Edwards-Penney Chapter 8, sections 8.1, 8.2 Topics, Definitions, Theorems, Examples ABOUT 2250. Details about generalized eigenpairs are skipped. This is caused by skipping section 7.5. Non-homogeneous systems are treated in Laplace theory, long before this section, so chapter 8 is enrichment. ADDED MATERIAL. The Putzer formula for a 2x2 matrix A is the fastest way to find exp(At). This formula keeps us with pencil and paper longer, before hitting the computer. We add scalar solution methods for triangular systems and MAPLE methods for vector-matrix integration. 8.1 Matrix Exponentials exp(A t) and Linear Systems u'=Au ==== DEF. Fundamental matrix. It is a matrix W(t) whose columns are independent solutions of the differential system u'=Au. Routinely W(t) is found by taking partials on symbols in the general solution, which itself is found by a variety of methods. THEOREM 1. Fundamental Matrix Solutions Let W(t) be a fundamental matrix of u'=Au. Then the following properties hold: 1. W'(t) = A W(t) [matrix multiply identity] 2. The unique solution of u'=Au, u(0)=u0 is given by the formula u(t) = W(t) W(0)^(-1) u0 CONSTRUCTION of a FUNDAMENTAL MATRIX Method 1. [Eigenvalue Method] If 2x2 matrix A is diagonal, then use thismethod. If A has n eigenpairs (lambda[i],v[i]) with real eigenvalues, i=1,...,n, then W(t)= where the columns are w1=exp(lambda[1] t) v[1], ... , wn=exp(lambda[n] t) v[n] Method 2. [Cayley-Hamilton Method, shortcut] If non-diagonal 2x2 matrix A has eigenvalues lambda[1], lambda[2] with corresponding Euler solution atoms A1(t), A2(t), then x1(t) = c1 A1(t) + c2 A2(t), and x2(t) = answer resolved from the Cayley-Hamilton shortcut (involves symbols c1, c2). The columns of W(t) are then the partial derivatives on symbols c1, c2 taken on the vector solution u(t)= EXAMPLE. Find fundamental matrix W(t) for the system x' = 4x + 3y, y' = 2x - y u'(t) = Au(t), A:=Matrix([[4,2],[3,-1]]) ANSWER: W(t)= | exp(-5t)*<1,1/2>> BOOK ANSWER: Phi(t)=|exp(-5t)*<2,1>>, which differs in column 2 from W(t): col(Phi,2)=2 col(W,2). DETAILS: Char equation r^2-3r-10=0 has roots r=-2,5 and Euler solution atoms exp(-2t), exp(5t). Then x(t) = c1 exp(-2t) + c2 exp(5t), y(t) = (1/2)(x'-4x) [from x'=4x+2y] = -3 c1 exp(-2t) + (1/2)c2 exp(5t) Then W(t) = <(d/dc1)u(t)|(d/dc2)u(t)> = | exp(-5t)*<1,1/2>> EXAMPLE. Find the solution of the system x' = 4x + 3y, y' = 2x - y, x(0)=1, y(0)=-1 ANSWER: x(t)=(1/7)( exp(-2t) + 4 exp(5t)), y(t)=(1/7)(-9 exp(-2t) + 2 exp(5t)) METHOD: Linear algebra. Solve for c1, c2 in the general solution u(t)=W(t) using <1,-1>=u(0)=W(0). EXPONENTIAL MATRICES DEF. The exponential matrix is defined by exp(A t)=W(t)W(0)^(-1) for any fundamental matrix W(t) of the system u'=Au. THEOREM. If W(t) and W1(t) are two fundamental matrices, then W(t)W(0)^(-1)=W1(t)w1(0)^(-1). Therefore, the exponential matrix exp(A t) is well-defined. THEOREM. The exponential matrix has these properties. 1. exp(A) = I + A + A^2/2 + A^3/3! + ... 2. exp(A t) = I at t=0. 