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A theorem in my textbook is confusing me:
For the functions [itex] p(t) \ \ \text{and} \ \ q(t) [/itex] continuous on an open inteval [itex] I [/itex] defined by [itex] \alpha < t < \beta [/itex]:
We have differential equation [itex] L[y] = 0 [/itex] where
[tex] L = (\frac{d^2}{dt^2} + p\frac{d}{dt} + q) [/tex]
The theorem attempts to prove that the general solution can be expressed in the form
[tex] y = c_{1} y_{1} + c_{2} y_{2} [/tex]
It does so by looking at an arbitrary solution [itex] y = \phi (t) [/itex]. Let's say that at some point [itex] t_0 [/itex], we let [itex] \phi (t_0) = y_0 \ \ \text{and} \ \ \phi^{\prime} (t_0) = y^{\prime}_0 [/itex].
Then [itex] \phi (t) [/itex] is certainly a solution to the initial value problem:
[tex] y^{\prime\prime} + py^{\prime} + qy = 0 [/tex]
[tex] y(t_0) = y_0 [/tex]
[tex] y^{\prime}(t_0) = y^{\prime}_0 [/tex]
But, as proven by another theorem, as long as the Wronskian of [itex] y_1 \ \ \text{and} \ \ y_2 [/itex] is not zero at [itex] t_0 [/itex], then we can find constants c1 and c2 such that
[tex] y = c_{1} y_{1} + c_{2} y_{2} [/tex]
is also a solution to this initial value problem. By the uniqueness theorem (which the textbook declines to prove), there is only one solution to the initial value problem i.e.:
[tex] y = \phi (t) = c_{1} y_{1} + c_{2} y_{2} [/tex]
In my mind, this constitutes an actual proof of the theorem if and only if it is true for all possible initial value problems i.e. if I can show this for every possible point [itex] t_0 [/itex] that I care to choose in [itex] I [/itex]. That condition demands that the Wronskian of [itex] y_1 \ \ \text{and} \ \ y_2 [/itex] is not zero at any point in the interval! Not only does the textbook not make this point clear, it says something entirely different!
"Not everywhere zero" is not the same thing as "nonzero everywhere"! As I've already said, I think it should be the latter. The former implies that it can be zero in certain places, as long as it isn't zero everywhere.
So which is wrong, my interpretation of the proof, or the statement in the textbook?
For the functions [itex] p(t) \ \ \text{and} \ \ q(t) [/itex] continuous on an open inteval [itex] I [/itex] defined by [itex] \alpha < t < \beta [/itex]:
We have differential equation [itex] L[y] = 0 [/itex] where
[tex] L = (\frac{d^2}{dt^2} + p\frac{d}{dt} + q) [/tex]
The theorem attempts to prove that the general solution can be expressed in the form
[tex] y = c_{1} y_{1} + c_{2} y_{2} [/tex]
It does so by looking at an arbitrary solution [itex] y = \phi (t) [/itex]. Let's say that at some point [itex] t_0 [/itex], we let [itex] \phi (t_0) = y_0 \ \ \text{and} \ \ \phi^{\prime} (t_0) = y^{\prime}_0 [/itex].
Then [itex] \phi (t) [/itex] is certainly a solution to the initial value problem:
[tex] y^{\prime\prime} + py^{\prime} + qy = 0 [/tex]
[tex] y(t_0) = y_0 [/tex]
[tex] y^{\prime}(t_0) = y^{\prime}_0 [/tex]
But, as proven by another theorem, as long as the Wronskian of [itex] y_1 \ \ \text{and} \ \ y_2 [/itex] is not zero at [itex] t_0 [/itex], then we can find constants c1 and c2 such that
[tex] y = c_{1} y_{1} + c_{2} y_{2} [/tex]
is also a solution to this initial value problem. By the uniqueness theorem (which the textbook declines to prove), there is only one solution to the initial value problem i.e.:
[tex] y = \phi (t) = c_{1} y_{1} + c_{2} y_{2} [/tex]
In my mind, this constitutes an actual proof of the theorem if and only if it is true for all possible initial value problems i.e. if I can show this for every possible point [itex] t_0 [/itex] that I care to choose in [itex] I [/itex]. That condition demands that the Wronskian of [itex] y_1 \ \ \text{and} \ \ y_2 [/itex] is not zero at any point in the interval! Not only does the textbook not make this point clear, it says something entirely different!
Boyce and DiPrima, 8th Ed. pg. 148:
"Theorem 3.2.4 of states that, as long as the Wronksian of y1 and y2 is not everywhere zero, the linear combination c1y1 + c2y2 contains all possible solutions of [the differential equation]."
"Not everywhere zero" is not the same thing as "nonzero everywhere"! As I've already said, I think it should be the latter. The former implies that it can be zero in certain places, as long as it isn't zero everywhere.
So which is wrong, my interpretation of the proof, or the statement in the textbook?
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