The heat kernel for gravitational potential
operators in one and two variables
Dedicated to Professor Joseph Kohn on the occasion of his 75th birthday and retirement
Ovidiu Calin, Eastern Michigan University, Department of Mathematics,
Ypsilanti, MI, 48197, USA, ocalin@emich.edu ∗
Der-Chen Chang, Georgetown University, Department of Mathematics,
Washington DC 20057, USA, chang@math.georgetown.edu†
January 8, 2008
Abstract
In this article we find the close form formulas for the heat kernels of the operators 1
2∂x2+xλ2 and 12(∂2x+ ∂y2) − λ
2
x2+y2. We are using a
geometric method and the expansion over eigenvalues method.
Key words: heat equation, Euler-Lagrange equation, Hamiltonian, Lagrangian, Van-Vleck determinant, Laguerre polynomials
MS Classification (2000): Primary: 53C17; Secondary: 34K10, 35H20
1
Introduction
In the first part of this paper we shall obtain the heat kernel for the operator
L = 1 2∂ 2 x+ λ x2, λ ∈ R, (1.1)
involving a geometric technique presented in [1]. An approach using path integrals can be found in [4]. An extension to a propagator involving time-dependent harmonic oscillator was done by means of path integrals in [5].
∗The first author is partially supported by NSF grant #0631541.
†The second author partially supported by a competitive research grant at Georgetown
The authors of this paper have been inspired to work on this problem after getting acquainted with the problem from [6].
This operator owes its physical significance to its relation to the two body problem. Consider two unit mass particles described by the coordinates
ξ1 and ξ2 on the real line with an interaction which varies as the inverse
square of the distance between the particles. The problem is characterized by a Lagrangian which is the difference between the kinetic energy and the gravitational potential L( ˙x, x) = 1 2( ˙ξ 2 1 + ˙ξ22) − 2λ (ξ1− ξ2)2, (1.2)
where the strength of the interaction λ > 0. With the change of coordinates
x = ξ1√− ξ2
2 , y =
ξ1+ ξ2
2 the Lagrangian (1.2) becomes
L(x, y, ˙x, ˙y) = ˙y2+³ 1 2˙x 2− λ x2 ´ = L0+ L1,
which is the sum of two Lagrangians.
The momenta associated with the coordinates x and y are
px = ∂L ∂ ˙x = ∂L1 ∂ ˙x = ˙x, py = ∂L ∂ ˙y = ∂L0 ∂ ˙x = 2 ˙y.
Then the associated Hamiltonian is obtained by applying the Legendre transform on the Lagrangian L
H(px, py, x, y) = px˙x + py˙y − L = (py˙y − L0) + (px˙x − L1) = H0(py, y) + H1(px, x), with H0(py, y) = p2 y 4 , H1(px, x) = 1 2p 2 x+ λ x2,
Quantizing the Hamiltonian H = H0 + H1, i.e., replacing px → ∂x and
py → ∂y yields the following operator
P = 1 4∂ 2 y + 1 2∂ 2 x+ λ x2. (1.3)
Due to separation of variables, the heat kernel of P can be written as etPδ(x0,y0) = et14∂y2+t ¡ 1 2∂x2+x2λ ¢ δ(x0,y0) = et14∂y2δ y0⊗ e t¡1 2∂x2+x2λ ¢ δx0 = √1 16πte −(y−y0)16t 2 ⊗ et¡1 2∂x+2 x2λ ¢ δx0, t > 0.
We need to compute the second term of the above convolution which is the heat kernel of the operator (1.1). We recall in the following section a geometric method for computing heat kernels.
In the second part of this paper we shall treat the heat kernel problem for the operator
L = 1 2(∂ 2 x+ ∂y2) − λ2 x2+ y2 (1.4)
The method is using Laguerre polynomials and the expansion over the eigen-functions. The exact summation is given by the Hille-Hardy’s formula.
