Bargmann transform and generalized heat Cauchy problems

TL;DR

This paper explicitly solves certain heat-type Cauchy problems linked to generalized Dirac, Euler, and Harmonic oscillator operators using the Bargmann transform, providing new analytical solutions.

Contribution

It introduces explicit solutions to generalized heat Cauchy problems using the Bargmann transform, expanding analytical methods for these operators.

Findings

Explicit solutions for heat Cauchy problems with Dirac, Euler, and Harmonic oscillator operators

Application of the Bargmann transform as a key analytical tool

Enhanced understanding of heat equations related to generalized operators

Abstract

In this article we solve explicitly some Cauchy problems of the heat type attached to the generalized real and complex Dirac, Euler and Harmonic oscillator operators. Our principal tool is the Bargmann transform.

Equations

$$\displaystyle\qquad\left\{\begin{array}[]{cc}\displaystyle\Delta u(t,x)=\frac{\partial}{\partial t}u(t,x);\quad(t,x)\in\mathbb{R}^{\ast}_{+}\times\mathbb{R}^{n},\vspace*{0.3cm}\\ u(0,x)=u_{0}(x);\quad u_{0}\in L^{2}(\mathbb{R}^{n}).\end{array}\right.$$

$$\displaystyle\qquad\left\{\begin{array}[]{cc}\displaystyle D_{z}^{a}U(t,z)=\frac{\partial}{\partial t}U(t,z);\quad(t,z)\in\mathbb{R}^{\ast}_{+}\times\mathbb{C},\vspace*{0.3cm}\\ U(0,z)=U_{0}(z);\quad U_{0}\in F_{\frac{a}{2}}^{2}.\end{array}\right.$$

$$\displaystyle\qquad\left\{\begin{array}[]{cc}\displaystyle d^{a}_{x}u(t,x)=\frac{\partial}{\partial t}u(t,x);\quad(t,x)\in\mathbb{R}^{\ast}_{+}\times\mathbb{R},\vspace*{0.3cm}\\ u(0,x)=u_{0}(x);\quad u_{0}\in L^{2}(\mathbb{R}).\end{array}\right.$$

$$D_z^a=\frac{1}{a}\frac{\partial }{\partial z}+\frac{z}{2},d_x^a=\frac{\partial }{\partial x}-ax.$$

$$\displaystyle\qquad\left\{\begin{array}[]{cc}\displaystyle e^{a}_{x}v(t,x)=\frac{\partial}{\partial t}v(t,x);\quad(t,x)\in\mathbb{R}^{\ast}_{+}\times\mathbb{R},\vspace*{0.3cm}\\ v(0,x)=v_{0}(x);\quad v_{0}\in L^{2}(\mathbb{R}).\end{array}\right.$$

$$\displaystyle\qquad\left\{\begin{array}[]{cc}\displaystyle E^{a}_{z}Y(t,w)=\frac{\partial}{\partial t}Y(t,w);\quad(t,w)\in\mathbb{R}^{\ast}_{+}\times\mathbb{C},\vspace*{0.3cm}\\ Y(0,w)=Y_{0}(w);\quad Y_{0}\in F_{\frac{a}{2}}^{2}.\end{array}\right.$$

$$E_z^a=-2az\frac{\partial }{\partial z}-a,e_x^a=ax\frac{\partial }{\partial x}.$$

$$\displaystyle\qquad\left\{\begin{array}[]{cc}\displaystyle h^{a}_{x}y(t,x)=\frac{\partial}{\partial t}y(t,x);\quad(t,x)\in\mathbb{R}^{\ast}_{+}\times\mathbb{C},\vspace*{0.3cm}\\ y(0,x)=y_{0}(x);\quad y_{0}\in L^{2}(\mathbb{R}).\end{array}\right.$$

