# Leads to the laplace trans the fundamental solution 1

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) , leads to the Laplace trans- the fundamental solution 1 . x Σ 1 (1 n ) 2 y . xy Σ t . (x + y) Σ 2 t 2 t t 2 y Now E Σ e λX t µ R t ds Σ = e λy p ( x, y , t ) dy 0 x . x Σ d Γ(α) 1 F 1 (α, β, x ) 2 t 2 t with α = d + n , β = 2d + n . Here 1 F 1 is Kummer’s confluent hyper- 2 2 geometric function. This is the Laplace transform of the joint density of (X , t ds ). See [8] for more on this example and applications of symmetries to the calculation of joint densities. Example 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 , A > 0 for m ex p t . Inversion gives 0 x d (1+2 λt ) 2 d + n/ λx 1+2 λ ex p . ,

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x 0 u(x, y, 0) = f(x, y). We convert the problem to polar coordinates. So we let u(x, y, t) = U ( x 2 + y 2 , tan −1 ( y ), t), where U (r, θ, t) satisfies U t = U rr + r U r + r 2 U θθ r 2 U, with U (r, θ, 0) = f(r cos θ, r sin θ) = F (r, θ). Taking the Fourier trans- form in θ, where f ^ (n) = 2 π f(θ)e inθ dθ, this becomes U ^ t = U ^ rr + r U ^ r n 2 + A r 2 U ^ . (3.12) With initial data U ^ ( r , n, 0) = F ^ ( r , n, 0), this has solution n 2 + A 1 + 4λt 1
^ 1 p(r, ρ, n, t) = I n 2 + F (ρ, n) e 4πt 4 t I n 2 + A ( 2t ) e The change of variables ρ 2 = z converts this to a Laplace transform and inversion gives 1 . r 2 + ρ 2 ) Σ . r ρ Σ 2 t 4 t 2 t 4 t Formally at least, this gives us the Fourier series expansion of the solution of (3.11) Σ ^ 0 n Z 1 r 2 + ρ 2 inθ 0 n Z Convergence of the series obviously depends on F . As in the one dime nsional case, we can find further solutions. If we take u (r) = r n 2 + A as a stationary solution, then we see that there is a fundamental solution of (3.12) such that ∫ ∞ e λρ 2 u (ρ)p(r, ρ, n, t)dρ = r n 2 + A exp . λr 2 . 1 0 (1 + 4λt) 1− n 2 + A 1 + 4λt . U (r, θ, t) = dρ. Σ

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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
• Complex number, Dirac delta function, Pierre-Simon Laplace

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