3. The columns of exp(A t) are independent solutions of u'=Au. 4. (d/dt)exp(A t) = A exp(A t) = exp(A t) A. 5. If AB=BA, then exp(A)exp(B)=exp(A+B). THEOREM 2. The solution of u'=Au, u(0)=u0 is u(t) = exp(A t) u0. EXAMPLE. If A:=Matrix([[a,0],[0,b]]), then exp(A):=Matrix([[exp(a),0],[0,exp(b)]]). THEOREM. exp(diag(a1,...,an))=diag(exp(a1),...,exp(an)) EXAMPLE. If A:=Matrix([[0,3,4],[0,0,6],[0,0,0]]), then exp(At)=I+At+A^2t^2/2 =Matrix([[1,3t,4t+9t^2],[0,11,6t],[0,0,0]]). EXAMPLE. Let B=Matrix([[2,3,4],[0,2,6],[0,0,2]]) Let B=Matrix([[0,3,4],[0,0,6],[0,0,0]])+ Matrix([[2.0.0],[0,2,0],[0,0,2]]) =A + 2I where A is from the previous example and I is the 3x3 identity matrix. Then exp(Bt)=exp(At+2It) =exp(At)exp(2It) =exp(2t)Matrix([[1,3t,4t+9t^2],[0,11,6t],[0,0,0]]) EXAMPLE. Let A:=Matrix([[4,2],[3,-1]]). Find a fundamental matrix and compute exp(At). ANSWER: By the Cayley-Hamilton shortcut, W:=, w1:=exp(-2t)<1,-3> and w2:=exp(5t)<2,1>. Then compute exp(At)=W(t)W(0)^(-1). MAPLE ANSWER CHECK w1:=exp(-2*t)*<1,-3>;w2:=exp(5*t)*<2,1>; W:=unapply(,t); W(t).W(0)^(-1); # Hand computation verified MAPLE ANSWER CHECK A:=Matrix([[4,2],[3,-1]]); LinearAlgebra[MatrixExponential](A,t); # Check final answer only EXAMPLE. Let B:=Matrix([[2,3,4],[0,2,6],[0,0,2]]), u0:=<19,29,39>. Solve u'=Bu, u(0)=u0 using exp(Bt). ANSWER: u(t):=exp(2t)*<19+243t+351t^2,29+234t,39> METHOD: Find exp(Bt) by previous example. Then u(t)=exp(Bt)u0. MAPLE ANSWER CHECK B:=Matrix([[2,3,4],[0,2,6],[0,0,2]]); expBt:=LinearAlgebra[MatrixExponential](B,t); u:=expBt.<19,29,39>; simplify(%); # Check final answer only GENERALIZED EIGENVECTORS We don't study section 7.5 and therefore skip the remarks in 8.1 about generalized eigenvectors. We'll do the examples by the simplest method possible. EXAMPLE. Find exp(At) for A:=Matrix([[3,4,5],[0,5,4],[0,0,3]]) ANSWER: exp(At)= where v1=, v2=<2exp(5t)-2exp(2t),exp(5t),0>, v3=<4exp(5t)-(4+3t)exp(3t),2exp(5t)-2exp(3t),exp(3t)> METHOD: Write the differential equation as x'=3x+4y+5z, y'=5y+4z, z'=3z The system is triangular and therefore it can be solved by the linear integrating factor method for first order equations. Method: solve for z(t)=c1 exp(3t), then back-substitute the answer for z into the second equation and solve for y = c2 exp(5t)- 2 c1 exp(3t). Back-substitute both z and y into the first equation to solve for x. Then x(t) = (-3 c3 t + c1) exp(3t) + 2 c2 exp(5t), y(t) = -2 c3 exp(3t) + c2 exp(5t), z(t) = c3 exp(3t) Take partial derivatives on symbols c1,c2,c3 to find 3 independent solutions v1=, v2=<2 exp(5t),exp(5t),0>, v3=<-3 t exp(3t),-2 exp(3t),exp(3t)> Then W= and exp(At)=W(t)W(0)^(-1). MAPLE ANSWER CHECK for HAND DETAILS v1:=; v2:=<2 *exp(5*t),exp(5*t),0>; v3:=<-3* t *exp(3*t),-2*exp(3*t),exp(3*t)>; W:=unapply(,t); expAt:=W(t).W(0)^(-1); MAPLE ANSWER CHECK A:=Matrix([[3,4,5],[0,5,4],[0,0,3]]); LinearAlgebra[MatrixExponential](A,t); 8.2 Non-Homogeneous Linear Systems u'=Au+F(t) ==== DEF. The non-homogeneous system is u'=Au+f(t), the homogeneous system is u'=Au. SUPERPOSITION The general solution of u'=Au+f(t) is of the form u=uh+up, where uh(t) is the general solutionof u'=Au and up(t) is a particular solution of u'=Au+f(t). UNDETERMINED COEFFICIENTS If the components of f(t) are linear combinations of Euler solution atoms, then the method of