Second Order Differential Equation Solved With Green's Function

[dltsmith]

Hi there,

I'm trying to solve the following second order differential equation using the greens function theorem and not sure where to start as I don't know the values of the constants and am not sure how to apply the boundary of dy/dt = 0.Thanks for any help

the equation and conditions given are:

(d^2y/dx^2) + 2B*(dy/dx) + (z^2)*(y) = f(x)

for y(0)=dy/dx=0

thanks

 

[HallsofIvy]

Have you determined the Green's function for this equation?

The "x" and "y" are the variables in this equation but what is "z"? Is it a constant?

 

[dltsmith]

I have not obtained the Green's function as of yet. Yes sorry the z is a constant, probably was not the best of choices.

 

[HallsofIvy]

Frankly, because your conditions are initial conditions, both given at x= 0, rather than boundary conditions, one given at each endpoint, I wouldn't consider "Green's function" to be appropriate here.

The "corresponding homogeneous equation" is $$\frac{d^2y}{dx^2}+ 2B\frac{dy}{dx}+ z^2y= 0$$ which has characteristic equation [mathr^2+ 2Br+ z^2= 0[/math] and that has roots $$r= \frac{-2B\pm\sqrt{4B^2- 4z^2}}{2}= -B\pm\sqrt{B^2- z^2}$$.

Whether that is real or not depends upon whether $$B^2- z^2$$ is non-negative. If it is non-negative then the general solution to the homogeneous equation is $$y= e^{-Bt}(C_1e^{(B^2- z^2)x}+ C_2e^{-(B^2- z^2)x}$$.

The initial conditions make this almost trivial! $$y(0)= C_1+ C_2= 0$$. $$y'= -Be^{-Bt}(C_1e^{(B^2- z^2)x}+C_2e^{-(B^2- z^2)x}+$$$$e^{-Bt}(C_1(B^2- z^2)e^{(B^2- z^2)}x- C_2(B^2- r^2)e^{-(B^2- r^2)x}$$ so $$y'(0)= (-1)(C_1+ C_2)+ (B^2- z^2)(C_1- C_2)= (B^2-z^2- 1)C_1+ (B^2-z^2+1)C_2= 0$$. Those give y(x)= 0!

 

[dltsmith]

Thanks for the help!