Series solution to coupled nonlinear ODEs

[daso]

Hi,

I have a system of three coupled nonlinear ODEs:

$$\frac{d}{du}(nu)=exp(-\phi),$$

$$u\frac{du}{dx}=\frac{d\phi}{dx}-\frac{u}{n}exp(-\phi),$$

$$\frac{d^2 \phi}{dx^2}=n-exp(-\phi),$$

with boundary conditions

$$\phi=\phi'=u=n=0 \text{ at } x=0$$

Does anyone know, or have references to, how to solve these with a series expansion? I am only interested in the solution for $$x \ll 1$$, so higher order terms can be neglected.

Thankful for any help!

/Daniel

 

[jvicens]

Hi Daniel,

Thank you for sharing your system of three coupled nonlinear ODEs. To solve this system using a series expansion, you can use the method of perturbation. This involves breaking down the system into a linear and nonlinear part, and then solving for the solution using a series expansion in terms of a small parameter, in this case x \ll 1.

First, we can rewrite the system as:

\frac{d}{du}(nu)=exp(-\phi),

\frac{d}{dx}(u^2)=\frac{d\phi}{dx}-\frac{u^2}{n}exp(-\phi),

\frac{d^2 \phi}{dx^2}=n-exp(-\phi),

with boundary conditions

\phi=\phi'=u=n=0 \text{ at } x=0

Next, we can make the substitution u = u_0 + u_1x + u_2x^2 + ..., \phi = \phi_0 + \phi_1x + \phi_2x^2 + ... and n = n_0 + n_1x + n_2x^2 + ..., where u_0, \phi_0 and n_0 are the initial conditions at x=0. We can then substitute these into the system and equate terms of the same order in x to obtain a set of equations for each order.

For the linear terms, we have:

\frac{d}{du}(nu_0)=1,

\frac{d}{dx}(u_0^2)=\frac{d\phi_0}{dx},

\frac{d^2 \phi_0}{dx^2}=n_0-1,

with boundary conditions

\phi_0=\phi_0'=u_0=n_0=0 \text{ at } x=0

These equations can be solved analytically to obtain the leading order solution for x \ll 1. For the nonlinear terms, we have:

\frac{d}{du}(nu_1)=exp(-\phi_0),

\frac{d}{dx}(2u_0u_1)=\frac{d\phi_1}{dx}-\frac{2u_0u_1}{n_0}exp(-\phi_0),

\frac{d^2 \phi_

 

[tacman]

Hi Daniel,

Thank you for sharing your problem with us. Solving coupled nonlinear ODEs can be challenging, but using a series expansion can simplify the process. The idea behind a series solution is to approximate the solution as a sum of infinite terms, each with a different power of x. In your case, you are interested in the solution for x \ll 1, which means that higher order terms can be neglected. This is known as a perturbation method, where the solution is found by adding small corrections to an approximate solution.

To solve your system of equations using a series expansion, you can follow these steps:

1. Rewrite your equations in a form that is suitable for a series solution. This usually involves isolating the highest derivative term on one side of the equation, with all other terms on the other side.

2. Assume a series solution of the form u=u_0+u_1x+u_2x^2+..., \phi=\phi_0+\phi_1x+\phi_2x^2+..., n=n_0+n_1x+n_2x^2+...

3. Substitute the series solution into your equations and equate coefficients of the same powers of x. This will give you a system of equations for the coefficients u_i, \phi_i, and n_i.

4. Solve the system of equations for the coefficients, starting with the lowest order terms and working your way up.

5. Once you have the coefficients, substitute them back into your series solution to get the approximate solution for x \ll 1.

Some references that may be helpful for solving coupled nonlinear ODEs using a series expansion are "Perturbation Methods" by E.J. Hinch and "Introduction to Perturbation Methods" by Mark H. Holmes. You can also find many online resources and examples that may be useful in understanding and solving your problem.

I hope this helps and good luck with your problem!

Best regards,