Wronskian and differential equations

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1. Feb 1, 2016

bubblewrap

1. The problem statement, all variables and given/known data
The problems are in the uploaded file. 18) satisfies the differential equation
y''+p(t)*y'+q(t)*y=0
p(t) and q(t) are continuous

2. Relevant equations
Wronskian of y1 and y2

3. The attempt at a solution

18) I don't really get this one

19) Solved most but at the end, where I got y2(t)=c*y1(t)
I got the value of c for when t=t* but here y2=0. Can I still use it here?

Last edited by a moderator: Feb 1, 2016
2. Feb 1, 2016

Samy_A

For 18, the hint given after the question is excellent.
Assume you have two consecutive zeros of $y_2$. Let's name these zeros $a$ and $b$.

Compute the derivative of $\frac{y_2}{y_1}$, and apply the theorem of Rolle to the function $\frac{y_2}{y_1}$ in the interval $[a,b]$.
To be allowed to do this you must make an assumption about $y_1$ in the interval $[a,b]$.

The goal is to get a contradiction.

Last edited: Feb 1, 2016
3. Feb 2, 2016

bubblewrap

when differentiated the function has square of y1 at the bottom and we have to find y1=0 So the top has to be 0 as well right?

4. Feb 2, 2016

Samy_A

We have to find a zero for $y_1$ in $[a,b]$. But if $y_1$ has a zero in $[a,b]$, we can't just work with the function $\frac{y_2}{y_1}$, because that would mean dividing by zero.
So, assume $y_1$ has no zeros in $[a,b]$, do what the hint suggests, and find a contradiction. That will show that the assumption that $y_1$ has no zeros in $[a,b]$ was wrong.
Next you will have to show why $y_1$ can only have 1 zero in $[a,b]$

5. Feb 2, 2016

bubblewrap

Ah so if y1 is not 0 then we can multiply the differentiated y2/y1 which is y'2/y1-y'1y2/y21=0 which makes the Wronskian of y1 and y2 0. Because this means that there is no fundamental set of solutions (y1 and y2 are linearly dependent) it contradicts the first line of the question(y1 and y2 are a fundamental set of solutions. But then how does this work when y1 is 0, can we not know about that?

6. Feb 2, 2016

Samy_A

You found a contradiction. That means that the assumption that y1 is never 0 in the interval [a,b] is wrong.
That's exactly the first part of what you had to prove: y1 has at least 1 zero in [a,b].

But what about the second part? Can y1 have more than 1 zero in [a,b]?

Hint: notice that you haven't yet used the fact that a and b are consecutive zeros of y2.

7. Feb 3, 2016

bubblewrap

If a and b are consecutive it means that there is no $y_2$ that's 0 between them. Which would mean for a continuous $y_2$, there exists only 1 value c that satisfies $f'(c)=0$, where $f(x)=\frac{y_2}{y_1}$. This means that there cannot exist two $y_1$s that are not 0.

Also you said in your previous post
So how does it work when $y_1$ is 0?

8. Feb 3, 2016

Samy_A

The whole "trick" with $\frac{y_2}{y_1}$ and Rolle's theorem only works if we assume that $y_1$ has no zeros in [a,b]. But, as you showed in post 5, this will lead to a contradiction with the fact that $y_1,y_2$ are a fundamental set of solutions to the differential equation.
The conclusion is that $y_1$ must be 0 somewhere in [a,b]. Actually, we even can say that $y_1$ must be 0 somewhere in ]a,b[, as functions in a fundamental set of solutions can't be 0 in the same point.

The question about what happens when $y_1$ is 0 is not relevant. That's what we wanted to prove in the first place, and that has been done.

Now for the second part: how can we prove that $y_1$ has only 1 zero in ]a,b[? We can't work with $\frac{y_2}{y_1}$, because we already know that that function is not well-defined in the zero(s) of $y_1$.

Hint: assume that $y_1$ has 2 zeros in ]a,b[. Use the first part of this exercise, and the "symmetry" between $y_1$ and $y_2$ to get a contradiction with the fact that a and b are consecutive zeros of $y_2$.

