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Chapter 4 Systems of Linear Differential Equations

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Chapter 4 Systems of Linear Differential Equations Chapter 4 Systems of Linear Differential Equations Introduction to Systems Up to this point the entries in a vector or matrix have been real numbers. In this section, and in the following sections, we will be dealing with vectors and matri- ces whose entries are functions. A vector whose components are functions is called a vector-valued function or vector function. Similarly, a matrix whose entries are functions is called a matrix function. The operations of vector and matrix addition, multiplication by a number and matrix multiplication for vector and matrix functions are exactly as defined in Chap- ter 5 so there is nothing new in terms of arithmetic. However, there are operations on functions other than arithmetic operations, e.g., limits, differentiation, and integra- tion, that we have to define for vector and matrix functions. These operation from calculus are defined in a natural way. Let v(t)=(f1(t),f2(t),...,fn(t)) be a vector function whose compo- nents are defined on an interval I. Limit: Let c ∈ I. If lim fi(t)=αi exists for i =1, 2,...n, then x→c lim v(t)=lim f1(t), lim f2(t), ..., lim fn(t) =(α1,α2,...,αn) . t→c t→c t→c t→c Limits of vector functions are calculated “component-wise.” Derivative: If f1,f2, ..., fn are differentiable on I, then v is 87 differentiable on I, and 0 0 0 0 v =((f1(t),f2(t), ..., fn(t)) . That is, v0 is the vector function whose components are the derivatives of the components of v. Integration: Since differentiation of vector functions is done component- wise, integration must also be component-wise. That is Z Z Z Z v(t) dt = f1(t) dt, f2(t) dt, . , fn(t) dt . Limits, differentiation and integration of matrix functions is done in ex- actly the same way, component-wise. 4.1. Systems of Linear Differential Equations Consider the third-order linear differential equation y000 + p(t)y00 + q(t)y0 + r(t)y = f(t) where p, q, r, f are continuous functions on some interval I. Solving the equation for y000, we get y000 = −r(t)y − q(t)y0 − p(t)y00 + f(t). Introduce new dependent variables x1,x2,x3, as follows: x1 = y 0 0 x2 = x1 (= y ) 0 00 x3 = x2 (= y ) Then 000 0 y = x3 = −r(t)x1 − q(t)x2 − p(t)x3 + f(t) and the third-order equation can be written equivalently as a system of three first- order equations: 0 x1 = x2 0 x2 = x3 0 x3 = −r(t)x1 − q(t)x2 − p(t)x3 + f(t) 88 Example 1. (a) Consider the third-order nonhomgeneous equation y000 − y00 − 8y0 +12y =2et. Solving the equation for y000, we have y000 = −12y +8y0 + y00 +2et. 0 0 0 00 Let x1 = y, x1 = x2 (= y ),x2 = x3 (= y ). Then 000 0 t y = x3 = −12x1 +8x2 + x3 +2e and the equation converts to the equivalent system: 0 x1 = x2 0 x2 = x3 0 t x3 = −12x1 +8x2 + x3 +2e Note: This system is just a very special case of the “general” system of three, first-order differential equations: 0 x1 = a11(t)x1 + a12(t)x2 + a13(t)x3(t)+b1(t) 0 x2 = a21(t)x1 + a22(t)x2 + a23(t)x3(t)+b2(t) 0 x3 = a31(t)x1 + a32(t)x2 + a33(t)x3(t)+b3(t) (b) Consider the second-order homogeneous equation t2y00 − ty0 − 3y =0. Solving this equation for y00, we get 3 1 y00 = y + y0. t2 t 0 0 To convert this equation to an equivalent system, we let x1 = y, x1 = x2 (= y ). Then we have 0 x1 = x2 3 1 x0 = x + x 2 t2 1 t 2 which is just a special case of the general system of two first-order differential equa- tions: 0 x1 = a11(t)x1 + a12(t)x2 + b1(t) 0 x2 = a21(t)x1 + a22(t)x2 + b2(t) 89 General Theory Let a11(t),a12(t),...,a1n(t),a21(t),...,ann(t),b1(t),b2(t), ..., bn(t) be contin- uous functions on some interval I. The system of n first-order differential equations 0 x1 = a11(t)x1 + a12(t)x2 + ···+ a1n(t)xn(t)+b1(t) 0 x2 = a21(t)x1 + a22(t)x2 + ···+ a2n(t)xn(t)+b2(t) . (S) . 