I've been playing with a boundary value problem consisting of a fourth-order, nonlinear, inhomogeneous differential equation (containing two parameters $P$ and $Q$) and four or five boundary conditions. By "four or five", I mean the following four:

  1. $f\left(1\right) = 1$
  2. $f'\left(1\right) = 0$
  3. $f''\left(1\right) = 0$
  4. $f\left(0\right) = 0$

together with an additional condition $f\left(0\right)f''\left(0\right) = 0$, which is not entirely distinct from boundary condition 4.

I've got solutions (for lots of values of the parameters $P$ and $Q$) to the differential equation with boundary conditions 1-4, obtained using a finite difference shooting method. Now I need to verify whether the additional condition is satisfied, by checking whether $f''\left(x\right)$ is singular at $x = 0$, and if so whether the singularity is strong enough to prevent $f\left(x\right)f''\left(x\right)$ vanishing. Visual inspection of the numerical results suggests that $f''\left(x\right)$ probably is singular at $x = 0$. In an attempt to assess the strength of that singularity, I've substituted an ansatz $f\left(x\right) \simeq F_0x^n$ as $x \to 0$ into the differential equation (as per first step of the Frobenius method) to get an indicial equation. The possibilities I found were $n = 0$ (ruled out by boundary condition 4), $n = 1$ (a doubly degenerate solution), $n = 9P/\left(9P-2\right)$ (consistent with boundary condition 4 only if $P > 2/9$, consistent with matching the inhomogeneity in the differential equation only if $P > 10/9$), and $n = 5/4$ (which directly matches the inhomogeneity in the differential equation, but needs a specific value of $F_0$ to do so).

Well, so far so good: $f\left(x\right)f''\left(x\right) \to 0$ as $x \to 0$ in any of those power-law forms. However, the degeneracy of the $n = 1$ solution suggests to me that there's some additional, non-power-law form that I haven't tested. How do I go about finding and testing it, please? I've tried the method of reduction of order, but just get an intractable-looking third-order equation.

If anyone really wants to know, the differential equation is

$$12P\left(f\left(x\right)\right)^3f''''\left(x\right)+12\left(3P-1\right)\left(f\left(x\right)\right)^2f'\left(x\right)f'''\left(x\right)+4\left(3P+2\right)\left(f\left(x\right)\right)^2\left(f''\left(x\right)\right)^2+12\left(P-1\right)f\left(x\right)\left(f'\left(x\right)\right)^2f''\left(x\right)-\frac{9P^2}{Q^4}x = 0$$ It comes from a fluid mechanics paper from the 1930s (doi:10.1098/rspa.1939.0089).

Additional information: I tried a more general ansatz $$f\left(x\right) \simeq F_0T^n\left(\ln\left(x\right)\right)^m$$ as $x \to 0$. This With this ansatz, the leading three orders of terms in the differential equation ($\sim x^{4n-4}\left(\ln\left(x\right)\right)^{4m}$, $\sim x^{4n-4}\left(\ln\left(x\right)\right)^{4m-1}$, $\sim x^{4n-4}\left(\ln\left(x\right)\right)^{4m-2}$) are balanced (only) if $n = 1$ and $m = 3/4$ (or with the $m = 0$ forms we already know about). This form is consistent with boundary condition 4, and does not satisfy the additional condition. However, it looks to me like this solution is not valid because terms $\sim x^{4n-4}\left(\ln\left(x\right)\right)^{4m-3}$ in the differential equation pose a solvability condition that cannot be satisfied by any straightforward "next term" in the asymptotic series.

Further additional information: Aha! Imagine that, instead of $f\left(0\right) = 0$, we have $f\left(x_{\star}\right) = 0$ at some small, non-zero value of $x_{\star}$, then we formulate the Frobenius indicial equation at $x = x_{\star}$. Thinking specifically about the $f\left(x\right) \simeq F_0\left(x-x_{\star}\right)\left(\ln\left(x-x_{\star}\right)\right)^{3/4}$ solution: the terms $\sim \left(x-x_{\star}\right)^{4n-4}\left(\ln\left(x-x_{\star}\right)\right)^{4m-3} = \left(x-x_{\star}\right)^0\left(\ln\left(x-x_{\star}\right)\right)^0$ no longer pose an unsolvable solvability condition; instead, they match the $\left(9P^2/Q^4\right)x_{\star}$ term within the inhomogeneity $-\left(9P^2/Q^4\right)\left(x-x_{\star}\right)$ in the differential equation, but only with a fixed value of $F_0$ which is $\sim x_{\star}^{1/4}$, so $F_0 \to 0$ as $x_{\star} \to 0$. I think that's a strong enough argument for my purposes that the leading order behaviour when the boundary condition is applied right at $x = 0$, $f\left(0\right) = 0$, is not the form with the logarithmic factor, which is all I need.



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