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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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Theorem 3.3.1 If f and g are differentiable functions on an open interval I and if W(f, g)(10) = 0 for some point t0 in I, then f and g are linearly independent on I. Moreover, if f and g are linearly dependent on I, then W(f, g)(1) = 0 for every 1 in I.
To prove the first statement in Theorem 3.3.1, consider a linear combination k2 f (1) + k2g(1), and suppose that this expression is zero throughout the interval. Evaluating the expression and its derivative at t0, we have
k1f (10) + k2 g(10) = > (5)
ki f (0) + k2g'(10) = 0.
The determinant of coefficients of Eqs. (5) is precisely W (f, g)(10), which is not zero by hypothesis. Therefore, the only solution of Eqs. (5) is k2 = k2 = 0, so f and g are linearly independent.
The second part of Theorem 3.3.1 follows immediately from the first. Let f and g be linearly dependent, and suppose that the conclusion is false, that is, W(f, g) is not everywhere zero in I. Then there is a point t0 such that W(f, g)(10) = 0; by the first part of Theorem 3.3.1 this implies that f and g are linearly independent, which is a contradiction, thus completing the proof.
3.3 Linear Independence and the Wronskian
149
Theorem 3.3.2
We can apply this result to the two functions f (1) = el and g(1) = e2t discussed in Example 2. For any point t0 we have
W( f, g)(L) =
ef0 e2t0
e*0 2e210
= e310 = 0. (6)
The functions el and e2t are therefore linearly independent on any interval.
You should be careful not to read too much into Theorem 3.3.1. In particular, two functions f and g may be linearly independent even though W(f, g)(1) = 0 for every
1 in the interval I. This is illustrated in Problem 28.
Now let us examine further the properties of the Wronskian of two solutions of a second order linear homogeneous differential equation. The following theorem, perhaps surprisingly, gives a simple explicit formula for the Wronskian of any two solutions of any such equation, even if the solutions themselves are not known.
(Abels Theorem)4 If y2 and y2 are solutions of the differential equation
L [y] = y + p(1) + q (1 )y = 0, (7)
where p and q are continuous on an open interval I, then the Wronskian W(y2, y2)(1) is given by
W(yi,y2)(1) = exp
/
- (1) d1
(8)
where is a certain constant that depends on y2 and y2, but not on 1. Further,
W(yi, y2)(1) is either zero for all 1 in I (if = 0) or else is never zero in I (if = 0).
To prove Abels theorem we start by noting that y2 and y2 satisfy
ӳ + p(1) yi + q(1) ӳ = 0 y2 + p(1) y2 + q(1) 2 = .
If we multiply the first equation by -y2, the second by y2, and add the resulting equations, we obtain
(ӳ2 - ӳ2) + p(1 )(ӳ2 - Ӳ2) = . (10)
Next, we let W(1) = W(yvy2)(1) and observe that
W' = ӳ2/- 12. (11)
Then we can write Eq. (10) in the form
W' + p(t )W = 0. (12)
4The result in Theorem 3.3.2 was derived by the Norwegian mathematician Niels Henrik Abel (1802-1829) in 1827 and is known as Abels formula. Abel also showed that there is no general formula for solving a quintic, or fifth degree, polynomial equation in terms of explicit algebraic operations on the coefficients, thereby resolving a question that had been open since the sixteenth century. His greatest contributions, however, were in analysis, particularly in the study of elliptic functions. Unfortunately, his work was not widely noticed until after his death. The distinguished French mathematician Legendre called it a monument more lasting than bronze.
150
Chapter 3. Second Order Linear Equations
EXAMPLE
3
Theorem 3.3.3
Equation (12) can be solved immediately since it is both a first order linear equation (Section 2.1) and a separable equation (Section 2.2). Thus
W (1) = exp
/
- p(1) d1
(13)
where is a constant. The value of depends on which pair of solutions of Eq. (7) is involved. However, since the exponential function is never zero, W(1) is not zero unless = 0, in which case W(1) is zero for all 1, which completes the proof of Theorem
3.3.2.
Note that the Wronskians of any two fundamental sets of solutions of the same differential equation can differ only by a multiplicative constant, and that the Wronskian of any fundamental set of solutions can be determined, up to a multiplicative constant, without solving the differential equation.
In Example 5 of Section 3.2 we verified that y2 (1) = 11/2 and y2(1) = 1 1 are solutions of the equation
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