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This may be very obvious but I am stuck trying to solve a boundary value problem. I am trying to solve the following differential equation : $$F^{3}F^{(5)}+F=1 \space (Eq.1)$$ I have been advised to write the parametrization : $\boxed{F=1+\phi}$ with $\phi(\infty)=0$ : I thus obtain an asymptotic behavior : $$\phi^{(5)}+\phi \approx 0$$ that have 5 exponential solutions of the form : $\boxed{\phi \approx exp(\omega_i \eta)}$ with $\omega_i^{5}=-1$.

I can write that : $$ \phi = A e^{\omega_1\eta} + B e^{\omega_2\eta}+C e^{\omega_3\eta}+D e^{\omega_4\eta}+E e^{\omega_5\eta} $$ Knowing that : $\phi(\infty)=0$ and that 2 of the 5 complex roots have positive real parts (giving growing exponentials) : the coefficients of these growing exponentials must be zero.

It is said that this remark is supposed to provide 2 boundary conditions to the $(Eq.1)$ but I don't see how it's related since these are "just" conditions on coefficients of $\phi$...

It seems very immediate but I don't see it for the moment.

Thank you in advance for your help.

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1 Answer 1

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With $F=1+ϕ$ the approximate equation for $ϕ\approx 0$ results from $$ ϕ^{(5)}+1+ϕ =(1+ϕ)^{-3}=1-3ϕ+6ϕ^2-10ϕ^3+...+\binom{n+2}{2}(-ϕ)^n+... $$ Thus the linear approximation is $$ ϕ^{(5)}+4ϕ=0 $$ As observed, the characteristic roots are $α<0$, $β,\barβ$ in the negative half-plane and $γ,\barγ$ in the positive half-plane, all on a regular pentagon around the origin.

This makes the equilibrium point a saddle point. Almost all solutions, and thus all numerical solutions except the constant one, which approach the saddle point will at some point turn away and exponentially diverge.

An approximation to a solution that is asymptotic to the saddle point can be obtained by using a stable solution of the linear approximation as far-field piece of a piece-wise defined function. This far-field approximation can be parametrized as $$ ϕ=c_1e^{αη}+c_2e^{βη}+c_3e^{\barβη} $$ or equivalently as a solution of the linear ODE $$ (D-α)(D-β)(D-\barβ)\phi=0. $$ In this form it is easy to obtain further linear conditions on the derivatives sequence as $$ (D-α)(D-β)(D-\barβ)\phi^{(k)}=0,~~~k=1,2,... $$ To "optimally" fit a numerical solution and a far-field approximation at some point $η=b$, one can impose these linear conditions on the derivatives $[ϕ(b),ϕ'(b),...,ϕ^{(4)}(b)]$ that make up the state space for the original equation. This results in 2 boundary conditions.

That using the differential equations as "connection condition" is easier than extending the system with the coefficients as free parameters and adding continuity conditions can also be seen in Non-unique solutions for ODE with boundary conditions at infinity

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  • $\begingroup$ Thank you for your answer @lutz. So the first equation should be verified on $\eta = \infty$ and the second one on $\eta = 0$ : is it correct ? Moreover, could you please explicit how did you obtain the second condition ? I am not sure to have understand the explanation. Thank you very much. $\endgroup$
    – Waxler
    Apr 18, 2022 at 9:14
  • $\begingroup$ You approximate the infinite interval $[a,\infty)$ by some large finite one $[a,b]$ where $b$ is large. If $F(a)=F_a> 0$ is one of the two left boundary conditions, everything is fine there. The two conditions I gave are for $η=b$, the point to be reached at infinity is a saddle point, so you have to zero-in to the stable manifold. This is exactly what these two conditions do. $\endgroup$ Apr 18, 2022 at 11:48
  • $\begingroup$ Thank you for your precision. However, it may seem obvious but I still didn't get the second condition : could you please write explicitly how did you came up with this deduction ? Thank you in advance. $\endgroup$
    – Waxler
    Apr 20, 2022 at 8:40
  • $\begingroup$ I'm not sure that there is a good justification. It forces the numerical solution to have its value and up to 4th derivative to be equal at $η=b$ to the far-field approximation (that is, a solution of the 3rd order DE). It is better adapted to the theoretical far-field behavior than just setting $F(b)=1$, $F'(b)=0$ for the very crude far-field approximation of a constant function. One could probably find better conditions by applying some perturbation argument to get a better far-field approximation. $\endgroup$ Apr 21, 2022 at 10:49
  • $\begingroup$ Thank you @LutzLehmann for your clarification : indeed the second condition was a bit intuitive for me but with your new comment and the reading of the associated post that you shared, I think that I got it. Thanks again ! $\endgroup$
    – Waxler
    Apr 22, 2022 at 14:07

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