Fourier Analysis, Distributions, and Constant-Coefficient Linear PDE

Abstract

Fourier analysis is perhaps the most important single tool in the study of linear partial differential equations. It serves in several ways, the most basic–and historically the first–being to give specific formulas for solutions to various linear PDE with constant coefficients, particularly the three classics, the Laplace, wave, and heat equations:

$$\Delta u = f,\quad\frac{{\partial }^{2}u} {\partial {t}^{2}} -\Delta u = f,\quad\frac{\partial u} {\partial t} -\Delta u = f,$$

(0.1)

with Δ = ∂²∕∂x₁² + ⋯ + ∂²∕∂x_n². The Fourier transform accomplishes this by transforming the operation of ∂∕∂x_j to the algebraic operation of multiplication by iξ_j. Thus the (0.1) are transformed to algebraic equations and to ODE with parameters.

A The mighty Gaussian and the sublime gamma function

The Gaussian function ${e}^{-\vert x{\vert }^{2} }$ on ${\mathbb{R}}^{n}$ is an object whose study yields many wonderful identities. We will use the identity

$${\int \limits _{{\mathbb{R}}^{n}}}{e}^{-\vert x{\vert }^{2} }dx = {\pi }^{n/2},$$

(A.1)

which was established in (3.18), to compute the area A_{n − 1} of the unit sphere S^{n − 1} in ${\mathbb{R}}^{n}$. This computation will bring in Euler’s gamma function, and other results will flow from this. Switching to polar coordinates for the right side of (A.1), we have

$$\begin{array}{rlrlrl}{\pi }^{n/2}& = {A}_{ n-1}{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-{r}^{2} }\ {r}^{n-1}\ dr& &\cr & =\frac{1} {2}{A}_{n-1}{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-t}\ {t}^{n/2-1}\mathit{dt}&\cr & =\frac{1} {2}{A}_{n-1}\Gamma {\Bigl (\frac{n} {2}\Bigr)},&\cr\end{array}$$

(A.2)

where the gamma function is defined by

$$\Gamma (z) ={\int \nolimits\nolimits }_{0}^{\infty }{e}^{-t}\ {t}^{z-1}\mathit{dt},$$

(A.3)

for Re z > 0. Thus we have the formula

$${A}_{n-1} =\frac{2{\pi }^{n/2}} {\Gamma\left (\frac{1} {2}n\right)}.$$

(A.4)

To be satisfied with this, we need an explicit evaluation of Γ(n∕2). This can be obtained from Γ(1∕2) and Γ(1) via the following identity:

$$\begin{array}{rlrlrl}\Gamma (z + 1)& ={\int \nolimits\nolimits }_{0}^{\infty }{e}^{-t}\ {t}^{z}\ \mathit{dt}& &\cr & = -{\int \nolimits\nolimits }_{0}^{\infty }\frac{d} {dt}\left ({e}^{-t}\right)\ {t}^{z}\ \mathit{dt}&\cr & = z\Gamma (z),&\cr\end{array}$$

(A.5)

for Re z > 0, where we used integration by parts. The definition (A.3) clearly gives

$$\Gamma (1) = 1.$$

(A.6)

Thus, for any integer k ≥ 1,

$$\Gamma (k) = (k - 1)\Gamma (k - 1) =\cdots = (k - 1)!\,.$$

(A.7)

Note that, for n = 2, we have A₁ = 2π∕Γ(1), so (A.6) agrees with the fact that the circumference of the unit circle is 2π (which, of course, figured into the proof of (3.18), via (3.20). In case n = 1, we have A₀ = 2, which by (A.4) is equal to 2π^1∕2∕Γ(1∕2), so

$$\Gamma\left (\frac{1} {2}\right) = {\pi }^{1/2}.$$

(A.8)

Again using (A.5), we see that, when k ≥ 1 is an integer,

$$\begin{array}{rlrlrl}\Gamma\left (k +\frac{1} {2}\right)& =\left (k -\frac{1} {2}\right)\Gamma\left (k -\frac{1} {2}\right) =\cdots & &\cr & =\left (k -\frac{1} {2}\right)\left (k -\frac{3} {2}\right)\cdots\left (\frac{1} {2}\right)\Gamma\left (\frac{1} {2}\right)&\cr & = {\pi }^{1/2}\left (k -\frac{1} {2}\right)\left (k -\frac{3} {2}\right)\cdots\left (\frac{1} {2}\right).&\cr\end{array}$$

(A.9)

In particular, Γ(3∕2) = (1∕2)Γ(1∕2) = π^1∕2∕2, so A₂ = 2π^3∕2∕(π^1∕2∕2) = 4π,which agrees with the well known formula for the area of the unit sphere in ${\mathbb{R}}^{3}$.

