ODE Proof (2nd order linear homogeneous equations)

[Daxin]

Homework Statement

Suppose u, v are two linearly independent solutions to the differential equation u''+p(x)u'+q(x)v=0. If x0,x1 are consecutive zeros of u, then v has a zero on the open interval (x0,x1)

Homework Equations

The Attempt at a Solution

I'm trying to use the Wronskian(u,v;x) to arrive at a contradiction.

I know that the W = uv'-u'v is never 0 since u,v are linearly independent. Furthermore, since it is a combination of C2 functions it must also be continuous on [x0,x1] = I. Also if I suppose that v is never zero on I, then it must be always positive or always negative. Same with u, since we know x0,x1 are consecutive zeros. From this I want to show that the Wronskian is positive at some point in I, and negative at another point, which would lead to a contradiction. Is this the right idea?

 

[HallsofIvy]

Homework Statement

Suppose u, v are two linearly independent solutions to the differential equation u''+p(x)u'+q(x)v=0. If x0,x1 are consecutive zeros of u, then v has a zero on the open interval (x0,x1)

Surely you don't mean to have both u and v in the equation itself? Since u and v are to be specific solutions, it would be better to say
"Suppose u, v are two linearly independent solutions to the differential equation y''+p(x)y'+q(x)y=0.

Homework Equations

The Attempt at a Solution

I'm trying to use the Wronskian(u,v;x) to arrive at a contradiction.

I know that the W = uv'-u'v is never 0 since u,v are linearly independent. Furthermore, since it is a combination of C2 functions it must also be continuous on [x0,x1] = I. Also if I suppose that v is never zero on I, then it must be always positive or always negative. Same with u, since we know x0,x1 are consecutive zeros. From this I want to show that the Wronskian is positive at some point in I, and negative at another point, which would lead to a contradiction. Is this the right idea?

Sounds like a good plan. How are you going to implement it?

 

[pasmith]

I'm trying to use the Wronskian(u,v;x) to arrive at a contradiction.

I know that the W = uv'-u'v is never 0 since u,v are linearly independent. Furthermore, since it is a combination of C2 functions it must also be continuous on [x0,x1] = I. Also if I suppose that v is never zero on I, then it must be always positive or always negative. Same with u, since we know x0,x1 are consecutive zeros. From this I want to show that the Wronskian is positive at some point in I, and negative at another point, which would lead to a contradiction. Is this the right idea?

I think there is a direct proof: You have
$$W(x_0) = u(x_0)v'(x_0) - u'(x_0)v(x_0) = -u'(x_0)v(x_0)$$
and similarly
$$W(x_1) = -u'(x_1)v(x_1)$$
and, since $W(x)$ vanishes nowhere, $W(x_0)$ and $W(x_1)$ must have the same sign. Thus
$$W(x_0)W(x_1) = u'(x_0)v(x_0)u'(x_1)v(x_1) > 0.$$

Now use the fact that $x_0$ and $x_1$ are consecutive zeroes of $u$ to show that $u'(x_0)u'(x_1) < 0$.

What does that require of $v(x_0)v(x_1)$ if the condition on $W(x_0)W(x_1)$ is to hold?

 

[Daxin]

Yeah thanks guys, I think I get it now.

v(x0)v(x1) would have to be negative. And since v is cts it must have an 0 between x0 and x1.