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

[galuoises]

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

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

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

 

[jackmell]

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

y''=y+y^3

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

x(y)=g(y,c_1,c_2)

then try and solve simultaneously the expressions:

0=g(0,c_1,c_2)

g(f_0,c_1,c_2)\to+\infty

for the constants c_1,c_2 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.

 

[JJacquelin]

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

 

[jackmell]

Nicely done Jacquelin and beautiful too. Thanks.

 

[galuoises]

Thank you so much!