- #1

galuoises

- 8

- 0

[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?

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In summary, the conversation is about solving a non-linear ODE with a boundary condition using an analytical approach. The ODE is given as \frac{d}{dx}f(x)=a f(x)+ b f^3 (x) and the boundary is f(0)=0 and f(+\infty)=f_0. The conversation also discusses integrating the ODE and using elliptic functions to solve it. There is a mention of a possible analytical solution x(y)=g(y,c_1,c_2) and a need for careful verification.

- #1

galuoises

- 8

- 0

[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?

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- #2

JJacquelin

- 801

- 35

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)

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)

Last edited:

- #3

galuoises

- 8

- 0

[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

jackmell

- 1,807

- 54

[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

JJacquelin

- 801

- 35

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.

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

- #7

jackmell

- 1,807

- 54

Nicely done Jacquelin and beautiful too. Thanks.

- #8

galuoises

- 8

- 0

Thank you so much!

A non linear ODE (Ordinary Differential Equation) is a type of mathematical equation that involves the derivatives of a function and its independent variable, where the function is not linear. This means that the dependent variable appears in a non-linear form, such as squared or cubed, and cannot be simplified to a linear equation.

This equation represents a second-order non linear ODE, where the second derivative of the function y is equal to the sum of two terms: a times y and b times y cubed. This equation is a common form of a non linear ODE and is used to model various physical and biological phenomena.

Unlike linear ODEs, there is no general method for solving non linear ODEs. In most cases, numerical methods are used to approximate the solution. These methods involve breaking the problem down into smaller, simpler parts and using iterative calculations to find a solution. Other techniques, such as perturbation methods and series solutions, may also be used in certain cases.

Non linear ODEs are used to model a wide range of phenomena in physics, chemistry, biology, and engineering. Some common applications include modeling population growth, chemical reactions, and oscillating systems. Non linear ODEs are also used in chaos theory to study complex systems and their behavior.

Non linear ODEs can be much more difficult to solve than linear ODEs because they do not have a general solution. This means that each problem must be approached individually and specific techniques must be used to find a solution. In addition, non linear ODEs can exhibit chaotic behavior, making it difficult to predict the long-term behavior of a system.

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