# Improper boundary in non-linear ODE (pseudospectral methods)

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In summary, the conversation discusses using pseudospectral/collocation methods to solve non-linear equations. The solution is expanded using Chebyshev polynomials on a finite domain, but there is a problem with boundary conditions at infinity. The suggestion is made to map the domain to [-1,1] and use Chebyshev collocation, which would result in a bounded solution. Another suggestion is to set the solution as a sum of the Chebyshev polynomials plus a constant, which would solve the boundary condition at infinity. However, it is uncertain if this method would have all the desired properties of collocation methods.
TL;DR Summary
I am not being able to determine the behavior of my solutions for improper boundaries if the behavior of the solution is expected to diverge.
Hello,

I am trying to compute some non-linear equations with pseudospectral/collocation methods. Basically I am expanding the solution as

$$y(x)=\sum_{n=0}^{N-1} a_n T_n(x),$$

Being the basis an Chebyshev polynomial with the mapping x in [0,inf].

Then we put this into a general differential equation

$$Ly(x)=f(x,y),$$

$$\sum_{n=0}^{N-1} L T_n(x) a_n = f(x,y).$$

This function is evaluated at the collocation points associated with the Chebyshev polynomials as usual, leading to N-1 non-linear equations. Also, there is also equations for the boundaries, i. e.,

$$\sum_{n=0}^{N-1}a_n T_n(0)=A,$$
$$\sum_{n=0}^{N-1}a_n T_n(inf)=B.$$

My problem is here. How can I treat a boundary condition which leads to infinity somehow? For example$$\sum_{n=0}^{N-1}a_n \frac{dT_n(inf)}{dx}=1,$$

or even,

$$\sum_{n=0}^{N-1}a_n T_n(inf)=inf.$$

I would suggest that you map $[0, \infty)$ to $[-1,1)$ by $x \mapsto z = 2 \tanh(x) - 1$ and use Chebyshev collocation on $[-1,1]$ as normal, rewriting your operator $L$ in terms of $z$ using $$\frac{d}{dx} = \frac{dz}{dx}\frac{d}{dz} = 2\,\mathrm{sech}^{2}(x) \frac{d}{dz} = \left(\tfrac32 - z - \tfrac12 z^2\right)\frac{d}{dz}$$ This has the advantage that you are working with functions which are bounded on a finite domain, so your solution will be also. It also avoids having most of your collocation points in $[0,1)$ and the other at infinity.

You then have \begin{align*} f(x) &= \sum_n a_n T_n(2 \tanh(x) - 1) \\ f(\infty) &= \sum_n a_n T_n(1) \end{align*}

pasmith said:
I would suggest that you map $[0, \infty)$ to $[-1,1)$ by $x \mapsto z = 2 \tanh(x) - 1$ and use Chebyshev collocation on $[-1,1]$ as normal, rewriting your operator $L$ in terms of $z$ using $$\frac{d}{dx} = \frac{dz}{dx}\frac{d}{dz} = 2\,\mathrm{sech}^{2}(x) \frac{d}{dz} = \left(\tfrac32 - z - \tfrac12 z^2\right)\frac{d}{dz}$$ This has the advantage that you are working with functions which are bounded on a finite domain, so your solution will be also. It also avoids having most of your collocation points in $[0,1)$ and the other at infinity.

You then have \begin{align*} f(x) &= \sum_n a_n T_n(2 \tanh(x) - 1) \\ f(\infty) &= \sum_n a_n T_n(1) \end{align*}

The problem is that the derivatives evaluated at infinity would still be zero aways, becase sech(infinity)=0. I mean,

$$\sum_n a_n \frac{dT_n(x)}{dx}=\sum_n a_n 2 sech(x) \frac{dT^*_n(z)}{dz}$$

evaluated at infinity still cannot be a finite number.

I think if a function has a non-zero derivative at infinity then it's not going to be finite there.

Perhaps you could set $f(x) = f'(\infty)x + g(x)$ with $g'(\infty) = 0$ and work with $g$ instead. The boundary condition at $x = \infty$ is then satisfied automatically. So in this case you don't need a collocation point at $z = 1$.

pasmith said:
I think if a function has a non-zero derivative at infinity then it's not going to be finite there.
Yes. Unfortunately this is part of this problem I am tr

pasmith said:
Perhaps you could set $f(x) = f'(\infty)x + g(x)$ with $g'(\infty) = 0$ and work with $g$ instead. The boundary condition at $x = \infty$ is then satisfied automatically. So in this case you don't need a collocation point at $z = 1$.

I was thinking about expanding the solution, just like you said, but with

$$y(x)=u(x)+r= \sum_n a_n T_n(x) + r.$$

This way of writting turns the condition

$$\frac{dy(infinity)}{dx}=1,$$

into

$$\frac{du(infinity)}{dx}=0,$$

which is easily solved. But I am not sure if it will have all the good properties of collocation methods. It is something that I must implement frist to judge.

## 1. What is an improper boundary in non-linear ODE?

An improper boundary in a non-linear ODE refers to a boundary condition that is not well-defined or does not satisfy the necessary conditions for the solution of the ODE. This can lead to difficulties in finding a solution or can result in an incorrect solution.

## 2. What causes improper boundaries in non-linear ODEs?

Improper boundaries can be caused by a variety of factors, such as incorrect boundary conditions, singularities in the ODE, or discontinuities in the solution. They can also arise from errors in the numerical methods used to solve the ODE.

## 3. How do pseudospectral methods handle improper boundaries in non-linear ODEs?

Pseudospectral methods use a spectral approach to solving ODEs, which means that they approximate the solution using a sum of basis functions. These methods can handle improper boundaries by adjusting the basis functions to better fit the solution near the boundary.

## 4. What are the limitations of using pseudospectral methods for solving ODEs with improper boundaries?

While pseudospectral methods are effective at handling improper boundaries, they can be computationally expensive and require a large number of basis functions to accurately represent the solution. Additionally, they may not be suitable for problems with highly irregular boundaries.

## 5. Are there other methods for solving ODEs with improper boundaries?

Yes, there are other methods such as finite difference methods, finite element methods, and boundary element methods that can also handle improper boundaries in ODEs. The choice of method will depend on the specific problem and the desired level of accuracy and efficiency.

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