Using the Chebyshev derivative matrix $D$, we can numerically approximate the first and second derivative of a function by doing matrix multiplication:

$${df(x) \over dx} = Df(x) \tag 1$$ $${d^2f(x) \over dx^2} = D^2f(x) \tag 2$$

where $x \in \{x_0,x_1,...x_n\}$, $x_j=\cos(j \pi/n)$, $j \in \{0, 1,...n\}$.

When we want to calculate the Laplacian of a function we can use the Kronecker product of matrices:

$$\nabla ^2f(x, y) = (I \otimes D^2 + D^2 \otimes I)\bar f \tag 3 $$

where $\bar f$ is a vector obtained by stacking the columns of matrix $f(x, y)$. My question is, how can we find the mixed partial derivatives of a function using the Chebyshev derivative matrix? For instance, how would we calculate:

$${\partial^2 f(x,y) \over \partial x \partial y}=? \tag 4$$ and $${\partial^3 f(x,y) \over \partial x^2 \partial y}=? \tag 5$$


1 Answer 1


Note: Your nomenclature is only valid on Cartesian elements. If you want to calculate derivatives on arbitrary shapes you also have to consider spatial metric terms.

Answer: To keep it simple, we stick to the special case. We can define a 2D differential operator $\mathbf{D}_2$ on a tensor-product element with

$\xi$-direction (or $x$):

$\mathbf{D}_2^{\xi} = \mathbf{D}_1 \otimes \mathbf{I}_1$

$\eta$-direction (or $y$):

$\mathbf{D}_2^{\eta} = \mathbf{I}_1 \otimes \mathbf{D}_1$

The Laplacian is defined as

$ \mathbf{\mathcal{L}}_2 := \mathbf{D}_2^{\xi}\mathbf{D}_2^{\xi} +\mathbf{D}_2^{\eta} \mathbf{D}_2^{\eta} \\ \hspace{0.75cm} = (\mathbf{D}_1\otimes\mathbf{I}_1) (\mathbf{D_1}\otimes\mathbf{I_1}) + (\mathbf{I_1}\otimes\mathbf{D_1}) (\mathbf{I}_1 \otimes\mathbf{D}_1)\\ \hspace{0.75cm} = (\mathbf{D}_1\mathbf{D}_1)\otimes (\mathbf{I_1}\mathbf{I_1}) + (\mathbf{I_1}\mathbf{I_1}) \otimes (\mathbf{D}_1 \mathbf{D}_1)$

For mixed derivatives you simply have to apply both operators one after the other

$ \partial_{\xi \eta}:= \mathbf{D}_2^{\xi}\mathbf{D}_2^{\eta} \\ \hspace{0.8cm} = (\mathbf{D}_1\otimes\mathbf{I}_1) (\mathbf{I}_1 \otimes\mathbf{D}_1)\\ \hspace{0.8cm} = (\mathbf{D}_1\mathbf{I}_1)\otimes (\mathbf{I_1}\mathbf{D_1}) \\ \hspace{0.8cm} =\mathbf{D_1} \otimes \mathbf{D}_1\equiv\partial_{\eta \xi}$

using the relation

$(\mathbf{A} \otimes \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{AC}) \otimes (\mathbf{BD})$

Your second example can be derived in a similar way.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.