0 xn = an1(t)x1 + an2(t)x2 + ···+ ann(t)xn(t)+bn(t) is called a first-order linear differential system. The system (S) is homogeneous if b1(t) ≡ b2(t) ≡···≡bn(t) ≡ 0onI. (S) is nonhomogeneous if the functions bi(t) are not all identically zero on I; that is, if there is at least one point a ∈ I and at least one function bi(t) such that bi(a) =0.6 Let A(t) be the n × n matrix a11(t) a12(t) ··· a1n(t) a21(t) a22(t) ··· a2n(t) A(t)= . . an1(t) an2(t) ··· ann(t) and let x and b be the vectors x1 b1 x2 b2 x = . , b = . . . . . xn bn Then (S) can be written in the vector-matrix form x0 = A(t) x + b. (S) The matrix A(t) is called the matrix of coefficients or the coefficient matrix. Example 2. The vector-matrix form of the system in Example 1(a) is: 010 0 0 x = 001x + 0 , −1281 2et 90 a nonhomogeneous system. The vector-matrix form of the system in Example 1(b) is: 0 01! 0 ! 01! x1 ! x = 2 x + = 2 x, where x = , 3/t 1/t 0 3/t 1/t x2 a homogeneous system. A solution of the linear differential system (S) is a differentiable vector function x1(t) x2(t) x(t)= . . . xn(t) that satisfies (S) on the interval I. 2t 1 t e 2 e Example 3. Verify that x(t)= 2t + 1 t is a solution of the nonho- 2e 2 e 2t 1 t 4e 2 e mogeneous system 010 0 0 x = 001x + 0 −1281 2et of Example 2. 91 SOLUTION 0 2t 1 t e 2 e x0 = 2t + 1 t 2e 2 e 2t 1 t 4e 2 e 2t 1 t 2e 2 e = 2t + 1 t 4e 2 e 2t 1 t 8e 2 e 2t 1 t 010 e 2 e 0 =? 001 2e2t + 1 et + 0 2 − 2t 1 t t 1281 4e 2 e 2e 2t 1 t 010 e 010 2 e 0 =? 2t + 1 t + 001 2e 001 2 e 0 − 2t − 1 t t 1281 4e 1281 2 e 2e 2t 1 t 2e 2 e 0 = 2t + 1 t + 4e 2 e 0 2t − 3 t t 8e 2 e 2e 2t 1 t 2e 2 e = 2t + 1 t . 4e 2 e 2t 1 t 8e 2 e x is a solution. THEOREM 1. (Existence and Uniqueness Theorem) Let a be any point on the interval I, and let α1,α2,...,αn be any n real numbers. Then the initial-value problem α1 0 α2 x = A(t) x + b(t), x(a)= . . . αn has a unique solution. Exercises 4.1 Convert the differential equation into a system of first-order equations. 1. y00 − ty0 +3y = sin 2t. 92 2. y00 + y =2e−2t. 3. y000 − y00 + y = et. 4. my00 + cy0 + ky = cos λt, m, c, k, λ are constants. Write the system in vector-matrix form. 5. 0 x1 = −2x1 + x2 + sin t 0 x2 = x1 − 3x2 − 2 cos t 6. 0 t 2t x1 = e x1 − e x2 0 −t t x2 = e x1 − 3e x2 7. 0 2t x1 =2x1 + x2 +3x3 +3e 0 x2 = x1 − 3x2 − 2 cos t 0 x3 =2x1 − x2 +4x3 + t 8. 0 2 x1 = t x1 + x2 − tx3 +3 0 t −2t x2 = −3e x2 +2x3 − 2e 0 2 x3 =2x1 + t x2 +4x3 t−1 ! 9. Verify that u(t)= is a solution of the system in Example 1 (b). −t−2 −3t 1 t e 2 e 10. Verify that u(t)= −3e−3t + 1 et is a solution of the system in 2 −3t 1 t 9e 2 e Example 1 (a). te2t 2t 2t 11. Verify that w(t)= e +2te is a solution of the homogeneous system 4e2t +4te2t associated with the system in Example 1 (a). 93 − sin t ! 12. Verify that x(t)= is a solution of the system − cos t − 2 sin t −21! 0 ! x0 = x + . −32 2 sin t −2e−2t 13. Verify that x(t)= 0 is a solution of the system 3e−2t 1 −32 0 x = 0 −10x. 0 −1 −2 4.2. Homogeneous Systems In this section we give the basic theory for linear homogeneous systems. This “theory” is simply a repetition results given in Sections 3.2 and 3.6, phrased this time in terms of the system 0 x1 = a11(t)x1 + a12(t)x2 + ···+ a1n(t)xn(t) 0 x2 = a21(t)x1 + a22(t)x2 + ···+ a2n(t)xn(t) . (H) . 0 xn = an1(t)x1 + an2(t)x2 + ···+ ann(t)xn(t) or x0 = A(t)x. (H) 0 0 Note first that the zero vector z(t) ≡ 0 = . is a solution of (H). As before, . . 0 this solution is called the trivial solution. Of course, we are interested in finding nontrivial solutions. THEOREM 1. If x1, x2,...,xk are solutions of (H), and if c1,c2,...,ck are real numbers, then c1x1 + c2x2 + ···+ ckxk is a solution of (H); any linear combination of solutions of (H) is also a solution of (H).
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