72 15. The Wronskian We know that a general second order homogeneous linear ODE, (1) y →→ + p(x)y → + q(x)y = 0, has a pair of independent solutions; and that if y1, y2 is any pair of independent solutions then the general solution is (2) y = c1y1 + c2y2. Suppose we wish to solve the initial value problem with y(x0) = a , y →(x0) = b. To solve for the constants c1 and c2 in (2), we get one equation by substituting x = x0 into this expression. We get another equation by first differentiating (2) and then setting x = x0 and using the value y→(x0) = b. We end up with the system of linear equations (3) y1(x0)c1 + y2(x0)c2 = a , y1→ (x0)c1 + y2→ (x0)c2 = b When you solve for the coefficients of these equations, you get (“Cramer’s rule”) = y2→ (x0)a − y2(x0)b, = −y1→ (x0)a + y1(x0)b c1 c2W (x0) W (x0) where W (x0) is the value at x0 of the Wronskian function ⎩ ⎪ → → y1 y2(4) W (x) = y1y2 − y2y1 = det → →y1 y2 determined by the pair of solutions y1, y2. You generally wouldn’t want to use these formulas for the coefficients; it’s better to compute them directly from (3) in the particular case you are looking at. But this calculation does draw attention to the Wronskian function. We can find a linear combination of y1 and y2 which solves the IVP for any given choice of initial conditions exactly when W (x0) = 0. ∈On the other hand it’s a theorem that one can solve the initial value problem at any x value using a linear combination of any linearly inde-pendent pair of solutions. A little thought then leads to the following conclusion: Theorem. Let y1, y2 be solutions of (1) and let W be the Wronskian formed from y1, y2. Either W is the zero function and one solution is a multiple of the other, or the Wronskian is nowhere zero, neither solution73 is a multiple of the other, and any solution is a linear combination of y1, y2. For example, if we compute the Wronskian of the pair of solutions {cos x, sin x} of y→→ + y = 0, we get the constant function 1, while the Wronskian of {cos x, 2 cos x} is the constant function 0. One can show (as most ODE textbooks do) that if W is the Wronskian of some linearly independent pair of solutions, then the Wronskian of any pair of solutions is a constant multiple of W . (That multiple is zero if the new pair happens to be linearly dependent.) Many references, including Edwards and Penney, encourage the im-pression that computing the Wronskian of a pair of functions is a good way to check whether or not they are linearly independent. This is silly. Two functions are linearly dependent if one is a multiple of the other; otherwise they are linearly independent. This is always easy to see by inspection. Nevertheless the Wronskian can teach us important things. To illus-trate one, let’s consider an example of a second order linear homoge-neous system with nonconstant coefficient: the Airy equation (5) y →→ + xy = 0. →→At least for x > 0, this is like the harmonic oscillator y + �n2 y = 0, except that the natural circular frequency �n keeps increasing with x: the x sits in the position where we expect to see �n2 , so near to a given value of x we expect solutions to behave like cos(≤xx) and sin(≤xx). I emphasize that these functions are not solutions to (5), but they give us a hint of what to expect. In fact the normalized pair (see Section 9) of solutions to (5), the “Airy cosine and sine functions,” have graphs as illustrated in Figure 9 One of the features this picture has in common with the graphs of cosine and sine is the following fact, which we state as a theorem. Theorem. Let {y1, y2} be any linearly independent pair of solutions of the second order linear ODE (1), and suppose that x0 and x1 are numbers such that x0 = x1 and y1(x0) = 0 = y1(x1). Then y2 becomes ∈zero somewhere between x0 and x1. This fact, that zeros of independent solutions interleave, is thus a completely general feature of second order linear equations. It doesn’t depend upon the solutions being normalized, and it doesn’t depend upon the coefficients being constant.74 0 2 4 6 8 10 12 −1 −0.5 0 0.5 1 1.5 Figure 9. Airy cosine and sine You can see why this must be true using the Wronskian. We might as well assume that y1 is not zero anywhere between x0 and x1. Since the two solutions are independent the associated Wronskian is nowhere zero, and thus has the same sign everywhere. Suppose first that the sign is positive. Then y1y→ 1y2 (x0) and y2(x0) have opposite signs, since y1(x0) = 0. Similarly, →everywhere. At x0 this says that > y 2 →y 1→(x1) and y2(x1) have opposite signs. But y→ 1(x0) and y→ 1(x1) must y 1have opposite signs as well, since x0 and x1 are neighboring zeros of y1. (These derivatives can’t be zero, since if they were both terms in the definition of the Wronskian would be zero, but W (x0) and W (x1) are nonzero.) It follows that y2(x0) and y2(x1) have opposite signs, and so y2 must vanish somewhere in between. The argument is very similar if the sign of the Wronskian is negative.MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations�� Spring 2010 For information about citing these materials or our Terms of Use, visit:
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