Orthogonal Functions Fourier Series Summary Chapter 11: Fourier Series Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw December 13, 2013 1 / 44 DE Lecture 13 王奕翔 王奕翔
Orthogonal Functions Fourier Series Summary Fourier Series is invented by Joseph Fourier, which basically asserts that most periodic functions can be represented by infinite sums of sine and cosine functions. Jean Baptiste Joseph Fourier, (1768 - 1830). 2 / 44 DE Lecture 13 王奕翔
Orthogonal Functions Boundary 3 / 44 how Fourier Series is motivated. Below, let’s try to follow Fourier’s steps in solving this problem and see the space) is general. Prior to Fourier, there is no known solution to the be derived from heat transfer theory. The above is called the Heat Equation , which can condition Initial Fourier Series condition DE Lecture 13 Summary Fourier’s Motivation: Solving the Heat Equation k ∂ 2 u ∂ x 2 = ∂ u Solve u ( x , t ) : 0 < x < L , t > 0 ∂ t , subject to : u (0 , t ) = 0 , u ( L , t ) = 0 , t > 0 u ( x , 0) = f ( x ) , 0 < x < L u = 0 u = 0 BVP if f ( x ) (initial temperature distribution over x 0 L 王奕翔
Orthogonal Functions Boundary 4 / 44 Step 2 : Convert the original PDE into the following: (This approach was also taken by other predecessors like D. Bernoulli.) Fourier Series Initial condition condition DE Lecture 13 Summary Fourier’s Motivation: Solving the Heat Equation k ∂ 2 u ∂ x 2 = ∂ u Solve u ( x , t ) : 0 < x < L , t > 0 ∂ t , subject to : u (0 , t ) = 0 , u ( L , t ) = 0 , t > 0 u ( x , 0) = f ( x ) , 0 < x < L Step 1 : Assume that the solution takes the form u ( x , t ) = X ( x ) T ( t ) . { X ′′ + λ X ⇒ X ′′ X = T ′ = 0 kX ′′ T = XT ′ = kT = − λ = ⇒ T ′ + λ kT = 0 . Boundary condition becomes X (0) T ( t ) = X ( L ) T ( t ) = 0 . Since we want non-trivial solutions, T ( t ) ̸ = 0 = ⇒ X (0) = X ( L ) = 0 . 王奕翔
Orthogonal Functions Boundary 5 / 44 condition Initial Fourier Series condition DE Lecture 13 Summary Fourier’s Motivation: Solving the Heat Equation { X ′′ + λ X = 0 Solve u ( x , t ) = X ( x ) T ( t ) : T ′ + λ kT = 0 . subject to : X (0) = X ( L ) = 0 , u ( x , 0) = f ( x ) , 0 < x < L Step 3 : λ remains to be determined. What values should λ take? 1 λ = 0 : X ( x ) = c 1 + c 2 x . X (0) = X ( L ) = 0 = ⇒ c 1 = c 2 = 0 . 2 λ = − α 2 < 0 : X ( x ) = c 1 e − α x + c 2 e α x . Plug in X (0) = X ( L ) = 0 , we get c 1 = c 2 = 0 . 3 λ = α 2 > 0 : X ( x ) = c 1 cos ( α x ) + c 2 sin ( α x ) . Plug in X (0) = X ( L ) = 0 , we get c 1 = 0 , and c 2 sin ( α L ) = 0 . Hence, c 2 ̸ = 0 only if α L = n π . To obtain a non-trivial solution, pick λ = n 2 π 2 L 2 , n = 1 , 2 , . . . . 王奕翔
