Asymptotic solutions to ODE

Question

Wondering where I can find resources to do more questions of the following type and also if you guys can help me answer this problem.

Consider the differential equation:

$$u'' + \left( 1-\frac{\gamma}{x^2} \right) u = 0$$

for $x > 4$. Obtain the first two terms of the asymptotic solution for each of the two real solutions of this equation.

We start by writing out $u = u_0+u_1$ and consider the zeroth order solution which is $$u_0''+u_0=0$$ This has solution $u_0=e^{ix}$. Basically how do I proceed from here? Also where can I find more problems requiring this method of solution. Thanks!

Answer 1

An expansion of $u $ derived from the perturbation theory would be $u=u_0 +\gamma u_1 + \dots$ The $0$th-order and $1$st-order equations (in terms of powers of $\gamma$) are respectively \begin{aligned} u''_0 + u_0 &= 0 \, ,\\ u''_1 + u_1 &= \frac{u_0}{x^2} \, . \end{aligned} The $0$th-order solution is $$u_0(x) = a_0\cos x + b_0\sin x \, ,$$ whereas there is more work to find the $1$st-order solution. Indeed, $$u_1(x) = a_1\cos x + b_1\sin x + u_p(x)\, ,$$ where $u_p$ is a particular solution to the non-homogeneous ODE satisfied by $u_1$. The initial conditions $u(4) = U$, $u'(4) = V$ are first applied on $u_0$, $u'_0$: \begin{aligned} a_0\cos 4 + b_0\sin 4 &= U \, ,\\ b_0\cos 4 - a_0\sin 4 &= V \, , \end{aligned} which yields $a_0 = U\cos 4 - V\sin 4$ and $b_0 = U\sin 4 + V\cos 4$. Then, they are applied on $u_0 +\gamma u_1$, $u'_0 + \gamma u'_1$, etc.

There is much more to read about such methods in the book Perturbation Methods by A.H. Nayfeh (Wiley, 2008).

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This linear ODE may be solved analytically. The general solution involves Bessel functions.

Answer 2

As the equation is linear. We can find a complex solution and then take the real and imaginary part to obtain the two independent real solutions.

As you have noticed the zeros order solution (for $x\to\infty$) is given by $u_0 = e^{ix}$. We start with the expansion $$u=e^{ix}( 1 + y_1 + y_2 + \cdots)$$ with $y_i \ll y_{i+1}$ and $y_1 \ll 1$ for $x\to\infty$. Plugging the first two terms into the differential equation, we obtain $$y_1''+2 i y_1' \sim \frac{\gamma }{x^2} ;\tag{1}$$ here, we have neglected the term $\gamma y_1/x^2$ as it is (by assumption) small compared to the term on the right hand side of (1).

The task is to solve (1) asymptotically for $x\to \infty$ with the condition $y_1 \ll 1$. We assert that $y_1'' \ll y_1'$ (to be checked after having found the solution) and integrate $2 i y_1' \sim \frac{\gamma }{x^2}$ with the result $$y_1 = \frac{i \gamma}{x};$$ indeed, we have that $y_1'' \ll y_1'$ so the solution is consistent.

In conclusion, we have found that $$ u \sim e^{ix} \left( 1 + \frac{i \gamma}{x} +o(x^{-1})\right)$$ solves the problem. Taking the real and imaginary part leads to the two real solutions of the problem.

You find more problems and information about asymptotic solutions to differential equations in Bender, Orszag: "Advanced mathematical methods for scientists and engineers".