$$h_x^a=\frac{{\partial }^2}{\partial x^2}-a^2{x}^2.$$

$$\displaystyle\qquad\left\{\begin{array}[]{cc}\displaystyle H^{a}_{z}V(t,z)=\frac{\partial}{\partial t}V(t,z);\quad(t,z)\in\mathbb{R}^{\ast}_{+}\times\mathbb{C},\vspace*{0.3cm}\\ V(0,z)=V_{0}(z);\quad V_{0}\in F_{\frac{a}{2}}^{2}.\end{array}\right.$$

$$H_z^a=\frac{{\partial }^2}{\partial z^2}-\frac{a^2{z}^2}{4}-\frac{a}{2},h_x^a=\frac{{\partial }^2}{\partial x^2}-a^2{x}^2.$$

$$d{\lambda }_a(z)=\frac{a}{\pi }{e}^{-a|z{|}^2}dz.$$

$$⟨f,g⟩={\int }_{\mathbb{C}}f(z)\overline{g(z)}d{\lambda }_a(z).$$

$$\forall z\in \mathbb{C},\forall f\in L^2(\mathbb{R}),\left[Bf\right](z)=2^{\frac{1}{4}}{\int }_{-\infty }^{\infty }f(x)e^{2\pi xz-\pi x^2-\frac{\pi z^2}{2}}dx.$$

$$\left[B_{a}f\right](z)={\left(\frac{2a}{\pi }\right)}^{\frac{1}{4}}{\int }_{-\infty }^{\infty }f(x)e^{2axz-a{x}^2-\frac{a{z}^2}{2}}dx.$$

$$\left[B_a^{-1}f\right](x)={\left(\frac{2a}{\pi }\right)}^{\frac{1}{4}}{\int }_{\mathbb{C}}f(z)e^{2ax\overline{z}-a{x}^2-\frac{a}{2}{\overline{z}}^2}d{\lambda }_a(z).$$

$$\left[B_{\frac{a}{2}}(h_x^{a}f)\right](z)=E_z^a\left[B_{\frac{a}{2}}f\right](z).$$

$$\left[B_{\frac{a}{2}}(e_x^{a}f)\right](z)=H_z^a\left[B_{\frac{a}{2}}f\right](z).$$

$$\left[B_{\frac{a}{2}}\left(e^{a\frac{x^2}{2}}{e}^{-b(x-s{)}^2}\right)\right](z)={\left(\frac{a}{\pi }\right)}^{\frac{1}{4}}{\int }_{-\infty }^{\infty }{e}^{-b(x-s{)}^2}{e}^{axz-\frac{a{z}^2}{4}}dx$$

$$={\left(\frac{a}{\pi }\right)}^{\frac{1}{4}}{\int }_{-\infty }^{\infty }{e}^{-b(x-s-\frac{az}{2b}{)}^2}{e}^{asz}{e}^{\frac{a^2}{4b}{z}^2}{e}^{-\frac{a}{4}{z}^2}dx.$$

$${\int }_{-\infty }^{\infty }{e}^{-b(x-z{)}^2}dx=\sqrt{\frac{\pi }{b}},$$

$$\left[B_{\frac{a}{2}}\left(e^{a\frac{x^2}{2}}{e}^{-b(x-s{)}^2}\right)\right](z)={\left(\frac{a}{\pi }\right)}^{\frac{1}{4}}\sqrt{\frac{\pi }{b}}{e}^{asz}{e}^{\frac{a}{4}(\frac{a}{b}-1)z^2}dx.$$

$$\left[F_{r}f\right](x)=\sqrt{\frac{ar}{\pi }}{\int }_{\mathbb{R}}f(t)e^{iarxt}dt,\text{ for all }a,r>0\text{ and }f\in L^2(\mathbb{R}),$$

$$\left[h_{r}f\right](x)=f(rx),\text{ for all }f\in L^2(\mathbb{R})\text{ and }x\in \mathbb{R}.$$