Orthogonal Functions Boundary 6 / 44 Step 5 : Plug in the initial condition = exp = Fourier Series condition Initial condition DE Lecture 13 Summary Fourier’s Motivation: Solving the Heat Equation X ′′ + λ X { = 0 Solve u ( x , t ) = X ( x ) T ( t ) : T ′ + λ kT = 0 . subject to : X (0) = X ( L ) = 0 , u ( x , 0) = f ( x ) , 0 < x < L Step 4 : Once we fix λ = n 2 π 2 L 2 , n = 1 , 2 , . . . , we obtain − kn 2 π 2 ( ) ( n π ) X ( x ) = c 2 sin T ( t ) = c 3 exp , L 2 t L x − kn 2 π 2 ( n π ( ) ) ⇒ u n ( x , t ) = A n sin , ( A n := c 2 c 3 ) L 2 t L x ( n π ) ⇒ f ( x ) = A n sin L x not true for general f ( x ) ! 王奕翔
Orthogonal Functions condition 7 / 44 Not likely N for any N exp Fourier Series N Step 6 : By the superposition principle, below satisfies the PDE. condition Initial DE Lecture 13 Boundary Summary Fourier’s Motivation: Solving the Heat Equation X ′′ + λ X { = 0 Solve u ( x , t ) = X ( x ) T ( t ) : T ′ + λ kT = 0 . subject to : X (0) = X ( L ) = 0 , u ( x , 0) = f ( x ) , 0 < x < L − kn 2 π 2 ( ) ( n π ) ∑ L 2 t A n sin L x n =1 ( n π ) ∑ The question is, can it satisfy u ( x , 0) = = f ( x ) ? A n sin L x n =1 王奕翔
Orthogonal Functions Fourier Series 8 / 44 This motivates the theory of Fourier Series . DE Lecture 13 Fourier’s Idea : How about an infinite series? If we can represent of sine functions. Summary Key Observation : f ( x ) is arbitrary and hence not necessarily a finite sum arbitrary f ( x ) by the infinite series (for 0 < x < L ) ∞ ( n π ) ∑ f ( x ) = , A n sin L x n =1 and we can find the values of { A n } , the problem is solved. 王奕翔
Orthogonal Functions Fourier Series Summary 1 Orthogonal Functions 2 Fourier Series 3 Summary 9 / 44 DE Lecture 13 王奕翔
Orthogonal Functions a 10 / 44 . a Fourier Series Once inner product is defined, we can accordingly define norm . Definition (Norm of a Function) Definition (Inner Product of Functions) Functions as Vectors: Inner Product Summary DE Lecture 13 The inner product of f 1 ( x ) and f 2 ( x ) on an interval [ a , b ] is defined as ∫ b ⟨ f 1 , f 2 ⟩ := f 1 ( x ) f 2 ( x ) dx The norm of a function f ( x ) on an interval [ a , b ] is √∫ b ( f ( x )) 2 dx √ || f ( x ) || := ⟨ f , f ⟩ = 王奕翔
Orthogonal Functions Fourier Series 11 / 44 Definition (Orthonormal Set) a DE Lecture 13 Definition (Orthogonal Set) Definition (Orthogonal Functions) Orthogonality of Functions Summary f 1 ( x ) and f 2 ( x ) are orthogonal on an interval [ a , b ] if ⟨ f 1 , f 2 ⟩ = 0 . { φ 0 ( x ) , φ 1 ( x ) , · · · } are orthogonal on an interval [ a , b ] if ∫ b ⟨ φ m , φ n ⟩ = φ m ( x ) φ n ( x ) dx = 0 , m ̸ = n . { φ 0 ( x ) , φ 1 ( x ) , · · · } are orthonomal on an interval [ a , b ] if they are orthogonal and || φ n ( x ) || = 1 for all n . 王奕翔
