for loop one I1R1 q2 C Vt 645 and for loop two I3R2 LdI3 dt q2 C 646 Vt R1 R2 L C Figure 621 A circuit with two loops containing several different circuit ele ments There are three unknown functions for the charge Once we know the charge functions differentiation will yield the three currents However we only have two equations We need a third equation This equation is found from Kirchoffs Point Junction Rule Vt R1 R2 L A C B I1 I3 I2 1 2 Figure 622 The previous parallel circuit with the directions indicated for travers ing the loops in Kirchoffs Laws Consider the points A and B in Figure 622 Any charge current entering these junctions must be the same as the total charge current leaving the junctions For point A we have I1 I2 I3 647 or q1 q2 q3 648 Equations 645 646 and 648 form a coupled system of differential equations for this problem There are both first and second order derivatives involved We can write the whole system in terms of charges as R1 q1 q2 C Vt R2 q3 L q3 q2 C q1 q2 q3 649 The question is whether or not we can write this as a system of first order differential equations Since there is only one second order derivative we can introduce the new variable q4 q3 The first equation can be solved for q1 The third equation can be solved for q2 with appropriate substitutions for the other terms q3 is gotten from the definition of q4 and the second equation can be solved for q3 and substitutions made to obtain the system q1 V R1 q2 R1C q2 V R1 q2 R1C q4 q3 q4 q4 q2 LC R2 L q4 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 229 So we have a nonhomogeneous first order system of differential equa tions 623 MixtureProbllems There are many types of mixture problems Such problems are standard in a first course on differential equations as examples of first order differential equations Typically these examples consist of a tank of brine water con taining a specific amount of salt with pure water entering and the mixture leaving or the flow of a pollutant into or out of a lake We first saw such problems in Chapter 1 In general one has a rate of flow of some concentration of mixture enter ing a region and a mixture leaving the region The goal is to determine how much stuff is in the region at a given time This is governed by the equation Rate of change of substance Rate In Rate Out This can be generalized to the case of two interconnected tanks We will pro vide an example but first we review the single tank problem from Chapter 1 Example 615 Single Tank Problem A 50 gallon tank of pure water has a brine mixture with concentra C tion of 2 pounds per gallon entering at the rate of 5 gallons per minute See Figure 623 At the same time the wellmixed contents drain out at the rate of 5 gallons per minute Find the amount of salt in the tank at time t In all such problems one assumes that the solution is well ee mixed at each instant of time Let xt be the amount of salt at time t Then the rate at which the salt in the tank increases is due to the amount of salt entering the tank X less that leaving the tank To figure out these rates one notes that a dxdt has units of pounds per minute The amount of salt entering per minute is given by the product of the entering concentration times Figure 623 A typical mixing problem the rate at which the brine enters This gives the correct units 2 5 joPounds gal min min Similarly one can determine the rate out as x pounds gal x pounds 50 gal min 10 min Thus we have dx x 10 dt 10 This equation is easily solved using the methods for first order equations 230 DIFFERENTIAL EQUATIONS Figure 624 The two tank problem C P i e co Ce X 7 NS 7 e Example 616 Double Tank Problem One has two tanks connected together labeled tank X and tank Y as shown in Figure 624 Let tank X initially have 100 gallons of brine made with 100 pounds of salt Tank Y initially has 100 gallons of pure water Pure water is pumped into tank X at a rate of 20 gallons per minute Some of the mixture of brine and pure water flows into tank Y at 3 gallons per minute To keep the tank levels the same one gallon of the Y mixture flows back into tank X at a rate of one gallon per minute and 20 gallons per minute drains out Find the amount of salt at any given time in the tanks What happens over a long period of time In this problem we set up two equations Let xt be the amount of salt in tank X and yt the amount of salt in tank Y Again we carefully look at the rates into and out of each tank in order to set up the system of differential equations We obtain the system dx y 3x dt 100 100 dy 3x 3y di 100 100 659 This is a linear homogenous constant coefficient system of two first order equations which we know how to solve The matrix form of the system is given by 100 30 1 x x 0 100 100 The eigenvalues for the problem are given by A 3 V3 and the eigenvectors are 1 3 Since the eigenvalues are real and distinct the general solution is easily written down 1 1 tc e3v3t Cc e3Vv3F x 1 V3 C2 V3 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 231 Finally we need to satisfy the initial conditions So 1 1 100 0 c c or cy c 100 cy cr V3 0 So C2 cy 50 The final solution is 1 1 t 50 3v3 3v3t xt 5 1 Je or xt 50 ev eV9 yt 50V3 e39t e3V91 651 624 Chemical Kinetics THERE ARE MANY PROBLEMS IN THE