SET 8

MATH 543: INTEGRAL REPERSENTATIONS

References: DK and Hildebrandt

Let L_{z} u(z) = 0 be a homogenous linear differential equation. Let the integral representation of the solution of this differential equation be given by

u(z) = ∫_{I} K(z, t) v(t) dt,  (1)

where K(z, t) is the kernel , v(t), t ∈ I is a continuous function and I is an interval to be determined. To determine these three quantities we follow the steps (i) Assume that there exists an operator such that

L_{z} K(z, t) = M_{t} K(z, t)  (2)

Hence the first step is to determine the operator M_{t} if the Kernel of the integral representation is known. (ii) If (2) holds then

L_{z}u(z) = ∫_{I} L_{z} K(z, t) v(t) dt,  (3)
= ∫_{I} [M_{t} K(z, t)] v(t) dt  (4)

Using the generalized Green’s identity we obtain

L_{z} u(z) = ∫_{I} K(z, t) M_{t}^{†} v(t) dt + Q(K, v)|_{I} = 0  (5)

where M_{t}^{†} is adjoint of M_{t} and Q is the surface term resulting from the identity

v(t)M_{t}K(z, t) − K(z, t)M_{t}^{†}v(t) = \frac{dQ(K, v)}{dt},  (6)

Then we let

M_{t}^{†}v(t) = 0, and Q(K, v)|_{I} = 0,  (7)

(iii) third step is to find v(t) from the first equation in (7) and (iv) fourth step is to determine I from the second equation in (7).

The choice of Kernels:

(1) Laplace kernel. K(z, t) = e^{−zt}. It is used when the differential equation Lu = 0 has coefficients linear in z

(2) Euler kernel: K(z, t) = (z − t)^{ν}. It is used when the coefficient of nth order derivative in Lu = 0 is a polynomial of degree n. Here ν is a constant to be determined.

(3) Bessel kernel: K(z, t) = (z/2)^{ν} e^{t−z^{2}/(4t)} where ν is a constant to be determined.

For other types of kernels see ( For instance Fourier, Melin, Hankel) Sadri Hassan Chapter 15. As a summary We fisrts decide on the kernel and use the above four steps.

Problems

1. Hypergeometric function.

L_{z}u(z) = [z(1 − z) \frac{d^{2}}{dz^{2}} + [c − (a + b + 1)\frac{d}{dz} − ab] u(z) = 0  (8)

Choose the Euler kernel K(z, t) = (z − t)^{ν} and prove that ν = −a (or −b) and (for the choice ν = −a) (a)

M_{t} = (t^{2} − t) \frac{d^{2}}{dt^{2}} − [(b − a − 1)t + a − c + 1]\frac{d}{dt}

(b)

v(t) = kt^{a−c} (t − 1)^{c−b−1}, k is a constant

(c)

Q(K, v) = akt^{a−c+1}(t − 1)^{c−b}(z − t)^{−a−1}

(d) There are several choices of the interval I. One of them is I = [1, ∞). Hence (e)

u(z) = k ∫_{1}^{∞} (t − z)^{−a}t^{−a−c}(t − 1)^{c−b−1}dt

Provided Rec > Reb > 0. One can find the constant k by letting first a = 0 and then z = 0 the last equation becomes

u(0) = 1 = k ∫_{1}^{∞} t^{−a−c} (t − 1)^{c−b−1}dt
= k \frac{Γ(b)Γ(c − b)}{Γ(c)}  (9)

then we find an integral representation of the hypergeometric function which is defined not only in the disc |z| < 1 but in a more largeer region 0 ≤ arg(1 − z) ≤ 2π (prove this claim).

u(z) = F (a, b; c; z) = \frac{Γ(c)}{Γ(b)Γ(c − b)} ∫_{1}^{∞} (t − z)^{−a} t^{a−c} (t − 1)^{a−c−1} dt

By making t → 1/t transformation one finds also that

u(z) = \frac{Γ(c)}{Γ(b)Γ(c − b)} ∫_{0}^{1} (1 − tz)^{−a} t^{b−1} (1 − t)^{c−b−1} dt

2. Confluent hypergeometric equation:

L_{z}u(z) = [z \frac{d^{2}}{dz^{2}} + (c − z)\frac{d}{dz} − a]u(z) = 0  (10)

Choose the Laplace kernel K(z, t) = e^{zt} and prove that (a)

M_{t} = (t^{2} − t)\frac{d^{2}}{dt^{2}} + ct − a

Hence (b)

v(t) = kt^{a−1} (1 − t)^{c−a−1}, Q(K, v) = −ke^{zt}t^{a}(1 − t)^{c−a}

One can choose I = [0, 1] provided Rec > Rea > 0. (c)

u(z) = Φ(a, c; z) = Γ(c)Γ(c − a)Γ(a) ∫_{0}^{1} e^{zt}t^{a−1}(1 − t)^{c−a−1}dt

3. Find the above integral representation from the above integral representation of the hypergeometric function in Problem 1 by using the formula

Φ(a, c; z) = lim_{b→∞} F (a, b; c; z/b)

4. Bessel’s Equation:

L_{z}u(z) = [\frac{d^{2}}{dz^{2}} + \frac{1}{z} \frac{d}{dz} + 1 − \frac{ν^{2}}{z^{2}} ]u = 0  (11)

Choose the kernel K(z, t) = (z/2)^{ν}e^{t−t^{2}/4t} and show that (a)

M_{t} = \frac{d}{dt} − \frac{ν + 1}{t}

and (b)

v(t) = kt^{−ν−1}, Q(K, v) = k(z/2)^{ν}e^{t−z^{2}/4t}

Then one finds and integral repersentation of the Besel function J_{ν}(z) given by

J_{ν}(z) = k(z/2)^{ν} ∫_{C} t^{−ν−1} e^{t−z^{2}/4t} dt

where k is a constant to be determined and C is a closed contour closing the axis Rez ≤ 0. We find k by letting first t = \frac{uz}{2} which leads to

J_{ν}(z) = k ∫_{C} u^{−ν−1} e^{z(u−1/u)/2}du

and letting z = 0 in the above equation with ν = 0 then

J_{0}(0) = 1 = k ∫_{C} \frac{dt}{t}

Hence k = 2πi.

5. Hermite equation:

[\frac{d^{2}}{dz^{2}} − z \frac{d}{dz} + n]u(z) = 0  (12)

where n is a positive integer. Prove that the Hermite function H_{n}(z) has an integral representation

H_{n}(z) = \frac{1}{2πiΓ(n + 1)} ∫_{C} t^{−n−1}s^{zt} e^{−t^{2}/2}dt

where C is closed contour closing the origin. Consider also the transformation s = t − z and find the Rodriquez formula for the Hermite polynomials.