Lectures on Fractional Calculus - CARMA

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Green’s functions (2)• Once you have the Green’s function for apoint source, you can get the solution for anarbitrary set of sources by summing theGreen’s function multiplied by the strength ofthe source• You know well the potential for a pointelectrostatic charge in 3D:G(x, y, z) = 14πr , r = p x 2 + y 2 + z 2
The Green’s function in 2DWe start with(2π) 2 δ 2 (k x ,k y )= R ∞−∞ dxdyei(k xx+k y y)and∇ R 2 ∞−∞ dxdyei(k xx+k y y) =− R ∞−∞ dxdy(k2 x + k 2 y)e i(k xx+k y y)HenceG(x, y) = 14π 2 R ∞−∞R ∞−∞e i(k xx+ky y)(k 2 x +k2 y )Theintegralisdoneinpolarcoordinates:k x = r cos(θ), k y = r sin(θ).
Page 1 and 2: Three Lectures on
Page 3 and 4: References• A child’s garden of
Page 5 and 6: Motivation, History (2)• Initial
Page 7 and 8: Fractional Differentiation (1)d n x
Page 9 and 10: Factorial FunctionKey equations for
Page 11 and 12: A Fractional DifferentiationConundr
Page 13 and 14: Integration as NegativeDifferentiat
Page 15 and 16: Differentiation as Negative Integra
Page 17 and 18: Adding up Harmonics• Fourier seri
Page 19 and 20: Momentum space• A lot of physics
Page 21 and 22: The Laplacian (2)Once:n times:∇ 2
Page 23 and 24: The Dirac delta function (2)• The
Page 25: Green’s functions• A Green’s
Page 29 and 30: R ∞0The Green’s function in 2D
Page 31 and 32: A Neat TrickWe know the Green’s f
Page 33: Technical details (2)We write down
Page 36: • So we have deduced:G 2s (r) =

Lectures on Fractional Calculus - CARMA

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