1、Ch 7.9: Nonhomogeneous Linear Systems,The general theory of a nonhomogeneous system of equationsparallels that of a single nth order linear equation. This system can be written as x = P(t)x + g(t), where,General Solution,The general solution of x = P(t)x + g(t) on I: t has the formwhereis the genera
2、l solution of the homogeneous system x = P(t)x and v(t) is a particular solution of the nonhomogeneous system x = P(t)x + g(t).,Diagonalization,Suppose x = Ax + g(t), where A is an n x n diagonalizable constant matrix. Let T be the nonsingular transform matrix whose columns are the eigenvectors of A
3、, and D the diagonal matrix whose diagonal entries are the corresponding eigenvalues of A. Suppose x satisfies x = Ax, let y be defined by x = Ty. Substituting x = Ty into x = Ax, we obtainTy = ATy + g(t). or y = T-1ATy + T-1g(t)or y = Dy + h(t), where h(t) = T-1g(t). Note that if we can solve diago
4、nal system y = Dy + h(t) for y, then x = Ty is a solution to the original system.,Solving Diagonal System,Now y = Dy + h(t) is a diagonal system of the formwhere r1, rn are the eigenvalues of A. Thus y = Dy + h(t) is an uncoupled system of n linear first order equations in the unknowns yk(t), which
5、can be isolated and solved separately, using methods of Section 2.1:,Solving Original System,The solution y to y = Dy + h(t) has componentsFor this solution vector y, the solution to the original system x = Ax + g(t) is then x = Ty. Recall that T is the nonsingular transform matrix whose columns are
6、 the eigenvectors of A. Thus, when multiplied by T, the second term on right side of yk produces general solution of homogeneous equation, while the integral term of yk produces a particular solution of nonhomogeneous system.,Example 1: General Solution of Homogeneous Case (1 of 5),Consider the nonh
7、omogeneous system x = Ax + g below.Note: A is a Hermitian matrix, since it is real and symmetric. The eigenvalues of A are r1 = -3 and r2 = -1, with corresponding eigenvectors The general solution of the homogeneous system is then,Example 1: Transformation Matrix (2 of 5),Consider next the transform
8、ation matrix T of eigenvectors. Using a Section 7.7 comment, and A Hermitian, we haveT-1 = T* = TT, provided we normalize (1)and (2) so that (1), (1) = 1 and (2), (2) = 1. Thus normalize as follows:Then for this choice of eigenvectors,Example 1: Diagonal System and its Solution (3 of 5),Under the tr
9、ansformation x = Ty, we obtain the diagonal system y = Dy + T-1g(t):Then, using methods of Section 2.1,Example 1: Transform Back to Original System (4 of 5),We next use the transformation x = Ty to obtain the solution to the original system x = Ax + g(t):,Example 1: Solution of Original System (5 of
10、 5),Simplifying further, the solution x can be written asNote that the first two terms on right side form the general solution to homogeneous system, while the remaining terms are a particular solution to nonhomogeneous system.,Nondiagonal Case,If A cannot be diagonalized, (repeated eigenvalues and
11、a shortage of eigenvectors), then it can be transformed to its Jordan form J, which is nearly diagonal. In this case the differential equations are not totally uncoupled, because some rows of J have two nonzero entries: an eigenvalue in diagonal position, and a 1 in adjacent position to the right of
12、 diagonal position. However, the equations for y1, yn can still be solved consecutively, starting with yn. Then the solution x to original system can be found using x = Ty.,Undetermined Coefficients,A second way of solving x = P(t)x + g(t) is the method of undetermined coefficients. Assume P is a co
13、nstant matrix, and that the components of g are polynomial, exponential or sinusoidal functions, or sums or products of these. The procedure for choosing the form of solution is usually directly analogous to that given in Section 3.6. The main difference arises when g(t) has the form uet, where is a
14、 simple eigenvalue of P. In this case, g(t) matches solution form of homogeneous system x = P(t)x, and as a result, it is necessary to take nonhomogeneous solution to be of the form atet + bet. This form differs from the Section 3.6 analog, atet.,Example 2: Undetermined Coefficients (1 of 5),Conside
15、r again the nonhomogeneous system x = Ax + g:Assume a particular solution of the formwhere the vector coefficents a, b, c, d are to be determined. Since r = -1 is an eigenvalue of A, it is necessary to include both ate-t and be-t, as mentioned on the previous slide.,Example 2: Matrix Equations for C
16、oefficients (2 of 5),Substitutingin for x in our nonhomogeneous system x = Ax + g,we obtainEquating coefficients, we conclude that,Example 2: Solving Matrix Equation for a (3 of 5),Our matrix equations for the coefficients are:From the first equation, we see that a is an eigenvector of A correspondi