$$U(t,z)=e^{\frac{zt}{2}}{e}^{\frac{t^2}{4a}}{U}_0\left(z+\frac{t}{a}\right).$$

$$\qquad\left\{\begin{array}[]{cc}\displaystyle B^{-1}_{\frac{a}{2}}\left(\left(\frac{1}{a}\frac{\partial}{\partial z}+\frac{z}{2}\right)U(t,z)\right)=\frac{\partial}{\partial t}B^{-1}_{\frac{a}{2}}(U(t,z));\quad(t,z)\in\mathbb{R}^{\ast}_{+}\times\mathbb{C},\vspace*{0.3cm}\\ B^{-1}_{\frac{a}{2}}(U(0,z))=B^{-1}_{\frac{a}{2}}U_{0};\quad U_{0}\in F_{\frac{a}{2}}^{2}.\end{array}\right.$$

$$\qquad\left\{\begin{array}[]{cc}\displaystyle x\,\left[B^{-1}_{\frac{a}{2}}U\right](t,x)=\frac{\partial}{\partial t}\left[B^{-1}_{\frac{a}{2}}U\right](t,x);\quad(t,x)\in\mathbb{R}^{\ast}_{+}\times\mathbb{R},\vspace*{0.3cm}\\ \left[B^{-1}_{\frac{a}{2}}U\right](0,x)=\left[B^{-1}_{\frac{a}{2}}U_{0}\right](x);\quad U_{0}\in F_{\frac{a}{2}}^{2}.\end{array}\right.$$

$$\left[B_{\frac{a}{2}}^{-1}U\right](t,x)=e^{tx}\left[B_{\frac{a}{2}}^{-1}{U}_0\right](x).$$

$$U(t,z)=\left[B_{\frac{a}{2}}\left(e^{tx}{B}_{\frac{a}{2}}^{-1}{U}_0\right)\right](z)$$

$$={\left(\frac{a}{\pi }\right)}^{\frac{1}{2}}{\int }_{\mathbb{R}}{e}^{axz-\frac{a}{2}{x}^2-\frac{a{z}^2}{4}}{e}^{tx}{\int }_{\mathbb{C}}{U}_0(v)e^{ax\overline{v}-\frac{a}{2}{x}^2-\frac{a}{4}{\overline{v}}^2}d{\lambda }_{\frac{a}{2}}(v)dx$$

Full text

Keywords : Bargmann transform, heat Cauchy problem, Dirac operator, Euler operator, harmonic oscillator.

Subject classification : Primary: 35C05 and 35C15; Secondary:42A38.

1 Introduction

The classical Cauchy problem for the heat equation is of the form

[TABLE]

where $\displaystyle\Delta=\sum_{j=1}^{n}\frac{\partial^{2}}{\partial x_{j}^{2}}$ is the classical Laplacian of $\mathbb{R}^{n}$. We can formally replace the operator $\Delta$ by any partial differential operator and speak about generalized heat Cauchy problem. If we replace $\Delta$ by the Dirac operator $\displaystyle\frac{\partial}{\partial x}$, we obtain the famous transport equation (see [6]).

In this work, we solve explicitly some type of generalized heat Cauchy problems. Our method is based on the Bargmann transform which is an isometry from $L^{2}(\mathbb{R})$ to the Fock space $F_{\frac{a}{2}}^{2}$ (see [11] ). The paper is outlined as follows

In section 2, old and new results concerning the Bargmann transform are given.

In section 3, we compute explicitly the exact solutions of the heat Cauchy problems associated to the generalized complex and real Dirac operators

[TABLE]

and

[TABLE]

where the generalized real and complex Dirac operators are given by

[TABLE]

In section 4, we solve explicitly the following heat Cauchy problems associated to the generalized real and complex Euler operators

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and

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where the generalized real and complex Euler operators are given by

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In section 5 the explicit solutions of the following heat Cauchy problem for the (real) harmonic oscillators is computed.