Note that while Γ(z) defined by (A.3) is a priori holomorphic for Re z positive, the equation (A.5) shows that Γ(z) has a meromorphic extension to the entire complex plane, with simple poles at $z = 0,-1,-2,\ldots\,$. It turns out that Γ(z) has no zeros, so 1∕Γ(z) is an entire analytic function. This is a consequence of the identity

$$\Gamma (z)\Gamma (1 - z) =\frac{\pi } {\sin\pi z},$$

(A.10)

which we now establish. From (A.4) we have (for 0 < Re z < 1)

$$\begin{array}{rlrlrl}\Gamma (z)\Gamma (1 - z)& ={\int \nolimits\nolimits }_{0}^{\infty }{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-(s+t)}{s}^{-z}{t}^{z-1}\mathit{ds}\ \mathit{dt}& &\cr & ={\int \nolimits\nolimits }_{0}^{\infty }{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-u}{v}^{z-1}{(1 + v)}^{-1}\mathit{du}\ \mathit{dv}&\cr & ={\int \nolimits\nolimits }_{0}^{\infty }{(1 + v)}^{-1}{v}^{z-1}\ \mathit{dv},&\cr\end{array}$$

(A.11)

where we have used the change of variables u = s + t, v = t∕s. With v = e^x, the last integral is

$${\int \nolimits\nolimits }_{-\infty }^{\infty }{\left (1 + {e}^{x}\right)}^{-1}\ {e}^{\mathit{xz}}\mathit{dx},$$

(A.12)

which is holomorphic for 0 < Re z < 1, and we want to show that it is equal to the right side of (A.10) on this strip. It suffices to prove identity on the line $z = 1/2 + i\xi, \xi \in\mathbb{R}$; then (A.12) is equal to the Fourier integral

$${\int \nolimits\nolimits }_{-\infty }^{\infty }{\left (2\cosh\frac{x} {2}\right)}^{-1}\ {e}^{\mathit{ix}\xi }\mathit{dx}.$$

(A.13)

To evaluate this, shift the contour of integration from the real line to the line Im x = −2π. There is a pole of the integrand at x = − πi, and we have (A.13) equal to

$$-{\int \nolimits\nolimits }_{-\infty }^{\infty }{\left (2\cosh\frac{x} {2}\right)}^{-1}{e}^{2\pi\xi }{e}^{ix\xi }\mathit{dx} -\text{ Residue}\cdot (2\pi i).$$

(A.14)

Consequently, (A.13) is equal to

$$-2\pi i\frac{\text{ Residue}} {1 + {e}^{2\pi\xi }} =\frac{\pi } {\cosh\pi\xi },$$

(A.15)

and since π/ sinπ(1∕2 + iξ) = π∕coshπξ, the demonstration of (A.10) is complete.

The integral (A.3) and also the last integral in (A.11) are special cases of the Mellin transform:

$$\mathcal{M}f(z) ={\int \nolimits\nolimits }_{0}^{\infty }f(t){t}^{z-1}\mathit{dt}.$$

(A.16)

If we evaluate this on the imaginary axis:

$${\mathcal{M}}^{\#}f(s) ={\int \nolimits\nolimits }_{0}^{\infty }f(t){t}^{is-1}\mathit{dt},$$

(A.17)

given appropriate growth restrictions on f, this is related to the Fourier transform by a change of variable:

$${\mathcal{M}}^{\#}f(s) ={\int \nolimits\nolimits }_{-\infty }^{\infty }f({e}^{x}){e}^{\mathit{isx}}\mathit{dx}.$$

(A.18)

The Fourier inversion formula and Plancherel formula imply

$$f(r) = {(2\pi)}^{-1}{\int \nolimits\nolimits }_{-\infty }^{\infty }({\mathcal{M}}^{\#}f)(s){r}^{-\mathit{is}}\mathit{ds}$$

(A.19)

and

$${\int \nolimits\nolimits }_{-\infty }^{\infty }\vert {\mathcal{M}}^{\#}f(s){\vert }^{2}\mathit{ds} = (2\pi){\int \nolimits\nolimits }_{0}^{\infty }\vert f(r){\vert }^{2}{r}^{-1}\ dr.$$

(A.20)

In some cases, as seen above, one evaluates ℳf(z) on a vertical line other than the imaginary axis, which introduces only a slight wrinkle.

