1
$\begingroup$

Theorem - General solution of $y''+p(x)y'+q(x)y=0, x \in I (\star)$

Let $y_1, y_2$ be linearly independent solutions of $(\star)$ in an interval $I$.

Then if $y$ is a solution of $(\star)$ in $I$, there are $c_1, c_2 \in \mathbb{R}$ such that $y(x)= c_1 y_1(x)+ c_2 y_2(x), x \in I$.

For the proof of the theorem:

If $\psi$ is a solution of the differential equation $(\star)$ in $I$ we want to find $c_1, c_2 \in \mathbb{R}$ such that $\psi(x)= c_1 y_1(x) + c_2 y_2(x), \ \forall x \in I$.

We choose an arbitrary $x_0 \in I$. We consider the initial value problem:

$\left\{\begin{matrix} y''+p(x)y'+q(x)y=0 &, x \in I \\ y(x_0)=\psi(x_0) & \\ y'(x_0)=\psi'(x_0) & \end{matrix}\right.$

that has a unique solution in $I$.

We want:

$\begin{bmatrix} \psi(x_0)=c_1 y_1(x_0)+c_2 y_2(x_0)\\ \\ \psi'(x_0)=c_1 y_1'(x_0)+c_2y_2'(x_0) \end{bmatrix} (A)$

I suffices to find $c_1, c_2 \in \mathbb{R} $ so that the system $(A)$ has a unique solution as for $c_1, c_2 \in \mathbb{R} $.

It suffices to show that $\begin{vmatrix} y_1(x_0) & y_2(x_0) \\ \\ y_1'(x_0) & y_2'(x_0) \end{vmatrix}=y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0) \neq 0$

We define $W(y_1, y_2)(x)=y_1(x) y_2'(x)-y_1'(x) y_2(x), x \in I$.

Then there is the following lemma:

If $y_1, y_2$ linearly independent solutions of the differential equation $(\star)$ in $I$, then:

  • either $W(y_1, y_2)(x) \neq 0 \ \ forall x \in I$
  • either $W(y_1, y_2)(x)=0 \forall x \in I$.

Proof:

$W(y_1,y_2)'(x)=y_1(x) y_2''(x)-y_1''(x) y_2(x) \ \ \forall x \in I$

We have:

$$(\star \star) y_2 y_1''+p(x) y_2 y_1'+q(x) y_1 y_2=0 (\star \star)$$

$$(\star \star \star) y_1 y_2''+p(x) y_1 y_2'+q(x) y_1 y_2=0 (\star \star)$$

$$(\star \star \star)-(\star \star) \Rightarrow y_1 y_2''-y_2 y_1''=-p W(y_1, y_2)$$

Thus, $W'(y_1, y_2)+pW(y_1, y_2)=0$ in $I$.

So if $x_0 \in I$ then:

$$W(y_1,y_2)(x)= c e^{- \int_{x_0}^x p(t) dt}, c \in \mathbb{R}$$

So the conclusion follows from the last relation.


So we have shown that

If $y_1, y_2$ linearly independent solutions of the differential equation $(\star)$ in $I$, then:

  • either $W(y_1, y_2)(x) \neq 0 \ \forall x \in I$
  • either $W(y_1, y_2)(x)=0 \forall x \in I$.

So how can we deduce that the initial value problem:

$\left\{\begin{matrix} y''+p(x)y'+q(x)y=0 &, x \in I \\ y(x_0)=\psi(x_0) & \\ y'(x_0)=\psi'(x_0) & \end{matrix}\right.$

has always a unique solution, although it is also possible that $W(y_1, y_2)(x)=0 \forall x \in I$?

$\endgroup$
1
$\begingroup$

I believe some part of this proof is messed up.

The correct statement of the lemma should be:

If $y_1,y_2$ are two solutions of the differential equation $(*)$ in $I$, then

  • either $W(y_1, y_2)(x) \neq 0 \ \forall x \in I$
  • or $W(y_1, y_2)(x) = 0 \ \forall x \in I$

You can see that the proof of the lemma did not use the fact that $y_1,y_2$ are linearly independent.

In fact, we should have:

$y_1,y_2$ are two linearly independent solutions of the differential equation $(*)$ in $I$, if and only if $W(y_1, y_2)(x) \neq 0 \ \forall x \in I$.

This would clarify the proof of the theorem.

$\endgroup$
  • $\begingroup$ I see... Thank you very much :) $\endgroup$ – evinda Jun 3 '15 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.