Given the following ode for $x\rightarrow\infty$:
$$\left(x f(x)^3 \left(\frac{(x f^\prime)^\prime}{x}\right)^\prime\right)^\prime=0,$$
in the sense of "asymptotics", the equal sign is equivalent to "$\sim$".
Assume the function $f(x)>0$ defined on $x>0$. Is it possible to obtain a symbolic solution with AsymptoticDSolveValue
?
AsymptoticDSolveValue[D[x*f[x]^3*D[D[x*f'[x], x]/x, x], x] == 0, f[x], {x, \[Infinity], 2}]
However, the above naive code produces an output the same as the input.
My questions:
Can
AsymptoticDSolveValue
compute asymptotic approximations of solutions of odes for its independent variable approaching infinity? Thank to @Michael E2's comment, I found some examples in the document that consider this sort of limit. (Answer: yes, but for some problems)I am wondering what kind of expansion/approximation has been used in
AsymptoticDSolveValue
when it computes asymptotic approximations as $x\to\infty$. I ask this question because a common power series expansion $$f(x)\sim\sum_{n=0}^\infty c_n x^n$$ usually is appropriate for small $x$.If
AsymptoticDSolveValue
cannot solve this problem, is it possible to solve it withSeries
?
AsymptoticDSolveValue
, but of course it does not solve every problem. $\endgroup$AsymptoticDSolveValue
instead! I also updated the post. $\endgroup$AsymptoticDSolveValue[y''[x] - x y[x] == 0, y[x], {x, \[Infinity], 1} ]
$c_1 e^{-\frac{2 x^{3/2}}{3}} \left(\frac{1}{\sqrt[4]{x}}-\frac{5}{48 x^{7/4}}\right)+c_2 e^{\frac{2 x^{3/2}}{3}} \left(\frac{5}{48 x^{7/4}}+\frac{1}{\sqrt[4]{x}}\right)$ $\endgroup$AsymptoticDSolveValue
for a given problem? Or, could you suggest which kind of expansion is suitable for this equation? $\endgroup${x, ∞, n}
is effectively a request to use an asymptotic series in the solution. $\endgroup$