Systems of Linear Differential Equations Chapter 7 Doreen De Leon Department of Mathematics, California State University, Fresno June 22, 25 Motivating Examples: Applications of Systems of First Order Differential Equations Taken from an example in A First Course in Differential Equations with Modeling Applications, 8th ed, by Dennis G Zill Radioactive Series In our earlier discussion of radioactive decay, we assumed that the rate of decay was proportional to the number y(t of nuclei of the substance present at time t When a substance decays by radioactivity, it usually does not just transmute into one stable substance and then the proces stops Rather, the first substance decays into another radioactive substance, this substance in turn decays into a third radioactive substance, and so on This process, called a radioactive decay series, continues until a stable element is reached For example, the uranium decay series is U-238 Th-234 Pb-26, where Pb-26 is a stable isotope of lead The half-lives of the various elements in a radioactive series can range from billions of years (eg, 45 9 years for U-238 to a fraction of a second Suppose a radioactive series is described schematically by X λ Y λ 2 Z, where k λ < and k 2 λ 2 < are the decay constants for substances X and Y, respectively, and Z is a stable element Suppose that x(t, y(t, and z(t denote the amounts of substances X, Y, and Z, respectively, remaining at time t The decay of element X is described by dx dt λ x However, since X decays into Y (so Y is gaining atoms and Y is, itself, decaying (thereby losing atoms, the rate at which element Y decays is the net rate dy dt λ x λ 2 y
Since Z is a stable element, it is simply gaining atoms from the decay of element Y, so dz dt λ 2y In other words, a model of the radioactive decay series for the three elements is the linear system of first-order differential equations, 2 Electrical Networks dx dt λ x dy dt λ x λ 2 y dz dt λ 2y An electrical network having more than one loop also gives rise to a system of differential equations For example, in the figure below the current i (t splits in the directions shown at point B, called a branch point of the network By Kirchhoff s first law, we have i (t i 2 (t + i 3 (t ( Applying Kirchhoff s second law to each loop gives the following For loop A B B 2 A 2 A, summing the voltage drops across each part of the loop gives Similarly, for loop A B C C 2 B 2 A 2 A, we see that E(t i R + L di 2 dt + i 2R 2 (2 E(t i R + L 2 di 3 dt (3 2
We can use Equation ( to eliminate i from Equations (2 and (3 to obtain the following system of equations: L di 2 dt + (R + R 2 i 2 + R 3 i 3 E(t (4 L 2 di 3 dt + R i 2 + R i 3 E(t (5 2 First Order Systems of Differential Equations Section 7 We can write a higher order linear differential equation as a system of first order equations Note: The number of equations in the system equals the order of the differential equation Example: Write as a system of first order equations: x + p(tx + q(tx f(t So, the solution is Let x x x x x 2 x 2 x x 2 x q(tx p(tx f(t x 2 q(tx p(tx 2 f(t x x 2 x 2 q(tx p(tx 2 + f(t Our goal in this chapter will be to solve systems of first order constant coefficient differential equations using eigenvalue-eigenvector theory First, we must write the systems of differential equations in matrix form 3 Matrices and Systems of Linear Differential Equations Section 72 We can write a system of first order differential equations in matrix form Examples: ( Write the system below in matrix form x 2x 3x 2 x 2 4x + 2x 2 3
(2 Write the system below in matrix form ( x 2 3 x x 2 4 2 x 2 dx ( 2 3 dt x 4 2 ( x x 2 x x 2 x 2 q(tx p(tx 2 + f(t x + q(t p(t x 2 ( f(t Consider the homogeneous differential equation dx dt P (tx(t (6 Theorem (Superposition Principle Given x (t, x 2 (t,, x n (t n solutions to (6 on the open interval I, if c, c 2,, c n are constants, then is also a solution of (6 on I Proof Since x (t, x 2 (t,, x n (t solve (6, So, x(t c x (t + c 2 x 2 (t + + c n x n (t x P (tx, x 2 P (tx 2,, x n P (tx n x c x + c 2 x 2 + + c n x n c P (tx + c 2 P (tx 2 + + c n P (tx n P (t (c x + c 2 x 2 + + c n x n P (tx We can check the linear independence of solutions using the Wronskian If x, x 2,, x n are solutions of (6, their Wronskian is the determinant of the matrix whose columns are the solution vectors Theorem 2 Suppose x, x 2,, x n are solutions of the n n system (6 on an open interval I Suppose P (t is continuous on I Let W W (t Then x, x 2,, x n are linearly independent if and only if W for all t I They are linearly dependent if W on I 4
