Proving the Constant Wronskian Theorem for Scalar ODEs

[heinerL]

Hello

I'm trying to solve the following problem: given the scalar ODE [tex]x''+q(t)x=0[/tex] with a continuous function q.

x(t) and y(t) are two solution of the ODE and the wronskian is:

[tex]W(t):=x(t)y'(t)-x'(t)y(t)[/tex]. x(t) and y(t) are linear independent if [tex]W(t)\neq 0[/tex].

I want to show that W(t) is constant and that if [tex]x(t_1)=0 \Rightarrow x'(t_1) \neq 0[/tex] and [tex]y(t_1) \neq 0[/tex].

I am completely lost, can you help me?

Thx

 

[HallsofIvy]

Differentiate x(t)y'(t)- x'(t)y(t) again:
x'y'+ xy''- x"y- x'y'= xy"- x"y
Now, y'= -qy and x"= -qx so that becomes x(-qy)- (-qy)x= 0 for all t. That implies that the Wronskian itself isa constant.

 

[heinerL]

Differentiate x(t)y'(t)- x'(t)y(t) again:
x'y'+ xy''- x"y- x'y'= xy"- x"y
Now, y'= -qy and x"= -qx so that becomes x(-qy)- (-qy)x= 0 for all t. That implies that the Wronskian itself isa constant.

Do you mean y''=-qy and x''=-qx? How do you get this?

And for the second part can i say: Because x(t) and y(t) are linear independent if [tex]W(t) \neq 0, \forall t[/tex]
So let's say x(t) and y(t) are linear independent and [tex]x(t_1)=0 \Rightarrow W(t_1)=-x(t_1)*y(t_1)[/tex] and suppose [tex]x'(t_1)=0 \Rightarrow W(t_1)=0[/tex] which is a contradiction, that means that if [tex] x(t_1)=0 \Rightarrow x'(t_1) \neq 0 \ \ and \ \ y(t_1) \neq 0[/tex].

Is this correct?

 

[Office_Shredder]

Do you mean y''=-qy and x''=-qx? How do you get this?

Yeah, this is what Halls meant. It's true of course because x and y satisfy the differential equation.

Your argument for linear independence is correct except when you wrote [tex]W(t_1)=-x(t_1)*y(t_1)
[/tex][/tex] you meant [tex]W(t_1)=-x'(t_1)*y(t_1)
[/tex]

 

[heinerL]

Ah okey thanks I get it! And yes i meant [tex]W(t_1)=-x'(t_1)*y(t_1)[/tex].

And i found another argument but without proof, do you know how to proof it:

If [tex] x(t_1)=x(t_2)=0 \ \ and \ \ x(t) \neq 0 \ \ for \ \ t \in (t_1,t_2)[/tex] then it follows that [tex]y(t)[/tex] has a exactly one root in [tex](t_1,t_2)[/tex]

 

[Office_Shredder]

I found a way. I'll give a brief sketch for you to try to work out
1) Compare the signs of x'(t₁) and x'(t₂)
2) Use this to conclude that y(t) has at least one root on (t₁, t₂)
3) If it has two roots, explain why x(t) has a root inside of (t₁, t₂)

 

[heinerL]

So i try to write down my thoughts:

[tex]t_1 & t_2[/tex] are roots of x
So
[tex]W(t_k)=-x'(t_k)y(t_k)<0 \ \ k =1,2 \Rightarrow y(t_k), x'(t_k) \neq 0[/tex]

Suppose [tex]x'(t_1)>0 \Rightarrow \ \ \ "mean value theorm" \ \ \ x(t)>0[/tex]
then if [tex]x'(t_2)>0 \Rightarrow x(t)<0[/tex] which is a contradiction because then x would have another root between t_1 and t_2 therefore [tex]x'(t_2)<0[/tex]

[tex]\Rightarrow x'(t_1)>0, x'(t_2)<0 \Rightarrow y(t_1)>0, y(t_2)<0[/tex]

So y has root in (t_1,t_2)

And there is just one root because if there would be another root one can switch x and y and with above show that there is just one root!

Is this correct?