CHEMISTRY of chemical reactions which lead to systems of differential equations The simplest reaction is when a chemical A turns into chemical B This happens at a certain rate k 0 This reaction can be represented by the chemical formula k In this case we have that the rates of change of the concentrations of A A and B B are given by The chemical reactions used in these ex amples are first order reactions Second dl Al order reactions have rates proportional k A to the square of the concentration dt aB k A Tt A 652 Think about this as it is a key to understanding the next reactions A more complicated reaction is given by ky ko Here there are three concentrations and two rates of change The system of equations governing the reaction is aA kAl o 4 aB kyA keB c BA kB aC kpB 6 i 2B 653 232 DIFFERENTIAL EQUATIONS The more complication rate of change is when B increases from A chang ing to B and decrease when B changes to C Thus there are two terms in the rate of change equation for concentration B One can further consider reactions in which a reverse reaction is possible Thus a further generalization occurs for the reaction kg ky ky The reverse reaction rates contribute to the reaction equations for A and B The resulting system of equations is aA kAk3B di 1A ksB aB AB ky A b2B ks B aC kpB TE 2B 654 Nonlinear chemical reactions will be discussed in the next chapter 625 Predator Prey Models ANOTHER COMMON POPULATION MODEL is that describing the coexistence of species For example we could consider a population of rabbits and foxes Left to themselves rabbits would tend to multiply thus dR aR de witha 0 In sucha model the rabbit population would grow exponentially Similarly a population of foxes would decay without the rabbits to feed on So we have that dF bF dt for b 0 Now if we put these populations together on a deserted island they would interact The more foxes the rabbit population would decrease However the more rabbits the foxes would have plenty to eat and the pop ulation would thrive Thus we could model the competing populations as aRcF dF bFAaR 6 7 655 where all of the constants are positive numbers Studying this coupled system would lead to a study of the dynamics of these populations The nonlinear version of this system the LotkaVolterra model will be discussed in the next chapter linear systems of differential equations 233 626 Love Affairs The next application is one that was introduced in 1988 by Stro gatz as a cute system involving relationships1 One considers what happens 1 Steven H Strogatz introduced this problem as an interesting example of systems of differential equations in Mathematics Magazine Vol 61 No 1 Feb 1988 p 35 He also describes it in his book Nonlinear Dynamics and Chaos 1994 to the affections that two people have for each other over time Let R de note the affection that Romeo has for Juliet and J be the affection that Juliet has for Romeo Positive values indicate love and negative values indicate dislike One possible model is given by dR dt bJ dJ dt cR 656 with b 0 and c 0 In this case Romeo loves Juliet the more she likes him But Juliet backs away when she finds his love for her increasing A typical system relating the combined changes in affection can be mod eled as dR dt aR bJ dJ dt cR dJ 657 Several scenarios are possible for various choices of the constants For ex ample if a 0 and b 0 Romeo gets more and more excited by Juliets love for him If c 0 and d 0 Juliet is being cautious about her relationship with Romeo For specific values of the parameters and initial conditions one can explore this match of an overly zealous lover with a cautious lover 627 Epidemics Another interesting area of application of differential equation is in predicting the spread of disease Typically one has a population of sus ceptible people or animals Several infected individuals are introduced into the population and one is interested in how the infection spreads and if the number of infected people drastically increases or dies off Such models are typically nonlinear and we will look at what is called the SIR model in the next chapter In this section we will model a simple linear model Let us break the population into three classes First we let St represent the healthy people who are susceptible to infection Let It be the number of infected people Of these infected people some will die from the infection and others could recover We will consider the case that initially there is one infected person and the rest say N are healthy Can we predict how many deaths have occurred by time t We model this problem using the compartmental analysis we had seen for mixing problems The total rate of change of any population would be 234 DIFFERENTIAL EQUATIONS due to those entering the group less those leaving the group For example the number of healthy people decreases due infection and can increase when some of the infected group recovers Lets assume that a the rate of infection is proportional to the number of healthy people aS and b the number who recover is proportional to the number of infected people rl Thus the rate of change of healthy people is found as aS 711 Let the number of deaths be Dt Then the death rate could be taken to be proportional to the number of infected