% problem notes Ch8, S2009 %24 Apr, 8.1: 4, 12, 38 %24 Apr, 8.2: 4, 19 8.1-4: [ 3 -1] A = [ ] [ 1 1] Cayley-Hamilton-Ziebur method: The eigenvalues are 2,2 and therefore x(t) = c1 exp(2t) + c2 t exp(2t) That is, Euler's theorem with a double root r=2 implies two atoms exp(2t), t exp(2t) and component x(t) must be a linear combination of these two atoms. The first DE x'=3x-y implies y = 3x-x' = 3c1 exp(2t)+3c2 t exp(2t) - 2c1 exp(2t) -c2 exp(2t) - 2c2 t exp(2t)= (c1-c2) exp(2t) + c2 t exp(2t). Then x(t) = c1 exp(2t) + c2 t exp(2t) y(t) = (c1-c2) exp(2t) + c2 t exp(2t) Take the partials on symbols c1, c2 to find a fundamental matrix [ exp(2t) t exp(2t) ] Phi = [ ] [ exp(2t) t exp(2t)-exp(2t)] The general solution of u'=Au is then u=(Phi)c where c is an arbitrary 2-vector. Solve u(0)=Phi(0)c for c and then report the answer. Laplace resolvent method: Start with u'=Au, then [ s-3 1 ][1] L(u)=inverse(sI - a)u(0)=inverse[ ][ ] [ -1 s-1][0] [(s-1)/(s^2-4*s+4)] [L(t*exp(2*t)+exp(2*t))] = [ ]=[ ] [1/(s^2-4*s+4) ] [L(t*exp(2*t)) ] Then x(t) = t exp(2t)+exp(2t) y(t) = t exp(2t) The fundamental matrix can be taken to be Phi=invlaplace(B) where [ s-3 1 ] B=inverse(sI - a) =inverse[ ] [ -1 s-1] [(s-1)/(s^2-4*s+4) -1/(s^2-4*s+4) ] = [ ] [1/(s^2-4*s+4) (s-3)/(s^2-4*s+4)] Then [t exp(2t)+exp(2t) -t exp(2t) ] Phi = [ ] = exp(At) = Exp Matrix [t exp(2*t) -t exp(2*t)+exp(2t)] Some maple help: with(linalg): with(inttrans): # Find eigenvalues of A, then use Cayley-Hamilton method A:=matrix([[3,-1],[1,1]]); eigenvals(A); # Laplace resolvent method to find u(t)=exp(At)u(0) B:=inverse(evalm(s*diag(1,1)-A)); evalm(B&*vector([1,0]));map(invlaplace,%,s,t); # Find exp(At) inverse(evalm(s*diag(1,1)-A)); map(invlaplace,%,s,t); # alternate is exponential(A,t); 8.1-12: [ 5 -4] A = [ ] [ 3 -2] The methods of 8.1-4 apply. In particular: # Find exp(At) with(linalg): with(inttrans): A:=matrix([[5,-4],[3,-2]]); inverse(evalm(s*diag(1,1)-A)); Phi:=map(invlaplace,%,s,t); Then [-3*exp(t)+4*exp(2*t), 4*exp(t)-4*exp(2*t) ] exp(At) = [ ] [-3*exp(t)+3*exp(2*t), 4*exp(Phi)-3*exp(2*t)] Putzer Algorithm for exp(At) exp(lambda_1 t) - exp(lambda_2 t) exp(At) = exp(lambda_1 t) I + --------------------------------- (A - lambda_1 I) lambda_1 - lambda_2 The above is for 2x2 matrices A only. There is a general Putzer algorithm [see the manuscript on systems]. When lambda_1 = lambda_2, then limit with L'Hopital's Rule on the fraction to get the replacement term t exp(lambda_1 t). When lambda_1 and lambda_2 are complex conjugates, then take the real part of the formula, because exp(At) is REAL [throw away all terms containing complex unit i]. For this problem, [ 1 0 ] [ [ 5 -4 ] [ 1 0 ] ] exp(At) = exp(2t) [ ] + t exp(2t) [ [ ] - 2 [ ] ] [ 0 1 ] [ [ 3 -2 ] [ 0 1 ] ] 8.1-38: The methods of 8.1-4 apply. In particular: # Find exp(At) with(linalg): with(inttrans): A:=matrix([[5,20,30],[0,10,20],[0,0,5]]); inverse(evalm(s*diag(1,1,1)-A)); Phi:=map(invlaplace,%,s,t); Then [exp(5*t) 4*exp(10*t)-4*exp(5*t) -50*t*exp(5*t)-16*exp(5*t)+16*exp(10*t)] Phi=[0 exp(10*t) 4*exp(10*t)-4*exp(5*t) ] [0 0 exp(5*t) ] The book suggests to use Theorem 