Chapter 14: Fourier Transforms and Boundary Value Problems in an - PowerPoint PPT Presentation

Fourier Transforms BVP’s in an Unbounded Region Chapter 14: Fourier Transforms and Boundary Value Problems in an Unbounded Region Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw December 25, 2013 1 / 27 DE Lecture 15 王奕翔 王奕翔

Fourier Transforms Exception : a BVP on semi-finite plate 2 / 27 cosine/sine series in x . Note : we are able to solve this via Fourier series BVP’s in an Unbounded Region DE Lecture 15 bounded region, where the boundary conditions are given at finite boundaries, e.g., So far we have seen how to solve boundary-value problems within a (one-dimensional) heat equation, wave equation: x ∈ [0 , L ] (two-dimensional) Laplace’s equation: ( x , y ) ∈ [0 , a ] × [0 , b ] y because the homogeneous boundary conditions are still at finite boundaries x = 0 , a . r 2 u = 0 From the homogeneous boundary conditions, we u = 0 u = 0 can conclude that the solution u ( x , t ) is a Fourier x u = f ( x ) 王奕翔

Fourier Transforms BVP’s in an Unbounded Region 3 / 27 Separation of variables: unable to use the homogeneous boundary DE Lecture 15 What if the homogeneous boundary conditions are given at infinite boundaries , i.e., ±∞ ? y y u ( x, ∞ ) = 0 u ( x, ∞ ) = 0 r 2 u = 0 u = F ( y ) u = 0 r 2 u = 0 u = 0 u = F ( y ) x u = 0 u ( x, −∞ ) = 0 x conditions to find the possible values of the separation constant λ . 王奕翔

Fourier Transforms BVP’s in an Unbounded Region We introduce Fourier Transforms to deal with this issue. 4 / 27 DE Lecture 15 王奕翔

Fourier Transforms BVP’s in an Unbounded Region 1 Fourier Transforms 2 BVP’s in an Unbounded Region 5 / 27 DE Lecture 15 王奕翔

Fourier Transforms p 6 / 27 lim BVP’s in an Unbounded Region DE Lecture 15 From Fourier Series to Fourier Integral Recall : For a function f ( x ) defined on ( − p , p ) , its Fourier series is ( 1 ∫ p ) ∞ ∑ f ( x ) e − i n π e i n π p x dx p x 2 p − p n = −∞ Let α n := n π and ∆ α := α n +1 − α n = π p . Taking p → ∞ , ∆ α → 0 , and the Fourier series becomes F p ( α n ) � �� � (∫ p ) ∞ 1 ∑ f ( x ) e − i α n x dx e i α n x ∆ α 2 π lim p →∞ − p n = −∞ ∫ ∞ { } = 1 ∆ α = π p →∞ F p ( α ) e i α x d α p → 0 when p → ∞ 2 π −∞ ∫ ∞ {∫ ∞ } 1 f ( x ) e − i α x dx = e i α x d α 2 π −∞ −∞ 王奕翔

Fourier Transforms BVP’s in an Unbounded Region 7 / 27 DE Lecture 15 Definition (Fourier Integral and Fourier Transform) Fourier Integral and Fourier Transform The Fourier integral of a function f ( x ) defined on ( −∞ , ∞ ) is ∫ ∞ ∫ ∞ f ( x ) = 1 f ( x ) e − i α x dx . F ( α ) e i α x d α, where F ( α ) = 2 π −∞ −∞ The Fourier transform of f ( x ) is ∫ ∞ f ( x ) e − i α x dx := F ( α ) . F { f ( x ) } = −∞ The inverse Fourier transform of a function F ( α ) is ∫ ∞ F − 1 { F ( α ) } = 1 F ( α ) e i α x d α := f ( x ) . 2 π −∞ 王奕翔

Fourier Transforms Fourier Coefficients and Fourier Series 8 / 27 c n BVP’s in an Unbounded Region DE Lecture 15 Fourier and Inverse Fourier Transforms ∫ ∞ F f ( x ) e − i α x dx f ( x ) − → F ( α ) = −∞ ∫ ∞ f ( x ) = 1 F − 1 F ( α ) − − − → F ( α ) e i α x d α 2 π −∞ c n = 1 ∫ p FS f ( x ) e − i n π f ( x ) − − → p x dx 2 p − p ∞ FS − 1 ∑ c n e i n π − − − − → f ( x ) = p x n = −∞ 王奕翔

