Solving d²Q/dn² - A(B-n)dQ/dn + (A + C/n)Q = 0

[jen81]

Hi,

I have a second order differential equation but I do not know how to solve it.

$$\frac{d^2Q}{dn^2}$$$$\left -A(B-n\right)\frac{dQ}{dn}+$$ $$\left(A + \frac{C}{n}\right)Q = 0$$

Appreciate if anyone could help me on this.

Thank you

 

[LCKurtz]

Probably you must solve with an infinite series. Maple gives a solution in terms of HeunB functions which are computed as a power series around the origin.

 

[HallsofIvy]

Since that is a linear equation with constant coefficients, I see no reason to resort to power series. The characteristic equation is the quadratic
$$r^2- A(B-n)r+ A+ \frac{C}{n}= 0$$

Solve that with the quadratic formula:
[tex]r= \frac{A(B-n)\pm\sqrt{A^2(B-n)^2- 4A- 4\frac{C}{n}}}{2}[/itex][/tex]

 

[LCKurtz]

Since that is a linear equation with constant coefficients, I see no reason to resort to power series. The characteristic equation is the quadratic
$$r^2- A(B-n)r+ A+ \frac{C}{n}= 0$$

Solve that with the quadratic formula:
[tex]r= \frac{A(B-n)\pm\sqrt{A^2(B-n)^2- 4A- 4\frac{C}{n}}}{2}[/itex][/tex]

$$<br /> ? The independent variable is n. Bad notation admittedly, but not constant coefficient.$$

 

[HallsofIvy]

Ah- I missed that. Thanks.

 

[gato_]

The solution given by maple is not very helpful... I found it useful to make the following standard transformation, to get rid of the first derivative (I have changed the independent variable's name to x):
$$Q(x)=exp(-A(x-B)^{2}/4)y(x)$$
By doing that, the resulting equation is
$$y''(x)+[C/x+A/2-A^{2}(x-B)^{2}/4]y(x)=0$$
So the full solution consists of a gaussian like envelope multiplying a carrier function with what we can consider a spatial dependent frequency (not so, as it turns negative at some point, and the wave becomes evanescent). Close to x=0, the equation:
$$y''(x)+y(x)C/x=0$$
Has
$$y=C_{1}\sqrt{2}J_{1}(2\sqrt{Cx})$$
as solution, which goes as
$$y\approx x^{1/4}Cos(2\sqrt{Cx}-3\pi/4)$$
as x increases. However, we know that the frequency can't increase indefinitely, because there is a turning point when the coefficient in front of y(x) approaches 0. The solution around that point can be expressed locally in terms of Airy functions (that is, it matches oscillation with pure decay). The full solution then looks like in the attached figure

I am sure there is a nicer solution... but I don't find a simple one.