Solving for Coefficient in Nonlinear 2nd-Order Differential Equation

Question

Alexander Herrod 2018-1-4

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评论： Alexander Herrod 2018-7-22

Hi,

I'm trying to numerically solve for the coefficient "a" in a nonlinear second-order differential equation of the form:

y''(r) = y(r)(a - b*abs(y(r))/r + (c*r^3 - d)*y(r)^2)

given boundary conditions:

y(0) = y(inf) = 0

integral from 0 to inf of y(r)^2 dr = 1

However I don't seem to be able to find the tools to do this neatly. I have another platform that can do this, but I'd very much like to cross-check the results in Matlab!

Any help is appreciated!

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Answer 1

Torsten 2018-1-5

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编辑：Torsten 2018-1-5

在 MATLAB Online 中打开

Use "fzero" or "fsolve" to solve the equation

(integral_{0}^{inf} y(a,r)^2 dr) - 1 = 0

for a.

To this end, in the function routine from "fzero" or "fsolve", you will be given a suggestion for a. With this value for a, you can use "bvp4c" to solve the boundary value problem

y''(r) = y(r)(a - b*abs(y(r))/r + (c*r^3 - d)*y(r)^2)

given boundary conditions:

y(0) = y(inf) = 0

Once you have its solution in discrete points, use "trapz" to approximate integral_{0}^{inf} y(r)^2 dr and return integral_{0}^{inf} y(r)^2 dr - 1 to "fzero" or "fsolve".

I think one problem with a numerical solution will be that your boundary value problem always has y = 0 for all r as solution.

Best wishes

Torsten.

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Alexander Herrod 2018-7-21

Thanks for the reply, and sorry it's taken so long!

I don't assume the dependency of y(r) on any of the constants. The integration requirement is only to be applied at the end.

If I try your suggestion, Matlab throws me an error that it can't find and solutions to that boundary condition (because of course there are infinite possible functions!)

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Answer 2

John BG 2018-1-5

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编辑：John BG 2018-1-15

在 MATLAB Online 中打开

Hi Alexander

this is John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

1.

One of the tools available is MuPAD, start MuPAD with

mupad

once open define the equation, but avoid the abs(), it only complicates the result and since r seems to be a cylindrical coordinate, if r is the radius, then there is no point that r takes negative values

.

the result after ode:solve(o) is what comes up if including the abs()

the resulting functions do not seem that useful.

2.

It turns out that the Jacobi elliptic functions are solution to a similar differential equation:

clear all;clc;close all
rspan=[-20 20];
dr=0.0001;
r=[rspan(1):dr:rspan(2)];
M = 0.7;
[S,C,D] = ellipj(r,M);
figure(1);plot(r,S,r,C,r,D);
legend('SN','CN','DN','Location','best');grid on;
title('Jacobi Elliptic Functions sn,cn,dn');

a=1;b=1;c=.1;d=.1;
vS=d2ydr2(r,S,a,b,c,d);
vC=d2ydr2(r,C,a,b,c,d);
vD=d2ydr2(r,D,a,b,c,d);
d2Sdt2=diff(diff(S));
d2Cdt2=diff(diff(C));
d2Ddt2=diff(diff(D));
errS=d2Sdt2-vS([3:end]);
errC=d2Cdt2-vC([3:end]);
errD=d2Ddt2-vD([3:end]);
figure(2);plot(r([3:end]),abs(errS),r([3:end]),abs(errC),r([3:end]),abs(errD));grid on;

3.

now M=.99 CN and DN (not the blue one) show far smaller error

rspan=[1 5];
dr=0.0001;
r=[rspan(1):dr:rspan(2)];
M = 0.99;
[S,C,D] = ellipj(r,M);
vS=d2ydr2(r,S,a,b,c,d);
vC=d2ydr2(r,C,a,b,c,d);
vD=d2ydr2(r,D,a,b,c,d);
d2Sdt2=diff(diff(S));
d2Cdt2=diff(diff(C));
d2Ddt2=diff(diff(D));
errS=d2Sdt2-vS([3:end]);
errC=d2Cdt2-vC([3:end]);
errD=d2Ddt2-vD([3:end]);
figure(3);plot(r([3:end]),abs(errS),r([3:end]),abs(errC),r([3:end]),abs(errD));grid on;