Fourier Transforms The Fourier transform 9 / 27 i The Fourier integral BVP’s in an Unbounded Region DE Lecture 15 Find the Fourier integral representation and the Fourier transform of the Examples Example { 1 , 0 < x < 2 function f ( x ) = otherwise . 0 , ∫ ∞ ∫ 2 1 e − 2 i α − 1 f ( x ) e − i α x dx = e − i α x dx = ( ) F ( α ) = F { f ( x ) } = − i α −∞ 0 ∫ ∞ ∫ ∞ f ( x ) = 1 F ( α ) e i α x d α = 1 e − 2 i α − 1 ( ) e i α x d α 2 π 2 π α −∞ −∞ 王奕翔

Fourier Transforms BVP’s in an Unbounded Region Sufficient Condition of Convergence Theorem (Convergence of Fourier Integral) its Fourier integral converges to 10 / 27 DE Lecture 15 Let f and f ′ be piecewise continuous on every finite interval and let f be ∫ ∞ absolutely integrable on ( −∞ , ∞ ) (i.e., −∞ | f ( x ) | dx converges). Then, f ( x ) at a point where f ( x ) is continuous 1 2 ( f ( x +) + f ( x − )) at a point where f ( x ) is discontinuous. 王奕翔

Fourier Transforms BVP’s in an Unbounded Region 11 / 27 DE Lecture 15 Alternative Form of the Fourier Integral ∫ ∞ ∫ ∞ f ( x ) = 1 F ( α ) e i α x d α = 1 F ( α ) ( cos α x + i sin α x ) d α 2 π 2 π −∞ −∞ ∫ 0 {∫ ∞ } = 1 F ( α ) ( cos α x + i sin α x ) d α + F ( α ) ( cos α x + i sin α x ) d α 2 π 0 −∞ {∫ ∞ ∫ ∞ } = 1 F ( α ) ( cos α x + i sin α x ) d α + F ( − α ) ( cos α x − i sin α x ) d α 2 π 0 0 ∫ ∞ = 1 { } [ F ( α ) + F ( − α )] cos α x + i [ F ( α ) − F ( − α )] sin α x d α 2 π 0 ∫ ∞ {[∫ ∞ ] [∫ ∞ ] } 1 = f ( x ) cos α x dx cos α x + f ( x ) sin α x dx sin α x d α π 0 −∞ −∞ ∫ ∞ ∫ ∞ ∫ ∞ f ( x ) e − i α x dx + F ( α ) + F ( − α ) = f ( x ) e i α x dx = 2 f ( x ) cos α x dx −∞ −∞ −∞ ∫ ∞ ∫ ∞ ∫ ∞ f ( x ) e − i α x dx − F ( α ) − F ( − α ) = f ( x ) e i α x dx = − 2 i f ( x ) sin α x dx −∞ −∞ −∞ 王奕翔

Fourier Transforms Fourier Integral of an Even Function 12 / 27 Fourier Integral of an Odd Function BVP’s in an Unbounded Region DE Lecture 15 Alternative Form of the Fourier Integral ∫ ∞ f ( x ) = 1 { A ( α ) cos α x + B ( α ) sin α x } d α, π 0 ∫ ∞ ∫ ∞ where A ( α ) = −∞ f ( x ) cos α x dx , B ( α ) = −∞ f ( x ) sin α x dx . ∫ ∞ f ( x ) = 1 { } ✘ A ( α ) cos α x + ✘✘✘✘ B ( α ) sin α x d α, π 0 ∫ ∞ ∫ ∞ where A ( α ) = 2 f ( x ) cos α x dx , B ( α ) = −∞ f ( x ) sin α x dx = 0 . 0 ∫ ∞ f ( x ) = 1 { ✭✭✭✭ A ( α ) cos α x + B ( α ) sin α x } d α, ✭ π 0 ∫ ∞ ∫ ∞ where A ( α ) = −∞ f ( x ) cos α x dx = 0 , B ( α ) = 2 f ( x ) sin α x dx . 0 王奕翔

