📂Partial Differential Equations

Differentiation Operators and Symbols

Definition¹

For a natural number $m \in \mathbb{N}$, a differential operator refers to the following map $P$.

$$ P = \sum\limits_{\left| \alpha \right| \le m} a_{\alpha}(x) D^{\alpha},\qquad x = (x_{1}, \dots, x_{n}) \tag{1} $$

Here, $\alpha = (\alpha_{1}, \dots, \alpha_{n})$ is a multi-index. $D^{\alpha}$ is as follows.

$$ \begin{align*} D^\alpha &:= \dfrac{\partial ^{|\alpha|} } {{\partial x_{1}}^{\alpha_{1}}\cdots {\partial x_{n}}^{\alpha_{n}}} \\ &= \left( \frac{ \partial }{ \partial x_{1}} \right)^{\alpha_{1}}\left( \frac{ \partial }{ \partial x_{2}} \right)^{\alpha_{2}}\cdots \left( \frac{ \partial }{ \partial x_{n}} \right)^{\alpha_{n}} \\ &= \partial^{\alpha_{1}}_{x_{1}}\cdots\partial^{\alpha_{n}}_{x_{n}} \end{align*} $$

Explanation

$P$ is a mapping between suitable function spaces $X$ and $Y$. Of course, elements of $X$ should be differentiable at least once.

$$ P : X \to Y $$

Symbol

The polynomial $p$, obtained by substituting variable $\xi = (\xi_{1}, \dots, \xi_{n})$ into $D$ of $(1)$, is called the total symbol of $P$.

$$ p(x,\xi) = \sum\limits_{\left| \alpha \right| \le m} a_{\alpha}(x) \xi^{\alpha},\qquad \xi^{\alpha} = \xi^{\alpha}_{1} \dots \xi^{\alpha}_{n} $$

Also, the following polynomial $\sigma (x, \xi)$ is called the principal symbol of $P$.

$$ \sigma (x, \xi) = \sum\limits_{\left| \alpha \right| = m} a_{\alpha}(x) \xi^{\alpha} $$

Explanation

The total symbol $p$ satisfies $(1)$ by definition, but conversely, a polynomial $p$ that satisfies $(1)$ can also be defined as the total symbol of $P$.

Relation with Fourier Transform

Due to the properties of the Fourier transform, the following holds.

$$ \mathcal{F}[Df] (\xi) = i\xi \mathcal{F}f (\xi) \implies Df (x) = \mathcal{F}^{-1} \left[ i\xi \mathcal{F}f (\xi) \right] (x) $$

Therefore, the following is obtained.

$$ \begin{align} P f(x) &= \mathcal{F}^{-1} \left[ p(\cdot, i\xi) \hat{f}(\xi) \right] (x) \nonumber \\ &= \dfrac{1}{(2 \pi)^{n}}\int_{\mathbb{R}^{n}} p(x,i\xi)\hat{f}(\xi) e^{i x \cdot \xi} d\xi \end{align} $$