A full study of this is beyond the scope of the

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A full study of this is beyond the scope of the current paper. Example 3.5. Let X = { X t : t 0 } be a squared Bessel pro- cess, where dX t = ndt + 2 X t dW t , X 0 = x. The joint density of ( X t , R t 0 ds X s ) arises in the pricing of Asian options and other prob- lems, see [6]. To obtain its Laplace transform we require a funda- mental solution for the PDE u t = 2 xu xx + nu x - μ x u. From Theorem 2.3 the reader may check that the stationary solution u 0 ( x ) = x d ,

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FUNDAMENTAL SOLUTIONS 19 where d = 1 4 (2 - n + p ( n - 2) 2 + 8 μ ) , leads to the Laplace trans- form R 0 y d p ( x, y, t ) e - λy dy = x d (1+2 λt ) 2 d + n/ 2 exp ' - λx 1+2 λt . Inversion gives the fundamental solution p ( x, y, t ) = 1 2 t x y 1 2 (1 - n 2 ) I 2 d + n 2 - 1 xy t exp - ( x + y ) 2 t . (3.9) Now E x h e - λX t - μ R t 0 ds Xs i = Z 0 e - λy p ( x, y, t ) dy = e - x 2 t x 2 t · d Γ( α ) 1 F 1 ( α, β, x 2 t +4 t 2 λ ) Γ( β )(1 + 2 λt ) α , with α = d + n 2 , β = 2 d + n 2 . Here 1 F 1 is Kummer’s confluent hyper- geometric function. This is the Laplace transform of the joint density of ( X t , R t 0 ds X s ). See [8] for more on this example and applications of symmetries to the calculation of joint densities. Example 3.6. Let us now consider a two dimensional problem. We will solve u t = u xx + u yy - A x 2 + y 2 u, ( x, y ) R 2 , A > 0 (3.10) u ( x, y, 0) = f ( x, y ) . We convert the problem to polar coordinates. So we let u ( x, y, t ) = U ( p x 2 + y 2 , tan - 1 ( y x ) , t ), where U ( r, θ, t ) satisfies U t = U rr + 1 r U r + 1 r 2 U θθ - A r 2 U, (3.11) with U ( r, θ, 0) = f ( r cos θ, r sin θ ) = F ( r, θ ). Taking the Fourier trans- form in θ , where b f ( n ) = R 2 π 0 f ( θ ) e - inθ , this becomes b U t = b U rr + 1 r b U r - n 2 + A r 2 b U. (3.12) With initial data b U ( r, n, 0) = b F ( r, n, 0), this has solution b U ( r, n, t ) = Z 0 b F ( ρ, n, 0) p ( r, ρ, n, t ) dρ, with p ( r, ρ, n, t ) a fundamental solution of (3.12). Stationary solutions are u 0 ( r ) = r n 2 + A and u 1 ( r ) = r - n 2 + A . Applying Theorem 2.3 with u 0 we find that there is a fundamental solution of (3.12) such that Z 0 e - λρ 2 u 0 ( ρ ) p ( r, ρ, n, t ) = r n 2 + A (1 + 4 λt ) 1+ n 2 + A exp - λr 2 1 + 4 λt .
20 MARK CRADDOCK The change of variables ρ 2 = z converts this to a Laplace transform and inversion gives p ( r, ρ, n, t ) = 1 2 t exp - r 2 + ρ 2 ) 4 t I n 2 + A 2 t · . (3.13) Formally at least, this gives us the Fourier series expansion of the solution of (3.11) U ( r, θ, t ) = X n Z Z 0 b F ( ρ, n ) 1 4 πt e - r 2 + ρ 2 4 t I n 2 + A ( 2 t ) e inθ dρ. Convergence of the series obviously depends on b F. As in the one dimensional case, we can find further solutions. If we take u 1 ( r ) = r - n 2 + A as a stationary solution, then we see that there is a fundamental solution of (3.12) such that Z 0 e - λρ 2 u 1 ( ρ ) p ( r, ρ, n, t ) = r - n 2 + A (1 + 4 λt ) 1 - n 2 + A exp - λr 2 1 + 4 λt . Inversion of these generalized Laplace transforms will require distribu- tions and the result depends on n and A . These kinds of distributions are discussed in [10]. As we have a different distribution for each n , the fundamental solutions will depend upon infinite sums of right sided distributions, rather than a single distribution as happens in the one dimensional case.

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• Fall '16
• Dr Salim Zahir
• Fourier Series, Dirac delta function, fundamental solution

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