2
Heat kernels for some operators with potential
The following method is variational and it is related to the geometry gener-ated by the differential operator. In the following we shall briefly describe the method. This shows the interrelation between the geometry of an oper-ator and its heat kernel.
In this section the differential operators will be of the form
L = 1 2 d dx 2 + U (x) (2.5)
with U (x) at most quadratic. We associate the Hamiltonian function
H(p, x) = 1
2p
2+ U (x). (2.6)
Hamilton’s equations are ˙x = Hp= p,
For any two given points x0 and x, the geodesic joining them is obtained solving the equation
¨ x = −U0(x), x(0) = x0, x(t) = x. (2.7)
We shall assume that U (x) is such that the system (2.7) has a unique solution
x(s). The classical action between x0 and x in time t is obtained integrating
the Lagrangian L = p ˙x − H = 1
2˙x2− U (x) along the above geodesic x(s)
Scl(x0, x; t) = Z t 0 L¡˙x(s), x(s)¢ds = Z t 0 1 2˙x 2(s) − U¡x(s)¢ds.
With these notations the heat kernel of the operator (2.5), which is a path integral, is given by the formula
K(x0, x; t) = V (t)e−Scl(x0,x;t). (2.8)
If the Lagrangian is at most quadratic, then V (t) = r det ³ −2π1 ∂2Scl ∂x ∂x0 ´ is the Van-Vleck determinant. Otherwise, the function V depends on t, x0and
x and satisfies a transport equation, see section 4.
3
Finding the classical action
The geodesic joining the points x0 and x in time t satisfies the Euler-Lagrange equation associated with the Lagrangian L1
¨
x = 2λ x3
x(0) = x0, x(t) = x.
Since the regions {x < 0} and {x > 0} are separated, in order to have connectivity, we have to assume that either x0, x > 0 or x0, x < 0. We can
show that the energy is a first integral of motion, so 1
2˙x
2(s) + λ
x2(s) = E, s ∈ [0, t].
Under the assumption x0, x > 0 the above relation becomes
Let u(s) = x2(s), u
0= x20, ut= x2(t) = x2. Then u(s) verifies the ODE
˙u = 2√2Eu − 2λ
u(0) = u0, u(t) = ut.
We note that ˙u > 0, so the right side must have positive sign. Integrating
yields Z ut u0 du √ 2Eu − 2λ = 2t ⇐⇒ p
2Eut− 2λ −p2Eu0− 2λ = 2Et.
Eliminating the square roots we obtain ³
(u0+ ut) − 2Et2
´2
= 4(u0ut− 2λt2), (3.9)
where we assume the condition
λ < x20x2
2t2 .
Solving for E in (3.9) yields
E = x20+ x2 2t2 − p x2 0x2− 2λt2 t2 . (3.10)
The classical action Scl satisfies the following Hamilton-Jacobi equation
∂tScl= −E,
with E given by (3.10). We can write Scl= S0+ S1, where
∂tS0 = −x 2 0+ x2 2t2 =⇒ S0 = x2 0+ x2 2t (3.11) ∂tS1 = p x2 0x2− 2λt2 t2 . (3.12)
We shall solve (3.12) as a homogeneous equation. Let τ = t
x0x, and define S2(τ ) = S1(x0tx). Then d dτS2(τ ) = √ 1 − 2λτ2 τ2 .
With the substitution τ = √1
2λsin φ, integrating yields
S2(τ ) = Z √ 1 − 2λτ2 τ2 dτ = √ 2λ Z cot2φ dφ = √2λ Z ¡ − 1 − cot0φ¢dφ = −√2λ(φ + cot φ) = −√2λ©sin−1(√2λτ ) + √ 1 − 2λτ2 √ 2λτ ª and hence S1(x0, x, τ ) = −2 p x2 0x2− λt2 2t − √ 2λ sin−1¡√2λ t x0x ¢ . (3.13)
From (3.13) and (3.11) we obtain the classical action
Scl(x0, x, t) = x 2 0+ x2 2t − 2px2 0x2− λt2 2t − √ 2λ sin−1¡√2λ t x0x ¢ . (3.14)
By the general theory, or by a direct computation, the classical action
Scl satisfies the Hamilton-Jacobi equation
∂tScl+
1
2(∂xScl)
2+ λ
x2 = 0. (3.15)
We also note that for λ → 0 we obtain the Euclidean action
Scl(x0, x, t) = (x − x0) 2
2t .
