Green function

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  • #1
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1.
The zero order bessel function J0(0+) = 1, J'0(0+) = 0. and J0(t) satisfies teh differential equation


2. ty''(t) + y'(t) + ty(t) = 0, t>0
Prove thata the Laplace transform of J0(t) is 1/(s + 1)^1/2

The Attempt at a Solution



Could anyone help on this. How to I go about with the question. Tried using a general solution that is J(t) = C1e^r1t + C2e^r2t..But im not sure if that right... I wasn't able to prove...Thx

Homework Statement





Homework Equations





The Attempt at a Solution

 

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  • #2
Integral
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Why not start by finding the Laplace transform of your DE?
 
  • #3
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thx for replying...But how do I find the Laplace transform of the differential equation? Do i convert it to general solution J(t) = C1e^r1t + C2e^r2t. Is that right? Because i tried using that general equation and found two values of t and substituted in the general equation and solve simultaneously..but still wasn't able to prove...
 
  • #4
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I've found websites and even read books. It all says the same thing about bessel equation having a general solution. http://eqworld.ipmnet.ru/en/solutions/ode/ode0213.pdf
But what i dont get is that the in the general solution what is my v in my DE. I dont see any relationship. I am so confused..
 
  • #5
HallsofIvy
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If you are working with Bessel functions and Green's functions (you titled this "Green function" but I don't see anything in the problem that has anything to do with Green's function) then you should have learned long ago that functions of the form [itex]C_1e^{r_1t}+ C_2e^{r_2t}[/itex] satisfy linear equations with constant coefficients. Bessel's equation, the equation you are given, obviously does NOT have integer coefficients. You are not expected to "find" a solution, you are expected to find the Laplace transform of the differential equation.

Do you know how to find the Laplace transform of something like f(x) dy/dx in terms of the Laplace transform of y?
 
  • #6
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I am not quite sure but is the Lapalce transform of f(x) dy/dx is sY(s) - Yo and the laplace transform of d2y/dx2 is s2Y(s) - sY0 - Y1.

If I am wrong. Please kindly correct me and if you could kindly give me some hints to go about with the question. Thx
 
  • #7
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ty''(t) + y'(t) + ty(t) = 0
Divide the equation by t gives

y''(t) + 1/ty'(t) + ty(t) = 0

Using Laplace Transform it gives

s2*Y - s*y0 - y1 + 1/t*(S*Y - y0) + t*Y = 0

(s2 + 1/t*s + t)*Y - s - 1/t = 0

so now Y = (s = 1/t)/(s2 + 1/t*s + t)

but i still cant prove the equation.....:(
 
  • #8
HallsofIvy
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The Laplace transform of tf(t) is NOT t times the Laplace transform of f(t)!
 
  • #9
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oh there was an error

ty''(t) + y'(t) + ty(t) = 0
Divide the equation by t gives

y''(t) + 1/t*y'(t) + y(t) = 0

Using Laplace Transform it gives

s2*Y - s*y0 - y1 + 1/t*(s*Y - y0) + Y = 0

(s2 + 1/t*s + 1)*Y - s - 1/t = 0

so now Y = (s + 1/t)/(s2 + 1/t*s + 1)

I have tried completing the square but it doesnt help.....Is my Y correct?
 

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