I am having trouble at the very last step in proving the following theorem:
Theorem A. Let $y_1(x)$ and $y_2(x)$ be linearly independent solutions of the homogeneous equation $$y''+P(x)y'+Q(x)y=0 \space\space\space\space\space\space (1)$$
on the interval $[a,b]$.
Then $$c_1y_1(x)+c_2y_2(x) \space\space\space\space\space\space\space\space\space\space\space\space (2)$$ is the general solution of equation (1) on $[a,b]$, in the sense that every solution of (1) on this interval can be obtained from (2) by suitable choice of the arbitrary constants $c_1$ and $c_2$.
Proof.
We must show that, if $y(x)$ is any solution then the constants $c_1$ and $c_2$ can be found so that: $$y(x)=c_1y_1(x)+c_2y_2(x) \space\space\space\forall x \in[a,b]$$
By the uniquess or existence theorem (which we don't need to prove here). It suffices to show that for some point $x_0$ in $[a,b]$ we can find $c_1$ and $c_2$ such that:
$$c_1y_1(x_0)+c_2y_2(x_0)=y(x_0)$$
$$c_1y_1'(x_0)+c_2y_2'(x_0)=y'(x_0)$$
For it to be solvable, the determinant should not be equal zero:
So, the Wronskian: $$W(y_1,y_2)=y_1y_2'-y_2y_1' \neq 0$$
The book then goes on to investigate when $$W(y_1,y_2)$$ is equal to zero:
And to not make the post too long, using a lemma we arrived at:
$$W=ce^{-\int Pdx}$$
And so, W is either identically zero on $[a,b]$ or never zero on $[a,b]$ (Which was the claim of the lemma, I.e. W is either identically zero or never zero on $[a,b]$, and that $x_0$ location is of no consequence)
Now the textbook said: "This results reduces our overall task of proving the theorem to that of showing the Wronskian of any two linearly independent solutions of (1) is not identically zero"
Lemma 2.
If $y_1(x)$ and $y_2(x)$ are two solutions of equation (1) on $[a,b]$, then they are linearly dependent on this interval if and only if their Wronskian is identically zero.
(This is the proof I am having difficulty with in the "interval" parts:)
Proof of lemma.
Assume $y_1$ and $y_2$ are linearly dependent first, then we show as a consequence of this that the Wronskian is zero. Also assume neither function is identically zero, as this is a trivial case. So we have $y_2=cy_1$ and substituting into the Wronskian indeed yields 0.
Now the other half of the lemma:
Assume the Wronskian is identically zero and we want to prove that the solutions must be linearly dependent. Assume $y_1$ does not vanish identically on $[a,b]$, from which it follows by continuity that $y_1$ does not vanish at all on some subinterval $[c,d]$ of $[a,b]$. Since the Wronskian is identically zero on $[a,b]$, we can divide by $y_1^2$ to get:
$$\frac{y_1y_2'-y_2y_1'}{y_1^2}=0 $$
On [c,d]
Then,
$$\int(\frac{y_2}{y_1})' = \int 0 dx $$
Yielding, $$ y_2=ky_1 \space\space\space \forall x \in[c,d] $$
Now, what I do not understand is the following line:
"Finally,
since $y_2(x)$ and $ky_1(x)$ have equal values in $[c,d]$, they have equal derivatives
there as well; and Theorem 14-A allows us to infer that
$y_2(x)=ky_1(x)$
for all $x$ in $[a,b]$, which concludes the argument.
With this lemma, the proof of Theorem A is complete."
Theorem 14-A
Let $P(x)$, $Q(x)$, and $R(x)$ be continuous functions on a closed interval
$[a,b]$ If $x_0$ is any point in $[a,b]$, and if $y_0$ and $y_0'$
are any numbers whatever, then the equation $y''+Py'+Qy=R$ has one and only one solution $y(x)$ on the entire interval such that $y(x_0)=y_0$
and $y'(x_0)=y_0'$
My Question: How did the author use theorem 14-A to "extend" the interval from $[c,d]$ to $[a,b]$? I mean we assumed that $y_1$ can't be zero on $[c,d]$ and it may have some parts equal to zero on [a,b] so how did the author reach the conclusion that $y_2=ky_1$ for all k in $[a,b]$ and not only in $[c,d]$?
Sorry for the long post.
(Note, reading previous posts on similar questions, the proofs used are elegant and more "direct" however, I was following the author's proof (From Differential Equations book by G. Simmons) here and I just want to understand the last bit of it. Thanks.
Best Answer
This is due to the uniqueness of the solutions. Linear DE naturally satisfy the Lipschitz condition (if the coefficients are continuous), so global solutions exist and are unique.
As $ky_1$ is a solution, and $y_2=ky_1$ on some non-trivial interval, one gets $y_2=ky_1$ everywhere.