ReviewLet A be n × n with independent eigenvectors x1,, xn.ThenA can be diag onalized as A = PDP−1.• the columns of P are the eigenvectors• the d i agonal matrix D has the eigenvalues on the diagonalWhy? We need to see thatA P = P D:Axi= λixiA| |x1xn| |=| |λ1x1λnxn| |=| |x1xn| |λ1λn• The diff erential equation y′= ay with initial condition y(0) = Cis s olved by y(t) = Cea t.Recall from Calculus the Taylor serieset= 1 + t +t22!+t33!+• Goal: sim ilar treatment of system s like:y′=2 0 0−1 3 1−1 1 3y , y(0) =121Definition 1. Let A be n × n. The matrix exponential i seA= I + A +12!A2+13!A3+Then:ddteA t= AeA tWhy?ddteA t=ddtI + A t +12!A2t2+13!A3t3+= A +11!A2t +12!A3t2+= A eA tThe solution to y′= Ay, y(0) = y0is y(t) = eA ty0.Why? Because y′(t) = A eA ty0= A y(t) and y(0) = e0Ay0= y0.Example 2. If A =2 00 5, then :eA=1 00 1+2 00 5+12!"2200 52#+="e200 e5#eA t=1 00 1+2t 00 5t+12!"(2t)200 (5t)2#+="e2t00 e5t#Clearly, this works to obtain eDfor any diagonal matrix D.Armin [email protected] 3. Suppose A = PDP−1. Then, what is An?Solution.First, note th at A2= (PDP−1)(PDP−1) = PD2P−1.Likewise, An= PDnP−1.(The point being thatDnis trivial to compute because D is diagonal.)Theorem 4. Suppose A = PDP−1. Then, eA= PeDP−1.Why? Recall that An= P DnP−1.eA= I + A +12!A2+13!A3+= I + P D P−1+12!P D2P−1+13!P D3P−1+= PI + D +12!D2+13!D3+P−1= P eDP−1Example 5. Solve the differen tial equationy′=0 11 0y , y(0) =10.Solution. The solution to y′= Ay, y(0) = y0is y(t) = eA ty0.• Diagonalize A =0 11 0:◦−λ 11 −λ= λ2− 1, so the eigenvalues are ±1◦ λ = 1 has eigenspace Nul−1 11 −1= spann11o◦ λ = −1 ha s eigenspace Nul1 11 1= spann−11o◦ Hence, A = PDP−1with P =1 −11 1and D =1 00 −1.• Compute the solution y = eA ty0:y = PeD tP−1y0=1 −11 1"et00 e−t#121 1−1 110=121 −11 1"et00 e−t#1−1=121 −11 1"et−e−t#=12"et+ e−tet− e−t#Armin [email protected] 6. Solve the differen tial equationy′=2 0 0−1 3 1−1 1 3y , y(0) =121.Solution.• Recall that the solution to y′= Ay, y(0) = y0is y = eA ty0.• A has eigenva lues 2 and 4. (W e did that in an earlier class!)◦ λ = 2:0 0 0−1 1 1−1 1 1eigenspace span(110,101)◦ λ = 4:−2 0 0−1 −1 1−1 1 −1eigenspace span(011)• A = PDP−1with P =1 1 01 0 10 1 1, D =224• Compute the solution y = eA ty0:y = eA ty0= PeD tP−1y0=1 1 01 0 10 1 1e2te2te4t1 1 01 0 10 1 1−1121=1 1 01 0 10 1 1e2te2te4t101=1 1 01 0 10 1 1e2t0e4t=e2te2t+ e4te4tCheck (optional) that y =e2te2t+ e4te4tindeed solves the original problem:y′=2e2t2e2t+ 4e4t4e4t@!2 0 0−1 3 1−1 1 3e2te2t+ e4te4tRemark 7. The matrix exponential shares many othe r prope r ties of th e usual exponen-tial:• eAis inv er tible and (eA)−1= e−A• eAeB= eA+B= eBeAif AB = BAArmin [email protected] the Halloween tortureperimeter = 4perimeter = πperimeter = 4perimeter = πperimeter = 4perimeter = πperimeter = 4perimeter = πperimeter = 4perimeter = π• Length of t he graph of y(x) on [a, b] isRab1 + y′(x)2pdx.• While the blue curve does con verge to the circle,its derivative does not converge!• In the language of functional analysis:The linear mapD: yy′is not continuous!(That is, two functions can be close without their derivatives being close.)Even more ext r eme examples are provided by fractals. The Koch snowflake:Armin [email protected]• Its perimeter is infinit e !Why? At each iteration, the peri meter gets multiplied by4/3.• Its boundary has dimension log3(4) ≈ 1.262!!the effect of zooming in by a factor of3×3 d = 1 = log3(3)×9 d = 2 = log3(3)×4 d = log3(4)• Such fr actal behaviour is also observed wh e n attempting to measure the length ofa coastline: the me asured length increa ses by a factor when using a smaller scal e .See: http://en.wikipedia.org/wiki/Coastline_paradoxArmin
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