4
The transport equation
We shall assume that the heat kernel of L is of the type
K(x0, x, t) = V (x, t)e−S(x0,x,t).
Then a computation shows
∂tK = e−S ³ ∂tV − V ∂tS ´ ∂x2K = e−S ³ ∂2xV − 2∂xV ∂xS + V (∂xS)2− V (∂x2S) ´
and hence (∂t− L)K = (∂t−12∂x2− λ x2)K = e−S n − V [∂tS +12(∂xS)2+xλ2 | {z } =0 ] (by (3.15)) +∂tV −12∂x2V + ∂xV ∂xS + 12V ∂2xS o
We shall ask V to satisfy the following transport equation
∂tV −1 2∂ 2 xV + ∂xV ∂xS + 1 2V ∂ 2 xS = 0. (4.16)
The equation (4.16) might be hard to solve since the action S and its derivatives are complicated. In the following we shall consider a shortcut for these computations. We note that the action S = S0+ S1, where the
term S1 is a function of x0tx. Then
e−S = W (x0x
t )e
−x20+x2 2t .
Then it makes sense to look now for a heat kernel of the type
K(x0, x, t) = V (x, t)e−S0 = V (x, t)e− x20+x2 2t , where V (x, t) = √1 tZ( x0x
t ) satisfies the extended transport equation
∂tV −12∂x2V +∂xV ∂xS0+12V (∂x2S0)−V [∂tS0+12(∂xS0)2+xλ2] = 0. (4.17)
In the following we shall solve the equation (4.17). Let τ = x0x
t . Then we have V = t−12Z(τ ) ∂tV = −t− 3 2³ 1 2Z(τ ) + τ Z 0(τ )´ ∂xV = t−32Z0(τ ) x0 ∂x2V = t−32Z00(τ )x 2 0 t .
Since ∂tS = −x 2 0+ x2 2t2 , ∂xS = x t, ∂ 2 xS = 1 t,
the equation (4.17) becomes after cancelations
−1 2t −1 2Z00(τ )³ x0 t ´2 − t−12Z(τ )h λ x2 − 1 2 ³ x0 t ´2i = 0. Multiplying by −2x2t12 yields τ2Z00(τ ) + Z(τ )[2λ − τ2] = 0. (4.18) Let U (τ ) = τ−12Z(τ ). A computation shows
τ2U00(τ ) = 3 4τ −1 2Z(τ ) − τ12Z0(τ ) + τ32Z00(τ ) τ U0(τ ) = −1 2τ −12Z(τ ) + τ12Z0(τ ),
and using (4.18) we have
τ2U00(τ ) + τ U0(τ ) = τ−12 ³ τ2Z00(τ ) + 1 4Z(τ ) ´ = τ−12 ³ Z(τ )¡τ2− 2λ¢+1 4Z(τ ) ´ = U (τ )¡τ2− 2λ +1 4 ¢ .
Hence U (τ ) satisfies the modified Bessel equation
τ2U00+ τ U0+ (−τ2− γ2)U = 0, with γ = 1
2
√
1 − 8λ, with λ < 1/8. The general solution can be written as a linear combination
U (τ ) = αIγ(τ ) + βKγ(τ ), α, β ∈ R
where Iγ(τ ) and Kγ(τ ) are the modified Bessel function of the first and
second type. Hence the general solution of (4.18) is
Z(τ ) =√τ U (τ ) = α√τ Iγ(τ ) + β√τ Kγ(τ ), where Iγ(τ ) ∼ r 1 2πτ e τ, K γ(τ ) ∼ r π 2τ e −τ as τ → ∞, (4.19)
see Haberman [3], p. 323.
