Proof of Second Order ODE Theorem

[STEMucator]

Homework Statement

I'm pretty sure this is a typo?

http://gyazo.com/802746486cc68852e5384d5a12aed596

Homework Equations

See the image ^.

The Attempt at a Solution

I believe the theorem they're talking about, is that you can write the general solution of a second order ODE :

$L[y] = y'' + p(t)y' + q(t)y = 0$ Where L is the differential operator on a solution y.

In the form : c₁y₁ + c₂y₂ ⇔ y₁, y₂ were both linearly independent solutions to the homogeneous system. That is :

$W(y_1, y_2) ≠ 0$ where W is the wronskian of y₁, y₂.

So a bit of analysis here, when x > 0, y₁ = y₂ = x³, so that W(y₁, y₂) = 0. Also when x < 0, y₁ = x³ and y₂ = -y₁. So W(y₁, y₂) = 0 as well in this case.

Now, in part a) I showed that a linear combination of these two was indeed a solution by taking the required derivatives and blah blah, the boring stuff.

Now it's mind boggling me that this can't be a typo because this is a counter-example to this theorem apparently, so I must be missing something crucial here.

Any pointers on this?

 

[Dick]

Homework Statement

I'm pretty sure this is a typo?

http://gyazo.com/802746486cc68852e5384d5a12aed596

Homework Equations

See the image ^.

The Attempt at a Solution

I believe the theorem they're talking about, is that you can write the general solution of a second order ODE :

$L[y] = y'' + p(t)y' + q(t)y = 0$ Where L is the differential operator on a solution y.

In the form : c₁y₁ + c₂y₂ ⇔ y₁, y₂ were both linearly independent solutions to the homogeneous system. That is :

$W(y_1, y_2) ≠ 0$ where W is the wronskian of y₁, y₂.

So a bit of analysis here, when x > 0, y₁ = y₂ = x³, so that W(y₁, y₂) = 0. Also when x < 0, y₁ = x³ and y₂ = -y₁. So W(y₁, y₂) = 0 as well in this case.

Now, in part a) I showed that a linear combination of these two was indeed a solution by taking the required derivatives and blah blah, the boring stuff.

Now it's mind boggling me that this can't be a typo because this is a counter-example to this theorem apparently, so I must be missing something crucial here.

Any pointers on this?

It is true that if the wronskian is nonzero then the functions must be linearly independent. But is it true that if the wronskian is zero then they must linearly dependent?? Does your theorem really say that?

 

[STEMucator]

It is true that if the wronskian is nonzero then the functions must be linearly independent. But is it true that if the wronskian is zero then they must linearly dependent?? Does your theorem really say that?

Indeed, the theorem stated that if the Wronskian was non-zero for two solutions y₁ and y₂, then the solution y = c₁y₁ + c₂y₂ creates a fundamental solution set over the interval a < t < b.

 

[STEMucator]

I read this quickly on wiki : http://en.wikipedia.org/wiki/Wronskian

The Wronskian and linear independence :

If the functions fi are linearly dependent, then so are the columns of the Wronskian as differentiation is a linear operation, so the Wronskian vanishes. So the Wronskian can be used to show that a set of differentiable functions is linearly independent on an interval by showing that it does not vanish identically.

A common misconception is that W = 0 everywhere implies linear dependence, but Peano (1889) pointed out that the functions x2 and |x|x have continuous derivatives and their Wronskian vanishes everywhere, yet they are not linearly dependent in any neighborhood of 0. There are several extra conditions which ensure that the vanishing of the Wronskian in an interval implies linear dependence. Peano (1889) observed that if the functions are analytic, then the vanishing of the Wronskian in an interval implies that they are linearly dependent. Bochner (1901) gave several other conditions for the vanishing of the Wronskian to imply linear dependence; for example, if the Wronskian of n functions is identically zero and the n Wronskians of n–1 of them do not all vanish at any point then the functions are linearly dependent. Wolsson (1989a) gave a more general condition that together with the vanishing of the Wronskian implies linear dependence.

The part that caught my eye is highlighted.

EDIT : Is it because we have a saddle point for |x| in the neighborhood of zero because they're not differentiable at zero?

 

[STEMucator]

I read this quickly on wiki : http://en.wikipedia.org/wiki/Wronskian

The Wronskian and linear independence :

If the functions fi are linearly dependent, then so are the columns of the Wronskian as differentiation is a linear operation, so the Wronskian vanishes. So the Wronskian can be used to show that a set of differentiable functions is linearly independent on an interval by showing that it does not vanish identically.