The dimension of the solution space of (6 equals the number of variables If x, x 2,, x n are n linearly independent solutions of the n n system (6, then the general solution is x(t c x (t + c 2 x 2 (t + + c n x n (t Initial Value Problems Example: Given the initial value problem 3 2 x x, x( : 3 4 5 e 3t 2e 2t a Verify that x and x 2 are solutions 3e 3t e 2t b Show that x, x 2 are linearly independent c Solve the initial value problem Solution: a 3e x 3t 3 2 e 3t 3e 3t 9e 3t 3 4 3e 3t 9e 3t x 4e x 2t 3 2 2e 2t 4e 2t 2 2e 2t 3 4 e 2t x 2 2e 2t b W (t c e3t 3e 3t 2e 2t e 2t et 6e t 5e t, so x and x 2 are linearly independent e 3t 2e 2t x c x + c 2 x 2 c 3e 3t + c 2 e 2t ( 2 x( c 5 + c 3 2 c + 2c 2 3c + c 2 5 c 2 c 2 So, ( e 3t x 2 3e 3t ( ( 2e 2t 2e 3t 2e 2t 6e 3t e 2t e 2t 5
4 The Eigenvalue Method for Systems of First Order Linear Differential Equations Section 73 Goal: Solve x Ax, < t < (A a constant matrix Idea: Search for solutions of the form x e λt v Why? Consider the separable equation y λy The solution to this equation is y ce λt Plugging x into the system gives λe λt v Ae λt v λv Av Therefore, λ and v are the eigenvalues and corresponding eigenvectors of the matrix A Possibilities: Distinct eigenvalues Section 73 2 Complex conjugate eigenvalues Section 73 3 Repeated eigenvalues Section 75 4 Distinct Eigenvalues Procedure: Find the eigenvalues of A Solve det(a λi 2 For each eigenvalue, find corresponding eigenvector(s Find a basis for the solution space of (A λiv 3 Linear combinations yield general solutions Examples: ( Find a general solution of ( 4 x x 3 8 det(a λi λ 4 3 8 λ (λ 2(λ 6 λ 2, λ 2 6 λ(8 λ + 2 λ 2 8λ + 2 6
λ 2 Find a basis for the solution space of (A 2Iv In matrix form, we have 2 4 v 3 6 v 2 2 4 r r 2 4 3 6 3 6 So, we have 2v + 4v 2 v 2v 2 ( r 2 r 2 3 2 r ( 2 4 v 2 is the only arbitrary variable, so we may choose a value for it Let v 2 v 2, and ( 2 v Then, one solution is ( 2 x e λt v e 2t λ 2 6 Find a basis for the solution space of (A 6Iu In matrix form, we have ( 6 4 u 3 2 u 2 ( 6 4 r 2 r 3 2 r 2 r 2 +2r 3 2 3 2 6 4 So, we have 3u + 2u 2 u 2 3 u 2 u 2 is the only arbitrary variable, so let u 2 3 u 2, and ( 2 u 3 Then, a second solution is The general solution is x c x + c 2 x 2 ( 2 x 2 e λ2t u e 6t 3 2 2 Therefore, x c e 2t + c 2 e 6t 3 7
(2 Solve the initial value problem 3 2 2 x x, x( 6 General solution: 3 λ 2 det(a λi λ 6 λ (3 λ( λ(6 λ λ,λ 2 3, λ 3 6 λ Find a basis for the solution space of (A Iv In matrix form, we have 2 2 v v 2 5 v 3 2 2 2 2 r 3 r 3 5r 2 5 So we have v 3 2v + 2v 2 + v 3 v v 2 v 2 arbitrary, so let v 2 v and v Then, one solution is x e t λ 2 3 Find a basis for the solution space of (A 3Iu In matrix form, we have 2 u 2 u 2 3 u 3 8
2 2 3 So we have 2 2 2 r 3 r 3 3 2 r 2 2 3 r 2 r 2 +r u 3 2u 2 + u 3 u 2 u is arbitrary, so let u u Then, a second solution is λ 3 6 Solve (A 6Iw x 2 e 3t In matrix form, solve 3 2 w 5 w 2 w 3 The coefficient matrix is already in echelon form, so we have 5w 2 + w 3 w 2 5 w 3 3w + 2w 2 + w 3 w 2 3 w 2 + 3 w 3 w 3 is arbitrary, so since w 3 will be divided by both 3 and 5, let w 3 5 (eliminates fractions 7 w 2 3 and w 2(3 + (5 7 Therefore, w 3 3 3 5 Then, a third solution is 7 x 3 e 6t 3 5 General solution: x c x + c 2 x 2 + c 3 x 3 7 x c e t + c 2 e 3t + c 3 e 6t 3 5 9
Find c, c 2, c 3 : So, Therefore, or 2 7 c + c 2 + c 3 3 5 c + c 2 + 7c 3 2 c 2 3 c + 3c 3 c 5c 3 c 3 x e t + 3e 3t, e t + 3e 3t x e t 42 Complex Eigenvalues Goal: Obtain real-valued solutions Suppose λ a + bi, λ a bi are two eigenvalues corresponding to λ, then If v u + iw is an eigenvector (A λiv (A λiv (A λiv In other words, v is an eigenvector corresponding to λ The solution associated with λ and v is x ve λt (u + iwe (a+bit (u + iwe at (cos(bt + i sin(bt e at (u cos(bt w sin(bt +i e at (w cos(bt + u sin(bt }{{}}{{} x x 2
