General Solutions of 2nd Order Linear Homogeneous Ordinary D.E.s

[cepheid]

A theorem in my textbook is confusing me:

For the functions $p(t) \ \ \text{and} \ \ q(t)$ continuous on an open inteval $I$ defined by $\alpha < t < \beta$:

We have differential equation $L[y] = 0$ where

$$L = (\frac{d^2}{dt^2} + p\frac{d}{dt} + q)$$

The theorem attempts to prove that the general solution can be expressed in the form

$$y = c_{1} y_{1} + c_{2} y_{2}$$


It does so by looking at an arbitrary solution $y = \phi (t)$. Let's say that at some point $t_0$, we let $\phi (t_0) = y_0 \ \ \text{and} \ \ \phi^{\prime} (t_0) = y^{\prime}_0$.

Then $\phi (t)$ is certainly a solution to the initial value problem:

$$y^{\prime\prime} + py^{\prime} + qy = 0$$
$$y(t_0) = y_0$$
$$y^{\prime}(t_0) = y^{\prime}_0$$

But, as proven by another theorem, as long as the Wronskian of $y_1 \ \ \text{and} \ \ y_2$ is not zero at $t_0$, then we can find constants c1 and c2 such that

$$y = c_{1} y_{1} + c_{2} y_{2}$$

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.:

$$y = \phi (t) = c_{1} y_{1} + c_{2} y_{2}$$

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 $t_0$ that I care to choose in $I$. That condition demands that the Wronskian of $y_1 \ \ \text{and} \ \ y_2$ 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?

 

[ReyChiquito]

"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.

Remember that these conditions are all local. If $W[y_{1},y_{2}](t_{0})\neq 0$, means that you can build a solution in the form $c_{1}y_{1}+c_{2}y_{2}$ locally (in a vecinity of $t_{0}$, not in the whole interval). The curve $\phi(t)$ happens to be the graph of that solution, but only locally. To extend the solution to the whole interval, you need to be more carefull.

Im sure you can find a proof for the existence and uniqueness theorem for 2 order ode in Brown or Coddington book.

If you can't find the proof, give me some time and ill post a detailed one.

 

[cepheid]

I don't mind the lack of proof for the existence/uniqueness theorem at the moment. It's just this theorem for now that still baffles me. I know now that I'm wrong and that the W need only be nonzero at one point, but I have no idea how this proves the theorem. Couldn't I equally well pick a point at which the Wronskian is zero for sure, set up an initial value problem to which phi is a solution, and show that phi cannot be expressed in the form c1y1 + c2y2?

 

[HallsofIvy]

It can be shown that, if y1 and y2 are both solutions to a second order, linear, homogeneous differental equation, then the Wronskian can be written as an exponential. It is either identically 0 (so y1 and y2 are dependent) or never 0 (so y1 and y2 are independent) on an interval.

 

[ReyChiquito]

also, you migh take in mind that if $W[y_{1},y_{2}](t)=0$ and y is non trivial

$$y_{1}\dot{y}_{2}-\dot{y}_{1}y_{2}=0$$

if $y_{2}\neq 0$ (if it is 0 then take $y_{1}$)

$$\frac{y_{1}\dot{y}_{2}-\dot{y}_{1}y_{2}}{{y_{2}}^{2}}}=0$$

$$\frac{d}{dt}(\frac{y_{1}}{y_{2}})=0$$

integrating

$$\frac{y_{1}}{y_{2}}=\lambda$$

$$y_{1}=\lambda y_{2}$$

so $y_{1}$ and $y_{2}$ are lineary dependent.