Consequently, the solution of the extended transport equation (4.17) will be given by V (x0, x, t) = t−1/2Z(τ ) = √ xx0 t ³ αIγ¡ xxt0 ¢ + βKγ¡ xxt0 ¢´ , (4.20) with α, β ∈ R.
5
The heat kernel for L =
12∂
2x
+
xλ2In this section we shall state and prove the first main result of the paper. Theorem 5.1 The heat kernel for the operator L = 1
2∂x2+xλ2, with λ < 1/8 and x0, x > 0 is K(x0, x; t) = √ x0x t Iγ ³ x0x t ´ e−x20+x 2 2t , t > 0, where Iγ is the nonsingular modified Bessel function of order γ = 12
√
1 − 8λ.
Proof: We have shown already in the previous section that
(∂t− L)K(x0, x, t) = 0, t > 0,
with
K(x0, x, t) = V (x0, x, t)e− x20+x2
2t ,
and V given by (4.20). We need to choose the constants α and β such that lim
t&0K(x0, x, t) = δx0
in the distributions sense. Let K = K1+ K2 with
K1(x0, x, t) = α √ xx0 t Iγ ¡ xx0 t ¢ e−x2+x22t0 K2(x0, x, t) = β √ xx0 t Kγ ¡ xx0 t ¢ e−x2+x22t 0. Then lim
t&0K1(x0, x, t) = limt&0α
√ xx0 t Iγ ¡ xx0 t ¢ e−xx0t e−(x−x0) 2 2t = α lim τ →∞ √ τ Iγ(τ )e−τ lim t&0 1 √ te −(x−x0)2t 2 = α√1 2πlimt&0 1 √ te −(x−x0)2t 2 = αδx0,
where we have used the first relation of (4.19). Hence we shall choose α = 1. A similar computation, using the second relation of (4.19) yields
lim
t&0K2(x0, x, t) = limt&0β
√ xx0 t Kγ ¡ xx0 t ¢ e−xx0t e−(x−x0) 2 2t = β lim τ →∞ √ τ Kγ(τ )e−τ lim t&0 1 √ te −(x−x0)2t 2 = β√π lim τ →∞e −2τ lim t&0 1 √ 2πte −(x−x0)2t 2 = 0. Hence lim
t&0K(x0, x, t) = limt&0K1(x0, x, t) + limt&0K2(x0, x, t) = αδx0,
so we need to choose α = 1. In order to find β we shall consider the limit
λ → 0, case in which we recover the Gaussian kernel
1 √ 2πte −(x−x0)2t 2 = lim g&0K(x0, x, t) = √1 t √ τ I1/2(τ )e−τe−(x−x0) 2 2t + β√1 t √ τ J1/2(τ )e−τe−(x−x0) 2 2t = √1 2πte −(x−x0)2t 2 + β√1 t √ τ J1/2(τ )e−τe−(x−x0) 2 2t , since we take γ = 1/2 in Iγ(τ ) = √1 2πτe τ −12 (γ2− 14 ) τ +O(1/τ2).
Hence we need to choose β = 0.
In the case λ = −1 we obtain:
Corollary 5.2 The heat kernel for the operator L = 12∂2
x−x12 is K(x0, x; t) = √ x0x t I3/2 ³ x0x t ´ e−x20+x 2 2t , t > 0, where I3/2 is the nonsingular modified Bessel function of order 3/2.
Remark 5.3 Since Iγ> 0 it follows that all the heat kernel K(x0, x; t) > 0 for t > 0.
6
The two dimensional case
We shall extend the above results in the case of two variables, where the operator becomes L = 1 2(∂ 2 x1 + ∂ 2 x2) − λ2 x2 1+ x22 .
