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X = 0.06 X + 0.01 Y
Y = 0.04 X + 0.05 Y
which is homogeneous and has the same coefficients as the system (6). In terms of X and Y, the initial conditions (7) become
X (0) = 0.1, Y (0) = 0.1
Although we have converted a nonhomogeneous system to a homogeneous system in this particular case, it isn’t always possible to do so.
It is useful here to introduce matrix notation: it saves space and it expresses system (4) in the form of a single equation. Let x be the vector with components xi, x2, ... , xn and let A be the matrix of the coefficients, where aij is the element in the ith row and jth column of A. The derivative of the vector x, written dx/dt or x;, is defined to be the vector with the components dx1/ dt,... , dxn/dt. Therefore we can write the system (4) in the compact form
x; = Ax, where A =
In vector notation, the initial conditions (5) become
x(t0) = à
where a is the vector with components a1,... , an.
? Find the linear system matrix for system (8).
A solution of the initial value problem (10) and (11) is a set of functions
x1 = x1 (t)
xn = xn (t)
that satisfy the differential equations and initial conditions. Using our new notation, if x(t) is the vector whose components are x1 (t), ... , xn (t), then x = x(t) is a solution of the corresponding vector IVP, (10) and (11). The system x; = Ax is homogeneous, while a nonhomogeneous system would have the form x; = Ax + F, where F is a vector function of t or else a constant vector.
Solution Formulas: Eigenvalues and Eigenvectors
¦ Solution Formulas: Eigenvalues and Eigenvectors
The determinant of a matrix is denoted by det.
The keys to finding a solution formula for X = Ax are the eigenvalues and eigenvectors of A.
Formula (17)iscalled the general solution formula of system (10) because every solution has the form of (17) for some choice of the constants Cj . The other way around, every choice of the constants yields a solution of system (10).
To find a solution formula for system (10) let’s look for an exponential solution of the form
X = ve
where X is a constant and v is a constant vector to be determined. Substituting x as given by (13) into the ODE (10), we find that v and X must satisfy the algebraic equation
Av = Xv
Equation (14) can also be written in the form
(A - XI)v = 0
where I is the identity matrix and 0 is the zero vector with zero for each component. Equation (15) has nonzero solutions if and only if X is a root of the nth-degree polynomial equation
det(A - XI) = 0
called the characteristic equation for the system (10). Such a root is called an eigenvalue of the matrix A. We will denote the eigenvalues by X1,... ,Xn. For each eigenvalue Xt there is a corresponding nonzero solution v(l), called an eigenvector. The eigenvectors are not determined uniquely but only up to an arbitrary multiplicative constant.
For each eigenvalue-eigenvector pair (Xl7 v(l)) there is a corresponding vector solution v(i)eXit of the ODE (10). If the eigenvalues X1,... ,Xn are all different, then there are n such solutions,
In this case the general solution of system (10) is the linear combination
X = C1v(1)eX11 + ••• + Cnv(n)eXnt (17)
The arbitrary constants C1, ... ,Cn can always be chosen to satisfy the initial conditions (11). If the eigenvalues are not distinct, then the general solution takes on a slightly different (but similar) form. The texts listed in the references give the formulas for this case. If some of the eigenvalues are complex, then the solution given by formula (17) is complex-valued. However, if all of the coefficients aij are real, then the complex eigenvalues and eigenvectors occur in complex conjugate pairs, and it is always possible to express the solution formula (17) in terms of real-valued functions. Look ahead to formulas (20) and (21) for a way to accomplish this feat.
¦ Calculating Eigenvalues and Eigenvectors
Here’s how to find the eigenvalues and eigenvectors of a 2 x 2 real matrix
a b c d
First define the trace of A (denoted by tr A) to be the sum a + d of the diagonal entries, and the determinant of A (denoted by det A) to be the number ad — bc. Then the characteristic equation for A is
det(A — AI) = det
a — A b c d— A
— A — (a + d) A + ad — bc