people So dD dl dt Finally the rate of change of infected people is due to healthy people getting infected and the infected people who either recover or die Using the corresponding terms in the other equations we can write the rate of change of infected people as a aSrIdl This linear system of differential equations can be written in matrix form d S a r 0 S it I a dr 0 I 658 D 0 d 0 D The reader can find the solutions of this system and determine if this is a realistic model 63 Matrix Formulation WE HAVE INVESTIGATED SEVERAL LINEAR SYSTEMS in the plane and in the next chapter we will use some of these ideas to investigate nonlinear systems We need a deeper insight into the solutions of planar systems So in this section we will recast the first order linear systems into matrix form This will lead to a better understanding of first order systems and allow for extensions to higher dimensions and the solution of nonhomogeneous equations later in this chapter We start with the usual homogeneous system in Equation 69 Let the unknowns be represented by the vector t xt 9 yt Then we have that v ax by 4 b x Ay y cx dy c d y LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 235 Here we have introduced the coefficient matrix A This is a first order vector differential equation x Ax Formerly we can write the solution as x xe You can verify that this is a solution by simply differentiating d d x07 e Axge Ax However there remains the question What does it mean to exponentiate a matrix The exponential of a matrix is defined using the Maclaurin series expansion oo x2 x3 x Syst a e ea deet 3 tate We define The exponential of a matrix is defined us oA K y aA 14 A4 a 4 fee 660 ing the Maclaurin series pansion k0 eso alex Gt gto In general it is difficult to sum this series but it is doable for some simple k0 examples So we define 1 0 apap ey Ay 659 Example 617 Evaluate e4 for A 2 3 oO 0 2 In general it is difficult computing e4 unless A is diagonal tA Pio Pas 2 3 1 o fto f10y fr oy Lol 0 2 2 0 2 3 0 2 190 f1 0 ff to Pfr 0 Lol 0 2 2o 4 3 0 8 14t4 545 0 0 1426 264 8 e 0 0 et 661 Example 618 Evaluate e for A 10 We first note that A2 0 1 O1 71 0 L 1 0 1 0 0 1 Therefore An A nodd I neven 236 DIFFERENTIAL EQUATIONS Then we have tA Pio PAs eo ItaA A 3A ae Pp ItaA sit aAt 1464h t4 04h 2 4 es ee cosh sinht 662 sinht cosht 662 Since summing these infinite series might be difficult we will now inves tigate the solutions of planar systems to see if we can find other approaches for solving linear systems using matrix methods We begin by recalling the solution to the problem in Example 616 We obtained the solution to this system as xt cye toe yt Zee one 663 3 2 This can be rewritten using matrix operations Namely we first write the solution in vector form t 0 yt cye coe fcyel heoe t 4t cye c2e loot Lan p4t 3c 1e o c2e 1 1 c etc 1 ett 664 3 2 We see that our solution is in the form of a linear combination of vectors of the form x ve with v a constant vector and A a constant number This is similar to how we began to find solutions to second order constant coefficient equations So for the general problem 63 we insert this guess Thus x Ax Ave Ave 665 For this to be true for all t we have that Av Av 666 This is an eigenvalue problem A is a 2 x 2 matrix for our problem but could easily be generalized to a system of n first order differential equa tions We will confine our remarks for now to planar systems However we LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 237 need to recall how to solve eigenvalue problems and then see how solutions of eigenvalue problems can be used to obtain solutions to our systems of differential equations 64 Eigenvalue Problems We seek nontrivial solutions to the eigenvalue problem Av Av 667 We note that v 0 is an obvious solution Furthermore it does not lead to anything useful So it is called a trivial solution Typically we are given the matrix A and have to determine the eigenvalues A and the associated eigenvectors v satisfying the above eigenvalue problem Later in the course we will explore other types of eigenvalue problems For now we begin to solve the eigenvalue problem for v 2 Inserting this into Equation 667 we obtain the homogeneous algebraic system a Av bv2 0 cv dAv2 0 668 The solution of such a system would be unique if the determinant of the system is not zero However this would give the trivial solution v 0 v2 0 To get a nontrivial solution we need to force the determinant to be zero This yields the eigenvalue equation aA b 0 aAdA be den aAdA This is a quadratic equation for the eigenvalues that would lead to nontrivial solutions If we expand the right side of the equation we find that M adAad be 0 This is the same equation as the characteristic equation 612 for the gen eral constant coefficient differential equation considered in the first chapter Thus the eigenvalues correspond to the solutions of the characteristic poly nomial for the system Once we find the eigenvalues then there are possibly an infinite number solutions to the algebraic system We will see this in the examples So the process is to a Write the coefficient matrix b Find the eigenvalues from the equation detA AI 0 and c Find the eigenvectors by solving the linear system A AIv 0 for