The Normalized Fundamental Matrix

The Normalized Fundamental Matrix In the previous note we saw two main facts about the fundamental matrix:     x1 x1 x2 = . (1) Φ(t) = x2 y1 y2 and x = Φ ( t ) Φ ( t 0 ) − 1 x0 . (2) Is there a “best” choice for fundamental matrix? There are two common choices, each with its advantages. If the ODE system has constant coefficients, and its eigenvalues are real and distinct, then a natural choice for the fundamental matrix would be the one whose columns are the normal modes — the solutions of the form xi = ~αi ełi t , i = 1, 2. There is another choice however which is suggested by (2) and which is particularly useful in showing how the solution depends on the initial conditions. Suppose we pick Φ(t) so that   1 0 Φ ( t0 ) = I = . (3) 0 1 Referring to the definition (1), this means the solutions x1 and x2 are picked so     1 0 x1 ( t 0 ) = , x2 (t0 ) = . (30 ) 0 1 Since the xi (t) are uniquely determined by these initial conditions, the fundamental matrix Φ(t) satisfying (3) is also unique; we give it a name. e t0 (t) satisfying Definition 2 The unique matrix Φ e 0t = A Φ e t0 , Φ 0 e t0 ( t 0 ) = I Φ is called the normalized fundamental matrix at t0 for A. For convenience in use, the definition uses Theorem 1 to e t0 will actually be a fundamental matrix. The guarantee Φ e t0 (t)| 6= 0 in Theorem 1 is satisfied, since the condition |Φ e t0 (t0 )| = 1. definition implies |Φ (4) To keep the notation simple, we will assume in the rest of this section e 0 is the normalized fundamental that t0 = 0, as it almost always is; then Φ e 0 (0) = I, we get from (2) the matrix form for the solution to matrix. Since Φ 0 the IVP: x = A(t) x, x(0) = x0 is e 0 (t)x0 . x( t ) = Φ (5) e 0 . One way is to find the two solutions in (30 ) and use Calculating Φ e 0 . This is fine if the two solutions can be deterthem as the columns of Φ mined by inspection. If not, a simpler method is this: find any fundamental matrix Φ(t); then e 0 ( t ) = Φ ( t ) Φ (0 ) −1 . Φ (6) To verify this, we have to see that the matrix on the right of (6) satisfies the two conditions in Definition 2. The second is trivial. The first is easy using the rule for matrix differentiation: If M = M(t) and B, C are constant matrices, then ( BM)0 = BM0 , ( MC )0 = M0 C, from which we see that since Φ is a fundamental matrix, (Φ(t)Φ(0)−1 )0 = Φ(t)0 Φ(0)−1 = AΦ(t)Φ(0)−1 = A(Φ(t)Φ(0)−1 ), showing that Φ(t)Φ(0)−1 also satisfies the first condition in Definition 2.    0 1 Example 2A. Find the solution to the IVP: x0 = x , x(0) = −1 0 x0 . Solution. Since the system is x 0 = y, y0 = − x, we can find by inspection the fundamental set of solutions satisfying (30 ) : x = cos t y = − sin t and x = sin t y = cos t . Thus by (5) the normalized fundamental matrix at 0 and solution to the IVP is        cos t sin t x0 cos t sin t e x = Φ x0 = = x0 + y0 . − sin t cos t y0 − sin t cos t 0 Example 2B.   Give the normalized fundamental matrix at 0 for x 1 3 x. 1 −1 = Solution. This time the solutions (30 ) cannot be obtained by inspection, so we use the second method. You can easily find the eigenvalues and eigenvectors for this system. Doing so produces the normal modes. Using them as the columns of a fundamental matrix gives us   2t −e−2t 3e . Φ(t) = e−2t e2t Using (6) and the formula for calculating the inverse matrix we get     1 1 1 3 −1 −1 , Φ (0) = , Φ (0) = −1 3 1 1 4 so that e (t) = 1 Φ 4  3e2t −e−2t e2t e−2t  1 1 −1 3  1 = 4  3e2t + e2t 3e2t − 3e−2t e2t − e−2t e2t + 3e−2t  .