AERONAUTIC
2004001553.pdf

# But u x 0 u x φ γ x this means that if we integrate

• 26

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But u ( x, 0) = u 0 ( x γ ( x ) . This means that if we integrate initial data u 0 Φ γ against p , the resulting function solves the Cauchy problem for (2.12), with this initial data, which proves that p is a fundamental solution. / Corollary 2.6. Suppose that the conditions of the previous theorem hold and that g = 0 . Take the stationary solution u 0 = 1 . The resulting fundamental solution p has the property that R 0 p ( x, y, t ) dy = 1 . Proof. We know that R 0 e - λy 2 - γ p ( x, y, t ) dy = U λ ( x, t ) . Observe that U 0 ( x, t ) = 1, since fl fl fl fl cosh( At 2 )+sinh( At 2 ) cosh( At 2 ) - sinh( At 2 ) fl fl fl fl B 2 σ A (2 - γ ) e - Bt σ (2 - γ ) = 1 . / Remark 2.7 . The extension of these results to the case A < 0 is achieved by replacing cosh( At ) with cos( p | A | t ) etc. The reader may check the details. For the final class of Riccati equations, we are able to establish results similar to Theorem 2.5, which give Laplace transforms under a change of variables. Theorem 2.8. Suppose that γ 6 = 2 and h ( x ) = x 1 - γ f ( x ) and g satisfy σxh 0 - σh + 1 2 h 2 +2 σx 2 - γ g ( x ) = Ax 4 - 2 γ 2(2 - γ ) 2 + Bx 3 - 3 2 γ 3 - 3 2 γ + Cx 2 - γ 2 - γ - κ, where κ = γ 8 ( γ - 4) σ 2 , γ 6 = 2 and A > 0 . Let u 0 be an analytic stationary solution of the PDE (2.2). Define the following constants: a = C 2 σ (2 - γ ) , b = (1 - γ ) A 2(2 - γ ) , k = 2(2 - γ ) B 3 A , d = B 2 9 , l = 3 Ak and s = a + d A - Ak 2 2 σ (2 - γ ) 2 . Let X ( ², x, t ) = x 1 - γ 2 + k q 1 + 2 ² 2 (cosh( At ) - 1) + 2 ² sinh( At ) - k 2 2 - γ ,

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12 MARK CRADDOCK and F 0 ( x ) = f ( x ) x γ . Then equation (2.12) has a solution of the form U ² ( x, t ) = x l (1 + 2 ² 2 (cosh( At ) - 1) + 2 ² sinh( At )) - 2 b A ( k + kx γ 2 (1 - q 1 + 2 ² 2 (cosh( At ) - 1) + 2 ² sinh( At ))) l × fl fl fl fl fl cosh( At 2 ) + (1 + 2 ² ) sinh( At 2 ) cosh( At 2 ) - (1 - 2 ² ) sinh( At 2 ) fl fl fl fl fl s e Ak 2 σ (2 - γ ) 2 - 2 s At × exp ( - ( x 1 - γ 2 + k ) 2 (cosh( At ) + ² sinh( At )) σ (2 - γ ) 2 (1 + 2 ² 2 (cosh( At ) - 1) + 2 ² sinh( At ) ) × exp 1 2 σ ( F ( X ( ², x, t ) - F ( x ))) u 0 ( X ( ², x, t )) . Further, (2.12) has a fundamental solution p ( x, y, t ) such that Z 0 e - λ ( y 2 - γ +2 ky 1 - γ 2 ) u 0 ( y ) p ( x, y, t ) dy = U λ ( x, t ) , (2.16) in which U λ ( x, t ) = U σ (2 - γ ) 2 λ A ( x, t ) . Proof. The proof is similar to the previous result. It is a matter of finding the right symmetry, which turns out to be v = x 2 - γ + 2 B 3 A x γ 2 sinh( At ) x + (cosh( At ) - 1) A t + N ( x, t ) 2 u∂ u , where N ( x, t ) = - A ( x 1 - γ 2 + k ) 2 σ (2 - γ ) 2 cosh( At ) + γB A x γ 2 - 1 sinh( At ) - f ( x ) σx γ x 2 - γ + 2 B 3 A x γ 2 sinh( At ) - 2 s cosh( At ) - 2 b A sinh( At ) . Exponentiation of the symmetry produces the solution U ² ( x, t ) . The rest of the proof proceeds as before. / Remark 2.9 . This theorem involves a somewhat different form of gen- eralized Laplace transform . Note however that the transform Φ( λ ) = R 0 e - λ ( y 2 - γ +2 ky 1 - γ 2 ) φ ( y ) dy reduces to a Laplace transform when we make the substitution z = y 2 - γ + 2 ky 1 - γ 2 . (This gives a quadratic for y 1 - γ 2 in terms of z and we take the positive root).
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• Fall '16
• Dr Salim Zahir
• Fourier Series, Dirac delta function, fundamental solution

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