
#1
Jan1213, 07:54 AM

P: 8

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? 



#2
Jan1213, 11:42 AM

P: 746

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) 



#3
Jan1313, 06:02 AM

P: 8

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] 



#4
Jan1313, 08:12 AM

P: 1,666

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



#5
Jan1313, 11:52 AM

P: 746

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. 



#6
Jan1413, 03:37 AM

P: 746

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




#7
Jan1413, 07:46 AM

P: 1,666

Nicely done Jacquelin and beautiful too. Thanks.




#8
Feb913, 06:28 AM

P: 8

Thank you so much!



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