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The Fourier transform of the heat kernel

Example

Assume Countable Choice and let $n \geq 1$ , and use the library's $2 π$ -normalised transform $F f (ξ) = \int_{R^{n}} f (x) e^{- 2 π i x \cdot ξ} d x$ of Fourier transform on complex L1 classes. Then for every $t > 0$ the heat kernel $Γ_{t}$ of The heat kernel on $R^{n}$ and its causal extension is the $L^{1}$ function with $F Γ_{t} (ξ) = e^{- 4 π^{2} t ∣ ξ ∣^{2}}, ξ \in R^{n} .$ In the unnormalised convention $G f (ξ) = \int_{R^{n}} f (x) e^{- i x \cdot ξ} d x$ the same computation reads $G Γ_{t} (ξ) = e^{- t ∣ ξ ∣^{2}}$ . The heat-flow multiplier $e^{- t ∣ ξ ∣^{2}}$ alone does not determine the forward normalization: Hunter uses $(2 π)^{- n} G$ , whose transform of $Γ_{t}$ is $(2 π)^{- n} e^{- t ∣ ξ ∣^{2}}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $t > 0$ and $ξ \in R^{n}$ .

[A1]

Countable Choice is the hypothesis of the Gaussian transform lemma below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $t > 0$ the heat kernel is $Γ_{t} (x) = (4 π t)^{- n / 2} \exp (- ∣ x ∣^{2} / (4 t)) > 0$ on $R^{n}$ (The heat kernel on $R^{n}$ and its causal extension).

[F2]

$Γ_{t} \in L^{1} (R^{n})$ with $∥ Γ_{t} ∥_{1} = 1$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F3]

For $f \in L^{1} (R^{n}; C)$ the $2 π$ -normalised Fourier transform is $F f (ξ) = \int_{R^{n}} f (x) e^{- 2 π i x \cdot ξ} d x$ , defined at every frequency (Fourier transform on complex L1 classes).

[F4]

Assume countable choice. For $n \geq 1$ , $s > 0$ and $ξ \in R^{n}$ , $F (e^{- π s ∣ x ∣^{2}}) (ξ) = s^{- n / 2} e^{- π ∣ ξ ∣^{2} / s}$ (Euclidean Gaussian transform with the 2π normalization).

Verification

technique · direct

1.1A1F1F2F3given

By [F1] and [F2] the kernel satisfies $Γ_{t} (x) = (4 π t)^{- n / 2} e^{- ∣ x ∣^{2} / (4 t)}$ with $Γ_{t} \in L^{1} (R^{n})$ , so its transform of [F3] is defined at every $ξ$ by the absolutely convergent integral $F Γ_{t} (ξ) = (4 π t)^{- n / 2} \int_{R^{n}} e^{- ∣ x ∣^{2} / (4 t)} e^{- 2 π i x \cdot ξ} d x$ .

2.1step 1.1givenalgebra

Writing $s : = 1 / (4 π t) > 0$ , the identity $- \frac{∣ x ∣^{2}}{4 t} = - π s ∣ x ∣^{2}$ holds, so $Γ_{t} (x) = (4 π t)^{- n / 2} e^{- π s ∣ x ∣^{2}}$ and $(4 π t)^{- n / 2} = s^{n / 2}$ .

3.1step 1.1step 2.1F3F4givenalgebra

Applying the Gaussian transform [F4] with this $s$ and factoring the constant out of the integral gives $F Γ_{t} (ξ) = (4 π t)^{- n / 2} s^{- n / 2} e^{- π ∣ ξ ∣^{2} / s} = e^{- π ∣ ξ ∣^{2} / s}$ , and substituting $s = 1 / (4 π t)$ yields $- π ∣ ξ ∣^{2} / s = - 4 π^{2} t ∣ ξ ∣^{2}$ , so $F Γ_{t} (ξ) = e^{- 4 π^{2} t ∣ ξ ∣^{2}}$ for every $ξ$ .

4.1step 3.1F3givenalgebra

For the unnormalised convention, the definition gives $G Γ_{t} (ξ) = \int Γ_{t} (x) e^{- i x \cdot ξ} d x = F Γ_{t} (ξ / (2 π))$ because $e^{- i x \cdot ξ} = e^{- 2 π i x \cdot (ξ / (2 π))}$ ; substituting $ξ / (2 π)$ in step 3.1 gives $G Γ_{t} (ξ) = e^{- 4 π^{2} t ∣ ξ / (2 π) ∣^{2}} = e^{- t ∣ ξ ∣^{2}}$ .

5.1step 1.1step 3.1step 4.1given∎

Steps 1.1, 2.1, 3.1 and 4.1 establish $F Γ_{t} (ξ) = e^{- 4 π^{2} t ∣ ξ ∣^{2}}$ in the normalisation of [F3] and $G Γ_{t} (ξ) = e^{- t ∣ ξ ∣^{2}}$ in the unnormalised convention, which is the whole example.

Depends on

Used by

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Dependency tree · two levels

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Sources

John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2025 archived author manuscript, AMS Graduate Studies in Mathematics) (standard reference, not scraped)
Victor Ivrii, Partial Differential Equations (University of Toronto, 2018, CC BY-SA) (standard reference, not scraped)