Wronskian and differential equations

  • #1
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Homework Statement


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
20160201_225904.jpg


Homework Equations


Wronskian of y1 and y2

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?
 
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Answers and Replies

  • #2
Samy_A
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Homework Statement


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

Homework Equations


Wronskian of y1 and y2

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?
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.
 
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  • #3
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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
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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?
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
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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
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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?
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
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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
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 how does it work when ##y_1## is 0?
 
  • #8
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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?
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##.
 
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  • #9
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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.
 
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  • #10
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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.
It's a proof by contradiction.

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[.

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.
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}##.
 
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  • #11
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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
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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!
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
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Thank you! But how about the second part? Question 19)?
 
  • #14
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Thank you! But how about the second part? Question 19)?
It is not clear to me what is known about ##y_1,\ y_2## in question 19.
 
  • #15
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Here's the proof related to Question 19)
 

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  • #16
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Here's the proof related to Question 19)
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
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I got to the point where I could find the value of ##c## but here it ##y_2=0## (at ##t*##)
 
  • #18
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I got to the point where I could find the value of ##c## but here it ##y_2=0## (at ##t*##)
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
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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
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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##
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.
 
  • #21
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I don't know what to do from this point onward.
 
  • #22
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I don't know what to do from this point onward.
It is not clear at what point you are.
Can you state clearly what you have proven so far? What precise value have you found for c?
 
  • #23
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I could only find the value of ##c## for the situation above.
 
  • #24
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I could only find the value of ##c## for the situation above.
Please state clearly what you have proven so far. Don't refer vaguely to "situation above" or something else. Double guessing you is difficult.
Just post the equations, equalities, ... that you have proven so far.
 
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