each A 238 DIFFERENTIAL EQUATIONS 65 Solving Constant Coefficient Systems in 2D Before proceeding to examples we first indicate the types of solutions that could result from the solution of a homogeneous constant coefficient system of first order differential equations We begin with the linear system of differential equations in matrix form dx a b Ra x ax 669 The type of behavior depends upon the eigenvalues of matrix A The pro cedure is to determine the eigenvalues and eigenvectors and use them to construct the general solution If we have an initial condition xt xo we can determine the two arbitrary constants in the general solution in order to obtain the particular Recall that linear independence means solution Thus if xt and x2t are two linearly independent solutions c1x1t coxot 0 if and only if sn ig Gi ct 0 The reader should derive the then the general solution is given as condition on the x for linear indepen dence xf c1x1t coxet Then setting t 0 we get two linear equations for c and c2 C1X1 0 C2x20 Xo The major work is in finding the linearly independent solutions This de pends upon the different types of eigenvalues that one obtains from solving the eigenvalue equation detA AI 0 The nature of these roots indicate the form of the general solution In Table 61 we summarize the classifica tion of solutions in terms of the eigenvalues of the coefficient matrix We first make some general remarks about the plausibility of these solutions and then provide examples in the following section to clarify the matrix methods for our two dimensional systems The construction of the general solution in Case I is straight forward However the other two cases need a little explanation Lets consider Case ITI Note that since the original system of equations does not have any is then we would expect real solutions S50 we look at the real and imaginary parts of the complex solution We have that the complex solution satisfies the equation d a LReyt imyt AReyt imyé Differentiating the sum and splitting the real and imaginary parts of the equation gives d d ap Rely t iz Imyt AReyt tAImyt Setting the real and imaginary parts equal we have d a Rely AlRey I LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 239 and J aly t Alimyt Therefore the real and imaginary parts each are linearly independent so lutions of the system and the general solution can be written as a linear combination of these expressions Classification of the Solutions for Two Table 61 Solutions Types for Planar Sys tems with Constant Coefficients Linear First Order Differential Equations 1 Case I Two real distinct roots Solve the eigenvalue problem Av Av for each eigenvalue obtaining two eigenvectors v1V2 Then write the general solution as a linear combination xt cyevy coe2v2 2 Case II One Repeated Root Solve the eigenvalue problem Av Av for one eigenvalue A obtaining the first eigenvector v One then needs a second linearly independent solution This is obtained by solving the nonhomogeneous problem Avo Avo vq for vo The general solution is then given by xt cyevy coev2 tvi 3 Case III Two complex conjugate roots Solve the eigenvalue problem Ax Ax for one eigenvalue A a if obtaining one eigenvector v Note that this eigenvector may have complex entries Thus one can write the vector yt ev e cos Bt isin Btv Now construct two linearly independent solu tions to the problem using the real and imaginary parts of yt yit Reyt and yt Imyt Then the general solution can be written as xf cyyit coyot We now turn to Case II Writing the system of first order equations as a second order equation for xt with the sole solution of the characteristic equation A 5a d we have that the general solution takes the form xt cy eate This suggests that the second linearly independent solution involves a term of the form vte It turns out that the guess that works is x tev ety Inserting this guess into the system x Ax yields tev evo A tes eva ety Ate v Ae Ate et Avo eM vi tAv2 eM Av 670 240 DIFFERENTIAL EQUATIONS Noting this is true for all t we find that vy Av2 Av2 671 Therefore A Alv2 Vi We know everything except for v2 So we just solve for it and obtain the second linearly independent solution 66 Examples of the Matrix Method Here we will give some examples for typical systems for the three cases mentioned in the last section 4 2 E le 619 A xample 619 3 3 Eigenvalues We first determine the eigenvalues 4A 2 0 6 3 BHA 672 Therefore 0 4A3A6 0 A7A6 0 A1DA6 673 The eigenvalues are then A 16 This is an example of Case I Eigenvectors Next we determine the eigenvectors associated with each of these eigenvalues We have to solve the system Av Av in each case Case A 1 4 2 O71 O71 6 3 2 C1 0 6 This gives 3v1 2v2 0 One possible solution yields an eigenvector of O71 2 02 3 Case A 6 4 2 1 6 7 676 33 02 02 2 2 O71 0 6 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 241 For this case we need to solve 2v1 2v2 0 This yields O71 1 02 1 General Solution We can now construct the general solution xt cyeyy cpe2vo 2 1 t 6t 2cye cpe 6 8 3cye cpe 678 35 E le 620 A xample 620 11 Eigenvalues Again one solves the eigenvalue equation 3A 5 0 6 1 1aA 679 Therefore 0 3A1A5 0 A242 2 4412 A Sa VER AOC 14i 680 The eigenvalues are then A 1i1i This is an example of Case Il