A common misconception is that W = 0 everywhere implies linear dependence, but Peano (1889) pointed out that the functions x2 and |x|x have continuous derivatives and their Wronskian vanishes everywhere, yet they are not linearly dependent in any neighborhood of 0. There are several extra conditions which ensure that the vanishing of the Wronskian in an interval implies linear dependence. Peano (1889) observed that if the functions are analytic, then the vanishing of the Wronskian in an interval implies that they are linearly dependent. Bochner (1901) gave several other conditions for the vanishing of the Wronskian to imply linear dependence; for example, if the Wronskian of n functions is identically zero and the n Wronskians of n–1 of them do not all vanish at any point then the functions are linearly dependent. Wolsson (1989a) gave a more general condition that together with the vanishing of the Wronskian implies linear dependence.

The part that caught my eye is highlighted.

EDIT : Is it because we have a saddle point for |x| in the neighborhood of zero because they're not differentiable at zero?

So for x > 0, the slope of |x| is positive ( Is 1 really ). For x < 0, the slope of |x| is negative ( -1 to be exact ). Thus we have a saddle point at x = 0 and |x| is not differentiable at 0.

Thus $|x^3| = |x^2||x| = x^2|x|$ is also not differentiable at the origin.

 

[Dick]

You might also note your equation doesn't start with y'', it starts with x^2*y''. It's singular at x=0. y=c*x is another linearly independent solution. You might want to check your theorems assumption about p and q.

 

[Dick]

So for x > 0, the slope of |x| is positive ( Is 1 really ). For x < 0, the slope of |x| is negative ( -1 to be exact ). Thus we have a saddle point at x = 0 and |x| is not differentiable at 0.

Thus $|x^3| = |x^2||x| = x^2|x|$ is also not differentiable at the origin.

|x^3| is differentiable at the origin. It's second derivative is also continuous.

 

[STEMucator]

Ahhhhh so take this from before :$L[y] = y'' + p(t)y' + q(t)y = 0$

If p and q are continuous, then I will be able to find a general solution. Now though, we have a co-efficient in front of y''. That is r(x) = x². So we would get :

$L[y] = x^2y'' + 3xy' + 3y = 0$
$L[y] = y'' + \frac{3}{x}y' + \frac{3}{x^2}y = 0$

If x > 0, y₁ and y₂ create my linear combination y which is a solution to this system.

If x < 0, we observe a similar case.

If x = 0, then our co-efficients p and q are discontinuous and we are not able to find a solution.

 

[Dick]

Ahhhhh so take this from before :$L[y] = y'' + p(t)y' + q(t)y = 0$

If p and q are continuous, then I will be able to find a general solution. Now though, we have a co-efficient in front of y''. That is r(x) = x². So we would get :

$L[y] = x^2y'' + 3xy' + 3y = 0$
$L[y] = y'' + \frac{3}{x}y' + \frac{3}{x^2}y = 0$

If x > 0, y₁ and y₂ create my linear combination y which is a solution to this system.

If x < 0, we observe a similar case.

If x = 0, then our co-efficients p and q are discontinuous and we are not able to find a solution.

If your p and q aren't continuous that just means your theorem doesn't apply. Doesn't mean you can't find a solution. I think we've found three linearly independent ones already.

 

[STEMucator]

I'm not seeing what you're getting at. Is this problem about showing there are other linearly independent solutions to this problem?

 

[Dick]

I'm not seeing what you're getting at. Is this problem about showing there are other linearly independent solutions to this problem?

Yes! y=x^3, y=|x^3| and y=x are all linearly independent solutions. You have too many! You just have to say why your theorem doesn't apply. It says you should have only two.

 

[STEMucator]

Yes! y=x^3, y=|x^3| and y=x are all linearly independent solutions. You have too many! You just have to say why your theorem doesn't apply.

Shortened version of the theorem :

If the Wronskian is non-zero, then the solutions are linearly independent.
If the Wronskian IS zero, then the solutions are linearly dependent.

So because there are other linearly independent solutions who's Wronskian's are non-zero say y₁ = x and y₂ = c/x then we know the theorem is not contradicted?

This seems weak to me?

 

[Dick]

If the Wronskian IS zero, then the solutions are linearly dependent.

That is FALSE. Isn't that the point here with x^3 and |x^3|?

