# 4 fourier transforms of fundamental solutions we have

• 26

This preview shows pages 20–23. Sign up to view the full content.

4. Fourier Transforms of Fundamental Solutions We have so far dealt only with PDEs defined over the positive half line. When the domain of the PDE is the whole of R , we can re- cover fundamental solutions by inverting a Fourier transform. We il- lustrate the procedure for an interesting subclass of the equations we have looked at. Consider as motivation the heat equation. This has a well known symmetry ˜ u ² ( x, t ) = 1 1 + 4 ²t exp - ²x 2 1 + 4 ²t u x 1 + 4 ²t , t 1 + 4 ²t . (4.1) Take u = 1 . The general method we have introduced implies that we should look for a fundamental solution K ( x, y, t ) with the property that Z -∞ e - ²y 2 K ( x, y, t ) dy = 1 1 + 4 ²t exp - ²x 2 1 + 4 ²t . (4.2) It is easy to verify that K ( x, y, t ) = 1 4 πt e - ( x - y ) 2 4 t is a solution of this equation. However, if we did not know K to begin with, it is not clear how it can be extracted from (4.2). This integral equation does not even have a unique solution, since for ² > 0 and any bounded,

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

FUNDAMENTAL SOLUTIONS 21 continuous odd function h we have R -∞ e - ²y 2 ( K ( x, y, t ) + h ( y )) dy = R -∞ e - ²y 2 K ( x, y, t ) dy. A method for extracting the heat kernel from (4.1) using two station- ary solutions is given in [7]. However, the heat equation also possesses a symmetry of the form ˜ u ² ( x, t ) = e - ²x + ² 2 t u ( x - 2 ²t, t ) . If we take ² = and let u = 1, then this symmetry leads to Z -∞ e - iλy K ( x, y, t ) = e - iλx - λ 2 t . (4.3) So we have obtained the Fourier transform of the heat kernel from a Lie group symmetry. The Fourier inversion theorem easily gives the one dimensional heat kernel. This method applies to more problems than the heat equation. When- ever we have a PDE u t = σu xx + f ( x ) u x - g ( x ) u, x R (4.4) with a six dimensional Lie algebra of symmetries, we may construct Fourier transforms of fundamental solutions. We have seen conditions on the drift f which guarantee the existence of a nontrivial symmetry group. Proposition 4.1 is a special case of our earlier results, which we have not yet exploited. It guarantees the existence of a six dimen- sional symmetry group for equations of the form (4.4) and provides the symmetries that will be used to produce Fourier transforms. Proposition 4.1. Let the drift function f in equation (4.4) satisfy the Riccati equation σf 0 + 1 2 f 2 + 2 σg = 1 2 Ax 2 + Bx + C (4.5) where A, B, C are arbitrary constants. Then equation (4.4) has a six dimensional Lie algebra of symmetries. Moreover if A 6 = 0 , then it has a symmetry of the form ˜ u ² ( x, t ) = e - σ cosh( At ) x + 2 2 σ sinh(2 At )+ σ A (1 - cosh( At )) × e 1 2 σ ( F ( x - 2 ² sinh( At )) - F ( x ) ) u ( x - 2 ² sinh( At ) , t ) . (4.6) If f satisfies the special case σf 0 + 1 2 f 2 + 2 σg = Ax + B (4.7) then it has a symmetry of the form ˜ u ² ( x, t ) = e - ²x 2 σ + ² 2 t 4 σ - 4 σ t 2 + 1 2 σ ( F ( x - ²t ) - F ( x )) u ( x - ²t, t ) . (4.8) In both cases F 0 ( x ) = f ( x ) .
22 MARK CRADDOCK Proof. Let v = ξ∂ x + τ∂ t + φ∂ u . Lie’s method shows that v generates symmetries if and only if ξ = 1 2 t + ρ , φ ( x, t, u ) = α ( x, t ) u where α = - x 2 8 σ τ tt - x 2 σ ρ t - 1 4 σ ( xf ( x )) τ t - 1 2 σ ρ + η (4.9) and - x 2 8 σ τ ttt - x 2 σ ρ tt + η t = - 1 4 τ tt - 1 2 σ ( σ ( xf ) 00 + f ( xf ) 2 + 2( σxg ) 0 ) τ t - 1 2 σ ( σf 00 + ff 0 + 2 σg 0 ) ρ. (4.10) The Lie algebra of symmetries is six dimensional if and only if ρ is nonzero. This occurs when f satisfies the given Riccati equations. In

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