An important identity for the gamma function follows from taking the Mellin transform with respect to y of both sides of the subordination identity

$${e}^{-yA} =\frac{1} {2}y{\pi }^{-1/2}{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-{y}^{2}/4t }{e}^{-t{A}^{2} }{t}^{-3/2}\mathit{dt}$$

(A.21)

(if y > 0, A > 0), established in §5; see (5.22). The Mellin transform of the left side is clearly Γ(z)A^{− z}. The Mellin transform of the right side is a double integral, which is readily converted to a product of two integrals, each defining gamma functions. After a few changes of variables, there results the identity

$${\pi }^{1/2}\Gamma (2z) = {2}^{2z-1}\Gamma (z)\Gamma\left (z +\frac{1} {2}\right),$$

(A.22)

known as the duplication formula for the gamma function. In view of the uniqueness of Mellin transforms, following from (A.18) and (A.19), the identity (A.22) conversely implies (A.21). In fact, (A.22) was obtained first (by Legendre) and this argument produces one of the standard proofs of the subordination identity (A.21).

There is one further identity, which, together with (A.5), (A.10), and (A.22), completes the list of the basic elementary identities for the gamma function. Namely, if the beta function is defined by

$$B(x,y) ={\int \nolimits\nolimits }_{0}^{1}{s}^{x-1}{(1 - s)}^{y-1}\mathit{ds} ={\int \nolimits\nolimits }_{0}^{1}{(1 + u)}^{-x-y}{u}^{x-1}\mathit{du}$$

(A.23)

(with u = s∕(1 − s)), then

$$B(x,y) =\frac{\Gamma (x)\Gamma (y)} {\Gamma (x + y)}.$$

(A.24)

To prove this, note that since

$$\Gamma (z){p}^{-z} ={\int \nolimits\nolimits }_{0}^{\infty }{e}^{-pt}{t}^{z-1}\mathit{dt},$$

(A.25)

we have

$${(1 + u)}^{-x-y} =\frac{1} {\Gamma (x + y)}{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-(1+u)t}{t}^{x+y-1}\mathit{dt},$$

so

$$\begin{array}{rlrlrl}B(x,y)& =\frac{1} {\Gamma (x + y)}{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-t}{t}^{x+y-1}{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-ut}{u}^{x-1}\mathit{du}\ \mathit{dt}& &\cr & =\frac{\Gamma (x)} {\Gamma (x + y)}{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-t}{t}^{y-1}\ \mathit{dt}&\cr & =\frac{\Gamma (x)\Gamma (y)} {\Gamma (x + y)},&\cr\end{array}$$

as asserted.

The four basic identities proved above are the workhorses for most applications involving gamma functions, but fundamental insight is provided by the identities (A.27) and (A.31) below. First, since 0 ≤ e^{− t} − (1 − t∕n)ⁿ ≤ e^{− t} ⋅t²∕n for 0 ≤ t ≤ n, we have, for Re z > 0,

$$\begin{array}{rlrlrl}\Gamma (z)& ={\int \nolimits\nolimits }_{0}^{\infty }{e}^{-t}{t}^{z-1}\ \mathit{dt}& &\cr & {=\lim \limits_{n\rightarrow\infty }}{\int \nolimits\nolimits }_{0}^{n}{\Bigl (1 -{\frac{t} {n}}\Bigr)}^{n}{t}^{z-1}\ \mathit{dt}&\cr & {=\lim \limits_{n\rightarrow\infty }}{n}^{z}{\int \nolimits\nolimits }_{0}^{1}{(1 - s)}^{n}{s}^{z-1}\mathit{ds}.&\cr\end{array}$$

(A.26)

Repeatedly integrating by parts gives

$$\Gamma (z) {=\lim \limits_{n\rightarrow\infty }}{n}^{z}\frac{n(n - 1)\cdots 1} {z(z + 1)\cdots (z + n - 1)}{\int \nolimits\nolimits }_{0}^{1}{s}^{z+n-1}\mathit{ds},$$

which yields the following result of Euler:

$$\Gamma (z) {=\lim \limits_{n\rightarrow\infty }}{n}^{z}\frac{1\cdot 2\cdots n} {z(z + 1)\cdots (z + n)}.$$

(A.27)

Using the identity (A.5), analytically continuing Γ(z), we have (A.27) for all z, other than $0,-1,-2,\ldots$. We can rewrite (A.27) as

$$\Gamma (z) {=\lim \limits_{n\rightarrow\infty }}{n}^{z}\ {z}^{-1}{(1 + z)}^{-1}{\left (1 +\frac{z} {2}\right)}^{-1}\cdots {\left (1 +\frac{z} {n}\right)}^{-1}.$$

(A.28)

If we denote by γ Euler’s constant:

$$\gamma\ {=\lim \limits_{n\rightarrow\infty }}\left (1 +\frac{1} {2} +\cdots +\frac{1} {n} -\log n\right),$$

