Applications of the Wronskian to ordinary linear differential equations

Transcription

1 Physics 116C Fall 2011 Applications of the Wronskian to orinary linear ifferential equations Consier a of n continuous functions y i (x) [i = 1,2,3,...,n], each of which is ifferentiable at least n times. Then if there exist a set of constants λ i that are not all zero such that λ i y 1 (x)+λ 2 y 2 (x)+ +λ n y n (x) = 0, (1) then we say that the set of functions {y i (x)} are linearly epenent. If the only solution to eq. (1) is λ i = 0 for all i, then the set of functions {y i (x)} are linearly inepenent. The Wronskian matrix is efine as: y 1 y 2 y n y 1 y 2 y n Φ[y i (x)] = y 1 y 2 y n, y (n 1) 1 y (n 1) 2 y n (n 1) where y i y i, y i 2 y i,, y(n 1) 2 i (n 1) y i. n 1 The Wronskian is efine to be the eterminant of the Wronskian matrix, W(x) et Φ[y i (x)]. (2) In Physics 116A, we learne that that if {y i (x)} is a linearly epenent set of functions then the Wronskian must vanish. However, the converse is not necessarily true, as one can fin cases in which the Wronskian vanishes without the functions being linearly inepenent. Nevertheless, if the y i (x) are solutions to an nth orer orinary linear ifferential equation, then the converse oes hol. That is, if the y i (x) are solutions to an nth orer orinary linear ifferential equation an the Wronskian of the y i (x) vanishes, then {y i (x)} is a linearly epenent set of functions. Moreover, if the Wronskian oes not vanish for some value of x, then it is oes not vanish for all values of x, in which case an arbitrary linear combination of the y i (x) constitutes the most general solution to the nth orer orinary linear ifferential equation. To emonstrate that the Wronskian either vanishes for all values of x or it is never equal to zero, if the y i (x) are solutions to an nth orer orinary linear ifferential equation, we shall erive a formula for the Wronskian. Consier the ifferential equation, a 0 (x)y (n) +a 1 (x)y (n 1) + +a n 1 (x)y +a n (x)y = 0. (3) 1

2 We are intereste in solving this equation over an interval of the real axis a < x < b in which a 0 (x) 0. We can rewrite eq. (3) as a first orer matrix ifferential equation. Defining the vector y y Y = y..., y (n 1) It is straightforwar to verify that eq. (3) is equivalent to Y = A(x) Y, where the matrix A(x) is given by A(x) = a n(x) a n 1(x) a n 2(x) a n 3(x) a 1(x) a 0 (x) a 0 (x) a 0 (x) a 0 (x) a 0 (x). (4) It immeiately follows that if the y i (x) are linearly inepenent solutions to eq. (3), then the Wronskian matrix satisfies the first orer matrix ifferential equation, Φ Using eq. (21) of Appenix I, it follows that etφ = etφtr = A(x)Φ. (5) ( Φ 1Φ ) = etφtr A(x), after using eq. (5) an the cyclicity property of the trace (i.e. the trace is unchange by cyclically permuting the matrices insie the trace). In terms of the Wronskian W efine in eq. (2), W = W Tr A(x). (6) This is a first orer ifferential equation for W that is easily integrate, { x } W(x) = W(x 0 )exp TrA(t)t. x 0 Using eq. (4), it follows that TrA(t) = a 1 (t)/a 0 (t). Hence, we arrive at Liouville s formula (also calle Abel s formula), { x } a 1 (t) W(x) = W(x 0 )exp x 0 a 0 (t) t. (7) 2

3 Note that if W(x 0 ) 0, then the result for W(x) is strictly positive or strictly negative epening on the sign of W(x 0 ). This confirms our assertion that the Wronskian either vanishes for all values of x or it is never equal to zero. Let us apply these results to an orinary secon orer linear ifferential equation, y +a(x)y +b(x)y = 0, (8) where for convenience, we have ivie out by the function that originally appeare multiplie by y. Then, eq. (7) yiels the Wronskian, which we shall write in the form: { } W(x) = cexp a(x), (9) where c is an arbitrary nonzero constant an a(x) is the inefinite integral of a(x). A simpler an more irect erivation of eq. (9) is presente in Appenix II. The Wronskian also appears in the following application. Suppose that one of the two solutions of eq. (8), enote by y 1 (x) is known. Then, one can erive a secon linearly inepenent solution y 2 (x) by the metho of variations of parameters. In this context, the iea of this metho is to efine a new variable v, y 2 (x) = v(x)y 1 (x) = y 1 (x) w(x), (10) where Then, we have v w. (11) y 2 = vy 1 +wy 1, y 2 = vy 1 +w y 1 +2wy 1. Since y 2 is a solution to eq. (8), it follows that w y 1 +w[2y 1 +a(x)y 1 ]+v[y 1 +a(x)y 1 +b(x)y 1 ] = 0, Using the fact that y 1 is a solution to eq. (8), the coefficient of v vanishes an we are left with a first orer ifferential equation for w w y 1 +w[2y 1 +a(x)y 1] = 0. After iviing this equation by y 1, we see that the solution to the resulting equation is { ( ) } 2y w(x) = c exp 1 (x) y 1 (x) +a(x) { } = ce 2lny1(x) exp a(x) = { c [y 1 (x)] exp 2 } a(x) = W(x) [y 1 (x)], (12) 2 More generally, if any non-trivial solution to eq. (3) is known, then this solution can be employe to reuce the orer of the ifferential equation by 1. This proceure is calle reuction of orer. 3

