MATH 304Linear AlgebraLecture 23:Matrix exponentials.Initial value problem for a system of linear PDEs:(dxdt= 4x + 3y,dydt= y,x(0) = 1, y(0) = 1.The system can be rewritten in vector form:dvdt= Av, where A =4 30 1, v =xy.Matrix A is diagonalizable: A = UBU−1, whereB =4 00 1, U =1 −10 1.Let w =w1w2be coordinates of the vector v relative to thebasis v1= (1, 0), v2= (−1, 1) of eigenvectors of A. Thenv = Uw =⇒ w = U−1v.It follows thatdwdt=ddt(U−1v) = U−1dvdt= U−1Av = U−1AUw.Hencedwdt= Bw ⇐⇒(dw1dt= 4w1,dw2dt= w2.General solution: w1(t) = c1e4t, w2(t) = c2et, where c1, c2∈ R.Initial condition:w(0) = U−1v(0) =1 −10 1−111=1 10 111=21.Thus w1(t) = 2e4t, w2(t) = et. Thenx(t)y(t)= Uw(t) =1 −10 12e4tet=2e4t−etet.• Initial value problem for a linear ODE:dydt= 2y, y(0) = 3.Solution: y(t) = 3e2t.• Initial value problem for a system of linear ODEs:(dxdt= 2x + 3y,dydt= x + 4y,x(0) = 2, y (0) = 1.The system can be rewritten in vector formddtxy= Axy, where A =2 31 4.Solution:x(t)y(t)= etA21.What is etA?Exponential functionExponential function: f (x) = exp x = ex, x ∈ R.Principal property: ex+y= ex· ey.Definition 1. ex= limn→∞1 +xnn.In particular, e = limn→∞1 +1nn≈ 2.7182818.Definition 2. ex= 1 + x +x22!+ · · · +xnn!+ · · ·Definition 3. f (x) = exis the unique solution ofthe initial value problem f′= f , f (0) = 1.Complex exponentialsDefinition. For any z ∈ C letez= 1 + z +z22!+ · · · +znn!+ · · ·Remark. A sequence of complex numbersz1= x1+ iy1, z2= x2+ iy2, . . . convergesto z = x + iy if xn→ x and yn→ y as n → ∞.Theorem 1 If z = x + iy , x, y ∈ R, th enez= ex(cos y + i sin y ).In particular, eiφ= cos φ + i sin φ, φ ∈ R.Theorem 2 ez+w= ez· ewfor all z, w ∈ C.Proposition eiφ= cos φ + i sin φ for all φ ∈ R.Proof: eiφ= 1 + iφ +(iφ)22!+ · · · +(iφ)nn!+ · · ·The sequence 1, i, i2, i3, . . . , in, . . . is periodic:1, i, −1, −i|{z }, 1, i, −1, −i| {z }, . . .It follows thateiφ= 1 −φ22!+φ44!− · · · + (−1)kφ2k(2k)!+ · · ·+ iφ −φ33!+φ55!− · · · + (−1)kφ2k+1(2k + 1)!+ · · ·= cos φ + i sin φ.Matrix exponentialsDefinition. For any square matrix A letexp A = eA= I + A +12!A2+ · · · +1n!An+ · · ·Matrix exponential is a limit of matrix polynomials.Remark. Let A(1), A(2), . . . be a sequence of n×nmatrices, A(n)= (a(n)ij). The sequence converges toan n×n matrix B = (bij) if a(n)ij→ bijas n → ∞,i.e., if each entry converges.Theorem The matrix exp A is well defined, i.e.,the series converges.Properties of matrix exponentialsTheorem 1 If AB = BA then eAeB= eBeA= eA+B.Corollary (a) etAesA= esAetA= e(t+s)A, t, s ∈ R;(b) eO= I ; (c) (eA)−1= e−A.Theorem 2ddtetA= AetA= etAA.Indeed, etA= I + tA +t22!A2+ · · · +tnn!An+ · · · ,and the series