Eigenvectors In order to find the general solution we need only find the eigenvector associated with 1 i 3 5 C1 1 4 i V1 1 1 02 02 2i 1 0 681 1 21 V2 0 We need to solve 2 iv 5v2 0 Thus 04 i 2 682 02 1 Complex Solution In order to get the two real linearly indepen dent solutions we need to compute the real and imaginary parts of vem edt 2i1 elit 21 1 1 242 DIFFERENTIAL EQUATIONS a4 ecostisint 4 2 icostisint cost isint 4 2cost sint icost2sint cost isint of 2cost sint iet cost 2sint cost sint General Solution Now we can construct the general solution 2 tsint t2sint xt cet cos sin oe cos sin cost sint 1 2cost sint c2 cos t 2sint 683 cy cost co sint Note This can be rewritten as xt e cost 2e1 C2 esint 202 41 C1 C2 7 1 E le 621 A xample 621 9 4 Eigenvalues 7A 1 0 68 9 1A 684 Therefore 0 7A1A9 0 A8A16 0 A4 685 There is only one real eigenvalue A 4 This is an example of Case Il Eigenvectors In this case we first solve for v and then get the second linearly independent vector 7 1 U1 4 U1 9 1 v2 02 31 O71 0 686 Therefore we have 1 02 3 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 243 Second Linearly Independent Solution Now we need to solve Avo Avo v1 7 1 uy 4 uy 1 9 1 ug ug 3 31 uy 1 68 Expanding the matrix product we obtain the system of equations 3uuU 1 9u 3u2 3 688 uy 1 The solution of this system is ug 2 General Solution We construct the general solution as yt cyedtvy cre v2 tv 1 1 1 At At ce t ot ateallt 689 3c c22 3t 661 Planar Systems Summary The reader should have noted by now that there is a connection between the behavior of the solutions obtained in Section 613 and the eigenvalues found from the coefficient matrices in the previous examples In Table 62 we summarize some of these cases Stability Table 62 List of typical behaviors in pla Node Real A same signs A 0 stable nar systems A 0 unstable Saddle Real A opposite signs Mostly Unstable A pure imaginary Focus Spiral Complex A ReA 0 ReA 0 stable ReA 0 unstable Degenerate Node A 0 stable Lines of Equilibria A 0 stable The connection as we have seen is that the characteristic equation for the associated second order differential equation is the same as the eigen value equation of the coefficient matrix for the linear system However one should be a little careful in cases in which the coefficient matrix in not diag onalizable In Table 63 are three examples of systems with repeated roots The reader should look at these systems and look at the commonalities and 244 DIFFERENTIAL EQUATIONS differences in these systems and their solutions In these cases one has un stable nodes though they are degenerate in that there is only one accessible eigenvector with a repeated root of A 2 coun an 2b060de2 crag jar Oba Le4dq4 poe ced an2be lem Odo coe SSA ESS SS DEERE eee REREREAAL ESeeaaal SEEDS Ly Cotes DoT ie PEEEER TL EEE DoS A yo rh BTR a Co ata ser aly he SIDES EG FH SS HST TELE CECII ISSEY NNR CIITA SSS OSSEPAIE SESSSSSAI LSS 2 0 0 1 2 1 x x x x x 0 2 4 4 0 2 Another way to look at the classification of these solution is to use the determinant and trace of the coefficient matrix Recall that the determinant a b and trace of A d are given by detA ad be and trA ad fa We note that the general eigenvalue equation adAadbe 0 can be written as A trAA detA 0 690 Therefore the eigenvalues are found from the quadratic formula as trA trA2 4detA tga AE VEAP eA 600 The solution behavior then depends on the sign of discriminant trA 4det A If we consider a plot of where the discriminant vanishes then we could plot trA 4detA in the detAtrAplane This is a parabolic cure as shown by the dashed line in Figure 625 The region inside the parabola have a negative discriminant leading to complex roots In these cases we have oscillatory solutions If trA 0 then one has centers If trA 0 the solutions are stable spirals otherwise they are unstable spirals If the discriminant is positive then the roots are real leading to nodes or saddles in the regions indicated 67 Theory of Homogeneous Constant Coefficient Systems There is a general theory for solving homogeneous constant coefficient sys tems of first order differential equations We begin by once again recalling LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 245 trA Figure 625 Solution Classification for Unstable Nodes we trAgdetA Planar Systems Y Unstable Spirals ra Saddles Centers detA N Stable Spirals Stable Nodes ss S the specific problem 616 We obtained the solution to this system as xt cye tone i yt gels 5028 692 This time we rewrite the solution as cye cpe xX Zcye cnet ef et Cy ze set C2 FC 693 Thus we can write the general solution as a 2 x 2 matrix times an arbi trary constant vector The matrix consists of two columns that are linearly independent solutions of the original system This matrix is an example of what we will define as the Fundamental Matrix of solutions of the system So determining the Fundamental Matrix will allow us to find the general solution of the system upon multiplication by a constant matrix In fact we will see that it will also lead to a simple representation of the solution of the initial value problem for our system We will outline the general theory Consider the homogeneous constant coefficient system of first order dif ferential equations dx Ge Mata A122 