- #1
saxen
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Hi all, I have my exam in differential equations in one week so I will probably post a lot of question. I hope you won't get tired of me!
This is Legendres differential equation of order n. Determine an interval [0 t_0] such that the basic existence theorem guarantees a solution.
(1-t^2) [itex]\frac{d^2y}{dt^2}[/itex] - 2t[itex]\frac{dy}{dt}[/itex]+n(n-1)y = 0
y(0)=0
y'(0)=0
Picard-Lindelöfs theorem
M = max |f(x)|
T_0 = min {T,δ/M}
I thought, Picards theorem is for first order equation so thought I should first rewrite the system as x'(t)=A(t)x(t)
Write
[itex]x_{1}[/itex] = y --- >[itex]dx_{1}/dt[/itex] = dy/dt
[itex]x_{2}[/itex] = dy/dt --- > [itex]dx_{2}/dt[/itex] = d^2y/dt^2
dx_1/dt = [itex]\frac{1}{1+t^2}[/itex]
dx_2/dt = [itex]\frac{-n(n+1)}{1+t^2}[/itex] + [itex]\frac{2t)}{1+t^2}[/itex]
Picards theorem require the function to be lipschitz, which it seems to be.
Picards theorem also states that you should pick the interval according to
M = max |f(x)|
T_0 = min {T,δ/M}
Here is where I am stuck, assuming I correct so far.
M = max |f(x)|
T_0 = min {T,δ/M}
I think M is equal to n(n+1) but I have NO idea on how to compute T_0.
As always, all help is greatly appreciated!
Homework Statement
This is Legendres differential equation of order n. Determine an interval [0 t_0] such that the basic existence theorem guarantees a solution.
(1-t^2) [itex]\frac{d^2y}{dt^2}[/itex] - 2t[itex]\frac{dy}{dt}[/itex]+n(n-1)y = 0
y(0)=0
y'(0)=0
Homework Equations
Picard-Lindelöfs theorem
M = max |f(x)|
T_0 = min {T,δ/M}
The Attempt at a Solution
I thought, Picards theorem is for first order equation so thought I should first rewrite the system as x'(t)=A(t)x(t)
Write
[itex]x_{1}[/itex] = y --- >[itex]dx_{1}/dt[/itex] = dy/dt
[itex]x_{2}[/itex] = dy/dt --- > [itex]dx_{2}/dt[/itex] = d^2y/dt^2
dx_1/dt = [itex]\frac{1}{1+t^2}[/itex]
dx_2/dt = [itex]\frac{-n(n+1)}{1+t^2}[/itex] + [itex]\frac{2t)}{1+t^2}[/itex]
Picards theorem require the function to be lipschitz, which it seems to be.
Picards theorem also states that you should pick the interval according to
M = max |f(x)|
T_0 = min {T,δ/M}
Here is where I am stuck, assuming I correct so far.
M = max |f(x)|
T_0 = min {T,δ/M}
I think M is equal to n(n+1) but I have NO idea on how to compute T_0.
As always, all help is greatly appreciated!