The discussion revolves around the proof of a theorem related to second order ordinary differential equations (ODEs), specifically focusing on the conditions under which solutions can be expressed as linear combinations of independent solutions. Participants are examining the implications of the Wronskian determinant in relation to linear independence and the continuity of coefficients in the ODE.
There is an active exploration of the assumptions underlying the theorem, particularly regarding the continuity of coefficients in the differential equation. Some participants suggest that the theorem may not apply if the coefficients are not continuous, while others propose that multiple linearly independent solutions may still exist despite this. The discussion is ongoing with no clear consensus reached.
Participants note that the original problem may contain a typo and discuss the implications of singularities at specific points, particularly at x = 0, where coefficients become discontinuous. This raises questions about the applicability of the theorem in such cases.
Zondrina said:Like i asked, is it because my equation is singular at x=0?
Zondrina said:Waiiiiit, if my equation is singular at x = 0, then there's no way my Wronskian is identically zero everywhere right?
Dick said: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?
Zondrina said:Yes of course : y=c*x is a L.I solution as well.
Dick said:Bingo. So?
Zondrina said: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.
Dick said:I said that. I think you should prove it. Once you've done that, does that save your theorem?
Zondrina said:So supposing I want to find some linear combination of x3 and |x3| which gives me cx, I want to satisfy this relation :
c1x3 + c2|x3| = cx.
The only way this is true is if all the constants c1, c2 and c are all zero?
Dick said:True. How would you prove that instead of just saying it's true?
Zondrina said:Hmmm okay so. I realize that I can take any straight line here including 0, so I'll use 0 since it's easy.
We have y1 = x3 and y2 = |x3| which are L.I solutions to our equation.
y'2 = 3x2 if x ≥ 0
y'2 = -3x2 if x < 0
Hence their Wronskian is zero everywhere for all x in our interval, which should mean they are linearly dependent, but this is not the case.
Let a and b be constants such that :
ax3 + b|x3| = 0
Then :
a(1)3 + b|13| = a + b = 0
a(-1)3 + b(-(-1)3) = a - b = 0
So it must be the case that a = b = 0 and thus y1 and y2 must be linearly independent.
Dick said:That's not very good. You already showed x^3 and |x^3| are linearly independent and you just repeated it. Now show me x^3 and |x^3| and x are linearly independent.
Zondrina said:Well first I note that their Wronskian is zero everywhere, which means they are linearly dependent, which is also not the case.
Following a similar proof style to before I form the equation :
ax3 + b|x3| + cx = 0
and show that a = b = c = 0 is the only possible solution.
Dick said:The wronskian doesn't have much to do with it. If you put x=1, x=(-1) and x=2 that should give you enough ammunition to show a=b=c=0, right?