(A.29)

then (A.28) is equivalent to

$$\Gamma (z) {=\lim \limits_{n\rightarrow\infty }}{e}^{-\gamma z}{e}^{z(1+1/2+\cdots +1/n)}{z}^{-1}{(1 + z)}^{-1}{\left (1 +\frac{z} {2}\right)}^{-1}\cdots {\left (1 +\frac{z} {n}\right)}^{-1},$$

(A.30)

that is, to the Euler product expansion

$$\frac{1} {\Gamma (z)} = z\ {e}^{\gamma z}{\prod \limits_{n=1}^{\infty }}\left (1 +\frac{z} {n}\right){e}^{-z/n}.$$

(A.31)

It follows that the entire analytic function 1∕Γ(z)Γ(− z) has the product expansion

$$\frac{1} {\Gamma (z)\Gamma (-z)} = -{z}^{2}{\prod \limits_{n=1}^{\infty }}\left (1 -\frac{{z}^{2}} {{n}^{2}}\right).$$

(A.32)

Since Γ(1 − z) = − zΓ(− z), by virtue of (A.10) this last identity is equivalent to the Euler product expansion

$$\sin\pi z =\pi z{\prod \limits_{n=1}^{\infty }}\left (1 -\frac{{z}^{2}} {{n}^{2}}\right).$$

(A.33)

It is quite easy to deduce the formula (A.5) from the Euler product expansion (A.31). Also, to deduce the duplication formula (A.22) from the Euler product formula is a fairly straightforward exercise.

Finally, we derive Stirling’s formula, for the asymptotic behavior of Γ(z) as z → + ∞. The approach uses the Laplace asymptotic method, which has many other applications. We begin by setting t = sz and then s = e^y in the integral formula (A.3), obtaining

$$\begin{array}{rlrlrl}\Gamma (z)& = {z}^{z}{\int \nolimits\nolimits }_{0}^{\infty }{e}^{-z(s-\log s)}{s}^{-1}ds& &\cr & = {z}^{z}{\int \nolimits\nolimits }_{-\infty }^{\infty }{e}^{-z({e}^{y}-y) }\,\mathit{dy}.&\cr\end{array}$$

(A.34)

The last integral is of the form

$${\int \nolimits\nolimits }_{-\infty }^{\infty }{e}^{-z\varphi (y)}\,\mathit{dy},$$

(A.35)

where φ(y) = e^y − y has a nondegenerate minimum at y = 0; φ(0) = 1, φ′(0) = 0, φ′(0) = 1. If we write 1=A(y) + B(y), A ∈ C₀^∞((−2,2)), A(y) = 1 for |y| ≤ 1, then the integral (A.35) is readily seen to be

$${\int \nolimits\nolimits }_{-\infty }^{\infty }A(y){e}^{-z\varphi (y)}dy + O({e}^{-(1+1/e)z}).$$

(A.36)

We can make a smooth change of variable x = ξ(y) such that ξ(y) = y + O(y²), φ(y) = 1 + x²∕2, and the integral in (A.36) becomes

$${e}^{-z}{\int \nolimits\nolimits }_{-\infty }^{\infty }{A}_{ 1}(x){e}^{-z{x}^{2}/2 }\,\mathit{dx},$$

(A.37)

where ${A}_{1}\in{C}_{0}^{\infty }(\mathbb{R}),\ {A}_{1}(0) = 1$, and it is easy to see that, as z → + ∞,

$${\int \nolimits\nolimits }_{-\infty }^{\infty }{A}_{ 1}(x){e}^{-z{x}^{2}/2 }\,dx\sim {\left (\frac{2\pi } {z}\right)}^{1/2}\left [1 +\frac{{a}_{1}} {z} +\cdots\,\right ].$$

(A.38)

In fact, if z = 1∕2t, then (A.38) is equal to (4πt)^1∕2 u(t,0), where u(t, x) solves the heat equation, u_t − u_xx = 0, u(0, x) = A₁(x). Returning to (A.34), we have Stirling’s formula:

$$\Gamma(z) =\left (\frac{2\pi }/ {z}\right)^{1/2}\ {z}^{z}\ {e}^{-z}\left [1 + O({z}^{-1})\right ].$$

(A.39)

Since n!=Γ(n + 1), we have in particular that

$$n! = {(2\pi n)}^{1/2}\ {n}^{n}\ {e}^{-n}\left [1 + O({n}^{-1})\right ]$$

(A.40)

as n → ∞.

Regarding this approach to the Laplace asymptotic method, compare the derivation of the stationary phase method in Appendix B of Chap. 6.