Solving ODE $g(\eta)''+\eta/2 \cdot g(\eta)'-n\cdot g(\eta)=0$; BC: $g(0)=1$ and $g(\infty)=0$

Question

I'm lost! How would I solve $$g(\eta)''+\eta/2 \cdot g(\eta)'-n\cdot g(\eta)=0$$

with BC $$g(0)=1$$ and $$g(\infty)=0$$

Answer 1

An attempt at solution.

Let's make a transformation for some $a$:

$$g(\eta)=f(\eta)e^{a\eta^2}$$

$$g'=(f'+ 2a \eta f)e^{a\eta^2}$$

$$g''=(f''+ 2a \eta f'+2a f+2a \eta f'+4a^2 \eta^2 f)e^{a\eta^2}=(f''+ 4a \eta f'+2a (1+2a\eta^2) f)e^{a\eta^2}$$

After substitution we obtain:

$$f''+ 4a \eta f'+2a (1+2a\eta^2) f+\frac{1}{2} \eta (f'+ 2a \eta f)-nf=0$$

Or:

$$f''+ (4a+1/2) \eta f'+(2a-n+a(1+4a)\eta^2) f=0$$

We can now pick $a$ to get rid of the first derivative:

$$a=-\frac{1}{8}$$

$$f(\eta)=g(\eta) e^{\eta^2/8}$$

Which gives us:

$$f''-\left(n+\frac{1}{4}+\frac{1}{16}\eta^2 \right) f=0$$

$$f''-\frac{1}{16}\left(4(4n+1)+\eta^2 \right) f=0 \tag{1}$$

This equation has a general solution in the form of parabolic cylinder functions.

So in principle, we have solved the problem, and now only need to apply the boundary conditions.

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However, the boundary conditions pose some challenge. The first one is simple enough.

$$f(0)=1$$

But the second one we can't really guess, because for $\eta \to \infty$ the exponential factor blows up, and we don't know enough about $g(x)$ to figure out what is the limit:

$$\lim_{\eta \to \infty} g(\eta) e^{\eta^2/8} = ?$$

Unless there's some way to apply this boundary condition correctly to (1), we are back to the drawing board.

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Update:

I tried to use the Mathematica ODE solver with the second initial condition:

$$f(\eta_m)=0, \qquad \eta_m \gg 1$$

For example $\eta_m =100$. The convergence is very good, i.e. the solutions for different but large $\eta_m$ agree very well.

So I think this may be the way to solve the original equation.