m2k_dfq_inhosy

m2k_dfq_inhosy - Inhomogeneous Linear Systems General...

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Unformatted text preview: Inhomogeneous Linear Systems General Properties tions has the form An inhomogeneous system of linear differential equa- dY = A(t) Y + G(t), dt where A(t) is a known n × n matrix function of t and G(t) is a known n- vector function of t. The familiar rules relating solutions of homogeneous and inhomogeneous equations apply. Thus if Y (t, C ) is the general solution of the homogeneous system, the general solution of the inhomogeneous system takes the form Y (t, C ) + Yp (t), where Yp (t) is any vector solution of the inhomogeneous system. That being the case, the next order of business must be to develop some techniques for obtaining particular solutions of inhomogeneous linear systems. The first of these is the Method of Variation of Parameters We have seen that, for any fundamental matrix solution Y(t), the general solution of the linear homogeneous system Y (t) = A(t) Y (t) may be taken to be Y(t) C where C is an arbitrary n-vector. Following our development of the variation of parameters formula in the first and second order scalar cases, it would now be reasonable to think we might find a particular solution of the inhomogeneous system shown above in the form Yp (t) = Y(t) C (t) where now C (t) is an n-vector function of t. Substituting this solution form into the inhomogeneous system we obtain Y (t) C (t) + Y(t) C (t) = A(t) Y(t) C (t) + G(t). Since Y (t) = A(t) Y(t), this gives Y(t) C (t) = G(t) ⇒ C (t) = Y(t)−1 G(t). Then, integrating, we have ˆ C (t ) = C + t Y(τ )G(τ ) dτ ⇒ 1 t ˆ Yp (t) = Y(t) C + Y(t) Y(τ )−1 G(τ ) dτ ˆ for an arbitrary constant n-vector C . We can obtain a particular solution for ˆ any choice of C and any choice of the lower limit of the integral. With the ˆ choice C = 0 we obtain the general solution of the inhomogeneous system in the form t Y (t, C ) = Y(t) C + Y(t) Y(τ )−1 G(τ ) dτ for an arbitrary n-vector C . If we want to solve an initial value problem Y (t) = Y0 we take the lower limit of the integral to be t0. Then Y0 = Y (t0, C ) = Y(t0 ) C + 0 ⇒ C = Y(t0 )−1 Y0 and we have the solution of the initial value problem in the form Y (t) = Y(t) Y(t0 )−1 Y0 + Y(t) t Y(τ )−1 G(τ ) dτ t0 t = Y(t, t0) Y0 + Y(t, τ ) G(τ ) dτ, t0 where we recall that Y(τ, τ ) = Y(t, t) = In . This is the variation of parameters formula for a linear inhomogeneous system with possibly time varying system matrix A(t). To check, we verify the initial condition t0 Y (t 0 ) = Y (t 0 , t 0 ) + t0 Y(t0 , τ ) G(τ ) dτ = I Y0 + 0 = Y0 and satisfaction of the differential system: t Y (t) = Y (t) Y0 + Y(t, t)G(t) + t0 d Y(t, τ ) G(τ ) dτ = A(t) Y(t) Y(t) Y0 dt t + I G(t) + A(t) Y(t, τ ) G(τ ) dτ = A(t) Y (t) + G(t). t0 Example 1 We recall the inhomogeneous system from Example 1 of the introductory linear systems section of the notes: d dt y1 y2 = 1/t 1 0 2/t 2 y1 y2 + t . t2 For the corresponding homogeneous system we obtained the fundamental matrix solution Y (t ) = t t3/2 − t/2 . 0 t2 This gives, for the inverse matrix at t = τ and Y(t, τ ) = Y(t)Y(τ )−1 , Y(τ )−1 = 1/(2τ 2 ) − 1/2 , Y(t, τ ) = 1/τ 2 1/τ 0 t/τ 0 t3/(2τ 2 ) − t/2 . t2/τ 2 Then the solution of the inhomogeneous system shown above with initial condition y 1(2) = 1, y 2(2) = −1 is given, using the variation of parameters formula, by Y (t ) = = t/2 t3/8 − t/2 0 t2 /4 t − t3/8 + −t 2/4 t 2 1 −1 t/τ 0 t + t + t3/2 − t τ 2/2 t2 2 dτ = t3 /(2τ 2 ) − t/2 t2/τ 2 τ τ2 dτ t4 /3 − 7t3 /8 + t2 − 2 t/3 . t3 − 9 t2 /4 For a linear inhomogeneous constant coefficient system and initial condition dY = A Y + G(t), Y (t0 ) = Y0 , dt the variation of parameters formula shown previously takes the form Y (t) = eA(t−t0) Y0 + t eA(t−τ ) G(τ ) dτ. t0 It can be used to compute solutions, quite usefully if G(t) has some nonstandard functional form, but it is mostly used as an abstract representation formula. More widely used is the Method of Undetermined Coefficients The method of undetermined coefficients, for an inhomogeneous system with constant n × n matrix A, as just shown, works much the same way for linear inhomogeneous systems as it does for scalar linear constant coefficient inhomogeneous differential equations, except that the ”undetermined coefficients” are now vectors of dimension n. 3 We consider the inhomogeneous system Example 1 d dt y1 y2 11 −1 2 = y1 y2 t . t2 + Since the finite family associated with t2 consists of that function together with t and 1 we try a solution y 1 (t ) y 2 (t ) = t2 a1 a2 b1 b2 +t c1 . c2 +1 Substituting this solution form into the system we obtain 2t 11 −1 2 a1 a2 t2 a1 a2 + b1 b2 +t b1 b2 +1 = c1 c2 1 0 +t + t2 0 . 