undetermined coefficients applies to find up(t). The form of the particular solution is changed from the scalar case, but it too can be learned from a few examples. EXAMPLE. Solve by undetermined coefficients u'=Au+f(t) where A:=Matrix([[3,2],[7,5]]) and f(t)=<3,2t>. ANSWER: uh(t)=exp(At) and up(t)=at+b where a= and b= are vectors of constants. Finally, up(t)=<4t+17,-6t-25> EXAMPLE. Three brine tanks, u'=Au+f(t), u(0)=zero vector Find u=uh+up An example from section 7.3, already studied. A:=Matrix([[-0.5,0,0],[0.5,-0.25,0],[0,0.25,-0.2]]) f:=<20,0,0> ANSWER: x1=40-40exp(-t/2), x2=80+80exp(-t/2)-160exp(-t/4), x3=100+(100/3)(-2exp(-t/2)+24exp(-t/4)-25exp(-t/5)) DETAILS: up(t)=<40,80,100> by undetermined coefficients using trial solution u=. By the work in the 7.3 example, there are three independent solutions v1, v2, v3 v1=exp(-t/2)<3,-6,5>, v2=exp(-t/4)<0,1,-5>, v3=exp(-t/5)<0,0,1> These make a fundamental matrix W= to give the general solution u=uh+up, uh=W(t). To find c1,c2,c3 apply the initial condition u(0)=<0,0,0> to obtain a linear system of equations W(0)+<40,80,100>=<0,0,0> <<3,-6,5>|<0,1,-5>|<0,0,1>>=<-40,-80,-100> EXAMPLE. Find the form of a trial solution for up(t), given the system u'=Au+f(t), A:=Matrix([[4,2],[3,-1]]), and f:=t exp(-2t) <15,4> ANSWER: u=vector linear combination of the Euler solution atoms exp(-2t), t exp(-2t), t^2 exp(-2t). DETAILS: The components of f(t) are linear combinations of Euler solution atoms exp(-2t), t exp(-2t), t^2 exp(-2t). So we take a vector linear combination of them for the trial solution. This has to be modified if it fails: multiply by t on the affected group, then add these atoms to those already found, and repeat. VARIATION of PARAMETERS The formula for up(t) in the equation u'=Au+f(t) is given by up(t) = exp(At)int(exp(-As)f(s),s=0..t) This formula is known as the VARIATION of PARAMETERS FORMULA. GENERAL THEORY THEOREM 1. If the equation u'=P(t)u+f(t) is considered instead and P(t) is not a constant matrix, then the variation of parameters formula is up(t) = W(t) int(W(s)^(-1)f(s),s=0..t) where W is a fundamental matrix for u'=P(t)u. USE: Such a formula is useful with a computer algebra system, which can erase integration difficulty and solve the technical problem of finding W(t). EXAMPLE. Solve u'=Au+f(t), u(0)=u0 by variation of parameters, A:=Matrix([[4,2],[3,-1]]), f(t) = t exp(-2t) <-15,-4>, u0:=<7,3> ANSWER: x1=(1/14)( (6+28t-7t^2)exp(-2t) + 92 exp(5t) ) x2=(1/14)( (-4+14t+21t^2)exp(-2t) + 46 exp(5t) ) METHOD: Let up(t)=W(t) int(W(s)^(-1)f(s),s=0..t), uh(t)=W(t)W(0)^(-1)u0 The fundamental matrix W(t) was found in an example in section 8.1 to be W(t)= with v1:=exp(-2t)<1,-3>, v2:=exp(5t)<2,1>. Integrations are by computer. Answer check by computer. MAPLE COMPUTATION DETAILS A:=Matrix([[4,2],[3,-1]]); f:=t->t*exp(-2*t)*<-15,-4>; u0:=<7,3>; v1:=exp(-2*t)*<1,-3>; v2:=exp(5*t)*<2,1>; W:=unapply(,t); up:=W(t).map(int,W(s)^(-1).f(s),s=0..t); uh:=W(t).W(0)^(-1).u0; u:=uh+up; MAPLE ANSWER CHECK de1:=diff(x(t),t)=4*x(t)+2*y(t)-15*t*exp(-2*t),x(0)=7; de2:=diff(y(t),t)=3*x(t)-1*y(t)- 4*t*exp(-2*t),y(0)=3; dsolve({de1,de2},{x(t),y(t)}); ====== end