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[TABLE]

Section 6 is devoted to the following generalized heat Cauchy problem

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associated to the operator

[TABLE]

called here the complex harmonic oscillator.

Note that the wave Cauchy and Poisson problems associated to the real Dirac, Euler and Harmonic oscillator are considered in [9, 10].

2 Bargmann transform

For $a>0$, we define a Gaussian measure on $\mathbb{C}$ as follows

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The Fock space, denoted $F_{a}^{2}$, is the subspace of all entire functions in $L^{2}(\mathbb{C},d\lambda_{a})$ which is a Hilbert space with the inner product

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V. Bargmann [1, 2] and G. Folland [7] have defined a mapping $B$, from the space of square integrable functions $L^{2}(\mathbb{R})$ to the Fock space $F_{\pi}^{2}$, called Bargmann transform

[TABLE]

We use in this paper a parametrized form of the Bargmann transform given by K. Zhu [11] as follows

[TABLE]

This mapping is an isometry from $L^{2}(\mathbb{R})$ to $F_{a}^{2}$, its inverse is given by

[TABLE]

In what follows, we give some formulas involving Bargmann transform.

Lemma 2.1**.**

(Appendix 1 of [8]) For $a>0$, we have

$\displaystyle\left[B_{\frac{a}{2}}(x\,f)\right](z)=\left(\frac{1}{a}\,\frac{\partial}{\partial z}+\frac{z}{2}\right)\left[B_{\frac{a}{2}}f\right](z)$. 2. 2.

$\displaystyle\left[B_{\frac{a}{2}}\left(\frac{\partial}{\partial x}\,f\right)\right](z)=\left(\frac{\partial}{\partial z}-\frac{a}{2}z\right)\left[B_{\frac{a}{2}}f\right](z)$. 3. 3.

$\displaystyle\left[B_{\frac{a}{2}}\left(\left(\frac{\partial}{\partial x}\,-ax\right)\,f\right)\right](z)=-az\,\left[B_{\frac{a}{2}}f\right](z)$. 4. 4.

$\displaystyle\left[B_{\frac{a}{2}}\left(\left(\frac{\partial}{\partial x}\,+ax\right)\,f\right)\right](z)=2\,\frac{\partial}{\partial z}\left[B_{\frac{a}{2}}f\right](z)$.

The following proposition is a direct consequence of lemma 2.1.

Proposition 2.1**.**

Let $E_{x}^{a}$ and $h_{x}^{a}$ be respectively the complex Euler and real harmonic operators given in (1.17) and (1.25). Then, we have

[TABLE] 2. 2.

Let $e_{x}^{a}$ and $H_{z}^{a}$ be respectively the real Euler and complex harmonic operators given in (1.17) and (1.25). Then, we obtain

[TABLE]

The following lemma gives a new result about the Bargmann transform of the Gaussian function.

Lemma 2.2**.**

Let $a$, $b$, $c$ and $s$ be real numbers such that $a>0$ and $b>\frac{a}{2}$, then the following lemma holds

$\displaystyle\left[B_{\frac{a}{2}}\left(e^{a\frac{x^{2}}{2}}\,e^{-b(x-s)^{2}}\right)\right](z)=\left(\frac{a}{\pi}\right)^{\frac{1}{4}}\sqrt{\frac{\pi}{b}}\,e^{asz}\,e^{\frac{a}{4}\left(\frac{a}{b}-1\right)z^{2}}$. 2. 2.

$\left[B_{\frac{a}{2}}\left(e^{-cx^{2}}\right)\right](z)=\left(\frac{a}{\pi}\right)^{\frac{1}{4}}\sqrt{\frac{\pi}{c+\frac{a}{2}}}\,e^{\frac{a}{4}z^{2}\left(\frac{a-2c}{a+2c}\right)}$*. *

Proof.  For $s\in\mathbb{R}$, $a>0$ and $b>\frac{a}{2}$ we can write

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[TABLE]

From the formula (see lemma 2 of [5])

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we obtain

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Thus we obtain the first assertion of the lemma. The second assertion is a consequence of the first one. $\square$

The following lemma is well known for the classical Bargmann transform $B_{\pi}$ (see [12]), we adapt its proof for the parametrized form $B_{a}$ in the appendix of this paper.