So, x and x 2 are also solutions to the system Therefore, we have two linearly independent real-valued solutions, x e at (u cos(bt w sin(bt and x 2 e at (w cos(bt + u sin(bt We get the same solutions by looking at ve λt So, we need only find one eigenvector corresponding to one of the complex conjugate eigenvalue pair in order to obtain a general solution Example: Solve ( 4 x x 5 det(a λi λ 4 5 λ ( λ( 5 λ + 4 λ 2 + 6λ + 45 λ 6 ± 36 4(45 2 6 ± 2i 2 3 ± 6i λ 3 + 6i Find a basis for the solution space of (A λiv In matrix form, we have ( ( 3 + 6i 4 ( v 5 ( 3 + 6i v 2 ( 2 6i 4 v 2 6i We can show that the augmented matrix can be reduced to ( 2 6i 4 Therefore, we have v 2 (2 6iv + 4v 2 ( v 2 is the only arbitrary variable, so let v 2 2 6i v 4 Thus, ( 4 v 2 6i
( ( 4 4 e λt v e ( 3+6it e 3t (cos(6t + i sin(6t 2 6i 2 6i ( e 3t 4 cos(6t + i( 4 sin(6t 2 cos(6t + 6 sin(6t + i(2 sin(6t 6 cos(6t ( ( e 3t 4 cos(6t +i e 3t 4 sin(6t General solution: x c x + c 2 x 2, or 2 cos(6t + 6 sin(6t }{{} x 2 sin(6t 6 cos(6t }{{} x 2 ( ( x c e 3t 4 cos(6t + c 2 cos(6t + 6 sin(6t 2 e 3t 4 sin(6t 2 sin(6y 6 cos(6t 43 Repeated Eigenvalues Section 75 Definitions: Algebraic multiplicity: the order of the root Example:(λ 3 λ has algebraic multiplicity 3 Geometric multiplicity: the dimension of the eigenspace associated with an eigenvalue Case : algebraic multiplicity geometric multiplicty In this case, we obtain an eigenvector for each instance of the eigenvalue same solution strategy as for distinct eigenvalues Example: Find a general solution of x ( 3 x 3 det(a λi 3 λ 3 λ (3 λ 2 λ 3 (multiplicity 2 λ 3 Find a basis for the solution space of (A 3Iv In matrix form, we have v v 2 ( 2
So, both v and v 2 are arbitrary we have two linearly independent eigenvectors Let v r, v 2 s Then ( r v s r + s r + s The basis for the eigenspace gives us the two eigenvectors we seek: {}, Therefore, the general solution is x c e 3t + c 2 e 3t Case 2: algebraic multiplicity > geometric multiplicity For the case of a double root λ, this means that we will only obtain one eigenvector v, giving us only one solution, x e λt v We need to find a second solution, x 2 To do this, we will try a product of an exponential with a polynomial Let x 2 e λt (tu + w, where u and w are determined so that x 2 is a solution of x Ax Plug x 2 into the system: Divide both sides by e λt to get: Equating coefficients of like terms gives: x 2 e λt tu + e λt w x 2 λe λt tu + e λt u + λe λt w x 2 Ax 2 λe λt tu + e λt u + λe λt w A ( e λt tu + e λt w e λt tau + e λt Aw e λt (λtu + (u + λw e λt (tau + Aw λtu + (u + λw tau + Aw Au λu Aw λw + u (A λiw u 3
The first equation tells us that u is an eigenvector corresponding to λ, so let u v To summarize, we do the following: Determine the eigenvalues 2 Given an eigenvalue λ of multiplicity 2: (a If the dimension of the associated eigenspace is : i Find an eigenvector, v, giving one solution, x e λt v ii Find a second solution x 2 e λt (tv + w, where w solves (A λiw v (7 (b If the dimension of the associated eigenspace is 2, follow the standard procedure 3 For eigenvalues of multiplicity, follow the standard procedure Example: Solve x ( 4 x, x( 4 8 General solution: det(a λi λ 4 4 8 λ λ(8 λ + 6 λ 2 8λ + 6 ( 2 (λ 4 2 λ 4 (multiplicity 2 λ 4 Find a basis for the solution space of (A 4Iv In matrix form, we have 4 4 v 4 4 This gives v 2 4v + 4v 2 v v 2 ( d v 2 is arbitrary, so let v 2 v So, ( ( v and x e 4t is one solution Next, find a second solution, x 2 e 4t (tv w, 4
where w solves (A 4Iw v, or 4 4 w 4 4 Solving, ( 4 4 4 4 So, w 2 r 2 r 2 +r 4w 4w 2 w 2 is arbitrary, so let w 2 w 4 So, and General solution: x c x + c 2 x 2, so Find c and c 2 : x( ( w 4, ( x 2 e 4t (tv + w ( ( e (t 4t + 4 ( t + e 4t 4 t x c e 4t ( ( 2 So, c 2 and c + 4 c 2 c 2 4c 8 So, x 2e 4t ( ( t + + 8e 4t 4 t ( 4 4 ( t + + c 2 e 4t 4 t ( c 2 + c 2 4 5