19: LECTURE SYSTEMS OF DIFFERENTIAL EQUATIONS
MATH 60 SPRING 2009
1. Systems of Differential Equations Suppose that we are confronted with an n n system of linear dierential equations (with constant coecients) of the form y1 = a11 y1 + a12 y2 + + a1n yn y2 = a21 y1 + a22 y2 + + a2n yn (1) . . . . . .. . . .. . . . . . . . . . . . . . . . . yn = an1 y1 + an2 y2 + + ann yn where y1 , y2 , . . . , yn are unknown functions of t and the aij s are constants. Such systems are usually written in terms of matrices and vectors: y = Ay where y1 . y = . , . yn y1 . y = . . yn (2)
and A denotes the coecient matrix (aij ). Let us consider a simple system of the form (1) to see what insights we might be able to gain. 2. Uncoupled Systems As one might expect, if the coecient matrix A is a diagonal matrix, the corresponding system of linear dierential equations is particularly easy to solve. In particular, the following lemma will be most helpful: Lemma 1. If a is a constant, then the general solution to y = ay is y(t) = ceat where c is an arbitrary constant. Proof. Since any function of the form y(t) = ceat is obviously a solution to y = ay, we need only prove that all solutions to y = ay must be of this form. Suppose that y = ay and consider the derivative of the function yeat : (yeat ) = y eat + y(aeat ) = (ay)eat ayeat = ayeat ayeat =0 (the zero function). This implies that yeat = c for some constant c. In particular, this means that y = ceat , as claimed.
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MATH 60 SPRING 2009
Example 1. By the preceding lemma, the system y1 = y1 y2 = 2y2 y3 = 3y3 has the general solution y1 y2 y3 = = = c1 e t c2 e2t c3 e3t
(3)
(4)
where c1 , c2 , c3 are arbitrary constants. This is due to the fact that each of the initial three dierential equations in (3) involved only one of the unknown functions y1 , y2 , y3 . In other words, the system (3) is uncoupled the unknown functions y1 , y2 , y3 are not linked together in a complicated way. Let us consider the same system in terms matrices of and vectors to gain some insights. We can rewrite the original system (3) in the form y = Ay where 1 A = 0 0 0 2 0 0 0 , 3 y1 y = y2 , y3 2 = 2, y1 y = y2 . y3
Since the coecient matrix A is diagonal, we see that its eigenvalues are 1 = 1, 3 = 3.
The corresponding eigenvectors for the matrix A are 1 0 0 v1 = 0 , v2 = 1 , v3 = 0 , 0 0 1 which happen to be the standard basis vectors for R3 . The general solution (4) can therefore be written in the following form: y1 y = y2 y3 c1 et = c2 e2t c3 e3t 1 0 0 = c1 0 et + c2 1 e2t + c3 0 e3t 0 0 1 = c1 v1 e1 t + c2 v2 e2 t + c3 v3 e3 t . The structure of our answer is of interest, for it indicates what we might expect to happen if we consider systems of the form y = Ay where A is diagonalizable (i.e., A is similar to a diagonal matrix).
LECTURE 19: SYSTEMS OF DIFFERENTIAL EQUATIONS
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3. More Observations Based upon our brief experience with diagonal systems, we conjecture that the solutions to a system of dierential equations of the form (2) are most likely linear combinations of expressions like y = vet (5) where v is some nonzero vector to be determined and is some number to be determined. If a solution to ...