Non linear ODE: y'' = a y + b y^3

galuoises

I would like to solve the non linear ODE
[itex] \frac{d}{dx}f(x)=a f(x)+ b f^3 (x)[/itex]
with the boundary
[itex]f(0)=0\quad f(+\infty)=f_0 [/itex]

How to find analitical solution?

JJacquelin

If it is f ' = a f+b f^3 then integrate dx = df/(a f+b f^3)
If it is y'' = a y+b y^3 then the first step is the integration of :
y'' y' = (a y+b y^3) y'
y'² = a y² + b (y^4)/2 +c
dx = dy/sqrt(a y² + b (y^4)/2 +c)
The integration will lead to x(y) on the form of elliptic functions. Very complicated to invert in order to obtain y(x)

galuoises

Pardon me, I write the uncorrect differential equation: the problem is at the second order

[itex] \frac{d^2}{dx^2}f(x)=a f(x) + b f^3 (x) [/itex]
with the boundary
[itex] f(0)=0,\ f(+\infty)=f_0[/itex]

jackmell

In my opinion, [itex]x(y)=g(y,c_1,c_2)[/itex] is an analytical expression for the solution but I think first, just scrap the a and b and look at:

[tex]y''=y+y^3[/tex]

then do what Jacq said and get the expression in terms of:

[tex]x(y)=g(y,c_1,c_2)[/tex]

then try and solve simultaneously the expressions:

[tex]0=g(0,c_1,c_2)[/tex]

[tex]g(f_0,c_1,c_2)\to+\infty[/tex]

for the constants [itex]c_1,c_2[/itex] and if need be, do so numerically for them just for starters.

JJacquelin

Hello !

It is a difficult problem on the analytical viewpoint.
A difficulty is that elliptic functions are involved. But the major difficulty is due to the condition f(x=+infinity)=finite constant.
I am not quite sure that the solution given in attachment is a correct answer to the problem, so it should be carefully verified.

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JJacquelin

In case of some complementary conditions specified in attachment, y(x) can be fully explicited thanks to a Jacobi elliptic function.

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jackmell

Nicely done Jacquelin and beautiful too. Thanks.

galuoises

Thank you so much!