17、ng to eigenvalue r = -1, and hence has the formWe will see on the next slide that = 1, and hence,Example 2: Solving Matrix Equation for b (4 of 5),Our matrix equations for the coefficients are:Substituting aT = (,) into second equation, Thus = 1, and solving for b, we obtain,Example 2: Particular So
18、lution (5 of 5),Our matrix equations for the coefficients are:Solving third equation for c, and then fourth equation for d, it is straightforward to obtain cT = (1, 2), dT = (-4/3, -5/3). Thus our particular solution of x = Ax + g isComparing this to the result obtained in Example 1, we see that bot
19、h particular solutions would be the same if we had chosen k = for b on previous slide, instead of k = 0.,Variation of Parameters: Preliminaries,A more general way of solving x = P(t)x + g(t) is the method of variation of parameters. Assume P(t) and g(t) are continuous on t , and let (t) be a fundame
20、ntal matrix for the homogeneous system. Recall that the columns of are linearly independent solutions of x = P(t)x, and hence (t) is invertible on the interval t , and also (t) = P(t)(t). Next, recall that the solution of the homogeneous system can be expressed as x = (t)c. Analogous to Section 3.7,
21、 assume the particular solution of the nonhomogeneous system has the form x = (t)u(t),where u(t) is a vector to be found.,Variation of Parameters: Solution,We assume a particular solution of the form x = (t)u(t). Substituting this into x = P(t)x + g(t), we obtain(t)u(t) + (t)u(t) = P(t)(t)u(t) + g(t
22、) Since (t) = P(t)(t), the above equation simplifies tou(t) = -1(t)g(t) Thuswhere the vector c is an arbitrary constant of integration. The general solution to x = P(t)x + g(t) is therefore,Variation of Parameters: Initial Value Problem,For an initial value problem x = P(t)x + g(t), x(t0) = x(0), th
23、e general solution to x = P(t)x + g(t) isAlternatively, recall that the fundamental matrix (t) satisfies (t0) = I, and hence the general solution isIn practice, it may be easier to row reduce matrices and solve necessary equations than to compute -1(t) and substitute into equations. See next example
24、.,Example 3: Variation of Parameters (1 of 3),Consider again the nonhomogeneous system x = Ax + g:We have previously found general solution to homogeneous case, with corresponding fundamental matrix: Using variation of parameters method, our solution is given by x = (t)u(t), where u(t) satisfies (t)
25、u(t) = g(t), or,Example 3: Solving for u(t) (2 of 3),Solving (t)u(t) = g(t) by row reduction,It follows that,Example 3: Solving for x(t) (3 of 3),Now x(t) = (t)u(t), and hence we multiplyto obtain, after collecting terms and simplifying,Note that this is the same solution as in Example 1.,Laplace Tr
26、ansforms,The Laplace transform can be used to solve systems of equations. Here, the transform of a vector is the vector of component transforms, denoted by X(s): Then by extending Theorem 6.2.1, we obtain,Example 4: Laplace Transform (1 of 5),Consider again the nonhomogeneous system x = Ax + g:Takin
27、g the Laplace transform of each term, we obtain where G(s) is the transform of g(t), and is given by,Example 4: Transfer Matrix (2 of 5),Our transformed equation is If we take x(0) = 0, then the above equation becomesorSolving for X(s), we obtainThe matrix (sI A)-1 is called the transfer matrix.,Exa
28、mple 4: Finding Transfer Matrix (3 of 5),Then Solving for (sI A)-1, we obtain,Example 4: Transfer Matrix (4 of 5),Next, X(s) = (sI A)-1G(s), and henceor,Example 4: Transfer Matrix (5 of 5),ThusTo solve for x(t) = L-1X(s), use partial fraction expansions of both components of X(s), and then Table 6.2
29、.1 to obtain:Since we assumed x(0) = 0, this solution differs slightly from the previous particular solutions.,Summary (1 of 2),The method of undetermined coefficients requires no integration but is limited in scope and may involve several sets of algebraic equations. Diagonalization requires findin
30、g inverse of transformation matrix and solving uncoupled first order linear equations. When coefficient matrix is Hermitian, the inverse of transformation matrix can be found without calculation, which is very helpful for large systems. The Laplace transform method involves matrix inversion, matrix
31、multiplication, and inverse transforms. This method is particularly useful for problems with discontinuous or impulsive forcing functions.,Summary (2 of 2),Variation of parameters is the most general method, but it involves solving linear algebraic equations with variable coefficients, integration, and matrix multiplication, and hence may be the most computationally complicated method. For many small systems with constant coefficients, all of these methods work well, and there may be little reason to select one over another.,
