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I have the following two theorems:

Theorem 1: Consider the initial value problem $$y''+p(t)y' + q(t)y = g(t), y(t_0) = y_0, y'(t_0) = y'_0$$ where $p, q,$ and $g$ are continuous on an open interval $I$. Then there is exactly one solution $y = \phi(t)$ of this problem, and the solution exists throughout the interval $I$.

Theorem 2: Suppose that $y_1$ and $y_2$ are solutions of $$L[y] = y'' + p(t)y' + q(t)y = 0,$$ and that the Wronskian $$W = y_1y_2' - y_1'y_2$$ is not zero at the point $t_0$ where the initial conditions, $$y(t_0) = y_0, y'(t_0) = y_0',$$ are assigned. Then there is a choice of the constants $c_1,c_2$ for which $y = c_1y_1(t) + c_2y_2(t)$ satisfies the differential equation and the initial conditions.

Question: How can these theorems exist at the same time? Theorem 1 says that there is only one solution of the given initial value problem, while theorem 2 says that for the same (I'm obviously mistaken) initial value problem you can choose the constants $c_1$ and $c_2$. The only difference I see here is that $g(t) = 0$ in the second theorem. But if $g(t) = 0$ then $g(t)$ is still continuous so this shouldn't imply that in that case theorem 1 doesn't hold right?

Thanks!

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    $\begingroup$ The choice of constants is unique, of course. :) $\endgroup$ – Pedro Tamaroff Feb 27 '18 at 12:01
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Given the initial conditions, both theorems provide a unique solution satisfying the differential equation and the initial conditions.

One is for homogenous and the other one for the more general case of inhomogenous equations.

There is no conflict here.

Continuity of coefficients in linear equations results in uniqueness of solutions passing through given initial values.

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