Fourier Transforms BVP’s in an Unbounded Region 13 / 27 c DE Lecture 15 Definition (Fourier Cosine Integral and Fourier Cosine Transform) Fourier Cosine Integral and Fourier Cosine Transform The Fourier cosine integral of a function f ( x ) defined on (0 , ∞ ) is ∫ ∞ ∫ ∞ f ( x ) = 2 F ( α ) cos α x d α, where F ( α ) = f ( x ) cos α x dx . π 0 0 The Fourier cosine transform of f ( x ) is ∫ ∞ F c { f ( x ) } = f ( x ) cos α x dx = F ( α ) . 0 The inverse Fourier cosine transform of a function F ( α ) is ∫ ∞ { F ( α ) } = 2 F − 1 F ( α ) cos α x d α = f ( x ) . π 0 王奕翔

Fourier Transforms BVP’s in an Unbounded Region 14 / 27 s DE Lecture 15 Definition (Fourier Sine Integral and Fourier Sine Transform) Fourier Sine Integral and Fourier Sine Transform The Fourier sine integral of a function f ( x ) defined on (0 , ∞ ) is ∫ ∞ ∫ ∞ f ( x ) = 2 F ( α ) sin α x d α, where F ( α ) = f ( x ) sin α x dx . π 0 0 The Fourier sine transform of f ( x ) is ∫ ∞ F s { f ( x ) } = f ( x ) sin α x dx = F ( α ) . 0 The inverse Fourier sine transform of a function F ( α ) is ∫ ∞ { F ( α ) } = 2 F − 1 F ( α ) sin α x d α = f ( x ) . π 0 王奕翔

Fourier Transforms otherwise 15 / 27 BVP’s in an Unbounded Region DE Lecture 15 Example Find the Fourier integral representation of the function Examples { 1 , − a < x < a f ( x ) = . 0 , f ( x ) is an even function. Hence, its Fourier integral is a cosine integral ∫ ∞ {∫ ∞ f ( x ) = 2 } f ( x ) cos α x dx cos α x d α π 0 0 ∫ ∞ {∫ a } = 2 cos α x dx cos α x d α π 0 0 ∫ ∞ = 2 sin α a cos α x d α π α 0 王奕翔

Fourier Transforms (a) 16 / 27 (b) BVP’s in an Unbounded Region DE Lecture 15 by a Fourier sine integral. Examples Example On (0 , ∞ ) , represent f ( x ) = e − x (a) by a Fourier cosine integral, and (b) ∫ ∞ {∫ ∞ ∫ ∞ f ( x ) = 2 } cos α x d α = 2 1 e − x cos α x dx 1 + α 2 cos α x d α π π 0 0 0 ∫ ∞ {∫ ∞ ∫ ∞ } f ( x ) = 2 sin α x d α = 2 α e − x sin α x dx 1 + α 2 sin α x d α π π 0 0 0 王奕翔

Fourier Transforms BVP’s in an Unbounded Region 17 / 27 DE Lecture 15 y y x x (b) sine integral (a) cosine integral 王奕翔

Fourier Transforms is useful in solving BVP’s of PDE’s. 18 / 27 BVP’s in an Unbounded Region Fact DE Lecture 15 In this lecture we only focus on the Fourier transform of derivatives, as it them resemble those of the Laplace transform. The Fourier transform has many operational properties, and many of Fourier Transforms of Derivatives If f ( x ) , f ′ ( x ) → 0 as x → ±∞ , then F { f ′ ( x ) } = i α F { f ( x ) } , F { f ′′ ( x ) } = − α 2 F { f ( x ) } F s { f ′ ( x ) } = − α F c { f ( x ) } , F s { f ′′ ( x ) } = − α 2 F s { f ( x ) } + α f (0) F c { f ′ ( x ) } = α F s { f ( x ) } − f (0) , F c { f ′′ ( x ) } = − α 2 F c { f ( x ) } − f ′ (0) 王奕翔