Orthogonal Functions a 12 / 44 a Fourier Series DE Lecture 13 Examples Summary Example (Orthogonal or Not Depends on the Inverval) The functions f 1 ( x ) = x and f 2 ( x ) = x 2 are orthogonal on the interval [ a , b ] , a < b , only if a = − b . Proof : When a < b , ∫ b [ 1 ] b = 1 a 4 − b 4 ) ⟨ x , x 2 ⟩ = x 3 dx = 4 x 4 ( = 0 ⇐ ⇒ a + b = 0 4 王奕翔
Orthogonal Functions a 13 / 44 since an exponential function is strictly monotone. a Fourier Series DE Lecture 13 Examples Example (Exponential Functions are Not Orthogonal) Summary For λ 1 , λ 2 ∈ R , f 1 ( x ) = e λ 1 x and f 2 ( x ) = e λ 2 x are not orthogonal on any interval [ a , b ] , a < b . Proof : If λ 1 = − λ 2 , ∫ b e ( λ 1 + λ 2 ) x dx = b − a ̸ = 0 . ⟨ e λ 1 x , e λ 2 x ⟩ = If λ 1 ̸ = − λ 2 , e ( λ 1 + λ 2 ) x dx = e ( λ 1 + λ 2 ) b − e ( λ 1 + λ 2 ) a ∫ b ⟨ e λ 1 x , e λ 2 x ⟩ = ̸ = 0 , λ 1 + λ 2 王奕翔
Orthogonal Functions dx dx Fourier Series cos L x L x L sin sin L x L sin L x 14 / 44 sin DE Lecture 13 The set of functions Example sin Summary Examples ( n π { ) } | n = 1 , 2 , . . . are orthogonal on [0 , L ] . L x ( n π ) Proof : Let φ n ( x ) := sin . For m ̸ = n , L x ∫ L ( m π ( n π ) ) ⟨ φ m , φ n ⟩ = L x L x 0 { ( ( m − n ) π ) ( ( m + n ) π )} ∫ L 1 = − cos 2 0 [ ( ( m − n ) π )] L = 2( m − n ) π 0 [ ( ( m + n ) π )] L − 2( m + n ) π 0 = 0 − 0 = 0 . 王奕翔
Orthogonal Functions We answer the former question later with a particular set of orthogonal 15 / 44 coefficient c m ! Fourier Series functions. DE Lecture 13 Summary Orthogonal Series Expansion Question : For a infinite orthogonal set { φ n ( x ) | n = 0 , 1 , . . . } on some interval [ a , b ] , can we expand an arbitrary function f ( x ) as ∞ ∑ f ( x ) = c n φ n ( x ) ? n =0 If so, how to find the coefficients { c n } ? For the latter, simply take the inner product ⟨ f , φ m ⟩ to find the ∞ ⇒ c m = ⟨ f , φ m ⟩ ∑ c n ⟨ φ n , φ m ⟩ = c m || φ m || 2 = ⟨ f , φ m ⟩ = || φ m || 2 . n =0 王奕翔
Orthogonal Functions . 16 / 44 exp dx , and L Fourier Series sin . DE Lecture 13 Coefficients in the Solution of the Heat Equation Recall in solving the Heat equation, the last step is to determine Summary ∞ ( n π ) ∑ { A n | n = 1 , 2 , . . . } such that f ( x ) = A n sin L x n =1 Based on the principle developed above, we obtain A n = ⟨ f ,φ n ⟩ || φ n || 2 , where ( n π ) φ n ( x ) := sin L x )) 2 dx = 1 { ( 2 n π )} ∫ L ( n π ∫ L || φ n || 2 = ( 1 − cos dx = L 2 . 2 L x L x 0 0 Hence, A n = 2 ∫ L ( n π ) f ( x ) sin L x 0 ∞ − kn 2 π 2 ( n π ( ) ) ∑ u ( x , t ) = L 2 t A n sin L x n =1 王奕翔
Orthogonal Functions Fourier Series 17 / 44 DE Lecture 13 Summary Remaining question: ∞ ( n π ) ∑ f ( x ) = A n sin L x n =1 Will the infinite series converge for x ∈ [0 , L ] ? Does it converge to the function f ( x ) for x ∈ [0 , L ] ? 王奕翔
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