Let $F_{r}$ be a general form of the Fourier transform defined by

[TABLE]

and let $h_{r}$ be the operator of $L^{2}(\mathbb{R})$ defined as

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Lemma 2.3**.**

For all function $f\in F_{\frac{a}{2}}^{2}(\mathbb{R})$, we have

$\displaystyle\left[B_{\frac{a}{2}}F_{r}B_{\frac{a}{2}}^{-1}f\right](z)=2\sqrt{\frac{r}{r^{2}+1}}e^{-\frac{a}{4}z^{2}\left(\frac{r^{2}-1}{r^{2}+1}\right)}\int_{\mathbb{C}}f(w)e^{\frac{a}{4}\overline{w}^{2}(\frac{1-r^{2}}{r^{2}+1})}e^{\frac{iarz\overline{w}}{r^{2}+1}}\,d\lambda_{\frac{a}{2}}(w).$** 2. 2.

$\displaystyle\left[B_{\frac{a}{2}}F_{1}B_{\frac{a}{2}}^{-1}f\right](z)=\sqrt{2}\,f(iz)$. 3. 3.

$\displaystyle\left[B_{\frac{a}{2}}F_{1}^{-1}B_{\frac{a}{2}}^{-1}f\right](z)=\frac{1}{\sqrt{2}}\,f(-iz)$. 4. 4.

$\displaystyle\left[B_{\frac{a}{2}}\,h_{r}\,B_{\frac{a}{2}}^{-1}f\right](x)=\sqrt{\frac{2}{r^{2}+1}}e^{-\frac{a}{4}z^{2}\left(\frac{r^{2}-1}{r^{2}+1}\right)}\int_{\mathbb{C}}f(-iw)e^{\frac{a}{4}\overline{w}^{2}(\frac{1-r^{2}}{r^{2}+1})}e^{\frac{iarz\overline{w}}{r^{2}+1}}\,d\lambda_{\frac{a}{2}}(w).$**

For recent works on Bargman transform see [4, 5].

3 Heat Cauchy problems for the complex and real Dirac operators

Theorem 3.1**.**

The solution of the heat Cauchy problem (1.6) associated to the complex Dirac operator is given by the formula

[TABLE]

Proof.  Let $U$ be a solution of (1.6). Then, by applying the inverse of the Bargmann transform $B^{-1}_{\frac{a}{2}}$, we obtain

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Using lemma 2.1 we get

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Thus $B^{-1}_{\frac{a}{2}}U$ satisfies the formula

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This implies that

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[TABLE]

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[TABLE]

$\square$

Theorem 3.2**.**

The solution of the heat Cauchy problem (1.9) associated to the real Dirac operator is given by the formula

[TABLE]

Proof.  Let $u$ be a solution of (1.9). Then, by applying the Bargmann transform $B_{\frac{a}{2}}$, we obtain

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Using lemma 2.1 we obtain

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Then $B_{\frac{a}{2}}u$ satisfies the formula

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This implies that

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[TABLE]

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By setting $s^{\prime}=s-t$, we obtain

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[TABLE]

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$\square$

4 Heat Cauchy problems for the complex and real Euler operators

Theorem 4.1**.**

The solution of the heat Cauchy problem (1.13) associated to the real Euler operator is given by the formula

[TABLE]

Proof.  Let $e$ be the solution of (1.13). We define the function $y(t,x)$ by

[TABLE]

We get easily $\frac{\partial}{\partial t}y(t,x)=0$ then $y(t,x)=y(0,x)=e(0,x)=e_{0}(x).$ Then

