1 Week 4 1. Fourier series: the definition and basics (continued) Fourier series can be written in a complex form. For 2π -periodic function, for example, where c 0, c 1, c –1, c 2, c –2, etc. are the complex Fourier coefficients. (1)

2 Theorem 1: The complex Fourier coefficients of a real function satisfy the condition Proof: Compare (1) with (1)* and take into account that f* = f – hence, On the l.-h.s. of this equality, change n → –n and, thus, obtain Theorem 1 follows from the above equality.

3 Theorem 2: The complex and real Fourier coefficients of the same function are related by Proof: This theorem follows from Euler’s formula: e iA = cos A + i sin A. The complex Fourier series of a 2L -periodic function is Theorem 3: (2)

4 Theorem 4: The complex Fourier coefficients are given by Proof (for simplicity, for L = π ): Multiply the Fourier series (2) by exp(–imx) and integrate the resulting equality with respect to x over (–π, π) : In this equality, subdivide the series on the r.-h.s. into three parts: –∞ < n ≤ m – 1, n = m, and m + 1 ≤ n < ∞, which yields...

5 Observe that Thus, both series on the r.-h.s. of (3) vanish and we can solve for c m (and obtain the desired expression for it). (3)

6 Theorem 5: Any doubly periodic function f(x, y) with periods L x and L y can be represented by a 2D Fourier series of the form (Replace the “...” with an appropriate expression at home.)

7 Theorem 6: The Fourier coefficients of a 2D Fourier series are given by (Replace the “...” with appropriate expressions at home.)

8 (4) (5) (6) Example 1: Consider the following (1+1)-dim. initial-boundary-value problem: 2. Solving PDEs using Fourier series

9 (7) To satisfy the BCs, we should eliminate the constant and cosines – hence, Seek the solution of the initial-boundary-value problem (4)-(6) as a Fourier series of a periodic function of period 2π : Substitution of solution (7) into Eq. (4) yields hence...

10 (8a) where A n and B n are undetermined constants. Next, extend the initial conditions H and H ’ to the interval (–π, π) as odd functions and represent them by their Fourier series, We’ll also need (8b) (9) Then, (7) yields

11 Comparing (8) and (9), we obtain and (8a) yields the ‘final answer’:

12 Comment: Replace BC (5) with In this case, the solution should be sought in the form and the initial conditions should be extended as even functions. Q: What happens if the BC is

13 (10) (11) Let the initial conditions be (12) Example 2: waves in a rectangular pond (from Week 2) Assume, for simplicity, that c = 1, L x = L y = π. Then the equation and BCs reduce to

14 Seek the solution of the initial-boundary-value problem (10)-(12) as a Fourier series of a doubly periodic function with both periods equal to 2π : To satisfy the BCs, we should eliminate all sines. (13)

15 To satisfy the ICs, we should eliminate all terms except those involving 1, cos 3x, cos 3y, and cos 3x cos 4y : (14) Substitution of solution (14) into Eq. (10) yields (15)

16 The initial conditions for these equations follow from (12): (16)

17 The solution of the initial-value problem (15)-(16) is (17) Summarising (14) and (17), we obtain