We note that in this case the constant λ2 ≥ 0. In order to find the heat
kernel of L we shall recall first a few basic properties of special functions. The Laguerre polynomial La
nof degree n and parameter a can be defined
either by the Rodrigues’ formula
Lan= exx−a n! dn dxn(e−xxn+a) = 1 n!(−x) n+ . . . ,
or equivalently, by the generating function formula X
n≥0
Lan(x)yn= 1 (1 − y)a+1e
xy/(y−1). (6.21)
Applying integration by parts several times yields
Z ∞ 0 Lan(x)Lam(x)e−xxadx = 0, if n 6= m; Γ(a + n + 1) n! , if n = m. Hence fn(x) = s n! Γ(a + n + 1)e −x/2xa/2La n(x), n = 0, 1, . . . (6.22)
form an orthogonal system for L2(0, ∞). One can show that this system is also complete. Using that y = La
n(x) verifies the Laguerre equation equation
xy00+ (α + a − x)y0+ ny = 0,
one can show that the functions (6.22) are eigenfunctions for the operator
L = x∂x2+ ∂x−14(x + a
2
x), (6.23)
with the corresponding eigenvalue λn= −
¡
The following result is known under the name of Hille-Hardy’s formula, see Erdelyi [2], p. 189:
If La
n denotes the Laguerre polynomial, and Ia is the modified Bessel
function, then for |z| < 1 we have
∞ X n=0 n! Γ(n + a + 1)z n(xyz)a/2La n(x)Lan(y) = 1 (1 − z)e −(x+y)z(1−z) I a ³ 2√xyz 1 − z ´ .
Proposition 6.1 Let a, b ∈ R. The heat kernel of the operator
Lb = x∂x2+ ∂x− 14(b2x +a 2 x ) is given by Kb(x0, x, t) = sinh(bt/2)b/2 e− b 2(x0+x) coth(bt/2)Ia³ b√x0x sinh(bt/2) ´ , t > 0. (6.24)
Proof: We shall first find the heat kernel of the operator
L1= x∂x2+ ∂x−14(x +a
2
x).
Then using the homogeneity property we shall determine the heat kernel for Lb.
In the following we shall denote z = e−t and use that
√ z 1 − z = 1 2 sinh(t/2), 1 + 2z 1 − z = 1 + e−t 1 − e−t = coth(t/2).
over the eigenvalues (6.22) K1(x0, x, t) = X n≥0 eλntf n(x0)fn(x) = X n≥0 e−nte−a2te− t 2 n! Γ(a + n + 1)L a n(x0)Lan(x)x0a/2e−x0/2xa/2e−x/2 = e−12(x0+x)√z X n≥0 n! Γ(a + n + 1)z n(x 0xz)a/2Lan(x0)Lan(x) = e−12(x0+x) √ z 1 − ze −(x0+x)1−zz I a ³ 2√x0x √ z 1 − z ´
(by Hille-Hardy’s formula)
= 1 2 sinh(t/2)e −1 2(x0+x)(1+ 2z 1−z)Ia³√x0x 1 sinh(t/2) ´ = 1 2 sinh(t/2)e −1 2(x0+x) coth(t/2)Ia³ √x0x sinh(t/2) ´ .
In order to compute the heat kernel of the operator Lb, we take ˜x = bx
as a new variable. A computation shows Lb = b˜L1. Then (etLbf )(x
0) = (etb˜L1f )(˜˜ x0) =
Z
K1(bx0, bx, bt)f (x0)bdx0,
and hence the heat kernel of etLb is bK
1(tx0, tx, tb), that is Kb(x0, x, t) = b/2 sinh(bt/2)e −b 2(x0+x) coth(bt/2)Ia³ b√x0x sinh(bt/2) ´ . (6.25)
Therefore the heat kernel of the operator Lb is given by formula (6.25).