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Consequently, $e(t,x)=e_{0}(e^{at}\,x)$ for all $(t,x)\in\mathbb{R}_{+}\times\mathbb{R}.$ $\square$

Theorem 4.2**.**

The solution of the heat Cauchy problem (1.16) associated to the complex Euler operator is given by the formula

[TABLE]

Proof.  The proof of the theorem is Mutatis mutandis the proof of theorem 4.1. $\square$

5 Heat Cauchy problems for the real harmonic oscillators

In this section, we study the heat Cauchy problem for the real harmonic oscillator $\displaystyle h^{a}_{x}=\frac{\partial^{2}}{\partial x^{2}}-a^{2}\,x^{2}.$ It is clear that

[TABLE]

Note that the heat kernel for the classical real harmonic oscillator has been known for a long time as Mehler kernel ([3] p.145) and is given by the formula

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[TABLE]

but the method used here and the obtained formula are new.

Theorem 5.1**.**

The heat Cauchy problem (1.20) associated to the real harmonic oscillator has the unique solution given by

[TABLE]

Where $K_{a}(x,s,t)=$

[TABLE]

Proof.  Let $\gamma$ be a solution of (1.20). Then, by applying the Bargmann transform $B_{\frac{a}{2}}$, we obtain

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Proposition 2.1 assures that

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Now, theorem 4.2 gives

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So

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[TABLE]

Using the change of variables $s^{\prime}=e^{-2at}s$, we get

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[TABLE]

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We deduce from lemma 2.2 (with $\frac{a}{b}-1=-e^{-4at}$) that

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So we obtain

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Set again $s=e^{2at}s^{\prime}$, we get

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[TABLE]

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$\square$

The formula (5.2) obtained in the theorem 5.1 agree with the well known formula given in (5.1). In fact,

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[TABLE]

[TABLE]

[TABLE]

Thus we obtain the formula (5.1). $\square$

6 Heat Cauchy problems for the complex harmonic oscillators

In this section, we study the heat Cauchy problem for the complex harmonic oscillator $\displaystyle H^{a}_{z}=\frac{\partial^{2}}{\partial z^{2}}-\frac{a^{2}z^{2}}{4}-\frac{a}{2}$. It is clear that

[TABLE]

Theorem 6.1**.**

The Cauchy problem (1.24) for the heat equation associated to the complex harmonic oscillator has the unique solution given by

[TABLE]

where

[TABLE]

Proof.  Let $E$ be the solution of (1.24). By applying the inverse of the Bargmann transform $B^{-1}_{\frac{a}{2}}$, we obtain

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Using Proposition 2.1 we obtain

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So theorem 4.1 assures that

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This implies that

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By Lemma 2.3 we obtain

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Then

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Using the change of variable $w^{\prime}=-iw$ we get

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$\square$

Appendix :

We give here the proof of lemma 2.3.

Proof. of lemma 2.3.

Firstly, we have

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[TABLE]

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Thus we obtain that  $(B_{\frac{a}{2}}F_{r}B_{\frac{a}{2}}^{-1}f)(z)=$

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[TABLE]

We denote $c=\frac{a}{2}+\frac{ar^{2}}{2}$. Using the change of variable $t=\sqrt{c}.\,x$ we get

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[TABLE]

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[TABLE] 2. 2.

In the case $r=1$, we obtain

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By the formula of reproducing kernel (see proposition 2.2 of [11]) this function is exactly equal to $\sqrt{2}\,f(iz)$. 3. 3.

On the other hand,

[TABLE]

which justifies the assertion 3. 4. 4.

A simple calculus gives $\sqrt{r}\,h_{r}\circ F_{1}=F_{r}\textrm{ and thus }h_{r}=\frac{1}{\sqrt{r}}\,F_{r}\circ F_{1}^{-1}$, then

[TABLE]

Using assertion 1 and 3, we deduce that

[TABLE]

$\square$