So because there are other linearly independent solutions who's Wronskian's are non-zero say y₁ = x and y₂ = c/x then we know the theorem is not contradicted?

This seems weak to me?

What theorem are you talking about? Can you quote it in full?

 

[STEMucator]

That is FALSE. Isn't that the point here with x^3 and |x^3|?



What theorem are you talking about? Can you quote it in full?

Literally the theorem I stated in my last post, the shortened version is the one he gave in class apparently.

 

[Dick]

Literally the theorem I stated in my last post, the shortened version is the one he gave in class apparently.

If you mean what you said in post 12, that's not generally true. Probably because it's 'shortened'.

 

[STEMucator]

If you mean what you said in post 12, that's not generally true. Probably because it's 'shortened'.

Waiiiiiiit a second here. I thought for awhile about this. Here's the theorem I think he's talking about from the book :

http://gyazo.com/fac8f81914efdedde890eeaef24d734a

So, it's easy to show the Wronskian of y₁ and y₂ is equal to zero for x > 0 and x < 0.

NOW, I didn't actually consider what happens on an interval CONTAINING x = 0.

Suppose that there exist real scalars c₁ and c₂ such that : c₁y₁ + c₂y₂ = 0

Then for all x in our interval containing 0 we have :

c₁x³ + c₂|x|³ = 0

Now for x > 0 , c₁x³ + c₂x³ = 0 which tells us x³(c₁ + c₂) = 0 which yields c₁ + c₂ = 0

Now a similar case for x < 0 yields c₂ - c₂ = 0.

Solving both of these equations yields c₁ = c₂ = 0 and hence y₁ and y₂ are linearly independent because there exists a trivial relation of linear dependence on our interval containing x = 0. Hence there is a point on our interval where W ≠ 0 and thus the theorem is not contradicted.

 

[Dick]

Waiiiiiiit a second here. I thought for awhile about this. Here's the theorem I think he's talking about from the book :

http://gyazo.com/fac8f81914efdedde890eeaef24d734a

So, it's easy to show the Wronskian of y₁ and y₂ is equal to zero for x > 0 and x < 0.

NOW, I didn't actually consider what happens on an interval CONTAINING x = 0.

Suppose that there exist real scalars c₁ and c₂ such that : c₁y₁ + c₂y₂ = 0

Then for all x in our interval containing 0 we have :

c₁x³ + c₂|x|³ = 0

Now for x > 0 , c₁x³ + c₂x³ = 0 which tells us x³(c₁ + c₂) = 0 which yields c₁ + c₂ = 0

Now a similar case for x < 0 yields c₂ - c₂ = 0.

Solving both of these equations yields c₁ = c₂ = 0 and hence y₁ and y₂ are linearly independent because there exists a trivial relation of linear dependence on our interval containing x = 0. Hence there is a point on our interval where W ≠ 0 and thus the theorem is not contradicted.

Yes, you have shown that x^3 and |x^3| are linearly independent over an interval containing positive and negative numbers. But their wronskian IS zero everywhere even at zero. Now why does that not contradict your theorem??

 

[STEMucator]

I think that's because my DE is not in the form y'' + py' + qy = 0. I would have to divide through by x^2 which would create discontinuities at zero so that my p and q are not continuous?

 

[Dick]

I think that's because my DE is not in the form y'' + py' + qy = 0. I would have to divide through by x^2 which would create discontinuities at zero so that my p and q are not continuous?

Yeah, that's one thing. But the theorem as quoted didn't give any conditions on p and q. Does it give them elsewhere? The other out would be to notice c1*x^3+c2*|x^3| doesn't span all possible solutions. Why not?

 

[STEMucator]

Ahhh there's a theorem wayyyy back stating that p and q have to be continuous functions on the interval. So because p and q are not continuous everywhere on the interval in this case, it follows that y₁ and y₂ are not solutions over the entire interval?

 

[Dick]

Ahhh there's a theorem wayyyy back stating that p and q have to be continuous functions on the interval. So because p and q are not continuous everywhere on the interval in this case, it follows that y₁ and y₂ are not solutions over the entire interval?

Don't go back to saying that again. It's wrong. y1 and y2 are solutions over all the reals in the original equation. I thought you checked that. p and q becoming discontinous after you divide by x^2 doesn't change that. It may change whether you can rely on theorems about the solutions of y''+py'+qy=0 with p and q continuous.

