求分布意义下v(x)=sign cos(πx)的一阶与二阶导数

Let’s break this down step by step—distributional derivatives can feel counterintuitive at first, especially with piecewise constant functions like this one. First, let’s map out what $v(x)$ behaves like, focusing on your interval of interest $(-1,1)$:

• When $\cos(\pi x) > 0$ (i.e., $x \in (-1/2, 1/2)$), $v(x) = 1$
• When $\cos(\pi x) < 0$ (i.e., $x \in (-1, -1/2) \cup (1/2, 1)$), $v(x) = -1$
• At points $x = k + 1/2$ (for any integer $k$), $\cos(\pi x) = 0$, so $v(x)$ has jump discontinuities here: jumping from $-1$ to $1$ at $x=-1/2$, and $1$ to $-1$ at $x=1/2$ in the $(-1,1)$ interval.

Distributional derivatives follow a core definition: for any smooth, compactly supported test function $\phi \in C_c^\infty(\mathbb{R})$, the derivative $v'$ satisfies:
$$\langle v', \phi \rangle = -\langle v, \phi' \rangle$$

First Derivative $v'(x)$ #

Let’s compute the integral for $\langle v', \phi \rangle$ by splitting it at the jump points of $v(x)$:
$$
\begin{align*}
\langle v', \phi \rangle &= -\int_{-\infty}^\infty v(x)\phi'(x)dx \
&= -\left( \int_{-\infty}^{-1/2} (-1)\phi'(x)dx + \int_{-1/2}^{1/2} 1 \cdot \phi'(x)dx + \int_{1/2}^\infty (-1)\phi'(x)dx \right)
\end{align*}
$$

Use the Fundamental Theorem of Calculus on each integral (remember $\phi$ vanishes at $\pm\infty$ because it’s compactly supported):

• $\int_{-\infty}^{-1/2} (-1)\phi'(x)dx = -\left( \phi(-1/2) - \lim_{x\to-\infty}\phi(x) \right) = -\phi(-1/2)$
• $\int_{-1/2}^{1/2} \phi'(x)dx = \phi(1/2) - \phi(-1/2)$
• $\int_{1/2}^\infty (-1)\phi'(x)dx = -\left( \lim_{x\to\infty}\phi(x) - \phi(1/2) \right) = \phi(1/2)$

Substitute these back into the expression:
$$
\begin{align*}
\langle v', \phi \rangle &= -\left( -\phi(-1/2) + \phi(1/2) - \phi(-1/2) + \phi(1/2) \right) \
&= -\left( -2\phi(-1/2) + 2\phi(1/2) \right) \
&= 2\phi(-1/2) - 2\phi(1/2)
\end{align*}
$$

Recall that the Dirac delta distribution $\delta(x - a)$ satisfies $\langle \delta(x - a), \phi \rangle = \phi(a)$. This means our expression matches:
$$\langle v', \phi \rangle = \langle 2\delta(x + 1/2) - 2\delta(x - 1/2), \phi \rangle$$

So in the distributional sense:
$$v'(x) = 2\delta\left(x + \frac{1}{2}\right) - 2\delta\left(x - \frac{1}{2}\right)$$

(Note: For the entire real line, since $v(x)$ is periodic with period 2, the derivative extends to a sum over all jump points: $v'(x) = 2\sum_{k\in\mathbb{Z}} \delta\left(x - (k + 1/2)\right) - 2\sum_{k\in\mathbb{Z}} \delta\left(x - (k - 1/2)\right)$, but only the two deltas above matter in $(-1,1)$.)

Second Derivative $v''(x)$ #

To find the second derivative, we apply the distributional derivative definition again to $v'$:
$$\langle v'', \phi \rangle = -\langle v', \phi' \rangle$$

We already know $\langle v', \phi' \rangle = 2\phi'(-1/2) - 2\phi'(1/2)$, so:
$$
\begin{align*}
\langle v'', \phi \rangle &= -\left( 2\phi'(-1/2) - 2\phi'(1/2) \right) \
&= -2\phi'(-1/2) + 2\phi'(1/2)
\end{align*}
$$

The derivative of the delta distribution $\delta'(x - a)$ has the property $\langle \delta'(x - a), \phi \rangle = -\phi'(a)$. Rearranging this gives $\phi'(a) = -\langle \delta'(x - a), \phi \rangle$. Substitute this in:

• $-2\phi'(-1/2) = -2\left( -\langle \delta'(x + 1/2), \phi \rangle \right) = 2\langle \delta'(x + 1/2), \phi \rangle$
• $2\phi'(1/2) = 2\left( -\langle \delta'(x - 1/2), \phi \rangle \right) = -2\langle \delta'(x - 1/2), \phi \rangle$

Combining these results:
$$\langle v'', \phi \rangle = \langle 2\delta'(x + 1/2) - 2\delta'(x - 1/2), \phi \rangle$$

So the second distributional derivative is:
$$v''(x) = 2\delta'\left(x + \frac{1}{2}\right) - 2\delta'\left(x - \frac{1}{2}\right)$$

(Extending to the entire real line, this becomes a periodic sum of delta derivatives at each jump point.)

Key Takeaway #

For piecewise constant functions, their distributional first derivatives are sums of delta distributions weighted by the size of the jump at each discontinuity. The second derivatives are then sums of delta derivatives, since differentiating a delta distribution gives a delta prime distribution.

内容的提问来源于stack exchange，提问作者muserock92