In the first case we find that there is an

• 26

This preview shows pages 22–25. Sign up to view the full content.

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)

This preview has intentionally blurred sections. Sign up to view the full version.

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
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) +

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Fall '16
• Dr Salim Zahir
• Fourier Series, Dirac delta function, fundamental solution

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern