Could Mathematica solve a differential equation asymptotically?

Question

Is there any possibility that Mathematica could give asymptotic behavior(s) of a differential equation as it independent variable tends to a certain value?

Because I didn't know how to decompose this hard problem, I just tried the naive code to find a trial solution to the 3rd-order nonlinear ODE:

$$H^2(H_{\eta\eta\eta}-\eta^{-2}H_\eta+\eta^{-1}H_{\eta\eta})-\frac{\eta}{10}=0,$$

which is subject to the boundary conditions: $H(0)=1$ and $H_{\eta\eta}(0)=-c$ with $c$ being a constant. If one wants to seek a solution $H(\eta)$ that is an even function in $\eta$.

DSolve[{H[\[Eta]]^2*(H'''[\[Eta]] - \[Eta]^-2 H'[\[Eta]] + 
   1/\[Eta] H''[\[Eta]]) - \[Eta]/10 == 0, H[0] == 1, 
   H''[0] == -c, H[\[Eta]] == H[-\[Eta]]}, H[\[Eta]], \[Eta]]

Mathematica complains that:

DSolve::litarg: To avoid possible ambiguity, the arguments of the dependent variable in {...} should literally match the independent variables.

Obviously, the error results from the even function condition, H[\[Eta]] == H[-\[Eta]]. If I comment this condition, the code just repeats itself without any solution. Another minor question is:

How to impose an even or odd function requirement in DSolve?

Motivation: The reason why I arise this problem is that the asymptotic behavior of $H(\eta)$ has been shown (I guess by perturbation method which is powerful math tool) as follows. That can be used to check the desired code/method with MMA. I really want to consult superiors of MMA here for some suggestion about how to solve an exactly unsolvable differential equation with the asymptotic method. Thank you in advance!

$H \sim 1-\frac{1}{2}c\eta^2 \quad \text{as} \quad \eta \to 0$

$H(\eta)=\tilde{h}F\left(\eta_0-\eta \right)$ near $\eta_0$, where $H(\eta)$ reaches its minmum value for a given constant $c$, denoted by $\tilde{h}=H(\eta_0)$, $F(x)\sim x^2$ as $x\ll0$ and $F(x)\sim x(\log x)^{1/3}$ for large $x\gg0$.

Answer 1

It is possible to roll one's own asymptotic solver with the help of Series. As a demonstration, here we show how to obtain an asymptotic power series solution around zero.

The equations in OP, except the even function condition H[η] == H[-η] (which will turned out to be redundant for this particular problem), can be rearranged to (omit the == 0 parts):

eqs = {
       H[η]^2*(H'''[η]-η^-2 H'[η]+1/η H''[η])-η/10,
       H[0] - 1,
       H''[0] + c
      };

If we assume the existence of power expansion about $x=0$:

$$H(x) = \sum_{k=0}^\infty h_k x^k$$

then a series representation of $H$ and its derivatives can be straightforwardly defined through following rules:

seriesRules = RightComposition[
    ReplaceAll[{
          H[x_] :>
                (Inactive[Series][H[η], {η, 0, max}] // Inactive[ReplaceAll][η -> x]),
          Derivative[s_Integer?Positive][H][x_] :>
                (Inactive[Series][Derivative[s][H][η], {η, 0, max}] // Inactive[ReplaceAll][η -> x])
        }],
    (* the odd/even function constraint can be described as following rule *)
    (* Inactive[ReplaceAll][{
          (* odd: *)(* H[0] :> 0,Derivative[s_Integer?EvenQ][H][0] :> 0 *)
          (* even: *)Derivative[s_Integer?OddQ][H][0] :> 0
        }], *)
    Inactive[ReplaceAll][{H[0] :> h[0], Derivative[s_Integer][H][0] :> h[s]}]
];

Applying it on eqs gives us its series version:

series = eqs // seriesRules;

For a given series order max, series can be Activated to become algebra equations serieseqs about $h_0,h_1,...$:

asymptoticOrder = 10;
serieseqs = series //
               ReplaceAll[max -> asymptoticOrder] // Activate //
               Map[
                   If[Head[#] === SeriesData, #[[3]], #] &
                  ] //
               Flatten // Thread[# == 0] & // DeleteCases[True];

Luckily the equations we got here are all nice and easy to solve:

seriessol = Inactive[Solve][serieseqs, 
                            Union[Cases[serieseqs, _h, ∞]][[;; UpTo[Length@serieseqs]]]
                ] // Activate;
seriessol // Apply[List, #, {2}] & // Map[Grid[#, Frame -> All] &] // Row

Thus the corresponding approximate solution for $H$

H[η] // seriesRules //
        ReplaceAll[max -> asymptoticOrder] // Activate //
        ReplaceAll[seriessol]

(But do be aware this is only a formal series solution. The convergence is yet to prove.)

Answer 2

Mathematica 11.3? Will have AsymptoticDSolve and support for these

And WKB also

It will also finally have series solution for DSolve

See Asymptotic Expansions at http://www.wolfram.com/broadcast/video.php?c=104&p=3&v=2091

Answer 3

It's not a answer of Yours question,only a info if Mathematica 11.3 can solve or not.

Using Mathematica 11.3,it seems can't find solution.