Corollary 6.2 The heat kernel for L0= x∂2x+ ∂x− a2
4x is given by K0(x0, x, t) = 1te− 1 t(x0+x)Ia³ 2√x0x t ´ , t > 0. (6.26)
Proof: The heat kernel of L0 is obtained making b → 0 in the relation
(6.25). Using lim
b→0
bt/2
sinh(bt/2) = 1 yields (6.26).
Proposition 6.3 The heat kernel of P = 1 2(∂ 2 r + 1 r∂r− a2 r2) is G(r0, r, τ ) = 2τ1 e− 1 2τ(r20+r2)Ia³ r0r τ ´ , τ > 0. (6.27)
Proof: Changing of variable x = r2 in L 0 yields L0= x∂2x+ ∂x− a 2 4x = 1 4 ¡ ∂r2+1 r∂r− a2 r2 ¢ .
Let τ = t/2. Since the following relation holds among the heat kernels
et(x∂x2+∂x−a24x) = e
t
4(∂r2+1r∂r−a2r2)= e τ
2(∂r2+1r∂r−a2r2),
then the heat kernel of 12(∂2
r + 1r∂r− a
2
r2) is obtained from the heat kernel
of x∂2
x+ ∂x− a
2
4x making the substitutions x0 = r20, x = r2 and τ = t/2 in
formula (6.26).
As an application of formula (6.27) we shall find the heat kernel of the operator L = 1 2(∂ 2 x1 + ∂ 2 x2) − λ2 x2 1+ x22 , (6.28)
which in polar coordinates becomes
L = 1 2(∂ 2 r+ 1 r∂r+ 1 r2∂θ2) − λ2 r2.
If K is the heat kernel of the above operator, then applying a partial Fourier transform with respect to θ, we obtain that ˆK = FθK is the heat
kernel of the operator
P = 1 2(∂ 2 r + 1 r∂r− 1 r2ξ2) − 2λ2 2r2 = 1 2(∂ 2 r + 1 r∂r− a2 r2),
with a2 = 2λ2 + ξ2. The heat kernel of etP can be obtained from formula (6.27) ˆ K(r0, r, τ ; ξ) = 2τ1 e− 1 2τ(r20+r2)Ia³ r0r τ ´ .
Applying the inverse Fourier transform yields the heat kernel of the operator (6.28) K(r0, θ0, r, θ, τ ) = 1 2π Z ei(θ−θ0)ξ 1 2τe −1 2τ(r02+r2)I (2λ2+ξ2)1/2 ³ r0r τ ´ dξ = 1 4πτe −2τ1(r2 0+r2) Z ei(θ−θ0)ξI (2λ2+ξ2)1/2 ³ r0r τ ´ dξ.
Proposition 6.4 The heat kernel of the operator (6.28) is given by K(x0, x, τ ) = 4πτ1 e− 1 2τ(|x0|2+|x|2)V ³ arg(x0, x),xτ0x ´ , τ > 0, where arg(x0, x) = cos−1³ x · x0
|x| |x0| ´ , and V (u, ρ) = Z eiuξI (2λ2+ξ2)1/2(ρ) dξ.
7
Directions to further studies
We suggest here a few open problems related to this paper. We first ask how can the method be extended to find the heat kernel for an elliptic operator with potential ³ ∂ ∂t− X i,j=1 aij(x)∂xi∂xj− λ2 P aij(x)xixj ´ K(x0, x, t) = 0, t > 0 lim t&0K(x0, x, t) = δx0.
In particular, we are interested in the role played by the geometry. Can we find the heat kernel for L = 1
2
P
∂2
xk −
λ2
|x|2, with λ 6= 0 and x ∈ Rn with n ≥ 3?
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3. Haberman, R. : Elementary Applied Partial Differential Equations, 3-rd ed., Prentice Hall, 1998.
4. Khandekar, D.C., Lawande, S. V. : Path integration of a three body
problem, J. Phys. A: Gen. Phys., Vol. 5, 1972.
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