1 nXny dxp a Ag1X1 A22X2 A2nXn 246 DIFFERENTIAL EQUATIONS dxn a7 7 Ani X1 ay2X2 aAnnXn 694 As we have seen this can be written in the matrix form x Ax where xy x2 x Xn and 41 412 An 421 422 dn le M71 42 Ann Now consider m vector solutions of this system 1t 2t mt These solutions are said to be linearly independent on some domain if C194 t Co2t CuPmt 0 for all f in the domain implies that cy cz Cm 0 Let 1 t f2t Pnt bea set of n linearly independent set of solutions of our system called a fundamental set of solutions We construct a matrix from these solutions using these solutions as the column of that matrix We define this matrix to be the fundamental matrix solution This matrix takes the form fir fiz Pin p21 p22 Pan e 4 om Doo Prt Pn2 vt Pun What do we mean by a matrix solution We have assumed that each gx is a solution of our system Therefore we have that Ady for k 1n We say that is a matrix solution because we can show that also satisfies the matrix formulation of the system of differential equations We can show this using the properties of matrices d ae o 1 Agi Agu A gi n A 695 Given a set of vector solutions of the system when are they linearly independent We consider a matrix solution Qt of the system in which we have n vector solutions Then we define the Wronskian of Qt to be W det Qt LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 247 If Wt 0 then Qt is a fundamental matrix solution Before continuing we list the fundamental matrix solutions for the set of examples in the last section Refer to the solutions from those examples Furthermore note that the fundamental matrix solutions are not unique as one can multiply any column by a nonzero constant and still have a fundamental matrix solution 4 2 Example 619 A rampteaga 3 5 et eet Pt 3et eet We should note in this case that the Wronskian is found as W det t et eet 3et bt 5e 0 696 Example 620 A 3 1 1 e2costsint ecost2sint t ot I cos t e sint 7 1 E le 621 A xample 621 9 4 At At 1 t Pt at 0 3e eo 2 3t So far we have only determined the general solution This is done by the following steps Procedure for Determining the General Solution 1 Solve the eigenvalue problem A AIv 0 2 Construct vector solutions from ve The method depends if one has real or complex conjugate eigenvalues 3 Form the fundamental solution matrix t from the vector solution 4 The general solution is given by xt tC for C an arbitrary con stant vector We are now ready to solve the initial value problem x Ax xto Xo 248 DIFFERENTIAL EQUATIONS Starting with the general solution we have that xo xfo toC As usual we need to solve for the cs Using matrix methods this is now easy Since the Wronskian is not zero then we can invert at any value of t So we have co toxo Putting C back into the general solution we obtain the solution to the initial value problem xt Btbtyxo You can easily verify that this is a solution of the system and satisfies the initial condition at t tg The matrix combination tto is useful So we will define the resulting product to be the principal matrix solution denoting it by Yt Bt tg Thus the solution of the initial value problem is xt Ytxo Further more we note that Yt is a solution to the matrix initial value problem x Ax xto I where I is the n x n identity matrix Matrix Solution of the Homogeneous Problem In summary the matrix solution of d Ax xfp x0o is given by xt txo ttoxo where f is the fundamental matrix solution and Yt is the principal matrix solution Example 622 Lets consider the matrix initial value problem x 5x43y yo 6x4y 697 satisfying x0 1 y0 2 Find the solution of this problem We first note that the coefficient matrix is A 5 8 6 4 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 249 The eigenvalue equation is easily found from 0 5A4A18 jA2 A2A1 698 So the eigenvalues are A 12 The corresponding eigenvectors are found to be y 1 vw 1 17 9 2 1 Now we construct the fundamental matrix solution The columns are obtained using the eigenvectors and the exponentials e 1 t 1 2 ti H pit 1 pit 1 So the fundamental matrix solution is et e2t Ot 2et e2t The general solution to our problem is then et e2t xt 2et e2t Cc for C is an arbitrary constant vector In order to find the particular solution of the initial value problem we need the principal matrix solution We first evaluate 0 then we invert it 1 1 1 1 0 10 o4 4 o3 3 The particular solution is then xt et t 1 1 1 2ef 2 2 1 2 et et 3 2e 4 3e 4e 6 6e 4e 699 Thus xt 3e 4e and yt 6e 4e 68 Nonhomogeneous Systems Before leaving the theory of systems of linear constant coefficient systems we will discuss nonhomogeneous systems We would like to solve systems of the form x Atx ft 6100 250 DIFFERENTIAL EQUATIONS We will assume that we have found the fundamental matrix solution of the homogeneous equation Furthermore we will assume that Af and ft are continuous on some common domain As with second order equations we can look for solutions that are a sum of the general solution to the homogeneous problem plus a particular so lution of the nonhomogeneous problem Namely we can write the general solution as xt tC xpt where C is an arbitrary constant vector t is the fundamental matrix solution of x Atx and x Atxp Such a representation is easily verified We need to find the particular solution