14 PRELIMINARIES0.3. systems of first order equations.(1) Linear differential equations can be reduced to first order.(2) First order equations are matrix equations.(3) Exponential of a matrix.(4) Eigenvalues and eigenvectors.(5) Diagonalization of a matrix.(6) Solution.0.3.1. reduction to first order.Theorem 0.8. Any nth order linear differential equation in m vari-ables can be reduced to a first order linear diffeq in nm variables.Here is an example with n = m = 2.x!!= 2x!+ 3y!+ 6x + 7y!!= x!+ 9y − 5.(Differentiation is with respect to time t.) The theorem says you neednm = 4 variables, i.e., 2 more. Call them v, w and letv = x!, w = y!Then we have four 1st order linear diffeq’s in 4 variables:v!= 2v + 3w + 6x + 7w!= v + 9y − 5x!= vy!= w.You always get derivatives of the variable on the left and linear com-binations of the variables on the right.This example is too complicated. So, I started over with a simplerexample.Example 0.9. Solve the following 2nd order differential equation:y!!+ y!− 6y = 0with initial conditions:y0= 0, y!0= 5.(Initial position at the origin with initial velocity 5.)Following the procedure, you introduce another variable z := y!.Then the 1st order equations are:y!= zz!= 6y − zMATH 56A SPRING 2008 STOCHASTIC PROCESSES 150.3.2. matrix form. In matrix form this is:ddt!yz"=!0 16 −1"!yz"Which we can write asddtY = AYwhereY =!yz", A =!0 16 −1".0.3.3. exponential of a matrix. The solution of this equations isY = etAY0where Y0=!y0z0"and the matrix exponential etAis defined by:etA:= I2+ tA +t2A22+t3A33!+ · · · =∞#k=0tkAkk!To prove that this is the solution, you differentiate the sum term byterm:ddtetA=ddt#tkAkk!=#ddttkAkk!=#ktk−1Akk!which simplifies to#tk−1Ak(k − 1)!= A#tk−1Ak−1(k − 1)!= AetA.So,ddtetA= AetA0.3.4. eigenvalues and eigenvectors. In order to compute etAwe needthe eigenvalues and eigenvectors of the matrix A. These are given bythe equationAX = λX = X(λ) .In this example we have:!0 16 −1"$ %& 'A!12"$%&'X=!24"= 2$%&'λ!12"$%&'X=!12"(2)(If λ is a scalar, it belongs on the left. If (λ) is a 1 × 1 matrix, it mustbe on the right by the rules of matrix multiplication.)16 PRELIMINARIESλ1= 2 is an eigenvalue of our matrix A with eigenvector X1=!12".The other eigenvalue is λ2= −3 with eigenvector X2=!1−3":!0 16 −1"!1−3"=!−39"=!1−3"(−3)0.3.5. diagonalization. Put these together to get:!0 16 −1"$%& 'A!1 12 −3"$ %& 'Q=(X1X2)=!2 −34 9"=!1 12 −3"$ %& 'Q=(X1X2)!2 00 −3"$ %& 'D.Or:AQ = QDwhere Q = (X1X2) is the matrix whose columns are the eigenvectorsof A and D =!λ100 λ2"is the diagonal matrix whose diagonal entriesare the eigenvalues. The equation AQ = QD should be rewritten as:A = QDQ−1.This is called the diagonalization of A.0.3.6. powers of A. After we successfully diagonalized A, we can raiseit to a power: First,A2= QD Q−1Q$ %& 'I2DQ−1= QDI2DQ−1= QD2Q−1.By induction we get:Ak= QDkQ−1.Divide by k! and sum over k to get:etA= QetDQ−1= Q!etλ100 etλ2"Q−1MATH 56A SPRING 2008 STOCHASTIC PROCESSES 170.3.7. solution. The inverse ofQ =!a bc d"=!1 12 −3"is given byQ−1=1ad − bc!d −b−c a"=1−5!−3 −1−2 1"=!3/5 1/52/5 −1/5".So,Y = etAY0= QetDQ−1Y0.Putting in all the numbers, including y0= 0, z0= y!0= 5, we get:!yz"=!1 12 −3"!e2t00 e−3t"!3/5 1/52/5 −1/5"!05"=!1 12 −3"!e2t00 e−3t"!1−1"=!1 12 −3"!e2t−e−3t"=!e2t− e−3t2e2t+ 3e−3t".In other words,y = e2t− e−3tz = y!= 2e2t+ 3e−3t.This algorithm works if the eigenvalues of A are distinct.0.3.8. review of linear algebra. The eigenvalues of a square matrix aredefined to be the solutions of the equationdet(A − λI) = 0but there is a trick to use for 2 × 2 matrices. The determinant of amatrix is always the product of its eigenvalues:det A = λ1λ2= −6The trace (sum of diagonal entries) is equal to the sum of the eigenval-ues:trA = λ1+ λ2= −1So, λ1= 2, λ2= −3. The eigenvectors are the solutions of the linearequationsAX = λXor(A − λI2)X =
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