Wronskian and two solutions being independent

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In the uploaded file, question 11 says that in b) the solutions y₁ and y₂ are linearly independent but the Wronskian equals 0. I think it said that they are independent because it's not a fixed constant times the other solution (-1 for -1<=t<=0 and 1 for 0<=t<=1) but it clearly says in the textbook that the two solutions are linearly independent if and only if Wronskian equals 0.

 

[S.G. Janssens]

If two (differentiable) functions are linearly dependent on an interval, then their Wronskian vanishes.

The converse is not true. So: If two functions are linearly independent on an interval, then it does not necessarily follow that their Wronskian does not vanish. The phrase

the two solutions are linearly independent if and only if Wronskian equals 0.

is wrong in any case, so I hope it is not really in your textbook like this.

 

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It says so here

 

[S.G. Janssens]

It says so here

No, it's not what is written there. It says "unequal".

In any case, if the Wronskian does not vanish, then the functions are linearly independent on the interval in question. This is just the contrapositive of what I wrote here:

If two (differentiable) functions are linearly dependent on an interval, then their Wronskian vanishes.

The converse is not true. Maybe you know more about the functions $y_1$ and $y_2$ that solve (3) in your book to be able to draw this conclusion, but in general it is false.

 

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I made a mistake at the end of my first post I meant 'if and only if ... unequals 0"

 

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This is the case where it says that they are linearly independent but the Wronskian is 0

Doesn't 'if and only if' mean necessary and sufficient condition? It means that if linearly independent then the Wronskian always does not equal zero, it never is another case and there are only two cases.

 

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Also in the differential equation, do y, dy/dt, dy2/d2t and so on have to be continuous?

 

[S.G. Janssens]

I believe the point of the exercise in post #1 is exactly to demonstrate that two differentiable functions (here: $y_1$ and $y_2$) may be linearly independent on an interval (here: $[-1,1]$) and yet their Wronskian may still vanish identically. This shows that the converse of

If two (differentiable) functions are linearly dependent on an interval, then their Wronskian vanishes.

is not true.

In part (d) of that exercise it is then asked to show that $y_1$ and $y_2$ cannot satisfy the premise of the corollary in post #3, although I have to guess here because I don't know what equation (3) in your book is. So there is no contradiction between the exercise and the corollary. The corollary is not about arbitrary differentiable functions, but about functions that satisfy additional properties.

If P and Q are two logical statements, then "P if and only if Q" means: "P implies Q" and "Q implies P". So: Either P and Q are both true, or they are both false.

 

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I believe the point of the exercise in post #1 is exactly to demonstrate that two differentiable functions (here: $y_1$ and $y_2$) may be linearly independent on an interval (here: $[-1,1]$) and yet their Wronskian may still vanish identically. This shows that the converse of

is not true.

In part (d) of that exercise it is then asked to show that $y_1$ and $y_2$ cannot satisfy the premise of the corollary in post #3, although I have to guess here because I don't know what equation (3) in your book is. So there is no contradiction between the exercise and the corollary. The corollary is not about arbitrary differentiable functions, but about functions that satisfy additional properties.

If P and Q are two logical statements, then "P if and only if Q" means: "P implies Q" and "Q implies P". So: Either P and Q are both true, or they are both false.

Does 'to vanish' mean being 0?
Like, does W vanishes mean W equals 0?

 

[S.G. Janssens]

Does 'to vanish' mean being 0?
Like, does W vanishes mean W equals 0?

Yes. In this context "$W$ vanishes" means that $W$ is zero not just in a point but on the entire interval.