xt We can do this by applying The Method of Variation of Parameters for Systems We consider a solution in the form of the solution of the homogeneous problem but replace the constant vector by unknown parameter functions Namely we assume that Xpt Btet Differentiating we have that Xp 0c Oc AGc Be or Xp Axp e But the left side is f So we have that c f or since is invertible why o In principle this can be integrated to give c Therefore the particular solu tion can be written as t xpt 8 1ss ds 6101 This is the variation of parameters formula The general solution of Equation 6100 has been found as t xt tC t 1sfs ds 6102 We can use the general solution to find the particular solution of an ini tial value problem consisting of Equation 6100 and the initial condition xto xo This condition is satisfied for a solution of the form t xt tC t 1ss ds 6103 fo LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 251 provided xo xto tgC This can be solved for C as in the last section Inserting the solution back into the general solution 6103 we have t xt t71tgxp t 1sfs ds 6104 to This solution can be written a little neater in terms of the principal matrix solution t t71to t xt txo t 1ss ds 6105 to Finally one further simplification occurs when A is a constant matrix which are the only types of problems we have solved in this chapter In this case we have that t t So computing Yt is relatively easy Example 623 x x 2cost x0 4 x0 0 This example can be solved using the Method of Undetermined Coefficients However we will use the matrix method described in this section First we write the problem in matrix form The system can be written as wey 6106 y x 2cost Thus we have a nonhomogeneous system of the form 1 x Axf 0 0 1 0 y 2cost Next we need the fundamental matrix of solutions of the homoge neous problem We have that A QT 1 0 The eigenvalues of this matrix are A i An eigenvector associated 1 with A is easily found as This leads to a complex solution 1 it cost isint 1 icostsint From this solution we can construct the fundamental solution matrix t in t 1 cost sin sint cost 252 DIFFERENTIAL EQUATIONS So the general solution to the homogeneous problem is t int x tC C1 cos c sin c sint c2 cost Next we seek a particular solution to the nonhomogeneous prob lem From Equation 6103 we see that we need sfs Thus we have 1ss cos s sins 0 sins coss 2 cos s 2 sins coss 610 2 cos s 6107 We now compute t t int tf 2si t sfsds cost sin sin sos ds to sint cost to 2 cos s cost sint sin t sint cost t 5 sin2t tsint 6108 sint tf cost therefore the general solution is x c cost cp sint 4 tsint cy sint cp cost sinttcost The solution to the initial value problem is cost sint 4 tsint xX y sint cost 0 sint tcost or x 4costtsint 3sinttcost LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 253 Problems 1 Consider the system x 4xy y x2y a Determine the second order differential equation satisfied by xt b Solve the differential equation for xt c Using this solution find yf d Verify your solutions for xt and yt e Find a particular solution to the system given the initial conditions x0 1 and y0 0 2 Consider the following systems Determine the families of orbits for each system and sketch several orbits in the phase plane and classify them by their type stable node etc a x 3x yo 2y b x y yo 5x c xi 2y y 3x d x xy yo e xo 2x3y yo 3x42y 3 Use the transformations relating polar and Cartesian coordinates to prove that dol Rae dx de at dtl 254 differential equations 4 Consider the system of equations in Example 613 a Derive the polar form of the system b Solve the radial equation r r1 r2 for the initial values r0 0 05 10 20 c Based upon these solutions plot and describe the behavior of all solutions to the original system in Cartesian coordinates 5 Consider the following systems For each system determine the coeffi cient matrix When possible solve the eigenvalue problem for each matrix and use the eigenvalues and eigenfunctions to provide solutions to the given systems Finally in the common cases which you investigated in Problem 2 make comparisons with your previous answers such as what type of eigenvalues correspond to stable nodes a x 3x y y 2x 2y b x y y 5x c x x y y y d x 2x 3y y 3x 2y e x 4x y y x 2y f x x y y x y LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 255 6 For the given matrix evaluate fA using the definition an Pio Bas A yi A ItArA AP ray 1 2 3 and simplifying 1 0 A b A 1 0 2 2 c A 01 0 1 d A 01 1 0 0 i 01 0 wa 3 tented A 0014 i 0 0 0 7 Find the fundamental matrix solution for the system x Ax where matrix A is given If an initial condition is provided find the solution of the initial value problem using the principal matrix a A 1 0 2 2 b A 12 15 x0 1 4 4 0 2 1 A 4 13 2 d A x0 4 2 A f A 3 8 1 1 8 5 1 A nh ai 2 2 3 5 4 2 i A 45 2 22 2 256 DIFFERENTIAL EQUATIONS 8 Solve the following initial value problems using Equation 6105 the solution of a nonhomogeneous system using the principal matrix solution 2 1 e 1 wx 3 Sar 4 xo 3 5 3 1 1 b x x0 2 1 cos t 0 x x0 3 ae St x0 g Add a third spring connected to mass two in the coupled system shown in Figure 62 to a wall on the far right Assume that the masses are the same and the springs are the same a Model this system with a set of first order differential