How do we deduce that the initial value problem has always a unique solution?

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 $W(y_1, y_2)(x)=0 \forall x \in I$?

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.

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