Chebyshev Differentiation Matrix

  • #1
Leonardo Machado
57
2
TL;DR Summary
It is a question about the method to obtain the Chebyshev coefficients for differential operators.
Hi everyone.

I am studying Chebyshev Polynomials to solve some differential equations. I found in the literature that if you have a function being expanded in Chebyshev polynomials such as
$$
u(x)=\sum_n a_n T_n(x),
$$

then you can also expand its derivatives as
$$
\frac{d^q u}{dx^q}=\sum_n a^{(q)}_n T_n(x),
$$

with the following relation
$$
a^{(q)}_{k-1}= \frac{1}{c_{k-1}} ( 2 k a^{(q-1)}_k+ a^{(q)}_{k+1}),
$$

being $c_k=2$ for k=0 and 1 if k>0.

It all together defines the Chebyshev differentiation matrix, which is $D$ in
$$
a^{(1)}_i=D_{ij} a^{(0)}_j.
$$

Now I would like to know if there is any way of doing

$$
x^l \frac{du}{dx}=\sum_n a^{(x)}_n T_n(x),
$$

I am looking for it everywhere in the literature but I can't find a way of dealing with this kind of operator that appears in the Laplacian. I can't describe every linear operator without it
 

Answers and Replies

  • #2
Filip Larsen
Gold Member
1,625
548
I haven't worked with Chebyshev polynomials since last millennia, but to me it looks like you are asking for the multiplication of two Chebyshev expansions. If so, that should be fairly straight forward.
 
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  • #3
Leonardo Machado
57
2
I have solved it today. As a computational problem i used a Gaussian integration to take inner products from both sides, as

$$
\sum_k a^{(0)}_k (T_n(x),x^l \frac{du}{dx})=a^{(x)}_n (T_n(x),T_n(x)).
$$

Using Chebyshev collocation points to solve the inner product integral.
 

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