Distributions & the Dirac Delta
Distributions & the Dirac Delta
Distributions (generalized functions) extend the notion of function to objects like the Dirac delta $\delta(x)$ that are not ordinary functions but act as linear functionals on test functions. The framework of Schwartz distributions makes it rigorous to differentiate any locally integrable function, and Fourier transforms of distributions extend to all tempered distributions. This gives meaning to $\hat{\delta}=1$, to the Fourier transform of $\cos(\omega_0 t)$, and to the Green's functions of PDEs.
Dirac Delta & Distributions
A distribution $T$ is a continuous linear functional on test functions $\phi\in\mathcal{D}(\mathbb{R})$ (smooth compactly supported). The Dirac delta: $\langle\delta,\phi\rangle=\phi(0)$. Delta is not a function: $\int\delta(x)\,dx$ is defined by $\langle\delta,1\rangle=1$, not by pointwise values. Differentiation of distributions: $\langle T',\phi\rangle=-\langle T,\phi'\rangle$ (integration by parts transposed). So $\delta'$ satisfies $\langle\delta',\phi\rangle=-\phi'(0)$. Every locally integrable function defines a distribution: $\langle T_f,\phi\rangle=\int f\phi\,dx$.
Fourier Transform of Distributions
For tempered distributions $T\in\mathcal{S}'(\mathbb{R})$, $\hat{T}$ is defined by $\langle\hat{T},\phi\rangle=\langle T,\hat{\phi}\rangle$. Key pairs: $\mathcal{F}\{\delta\}=1$ (flat spectrum); $\mathcal{F}\{1\}=\delta$ (duality); $\mathcal{F}\{e^{2\pi i\xi_0 x}\}=\delta(\xi-\xi_0)$ (pure frequency); $\mathcal{F}\{\cos(2\pi\xi_0 x)\}=\frac{1}{2}[\delta(\xi-\xi_0)+\delta(\xi+\xi_0)]$; $\mathcal{F}\{\text{comb}_T\}=\frac{1}{T}\text{comb}_{1/T}$ (Poisson summation formula: $\sum\delta(x-nT)\leftrightarrow\frac{1}{T}\sum\delta(\xi-k/T)$).
Example 1
Verify $\delta$ is the distributional derivative of the Heaviside function $H(x)=1_{x>0}$.
Solution: $\langle H',\phi\rangle=-\langle H,\phi'\rangle=-\int_0^\infty\phi'(x)\,dx=-[\phi(x)]_0^\infty=\phi(0)=\langle\delta,\phi\rangle$. So $H'=\delta$ in the distributional sense, consistent with the intuition that $H$ jumps from 0 to 1 at $x=0$.
Example 2
Solve $y''+\omega_0^2 y=\delta(t)$ using the Fourier transform.
Solution: $\mathcal{F}$: $(-\omega^2+\omega_0^2)\hat{y}=1$, so $\hat{y}(\omega)=\frac{1}{\omega_0^2-\omega^2}=-\frac{1}{(\omega-\omega_0)(\omega+\omega_0)}$. Partial fractions + inverse Fourier (careful with causal Green's function): $G(t)=\frac{\sin(\omega_0 t)}{\omega_0}H(t)$. This is the causal Green's function of the harmonic oscillator, describing the impulse response.
Practice
- Show that $x\delta'(x)=-\delta(x)$ as distributions.
- Use the Poisson summation formula $\sum_n f(n)=\sum_k\hat{f}(k)$ to prove the periodization formula.
- Find the distributional Fourier series of the Dirac comb $\text{III}(x)=\sum_{n\in\mathbb{Z}}\delta(x-n)$.
- Show $p.v.(1/x)$ (principal value distribution) has Fourier transform $-i\pi\text{sgn}(\xi)$.
Show Answer Key
1. For test function $\phi$: $\langle x\delta'(x),\phi\rangle=-\langle\delta',x\phi\rangle=\langle\delta,(x\phi)'\rangle=\langle\delta,\phi+x\phi'\rangle=\phi(0)$. But $\langle-\delta,\phi\rangle=-\phi(0)$. Check: $(x\phi)'|_{x=0}=\phi(0)$, so $\langle x\delta',\phi\rangle=-\phi(0)=\langle-\delta,\phi\rangle$. ✓
2. Poisson summation: $\sum_{n\in\mathbb{Z}}f(n)=\sum_{k\in\mathbb{Z}}\hat{f}(k)$. Apply to the periodization $f_P(x)=\sum_n f(x+n)$. Its Fourier coefficients are $\hat{f}(k)$, so $f_P(x)=\sum_k\hat{f}(k)e^{2\pi ikx}$. Setting $x=0$ gives the formula.
3. $\widehat{\text{III}}(\xi)=\sum_n e^{-2\pi in\xi}=\sum_k\delta(\xi-k)=\text{III}(\xi)$. The Dirac comb is its own Fourier transform.
4. The Hilbert transform of $1/(\pi x)$ has $\widehat{\text{p.v.}(1/x)}(\xi)=-i\pi\text{sgn}(\xi)$. Proof: compute $\lim_{\epsilon\to0}\int_{|x|>\epsilon}\frac{e^{-2\pi i\xi x}}{\pi x}dx$ using a contour deformation around $x=0$.