equations b If the masses are all 20 kg and the spring constants are all 100 Nm then find the general solution for the system c Move mass one to the left of equilibrium 100 cm and mass two to the right 50 cm Let them go find the solution and plot it as a function of time Where is each mass at 50 seconds d Model this initial value problem with a set of two second order differential equations Set up the system in the form Mx Kx and solve using the values in part b 1o In Example 614 we investigated a couple massspring system as a pair of second order differential equations a In that problem we used 3V5 v5 Prove this result b Rewrite the system as a system of four first order equations c Find the eigenvalues and eigenfunctions for the system of equa tions in part b to arrive at the solution found in Example 614 d Let k 500 Nm and m 0250 kg Assume that the masses are initially at rest and plot the positions as a function of time if initially i x10 x20 100 cm and i x10 x20 100 cm Describe the resulting motion 11 Consider the series circuit in Figure 24 with L 100 H R 100 x 107 QO C 100 x 104 F and Vo 100 x 108 V a Set up the problem as a system of two first order differential equa tions for the charge and the current b Suppose that no charge is present and no current is flowing at time t 0 when V is applied Find the current and the charge on the capacitor as functions of time c Plot your solutions and describe how the system behaves over time linear systems of differential equations 257 12 Consider the series circuit in Figure 621 with L 100 H R1 R2 100 102 Ω C 100 104 F and V0 100 103 V a Set up the problem as a system of first order differential equations for the charges and the currents in each loop b Suppose that no charge is present and no current is flowing at time t 0 when V0 is applied Find the current and the charge on the capacitor as functions of time c Plot your solutions and describe how the system behaves over time 13 Initially a 100 gallon tank is filled with pure water At time t 0 water with a half a pound of salt per two gallons is added to the container at the rate of 3 gallons per minute and the wellstirred mixture is drained from the container at the same rate a Find the number of pounds of salt in the container as a function of time b How many minutes does it take for the concentration to reach 2 pounds per gallon c What does the concentration in the container approach for large values of time Does this agree with your intuition 14 You make two quarts of salsa for a party The recipe calls for five teaspoons of lime juice per quart but you had accidentally put in five table spoons per quart You decide to feed your guests the salsa anyway Assume that the guests take a quarter cup of salsa per minute and that you replace what was taken with chopped tomatoes and onions without any lime juice 1 quart 4 cups and 1 Tb 3 tsp a Write down the differential equation and initial condition for the amount of lime juice as a function of time in this mixturetype problem b Solve this initial value problem c How long will it take to get the salsa back to the recipes suggested concentration 15 Consider the chemical reaction leading to the system in 654 Let the rate constants be k1 020 ms1 k2 005 ms1 and k3 010 ms1 What do the eigenvalues of the coefficient matrix say about the behavior of the system Find the solution of the system assuming A0 A0 10 µmol B0 0 and C0 0 Plot the solutions for t 00 to 500 ms and describe what is happening over this time 16 Find and classify any equilibrium points in the Romeo and Juliet prob lem for the following cases Solve the systems and describe their affections as a function of time a a 0 b 2 c 1 d 0 R0 1 J0 1 258 differential equations b a 0 b 2 c 1 d 0 R0 1 J0 1 c a 1 b 2 c 1 d 0 R0 1 J0 1 Figure 626 Figure for Problem 17 A 500L B 1000L 10Lmin 10Lmin 5Lmin 15Lmin 17 Two tanks contain a mixture of water and alcohol with tank A contain ing 500 L and tank B 1000L Initially the concentration of alcohol in Tank A is 0 and that of tank B is 80 Solution leaves tank A into B at a rate of 15 litermin and the solution in tank B returns to A at a rate of 5 Lmin while well mixed solution also leaves the system at 10 litermin through an outlet A mixture of water and alcohol enters tank A at the rate of 10 litermin with the concentration of 10 through an inlet What will be the concentration of the alcohol of the solution in each tank after 10 mins 18 Consider the tank system in Problem 17 Add a third tank C to tank B with a volume of 300 L Connect C with 8 Lmin from tank B and 2 Lmin flow back Let 10 Lmin flow out of the system If the initial concentration is 10 in each tank and a mixture of water and alcohol enters tank A at the rate of 10 litermin with the concentration of 20 through an inlet what will be the concentration of the alcohol in each of the tanks after an hour 19 Consider the epidemic model leading to the system in 658 Choose the constants as a 20 days1 d 30 days1 and r 10 days1 What are the eigenvalues of the coefficient matrix Find the solution of the system assuming an initial population of 1 000 and one infected individual Plot the solutions for t 00 to 50 days and describe what is happening over this time Is this model realistic