In the first case we find that there is an

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satisfies the given Riccati equations. In the first case, we find that there is an infinitesimal symmetry of the form v 1 = sinh( At ) x - (( Ax + B A ) cosh( At ) + f ( x ) sinh( At )) u 2 σ u . Exponentiating 2 v 1 and multiplying the result by the constant e σ A produces (4.6). In the second case there is an infinitesimal symmetry v = t∂ x - 1 2 σ ( x + tf ( x ) + A 2 t 2 ) u∂ u . This leads to (4.8). / Observe that in both cases ˜ u ² ( x, 0) = e - c²x u ( x, 0) for constant c. So we can seek a fundamental solution with the property that Z -∞ e - iλy u ( y, 0) p ( x, y, t ) dy = ˜ u c ( x, t ) . (4.11) In other words, we can obtain Fourier transforms of fundamental solutions. We present some examples. Example 4.1. We will obtain the transition density of the process X = { X t : t 0 } where dX t = 2 aX t ce 2 aX t - 1 ce 2 aX t + 1 dt + 2 dW t , (4.12) for a, c R . The Kolmogorov equation is u t = u xx + 2 ax ce 2 ax - 1 ce 2 ax +1 · u x and we seek a fundamental solution which integrates to 1. Observe that the drift satisfies f 0 + 1 2 f 2 = 2 a 2 . Using the stationary solution u 0 = 1 and Proposition 4.1 leads to the symmetry solution U λ ( x, t ) = e - λ 2 t - ( x - 2 at ) 1 + ce 2 ax - 4 aiλt 1 + ce 2 ax . (4.13)
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FUNDAMENTAL SOLUTIONS 23 Since U λ ( x, 0) = e - iλx we seek a fundamental solution p ( x, y, t ) such that R -∞ e - iλy p ( x, y, t ) dy = U λ ( x, t ) . The Fourier inversion immedi- ately gives the result. The integrals are standard Gaussians and we have p ( x, y, t ) = 1 2 π Z -∞ e iλy e - λ 2 t - ( x - 2 at ) 1 + ce 2 ax - 4 aiλt 1 + ce 2 ax = K ( x - 2 at - y, t ) + ce 2 ax K ( x + 2 at - y, t ) 1 + ce 2 ax , (4.14) where K ( x - y, t ) = 1 4 πt e - ( x - y ) 2 4 t . The reader can check that this is a fundamental solution and since R -∞ p ( x, y, t ) dy = U 0 ( x, t ) = 1, it is a density. In fact it is the transition density for the process. An interesting feature of this Fourier transform analysis is that it is not always optimal to use stationary solutions. In the next set of examples, we use nonstationary solutions, for the simple reason that they lead to Fourier transforms which are easier to invert. Example 4.2. We will obtain the transition density of a mean revert- ing Ornstein-Uhlenbeck process X = { X t : t 0 } , dX t = ( a - bX t ) dt + 2 σdW t , (4.15) with a 0 , b > 0. If a = 0, then we have the regular Ornstein- Uhlenbeck process. The Kolmogorov forward equation is u t = σu xx + ( a - bx ) u x , x R . (4.16) Observe that (4.16) has a symmetry of the form ˜ u ² ( x, t ) = e a b - a ² σ + σ 2 b + ² ( a - b x + b ² sinh( b t )) e b t σ u ( x - 2 ² sinh( b t ) , t ) . (4.17) Now ˜ u ² ( x, 0) = e x - b x ² σ u ( x, 0). We will let ² iσλ b . Then we look for a fundamental solution p ( x, y, t ) and a solution u such that Z -∞ e - iλy u ( y, 0) e y p ( x, y, t ) dy = ˜ u iσλ b ( x, t ) . (4.18) We could use u = 1, but as illustration, instead we seek a solution of (4.16) of the form u ( x, t ) = e m ( t ) x + z ( t ) . Substitution into the PDE gives u ( x, t ) = e - 2 e b t ( a - b x )+ σ 2 b e 2 b t . Using this solution and the given symmetry, we seek a fundamental solution such that Z -∞ e - iλy e y p ( x, y, t ) dy = e - λ 2 σ 2 sinh( b t ) b + i λ ( σ + e b t ( a - b x - e b t ( a + σ ) )) b e 2 b t + r ( x,t ) with r ( x, t ) = - 2 e b t ( a - b x ) - σ 2 b e 2 b t . We can now recover p by taking the inverse Fourier transform. The integrals are standard Gaussians and we leave
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24 MARK CRADDOCK the details to the reader. The result is that p ( x, y, t ) = be bt 2 p 4 πσ sinh( bt ) exp - e - bt ( a ( e bt - 1) +
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