can be differentiated term by term.ddttnn!An=ddttnn!An=tn−1(n − 1)!An.Lemma Let A be an n×n matrix and x ∈ Rn.Then the vector function v(t) = etAx satisfiesv′= Av.Proof:dvdt=ddtetAx =AetAx = AetAx= Av.Theorem For any t0∈ R and x0∈ Rnthe initialvalue problemdvdt= Av, v(t0) = x0has a unique solution v(t) = e(t−t0)Ax0.Indeed, v(t) = e(t−t0)Ax0= etAe−t0Ax0= etAx,where x = e−t0Ax0is a constant vector.Evaluation of matrix exponentialsExample. A = diag(a1, a2, . . . , ak).An= diag(an1, an2, . . . , ank), n = 1, 2, 3, . . .eA= I + A +12!A2+ · · · +1n!An+ · · ·= diag(b1, b2, . . . , bk),where bi= 1 + ai+12!a2i+13!a3i+ · · · = eai.Theorem If A = diag(a1, a2, . . . , ak) theneA= diag(ea1, ea2, . . . , eak),etA= diag(ea1t, ea2t, . . . , eakt).Let A be an n-by-n matrix and suppose there existsa basis v1, . . . , vnfor Rnconsisting of eigenvectorsof A. That is, Avk= λkvk, where λk∈ R.Then A = UBU−1, where B = diag(λ1, λ2, . . . , λn)and U is a transition matrix whose columns arevectors v1, v2, . . . , vn.A2= UBU−1UBU−1= UB2U−1,A3= A2A = UB2U−1UBU−1= UB3U−1.Likewise, An= UBnU−1for any n ≥ 1.I + 2A − 3A2= UIU−1+ 2UBU−1− 3UB2U−1== U(I + 2B − 3B2)U−1.=⇒ p(A) = Up(B)U−1for any polynomial p(x).=⇒ etA= UetBU−1for all t ∈ R.Example. A =2 31 4.The eigenvalues of A: λ1= 1, λ2= 5.Eigenvectors: v1= (3, −1), v2= (1, 1).Therefore A = UBU−1, whereB =1 00 5, U =3 1−1 1.ThenetA= UetBU−1=3 1−1 1et00 e5t3 1−1 1−1=3ete5t−ete5t141 −11 3=143et+ e5t−3et+ 3e5t−et+ e5tet+ 3e5t.Problem. Solve a system of differential equations(dxdt= 2x + 3y,dydt= x + 4ysubject to initial conditions x(0) = 2, y (0) = 1.The unique solution:x(t)y(t)= etAx0, where A =2 31 4, x0=21.etA=143et+ e5t−3et+ 3e5t−et+ e5tet+ 3e5t=⇒(x(t) =34et+54e5t,y(t) = −14et+54e5t.Example. A =0 1 00 0 10 0 0, a Jordan block.A2=0 0 10 0 00 0 0, A3= O, An= O for n ≥ 3.eA= I + A +12A2=1 1120 1 10 0 1,etA= I + tA +t22A2=1 t12t20 1 t0 0 1.Example. A =1 10 1, another Jordan block.A2=1 20 1, A3=1 30 1, An=1 n0 1.etA= I +tA+t22!A2+ · · · +tnn!An+ · · · =a(t) b(t)0 a(t),where a(t) = 1 + t +t22!+t33!+ · · · = et,b(t) = t + 2t22!+ 3t33!+ · · · = tet.Thus etA=ettet0 et.Example. A =λ 10 λ, a general Jordan block.We have that A = λI + B, where B =0 10 0.Since (λI )B = B(λI ), it follows that eA= eλIeB.Similarly, etA= etλIetB.etλI=eλt00 eλt= eλtI ,B2= O =⇒ etB= I + tB =1 t0 1.Thus etA= etλIetB=eλtteλt0 eλt.Problem. Solve a system of differential equations(dxdt= 2x + y ,dydt= 2ysubject to initial conditions x(0) = y (0) = 1.The unique solution:x(t)y(t)= etAx0, where A =2 10 2, x0=11.etA=e2tte2t0 e2t= e2t1 t0 1=⇒(x(t) = e2t(1 + t),y(t) =
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