Theory of Linear Systems

Theory of Linear Systems This proves the two fundamental results concerning the theory of homogeneous linear systems X 0 = A(t) X + f (t) . The coefficient A(t) is given as a continuous N × N matrix valued function on this interval and f is continuous both for 0 ≤ t < ∞. 1 No blow up theorem Extending A to be independent of t for −∞ < t ≤ 0 and for T ≤ t < ∞ we may suppose that A is defined and continuous for all t. The equation then has the form X 0 = F (t, X) with F continuous in t and continuously differentiable in X on all of R × RN . The fundamental existence and uniqueness theorem implies that for any X0 ∈ RN , the there is a unique maximal solution X(t) defined for 0 ≤ t ≤ T∗ satisfying the initial condition X(0) = X0 . In the nonlinear case one can have T∗ < ∞. For linear equations this cannot occur unless the coefficients are singular. Theorem 1.1 T∗ = ∞. In particular the solution exists throughout the interval of interest 0 ≤ t ≤ T. Proof. The fundamental existence theorem shows that if T∗ < ∞ then for any closed bounded set Γ ∈ RN there is a t < T∗ so that the solution takes values outside Γ for t ∈ [t, T∗ [. We show that this is not possible by finding an R so that for 0 ≤ t < T∗ one has kX(t)k ≤ R. With Γ := {|X| ≤ R} the blow up criterion is violated. The strategy is to bound the growth of the square of the norm. Toward that end compute the derivative, 0 X(t) , X(t) = X 0 (t) , X(t) + X(t) , X 0 (t) . Apply the Cauchy-Schwartz inequality twice to find X(t) , X(t) 0 ≤ 2 kXk kX 0 k . Estimate kX 0 k = kAX + f k ≤ kAXk + kf k ≤ kA(t)kkXk + kf (t)k . Define a := max kA(t)k , b := max kf (t)k [0,T∗ ] Combining yields X(t) , X(t) [0,T∗ ] 0 ≤ 2akX(t)k2 + 2bkXk . Estimate 2bkXk ≤ b2 + kXk2 so ϕ(t) := X(t) , X(t) satisfies ϕ0 ≤ (2a + 1)ϕ + b2 . 1 Employing the method of integrating factors,  
0 e−(2a+1)t ϕ = e−(2a+1)t ϕ0 − (2a + 1)e−(2a+1)t ϕ = e−(2a+1)t ϕ0 − (2a + 1)ϕ ≤ b2 e−(2a+1)t . Integrate from t = 0 to t to find −(2a+1)t e Z t ϕ(t) − ϕ(0) ≤ b2 e−(2a+1)t dt . 0 Therefore for 0 ≤ t < T∗ multiplying by e(2a+1)t shows that Z T∗   2 (2a+1)T∗ b2 e−(2a+1)t dt := R2 . kX(t)k = ϕ(t) ≤ e ϕ(0) + 0 completing the proof. Alternate Proof. The fundamental existence theorem shows that if T∗ < ∞ then for any closed bounded set Γ ∈ RN there is a t < T∗ so that the solution takes values outside Γ for t ∈ [t, T∗ [. We show that this is not possible by finding an R so that for 0 ≤ t < T∗ one has kX(t)k ≤ R. With Γ := {|X| ≤ R} the blow up criterion is violated. The fundamental theorem of calculus implies that for 0 ≤ t < T∗ Z t Z t 0 X(t) − X(0) = X (s) ds = A(s) X(s) ds . 0 0 The triangle inequality then implies that Z t Z t kX(t)k ≤ kX0 k + k A(s) X(s) dsk ≤ kX0 k + kA(s) X(s)k ds . 0 0 Let K := max kA(t)k < ∞, and 0≤t≤T C := kX0 k . K is finite because of the continuity of A and the fact that [0, T ] is closed and bounded. Then kA(s)X(s)k ≤ KkX(s)k so u(t) := kX(t)k satisfies for 0 ≤ t < T∗ Z t u(t) ≤ C + K u(s) ds . 0 Since u is continuous and nonnegative Gronwall’s lemma implies that for 0 ≤ t < T∗ one has u(t) ≤ C eKt . Setting R := C eKT∗ proves the desired conclusion. 2 Theorem on general solutions The superposition principal asserts that if Φj (t) for 1 ≤ j ≤ k are k solutions and cj are k scalars then c1 Φ1 (t) + c2 Φ2 (t) + · · · + ck Φk (t) is a k parameter family of solutions. The general solution is determined by its initial condition so is an N parameter family of solutions suggesting that for k = N one might get the general solution. The next result tells under what conditions, k = N solutions yield the general solution of the homogeneous linear system. 2 Theorem 2.1 If Φj (t) is a solution for 1 ≤ j ≤ N then the following are equivalent. 1. For some t0 ∈ [0, T ] the N vectors Φ1 (t0 ), Φ2 (t0 ), . . . , ΦN (t0 ) are linearly independent in RN . P 2. The general solution is cj Φj (t) with scalar cj . 3. For all t ∈ [0, T ], the vectors Φ1 (t), Φ2 (t), . . . , ΦN (t) are linearly independent in RN . P Remark 2.1 When the conditions hold, every solution is of the form P cj Φj for a unique choice of cj . The uniqueness of the cj follows upon representing X(t0 ) = cj Φj (t0 ). Thanks to 1, the vector X(t0 ) has only one such representation. Proof. Prove that 3 ⇒ 1 ⇒ 2 ⇒ 3. That 3 ⇒ 1 is trivial. P 1 ⇒ 2. The superposition principal shows that each cj Φj is a solution. Conversely if X is a solution consider the vector X(t0 ). Since the Φj (t0 ) are N linearly independent vectors in RN they span so there are scalars cj so that X(t0 ) = c1 ϕ1 (t0 ) + c2 Φ2 (t0 ) + · · · + cN Φn (t0 ) . Define Z(t) := c1 ϕ1 (t) + c2 Φ2 (t) + · · · + cN Φn (t) . Then Z is a solution and Z(t0 ) = P X(t0 ) because of the choice of the cj . Uniqueness implies that Z = X so X is one the solutions cj Φj . This completes the proof of 2. 2 ⇒ 3. Fix t ∈ [0, T ]. For any W ∈ RN let X(t) be the Punique solution with X(t) = W . Property 2 implies that there are constants cj so that X = j cj Φj . P Evaluating at t = t implies that j cj Φ(t) = X(t) = W . Since W is arbitrary this shows that the set of vectors Φj (t) spans RN . A set of N vectors in RN that spans is independent. This proves 3. 3