1 Equating the vector coefficients of t2, t and 1, respectively, we obtain the equations 11 a1 0 0 = ; + −1 2 a2 1 0 11 −1 2 b1 b2 1 0 + 11 −1 2 c1 c2 −2 a1 a2 b1 b2 = 1 3 = 1 3 2 −1 11 2 −1 11 = 0 . 0 11 −1 2 We can see easily that the inverse of we have = b1 b2 − a1 a2 1 −1 2 3 0 −1 − is = 1 3 1 0 1 3 0 ; 0 2 −1 . Using that 11 1 ; −1 = 1 3 0 ; −1 1 2 −1 1 0 1 1 = . 3 −1 311 9 −1 Using these vectors in the original solution form, we see that we have the particular solution c1 c2 y 1 (t ) y 2 (t ) = t2 3 = 1 −1 + t 3 0 −1 + 4 1 9 1 −1 = t2/3 + 1/9 . −t2/3 − t/3 − 1/9 This method works with inhomogeneous vector functions G(t) which are polynomials in t as long as the system matrix A is nonsingular; i.e., has determinant different from zero and is therefore invertible. The vector version of the method of undetermined coefficients has a particularly neat form when the inhomogeneous term is a constant vector times an exponential. Thus, if G(t) = ert B , where B is a given constant vector, the general linear inhomogeneous constant coefficient system becomes dY = A Y + ert B. dt The method of undetermined coefficients in this case corresponds to trying for a particular vector solution of the form, for some constant vector C , Yp (t) = ert C. Substituting this solution form into the system we have r ert C = A ert C + ert B ⇒ (r I − A) C = B. This system has the unique solution C = (r I − A)−1 B if and only if det(r I − A) = 0, that is, if and only if r is not an eigenvalue of A, just as in the scalar inhomogeneous cases where r is not a root of the characteristic polynomial associated with the homogeneous equation. The particular solution in the present case is then Yp (t) = ert (r I − A) Example 2 −1 B. We consider a system with the same homogeneous part as in the first example; namely: d dt y1 y2 = 11 −1 2 5 y1 y2 + ert 1 , 2 where r is any real number. We can verify quite immediately that the eigen√ values of the matrix appearing here are −3 ± i 3 /2, so a real r will never be an eigenvalue of the matrix. We try for a particular solution y 1 (t ) y 2 (t ) = ert c1 c2 c1 c2 = and obtain the equation r − 1 −1 1 r−2 1 . 2 Solving for c1 and c2 we have c1 = r2 r 2r − 3 , c2 = 2 − 3r + 3 r − 3r + 3 and thus the particular solution is y 1 (t ) y 2 (t ) ert =2 r − 3r + 3 r . 2r − 3 This is then added to the general solution of the corresponding homogeneous system to obtain the general solution of the inhomogeneous system. Linear constant coefficient systems with inhomogeneous terms of the form G(t) = eρ t cos σ t B or G(t) = eρ t sin σ t B can be treated in two equivalent ways. The first is to try for a particular solution in the form Yp (t) = eρt cos σt C + eρt sin σt D, Where C and D are vectors to be determined. We substitute this form into the inhomogeneous system and, assuming ρ ± i σ are not eigenvalues of A, identify C and D. The second way is to realize that eρt cos σt B and eρt sin σt B are the real and imaginary parts, respectively, of e(ρ+iσ)t B . We would then replace the original inhomogeneous system by dZ = A Z + e(ρ+iσ)t B dt and, proceeding as before, obtain the complex solution Z (t) = e(ρ+iσ)t (r I − A) 6 −1 B. The desired particular solution Yp (t) is then the real part of Z (t) in the cos σt case and the imaginary part of this solution in the sin σt case. The arithmetic is much the same either way but it is a little easier to organize the work following the complex solution route. Example 3 Consider the inhomogeneous system d dt y1 y2 11 −1 2 = y1 y2 + sin ωt 1 . 1 We replace this system by the complex inhomogeneous system d dt z1 z2 = 11 −1 2 z1 z2 1 . 1 + eiωt Using the formula developed earlier a solution is then Z (t) = eiωt where c1 c2 = (i ω )2 = 1 − 3ω i + 3 = = iω − 1 −1 1 iω − 2 1 (3 − ω 2 )2 + 9 ω 2 −1 1 1 iω − 2 1 −1 iω − 1 3 − ω 2 + 3ω i (3 − ω 2 )2 + 9 ω 2 c1 , c2 1 1 iω − 1 iω − 2 − (3 − ω 2 )2 − 3 ω 2 − 3 ω i + (3 − ω 2)ω i . −2 (3 − ω 2 ) − 3 ω 2 − 6 ω i + (3 − ω 2 )ω i Multiplying this vector by eiωt and taking the imaginary part of the result we obtain a particular solution of the original system in the form Yp (t) = y 1 (t ) y 2 (t ) − = cos ωt (3 − ω 2)2 + 9 ω 2 − 3 ω + (3 − ω 2 )ω −6 ω + (3 − ω 2 )ω (3 − ω 2 ) + 3 ω 2 2(3 − ω 2 ) + 3 ω 2 sin ωt 2 (3 − ω 2 ) + 9 ω 2 . One would again add this to the general solution of the corresponding homogeneous system to obtain the general solution Y (t, C ) of the inhomogeneous system. 7 ...
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