U ˆ x 1 2 ² 2 cosh at 1 2 ² sinh at 1 2 γ 213

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u 0 ˆ x (1 + 2 ² 2 (cosh( At ) - 1) + 2 ² sinh( At )) 1 2 - γ ! , (2.13) where F 0 ( x ) = f ( x ) /x γ and c = (1 - γ ) 2 - γ . Further, there exists a funda- mental solution p ( x, y, t ) of (2.12) such that Z 0 e - λy 2 - γ u 0 ( y ) p ( x, y, t ) dy = U λ ( x, t ) , (2.14) where U λ ( x, t ) = U σ (2 - γ ) 2 λ A ( x, t ) .
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10 MARK CRADDOCK Proof. Applying Lie’s algorithm shows that if f satisfies the given con- dition, then there are infinitesimal symmetries v 1 = Ax 2 - γ e At x + e At t - ˆ Ax 2 - γ 2 σ (2 - γ ) 2 + Ax 1 - γ 2 σ (2 - γ ) f ( x ) ! e At u∂ u - αue At u , v 2 = - Ax 2 - γ e - At x + e - At t - ˆ Ax 2 - γ 2 σ (2 - γ ) 2 - Ax 1 - γ 2 σ (2 - γ ) f ( x ) + β ! e - At u∂ u , v 3 = t , v 4 = u∂ u . Here α = 1 - γ 2(2 - γ ) A + B 2 σ (2 - γ ) , β = - 1 - γ 2(2 - γ ) A + B 2 σ (2 - γ ) . The key is to find the right symmetry to generate a generalized Laplace trans- form. This turns out to be given by using v 1 , v 2 , v 3 to produce the infinitesimal symmetry v = 2 Ax 2 - γ sinh( At ) x + 2(cosh( At ) - 1) t - 1 σ g ( x, t ) u∂ u , where g ( x, t ) = ( Ax 2 - γ (2 - γ ) 2 + B 2 - γ ) cosh( At ) + A sinh( At )( x 1 - γ 2 - γ f ( x ) + σc ) . Exponentiation of this symmetry shows that U ² ( x, t ) is a solution of (2.12). The change of parameters ² σ (2 - γ ) 2 λ A shows that U λ ( x, 0) = e - λx 2 - γ u 0 ( x ) . The basic principle is that if p ( x, y, t ) is a fundamental solution of (2.12), then we should have Z 0 e - λy 2 - γ u 0 ( y ) p ( x, y, t ) dy = U λ ( x, t ) . (2.15) That is, U λ ( x, t ) is the generalized Laplace transform of u 0 p. To prove this, we first observe that U λ is a generalized Laplace trans- form if it is a Laplace transform. This follows from the obvious change of variables reducing it to a Laplace transform. Any function which can be written as a product of Laplace transforms is a Laplace transform. The drift functions can be expressed in terms of hypergeometric func- tions, which are analytic. So a straightforward argument shows that U λ can be written as the product of functions analytic in 1 λ and any function analytic in 1 λ is automatically a Laplace transform. Thus U λ is a generalized Laplace transform of some distribution u 0 p. (Chapter Ten of [20] contains the most general conditions under which a distribution may be written as a Laplace transform). Now suppose that (2.15) holds for some distribution p . We prove that p is a fundamental solution of the PDE. The fact that U λ ( x, t ) is a solution and U λ ( x, t ) = R 0 e - λy 2 - γ u 0 ( y ) p ( x, y, t ) dy implies that p is a solution for each fixed y . Now integrate a test function ϕ ( λ )
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FUNDAMENTAL SOLUTIONS 11 with sufficiently rapid decay against U λ . By Lemma 1.1 the function u ( x, t ) = R 0 U λ ( x, t ) ϕ ( λ ) is a solution of (2.2). We also have u ( x, 0) = Z 0 U λ ( x, 0) ϕ ( λ ) = Z 0 u 0 ( x ) e - λx 2 - γ ϕ ( λ ) = u 0 ( x γ ( x ) , where Φ γ is the generalized Laplace transform of ϕ. Next observe that by Fubini’s Theorem Z 0 u 0 ( y γ ( y ) p ( x, y, t ) dy = Z 0 Z 0 u 0 ( y ) ϕ ( λ ) p ( x, y, t ) e - λy 2 - γ dλdy = Z 0 Z 0 u 0 ( y ) ϕ ( λ ) p ( x, y, t ) e - λy 2 - γ dydλ = Z 0 ϕ ( λ ) U λ ( x, t ) dx = u ( x, t ) .
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  • Fall '16
  • Dr Salim Zahir
  • Fourier Series, Dirac delta function, fundamental solution

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