- #1
member 428835
Hi PF!
Given the following ODE $$(p(x)y')' + q(x)y = 0$$ where ##p(x) = 1-x^2## and ##q(x) = 2-1/(1-x^2)## subject to $$y'(a) + \sec(a)\tan(a)y(a) = 0$$ and $$|y(b)| < \infty,$$ where ##a = \sqrt{1-\cos^2\alpha} : \alpha \in (0,\pi)## and ##b = 1##, what is the Green's function?
This is the associated Legendre ODE, which admits two linearly independent solutions: $$y_1 = P_1^1, \,\,\, y_1 = Q_1^1$$ where ##P_1^1## and ##Q_1^1## are associated Legendre polynomials first and second kind respectively. Things now get a little murky: notice ##y_1## automatically satisfies ##y_1'(a) + \sec(a)\tan(a)y_1(a) = 0##, but ##y_2(b) = \infty##. If ##y_2(b)## was bounded, the Green's function would be very simple to calculate. But it's not bounded, so I'm confused how to proceed?
Any help is greatly appreciated.
Given the following ODE $$(p(x)y')' + q(x)y = 0$$ where ##p(x) = 1-x^2## and ##q(x) = 2-1/(1-x^2)## subject to $$y'(a) + \sec(a)\tan(a)y(a) = 0$$ and $$|y(b)| < \infty,$$ where ##a = \sqrt{1-\cos^2\alpha} : \alpha \in (0,\pi)## and ##b = 1##, what is the Green's function?
This is the associated Legendre ODE, which admits two linearly independent solutions: $$y_1 = P_1^1, \,\,\, y_1 = Q_1^1$$ where ##P_1^1## and ##Q_1^1## are associated Legendre polynomials first and second kind respectively. Things now get a little murky: notice ##y_1## automatically satisfies ##y_1'(a) + \sec(a)\tan(a)y_1(a) = 0##, but ##y_2(b) = \infty##. If ##y_2(b)## was bounded, the Green's function would be very simple to calculate. But it's not bounded, so I'm confused how to proceed?
Any help is greatly appreciated.