What does the Chebyshev differentiation matrix look like for third and fourth derivative?

Question

I have a PDE that contains both the 3rd derivative and 4th derivative. Example shown below

$$ \frac{\partial u}{\partial t} =\frac{\partial}{\partial x}(u^3\frac{\partial^3u}{\partial x^3}) $$

$$ \frac{\partial u}{\partial t} =(u^3\frac{\partial^4u}{\partial x^4} + 3u^2\frac{\partial u}{\partial x}\frac{\partial^3u}{\partial x^3}) $$

From the material "Spectral Methods in Matlab", it says that for the second derivative it takes the Chebyshev differentiation matrix as $D^2$. Does that mean can I consider the Chebyshev differentiation matrix for the 3rd derivative and 4th derivative as $D^3$ and $D^4$ respectively? Is the below equation correct for me to use ode15s to solve it?

$$ \frac{\partial u}{\partial t} =(u^3*(D^4*u) + 3u^2*(D*u)*(D^3u)) $$

Trefethen, Lloyd N., Spectral methods in Matlab, Software - Environments - Tools. 10. Philadelphia, PA: SIAM, Society for Industrial and Applied Mathematics. 184 p. (2000). ZBL0953.68643.

Answer 1

Yes, you can take the third and fourth power of the Chebyshev differentiation matrix for approximating the third and fourh derivative.

Why is this so? Because differentiation is an associative operation, as is matrix multiplication, and thus one gets the same result by applying the second derivative to the original function or the first derivative to its derivative. The same holds for higher derivatives: so, if $D^2*u$ is the Chebyshev expansion coefficients of the second derivative $u''(x)$, then $D(D^2*u)=D^3*u$ gives the coefficients of the approximation of the third derivative $u^{(3)}(x)$. And so on.