- #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

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