 

[STEMucator]

Okay I'm fairly confident now that these two theorems :

http://gyazo.com/6212935f0c5e359748a623a88b309b2e
http://gyazo.com/428dbfb9ce40f8633cbc588e5c629953

Are the ones being called into question and are the only theorems relevant to this question at all. Every other theorem involves initial values which we do not have here.

Now, Theorem 3.2.4 states that, if and only if the Wronskian of y₁ and y₂ is not zero everywhere, then the linear combination c₁y₁ + c₂y₂ contains all solutions of (2).

So, is it that the linear combination here does not contain all the possible solutions since the Wronskian is zero everywhere?

 

[Dick]

Okay I'm fairly confident now that these two theorems :

http://gyazo.com/6212935f0c5e359748a623a88b309b2e
http://gyazo.com/428dbfb9ce40f8633cbc588e5c629953

Are the ones being called into question and are the only theorems relevant to this question at all. Every other theorem involves initial values which we do not have here.

Now, Theorem 3.2.4 states that, if and only if the Wronskian of y₁ and y₂ is not zero everywhere, then the linear combination c₁y₁ + c₂y₂ contains all solutions of (2).

So, is it that the linear combination here does not contain all the possible solutions since the Wronskian is zero everywhere?

Yes! The Wronskian is zero everywhere. But c₁y₁ + c₂y₂ does not contain all solutions. Why? That would save the theorem from contradiction.

 

[STEMucator]

This is confusing me for some reason. I took the required derivatives and did all the substituting to show that y₁ and y₂ were both solutions to the DE and then I showed their combination y = y₁ + y₂ was also a solution.

So from part (a) I deduced that y₁ and y₂ were linearly independent solutions to my DE.

Then when I went to part (b), I found the Wronskian for both cases when x ≥ 0 and x < 0. Turns out W is identically zero just as desired which WOULD mean that y₁ and y₂ were linearly dependent.

I think what saves the theorem from contradiction is that one solution is a multiple of the other? I'm a bit lost about this really.

 

[Dick]

This is confusing me for some reason. I took the required derivatives and did all the substituting to show that y₁ and y₂ were both solutions to the DE and then I showed their combination y = y₁ + y₂ was also a solution.

So from part (a) I deduced that y₁ and y₂ were linearly independent solutions to my DE.

Then when I went to part (b), I found the Wronskian for both cases when x ≥ 0 and x < 0. Turns out W is identically zero just as desired which WOULD mean that y₁ and y₂ were linearly dependent.

I think what saves the theorem from contradiction is that one solution is a multiple of the other? I'm a bit lost about this really.

YOU have proved that the solutions are linearly independent. That would rule out them being multiples of each other, right? YOU also showed that the wronskian is zero everywhere. The only way out is to show that c1*x^3+c2*|x^3| does not describe ALL solutions. I gave you a HUGE hint a number of posts back. Where is it?

 

[STEMucator]

EDIT :

Your hint was this yes : The other out would be to notice c1*x^3+c2*|x^3| doesn't span all possible solutions. Why not?

 

[STEMucator]

You might also note your equation doesn't start with y'', it starts with x^2*y''. It's singular at x=0. y=c*x is another linearly independent solution. You might want to check your theorems assumption about p and q.

If that wasn't your hint, it MUST be that ^.

 

[Dick]

EDIT :

Your hint was this yes : The other out would be to notice c1*x^3+c2*|x^3| doesn't span all possible solutions. Why not?

That was it. I gave you one. Don't you remember?

 

[STEMucator]

Like i asked, is it because my equation is singular at x=0?

 

[STEMucator]

Waiiiiit, if my equation is singular at x = 0, then there's no way my Wronskian is identically zero everywhere right?

 

[Dick]

Like i asked, is it because my equation is singular at x=0?

You've asked that more than once and that's part of it. More to the point, I gave you a solution that's not in the span of x^3 and |x^3|. Can you find it?

 

[Dick]

Waiiiiit, if my equation is singular at x = 0, then there's no way my Wronskian is identically zero everywhere right?

yes, it is. please stop denying things you already know are true.

 

[STEMucator]

You've asked that more than once and that's part of it. More to the point, I gave you a solution that's not in the span of x^3 and |x^3|. Can you find it?

Yes of course : y=c*x is a L.I solution as well.

 

[Dick]

Yes of course : y=c*x is a L.I solution as well.

Bingo. So?

 

[STEMucator]

Bingo. So?

Well you said that it's not in the span of x³ and |x³| so that means there is no linear combination of them which will give me c*x.