# Although these so called whittaker transforms have

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Although these so called Whittaker transforms have known inversion integrals, explicit inversion is usually not possible as few Whittaker transforms have been calculated. In this paper we significantly improve on previous results. We con- sider a class of equations of the form u t = σx γ u xx + f ( x ) u x - g ( x ) u . Suppose that g is fixed and given and γ 6 = 2. We show that if h ( x ) = x 1 - γ f ( x ) is a solution of any one of three families of Riccati equations, then by using the full group of symmetries, it is always possible to obtain a generalized Laplace transform or Fourier transform of funda- mental solutions of the PDE, purely by symmetry. Our method has natural advantages over many other techniques and we will show how it can actually be used to rectify a shortcoming of the method of re- duction to canonical form. We will apply the theory to a number of examples and problems in stochastic analysis. The fact that we may compute Laplace and Fourier transforms, purely through a Lie algebra calculation, suggests a close relationship between Lie symmetry analysis and harmonic analysis. The symmetry group itself gives us these transforms absolutely explicitly as functions of the coefficients of the derivatives in the PDE. This is a consequence of the relationship between the representations of the underlying Lie group and the symmetry transformations, which was first developed in [3] and [4]. We believe that the connections between Lie symmetry analysis and harmonic analysis needs to be explored further. It is hoped that this work will stimulate further investigations in this direction. 1.0.1. Integrating symmetries. Suppose that we have a linear PDE P ( x, D α ) u = X | α |≤ n a α ( x ) D α u, x Ω R m , (1.2)

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FUNDAMENTAL SOLUTIONS 3 with α = ( α 1 , ..., α m ) , α i N , | α | = α 1 + · · · + α m and D α = | α | ∂x α 1 1 ··· ∂x α m m . Let G be a one parameter group of symmetries, generated by some vector field v = m i =1 ξ ( x ) x i + φ ( x, u ) u , where x = ∂x etc. Denote the action of G on a solution u by σ (exp ² v ) u ( x ) = U ² ( x ) . By the Lie symmetry property, there is an interval I R containing zero such that U ² ( x ) is a continuous one parameter family of solutions of (1.2), for all ² I . Continuity and linearity imply the following. Lemma 1.1. Suppose that U ² ( x ) is a continuous one parameter family of solutions of the PDE (1.2), which holds for ² I R . Suppose further that ϕ : I R is a function with sufficiently rapid decay. Then u ( x ) = Z I ϕ ( ² ) U ² ( x ) (1.3) is a solution of the PDE (1.2). Further, if the PDE is time depen- dent and U ² ( x, t ) is the family of symmetry solutions, then u ( x, t ) = R I ϕ ( ² ) U ² ( x, t ) and u ( x, 0) = R I ϕ ( ² ) U ² ( x, 0) d². Further, d n U ² ( x ) n is also a solution for all n = 0 , 1 , 2 , 3 .. Proof. To prove that u is a solution, we simply differentiate under the integral sign. If I is unbounded, we may take ϕ to have compact support to achieve convergence of the integral. The final claim follows from the fact that the PDE does not depend on ², so the order of differentiation may be reversed. / This simple result lies at the heart of the results we obtain here.
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• Fall '16
• Dr Salim Zahir
• Fourier Series, Dirac delta function, fundamental solution

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