FOURIER-LIKE KERNELS AS SOLUTIONS OF ODE ’ S 711

In this paper, we generate asymmetric Fourier kernels as solutions of ODE's. These kernels give many previously known kernels as special cases. Several applications are considered.


INTRODUCTION.
In a previous paper [1], we indicated how Fourier kernels could be generated as solutions of ordinary differential equations and thus, we generated a large number of hitherto unknown Fourier kernels.In this paper we pursue the same idea and generate some more kernels of a different kind.
1 in equation ( 5) is a normalizing factor.The kernels in equations ( 6) and ( 7) were noted by Guinand [2], though his arguments were quite different.
We notice that the eigenfunction in equation ( 5) is symmetric in x and A. In this paper we consider eigenfunctions which are not symmetric. 3.
We shall show that in each one of the above cases, the corresponding solutions of equation (1), which are bounded at infinity, generate Fourier-like kernels.Specifically, taking the normalization factors into account, we shall show that for suitable functions f(x) and A(A), where k(A,x) takes any one of the following values (corresponding respectively to the five cases in equations ( 8)); and ( 5) ks(A,x ,/ [(A4 + Aa + ,/) + (A4 + A3 + ,)]'/ It may be noted that, if we put a 0 in k(A,x), we get the kernel in equations Also, if we let a (R) in k(A,x), we get the kernel in equations (6).It may also be noted that k, k, k and k are all special cases of ks(A,x).It may also be noted from equation (3) that the right hand side of this equation vanishes if u and v satisfy the following conditions: In this case k is not a self conjugate kernel.However, we get the pair where .
[ A(B-a)e-X + (2a+ Aa+ Afl)sinAx +(a+ [(++) + (a++)] '/ k,(,x)* 2 A(a-)e-Xx + (2a+ Aa+ A)sinAx + (a+ fl+,A)AcosAx_ and It may be noted that if we put fl 0 in ks, we get ,'( ,,( [ + ( + )] as a pr of conjugate kernels.If we now didde l through by a and let a go to ity, we get the known pr [3] [-x cosAx] k ,(,) ,(  Since the arguments for showing the validity of equations ( 9) (or equations ( 16)) are the same in each case, we shall concentrate on the simplest case, namely kl(A,x).Proof of Equations ( 9) for k kl(A,x We shall first show that f(x) f A(A) kl(A,x dA (9a) We shall assume that f(x) is in C l[O,(R)) and appropriately well-behaved at infinity.
Since now the integral (gb) exists, we may only show that A() Lim f e -sx f(x) k,(,x) dx.The change in the order of integration in equation ( 22) is justified because of the presence of the term e-sx, > O.

THs shows that
Lim G(A,#,s) where is the (generalized) Dirac delta function, and we get gim f A(#)G(A,#,s)d# A(A), A > 0, s-0 0 as desired.
In order to show that the converse is true, i.e. (9b) (9a), we need to show that f d g(x-), x > 0, > 0. (9c) Alternatively [5], we may show that the Laplace Transform of the left hand side where xp, q is equal to 1/(p+q).This is easily shown, since the product of f -'k,(,) d f kl(,) d is a rational function of A. Taking the Laplace Transform of (9c), changing the order of integration, and substituting, we get the integral of a rational function of A, from zero to infinity.Integrating, and simplifying on Mathematica, .weeasily get the desired result.The arguments for other kernels are the same.
These kernels kt, k,...,k would arise if we try to solve the problem of vibrations of a semi-infinite beam whose end (x 0) is subject to appropriate conditions.We try to solve, e.g., with and b4u + b:u 0 in 0 < x < (R), > 0 (26a where the subscript denotes partial derivative w.r.t, that variable.This problem gives u as the deflection in the problem of vibrations of an elastic beam whose end (x 0) is elastically supported, so that the deflection u is zero at x 0 in > 0, and the bending moment at x 0 is proportional to the slope at x 0. Physical considerations here would require a > 0.
An appropriate representation of u in this case would be u(x,t) f kt(A,x)[A(A)cosA2t + sinAt] dA A and we would require and f(x) f A(A) k,(A, These equations are easily inverted with the help of equtions ( 9) and then, boundary conditions u f2(y) on x 0 in j fs(Y) on x 0 in u h(x) on y 0 in 0 < y < L (31a) 0 < y < e (31b) x > 0 (31c) substitution gives u. kl(x,y is given by equation (10).

2.
The equation -=D lu Vu-D V4u + m4u-fl(x,y) ( where u denotes the cell density at a point, occurs in Mathematical Biology.The corresponding steady state equation is m is a known constant, depending upon the rate at which the cells multiply.D here accounts for the short range effects in the diffusion process while D accounts for the long range ones [6].If these effects are not isotropic, one may encounter a situation in which the short range effects are dominant in the y-direction while the long range ones are dominant in the x-direction.In such a case, after re-scaling, we would get the equation Ou O4u b m4u f(x,y).
(30) y 4 We look for solutions of this equation in 0 < y < L To solve this problem, we write f, fi(A,y) (e -Ax-cosAx + sinAx) dx (32) u(x,y) and look for fi(A,y).We get 1 f u(x,y)(e -Ax-cosAx + sinAx) dx. (33) The kernel in equation ( 33) is the same as in equation ( 6).We shall call fi(A,y) the F-Transform (x A) of u(x,y).
Taking the F-Transform of equation ( 30 with (,0) (A) and dfi t 0 on y=L i' and denote F-Transforms of f and h respectively.
solvable.If g 0, we get ( This problem in fi(A,y) is easily and a(A,y) _jY-- and then u(x,y) is obtained from equation (32).Equation (36) suggests that we shofld take L < r/(2m).( We consider the bending of an anisotropic plate whose deflection u(x,y) is given by 04u + 2b 04U q-Oq4u f(x,y).

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(37) so that equation (43) may be written as a pair of equations W g o and W d(x).