Last edited: Feb 3, 2016
9. Feb 3, 2016

bubblewrap

So my previous answer wasn't correct?

And we found out that $\frac{y_2}{y_1}$ is not well defined; not continuous in the interval $[a,b]$ but if the function is not continuous, then we can't use Rolle's theorem, so shouldn't the proof of the first part be wrong? This is connected with the first question of this post; if the reason why my previous answer is not correct is because the function $\frac{y_2}{y_1}$ is not continuous then the proof for the first part of the question should be wrong for the same reason.

For the second part of the question, I still think that if $y_1$ has 2 $0$s or more then $f'(c)=0$ for more than one value of $c$ between [a,b] it contradicts $a$ and $b$ being consecutive 0s since more than one value of c satisfying $f'(c)=0$ would mean that there is another value $d$ in [a,b] that makes $y_2$ 0. I am not quite sure what is wrong with this proof.

Last edited: Feb 3, 2016
10. Feb 3, 2016

Samy_A

You have to prove that $y_1$ has zeros in [a,b].
So let's assume that $y_1$ does not have zero's in [a,b].
In that case, $f=\frac{y_2}{y_1}$ is a well-defined differentiable function. You can calculate $f'$, you can apply Rolle's theorem.That gives you a point in [a,b] were the Wronskian is 0. But, this can't be, because $y_1,\ y_2$ form a fundamental set of solutions. We have reached a contradiction.
The way out is to conclude that our assumption that $y_1$ doesn't have zero's in [a,b] was wrong.

Conclusion: $y_1$ has at least one zero in ]a,b[.

As $y_1$ has zero(s) in ]a,b[, you can't work with the function $f=\frac{y_2}{y_1}$, as it is not defined in the whole interval. You can't apply Rolle's theorem to $f=\frac{y_2}{y_1}$.

Last edited: Feb 3, 2016
11. Feb 3, 2016

bubblewrap

Oh, if there are two 0s of $y_1$ between a and b then by the same logic (applying the Rolle's theorem) there should exist a 0 for $y_2$ between the two 0s of $y_1$, applying the same logic to more than 2 0s of $y_1$ we get that there can only exist one 0!

12. Feb 3, 2016

Samy_A

Exactly: it would contradict the given that a and b are consecutive zeros of $y_2$. You don't even need the "more than 2" part.

13. Feb 3, 2016

bubblewrap

Thank you! But how about the second part? Question 19)?

14. Feb 3, 2016

Samy_A

It is not clear to me what is known about $y_1,\ y_2$ in question 19.

15. Feb 3, 2016

bubblewrap

Here's the proof related to Question 19)

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16. Feb 3, 2016

Samy_A

Ok, So I assume that $y_1, \ y_2$ are solutions of a second order linear differential equation.
You already have found that $y_2(t)=cy_1(t)$ (see first post).
To find the requested formula for c, you could differentiate this equation. That should give you almost what you need.

17. Feb 3, 2016

bubblewrap

I got to the point where I could find the value of $c$ but here it $y_2=0$ (at $t*$)

18. Feb 3, 2016

Samy_A

I don't see why this should be a problem. I you have the value of c, that's what is asked for.
Anyway, in case you haven't yet deduced the requested value for c, two questions:
1) What do you know about $y'_1(t*)$?
2) What can you conclude from $y'_2(t)=cy'_1(t)$?

19. Feb 3, 2016

bubblewrap

I did get the requested value of c but if $y_2=0$ at $t*$ then it means that c too is 0. Is this okay? because that would mean that $y_2$ is just 0 and not a multiple of $y_1$

20. Feb 3, 2016

Samy_A

So you have proved
(1) $y_2(t)=cy_1(t)$,
where $c=\frac{y'_2(t*)}{y'_1(t*)}$. (That is unless we have $y_1 \equiv 0$ , that was the other possibility in the exercise).
As we know that $y_1(t*)=0, y_2(t*)=0$, equation (1) applied to $t*$ simply gives $0=c.0$. That's fine, and doesn't tell anything about c.