3, exp(At)=Phi(t)inverse(Phi(0)). The matrix Phi in this case is the augmented matrix of eigenvector solutions exp(lambda_k t) v_k which appear Fourier's differential system theorem [THEOREM 3 in the book]: u(t) =c1 exp(r1 t) v1 + c2 exp(r2 t) v2 + c3 exp(r3 t) v3. Then Phi is the agumented matrix of exp(r1 t) v1, exp(r2 t) v2, exp(r3 t) v3. Engineers should learn the Laplace Resolvent method and Putzer's method, because the eigenanalysis method of Theorem 3 is limited in normal use to matrices A with n distinct real eigenvalues. ANSWER CHECK: There is a direct route to finding exp(At) in maple: with(linalg):A:=matrix([[5,20,30],[0,10,20],[0,0,5]]); exponential(A,t); 8.2-4: [ 4 1] [ exp(t)] [1] A = [ ] F = [ ] u(0) = [ ] [ 6 -1] [-exp(t)] [1] FIRST SOLUTION. Superposition says u=uh+up. Find exp(At) for example by using the Cayley-Hamilton-Ziebur method for Phi followed by exp(At)=Phi(t)inverse(Phi(0)). See also the MAPLE SOLUTION below. [exp(-2t)+6 exp(5t) -exp(-2t)+exp(5t) ] exp(At)= (1/7)[ ] [-6 exp(-2t)+6 exp(5t) 6 exp(-2t)+exp(5t)] Then uh=exp(At)u0, with u0 unknown just now, to be found later. Undetermined coefficients suggests to find up=exp(t)c where c is a 2-vector with component c1, c2. Substitution into u'=Au+F gives exp(t)c = exp(t)Ac + F Then [ 1] c = Ac + b, b = [ ] [-1] has unique solution c1=-1/12, c2=-3/4 The result is [-1/12] up = exp(t)[ ] [-3/4 ] Add u=uh+up and determine u0 from the formula u0 = uh(0)+up(0) u0 = u(0)+vector([-1/12,-3/4]) = vector([11/12,-7/4]) Then u=uh+up=exp(At)u0+up gives [33/28*exp(5*t)-1/12*exp(t)-2/21*exp(-2*t)] u = [ ] [33/28*exp(5*t)-3/4*exp(t)+4/7*exp(-2*t) ] MAPLE SOLUTION. We will use maple assist to find exp(At), then proceed in maple to do the vector integrations required in variation of parameters. # Find exp(At) with(linalg): with(inttrans): A:=matrix([[4,1],[6,-1]]); inverse(evalm(s*diag(1,1)-A)); Phi:=map(invlaplace,%,s,t); # Integrate var of param formula up=int(G,t=0..x); F:=vector([exp(t),-exp(t)]); u0:=vector([1,1]); G:=evalm(inverse(Phi)&*F); H:=subs(x=t,map(int,G,t=0..x)); up:=evalm(Phi&*H); # Find uh from the initial condition uh:=evalm(Phi&*u0); # Superposition u=uh+up u:=evalm(uh+up); map(simplify,%); [33/28*exp(5*t)-1/12*exp(t)-2/21*exp(-2*t)] u = [ ] [33/28*exp(5*t)-3/4*exp(t)+4/7*exp(-2*t) ] 8.2-19: [1 2] [180t] [0] A = [ ] F = [ ] u(0) = [ ] [2 -2] [90 ] [0] We will use maple assist to find exp(At), then proceed in maple to do the vector integrations required in variation of parameters. The code is modified from the solution of the previous problem. # Find exp(At) with(linalg): with(inttrans): A:=matrix([[1,2],[2,-2]]); inverse(evalm(s*diag(1,1)-A)); Phi:=map(invlaplace,%,s,t); # Integrate var of param formula up=int(G,t=0..x); F:=vector([180*t,90]); u0:=vector([0,0]); G:=evalm(inverse(Phi)&*F); H:=subs(x=t,map(int,G,t=0..x)); up:=evalm(Phi&*H); # Find uh from the initial condition # uh:=evalm(Phi&*u0); uh:=vector([0,0]); # because of zero ic # Superposition u=uh+up u:=evalm(uh+up); map(simplify,%); [-60*t-70+16*exp(-3*t)+54*exp(2*t)] u = [ ] [-60*t+5-32*exp(-3*t)+27*exp(2*t) ] %%===end of Ch8 problem notes======