4 after using eq. (9) for the Wronskian. The secon solution to eq. (8) efine by eq. (10) is then given by W(x) y 2 (x) = y 1 (x) [y 1 (x)] 2 after employing eq. (12). Finally, we note that the Wronskian also appears in solutions to inhomogeneous linear ifferential equations. For example, consier y +a(x)y +b(x)y = f(x), (13) an assume that the solutions to the homogeneous equation [eq. (8)], enote by y 1 (x) an y 2 (x) are known. Then the general solution to eq. (13) is given by y(x) = c 1 y 1 (x)+c 2 y 2 (x)+y p (x), where y p (x), calle the particular solution, is etermine by the following formula, y2 (x)f(x) y1 (x)f(x) y p (x) = y 1 (x) +y 2 (x). (14) W(x) W(x) This result is erive using the technique of variation of parameters. Namely, one writes y p (x) = v 1 (x)y 1 (x)+v 2 (x)y 2 (x), (15) subject to the conition (which is chosen entirely for convenience): With this choice, it follows that v 1 y 1 +v 2 y 2 = 0. (16) y p = v 1 y 1 +v 2 y 2. Differentiating once more an plugging back into eq. (13), one obtains [after using eq. (16)]: v 1y 1 +v 2y 2 = f(x). (17) We now have two equations, eqs. (16) an (17), which constitute two algebraic equations for v 1 an v 2. The solutions to these equations yiel v 1 = y 2(x)f(x) W(x), v 2 = y 1(x)f(x) W(x) where W(x) is the Wronskian. We now integrate to get v 1 an v 2 an plug back into eq. (15) to obtain eq. (14). The erivation is complete. Reference: Daniel Zwillinger, Hanbook of Differential Equations, Thir Eition (Acaemic Press, San Diego, CA, 1998). 4,

5 APPENDIX I: Derivative of the eterminant of a matrix Recall that for any matrix A, the eterminant can be compute by the cofactor expansion. The ajugate of A, enote by aj A is equal to the transpose of the matrix of cofactors. In particular, eta = j a ij (aj A) ji, for any fixe i, (18) where the a ij are elements of the matrix A an (aj A) ji = ( 1) i+j M ij where the minor M ij is the eterminant of the matrix obtaine by eleting the ith row an jth column of A. Suppose that the elements a ij epen on a variable x. Then, by the chain rule, eta = i,j eta a ij a ij. (19) Using eq. (18), an noting that (aj A) ji oes not epen on a ij (since the ith row an jth column are remove before computing the minor eterminant), Hence, eq. (19) yiels Jacobi s formula: eta = i,j eta a ij = (aj A) ji. If A is invertible, then we can use the formula to rewrite eq. (20) as which is the esire result. Reference: [ a ij (aj A) ji = Tr (aj A) A ]. (20) A 1 eta = Aj A, ( eta = etatr A 1A ), (21) M.A. Golberg, The erivative of a eterminant, The American Mathematical Monthly, Vol. 79, No. 10 (Dec. 1972) pp Recall that if A = [a ij ] an B = [b ij ], then the ij matrix element of AB are given by k a ikb kj. The trace of AB is equal to the sum of its iagonal elements, or equivalently Tr(AB) = jk a jk b kj. Note that Tr (cb) = ctr B for any number c an matrix B. In eriving eq. (21), c = eta. 5

6 APPENDIX II: Derivation of Abel s equation Eq. (9) is usually calle Abel s formula, an can be erive irectly without using any fancy techniques. Consier the orinary linear secon orer ifferential equation, y +a(x)y +b(x)y = 0. (22) Suppose that y 1 (x) an y 2 (x) are linearly inepenent solutions of eq. (22). Then the Wronskian is non-vanishing, ( ) y1 y W = et 2 y 1 y 2 = y 1 y 2 y 1 y 2 0. Taking the erivative of the above equation, W = (y 1y 2 y 1y 2 ) = y 1 y 2 y 1y 2, since the terms proportional to y 1 y 2 exactly cancel. Using the fact that y 1 an y 2 are solutions to eq. (22), we have y 1 1 +b(x)y 1 = 0, (23) y 2 +a(x)y 2 +b(x)y 2 = 0. (24) Next, we multiply eq. (24) by y 1 an multiply eq. (23) by y 2, an subtract the resulting equations. The en result is: y 1 y 2 y 1 y 2 +a(x)[y 1 y 2 y 1 y 2] = 0. or equivalently [cf. eq. (6)], W +a(x)w = 0, (25) The solution to this first orer ifferential equation is Abel s formula, { } W(x) = c exp a(x), (26) where c is an arbitrary constant. Eq. (26) is a special case of Liouville s formula given in eq. (7) for orinary secon-orer linear ifferential equations (n = 2). 6