ENGG 2420: Fourier series and partial differential equations

ENGG 2420: Fourier series and artial differential equations Prof. Thierry Blu e-mail: tblu@ee.cuhk.edu.hk web: www.ee.cuhk.edu.hk/~tblu/ Deartment of Electronic Engineering The Chinese University of Hong Kong August 2014 Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 1/50

Outline 1 Fourier Series Trigonometric and Exonential Series Fourier series aroximation 2 General considerations on PDE s The Wave Equation The Heat Equation Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 2/50

Periodicity Fourier Series Trigonometric and Exonential Series Fourier series aroximation Definition A function f(x) is said to be -eriodic iff f(x+) = f(x) for all x R. The real number > 0 is the eriod of f. Proerties Linearity: if f(x) and g(x) are eriodic with same eriod, then af(x)+bg(x) is also -eriodic Product: if f(x) and g(x) are eriodic with same eriod, then f(x)g(x) is also -eriodic NOTE: if f(x) and g(x) are eriodic, but with a different eriod, then in general af(x)+bg(x) is not eriodic. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 3/50

Periodicity Fourier Series Trigonometric and Exonential Series Fourier series aroximation Examles of eriodic functions f(x) = cos(ωx) and sin(ωx) are eriodic with = 2π ω f(x) = tan(ωx) is eriodic with = π ω f(x) = x x is eriodic with = 1 Exercise: show that, if f(x) is l -eriodic and g(x) is m -eriodic where l and m are integers, then af(x)+bg(x) is n -eriodic, where n is the least common multile of l and m. x denotes the integer art of x; i.e., the largest integer that is smaller than x. Examles: 1.5 = 1, 1.5 = 2, 1 = 1, etc. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 4/50

Comlex Fourier series Trigonometric and Exonential Series Fourier series aroximation Exonential Fourier series Comlex exonentials are eriodic functions; e.g., e 2iπx/ and in general e 2inπx/ (n integer) are -eriodic functions. Thus, series of the form f 0 (x) = + n= define functions that are -eriodic as well. c n e i 2πnx, with c n comlex-valued NOTE: Kreyszig 10e does not mention exonential Fourier series anymore. However, it is usually much simler to work with exonentials (differentiation, addition) than with the equivalent trigonometric functions. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 5/50

Real Fourier series Fourier Series Trigonometric and Exonential Series Fourier series aroximation Trigonometric Fourier series Using Euler s relation e iθ = cosθ+isinθ and the symmetries of cos and sin, the exonential Fourier series can be exressed as (exercise: do it!) + ( ) 2πnx + ( ) 2πnx f 0 (x) = a 0 + a n cos + b n sin n=1 n=1 where a 0 = c 0, and when n 0, a n = c n +c n and b n = i(c n c n ). This series is an examle of a trigonometric series. NOTE: The trigonometric reresentation is useful to outline the arity or real-valuedness of the function. For most comutations, however, the exonential Fourier series is simler to maniulate. cos( θ) = cos(θ) and sin( θ) = sin(θ). Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 6/50

Functions defined by a Fourier series Trigonometric and Exonential Series Fourier series aroximation A Fourier series generates a well-defined, continuous function whenever the series is absolutely convergent exonential Fourier series: n c n < trigonometric Fourier series: n a n < and n b n < NOTE: This condition is sufficient, but by no means necessary. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 7/50

Functions defined by a Fourier series Trigonometric and Exonential Series Fourier series aroximation Examle: for = 1, a n = 0, and b n = sin(n2 ) n α for various values of α Slower convergence usually leads to more irregular functions. 1.5 1 0.5 0 0.5 1 1.5 1.4 1.6 0.8 1 1.2 α 0 0.2 0.4 x 0.6 0.8 1 1.8 NOTE: For 0 < α 1, the series is not absolutely convergent. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 8/50

Fourier series Fourier Series Trigonometric and Exonential Series Fourier series aroximation Proerties of the exonential Fourier series The exonential Fourier series f 0 (x), entirely characterized by its Fourier coefficients, c n, is such that f 0 (x) is real-valued iff c n = c n (Hermitian symmetry) Shift of f 0 (x) by x 0 amounts to multilication of c n by e 2iπnx 0 Differentiation of f 0 (x) amounts to multilication of c n by 2iπn Integration of f 0 (x) amounts to multilication of c n by 2iπn (n 0) What haens to the coefficient c 0? Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 9/50

Fourier series Fourier Series Trigonometric and Exonential Series Fourier series aroximation Orthogonality The Fourier series decomositions are orthogonal for the scalar roduct f,g = 1 f(x)g(x)dx. This means that 0 Exonential series: { e 2in πx,e 2inπx 1 if n = n = 0 if n n Trigonometric series: cos ( ) ( 2n πx,cos 2nπx ) = { ( sin 2n ) ( πx,sin 2nπx ) 1/2 if n = n = and cos ( 2n πx ),sin ( 2nπx ) = 0, for all n,n 0 0 if n n Question Is it ossible to exress any -eriodic function as a Fourier series? Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 10/50

Trigonometric and Exonential Series Fourier series aroximation Decomosition theorem (exonential Fourier series) Let f(x) be a -eriodic function that is square integrable over [0,]. If c n = 1 f(x)e i2πnx dx 0 then the exonential Fourier series f 0 (x) = equals f(x) almost everywhere; i.e., + n= c n e i2πnx 0 f(x) f 0(x) 2 dx = 0. NOTE: almost everywhere equality is not ointwise convergence. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 11/50

Trigonometric and Exonential Series Fourier series aroximation No comlete roof given but an indication why c n should be of this form. If f(x) = f 0 (x) then: 1 0 2πnx f(x)e i dx = 1 = 1 = f 0(x)e 0 ( + 0 + k= = c n = c n k= c k 0 dx i 2πnx dx ) c k e i2πkx 2πkx ei e i2πnx e i2πnx dx 0 }{{} = 0 if k n dx Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 12/50

Trigonometric and Exonential Series Fourier series aroximation Decomosition theorem (trigonometric Fourier series) Let f(x) be a -eriodic function that is square integrable over [0,]. If a 0 = 1 0 f(x)dx a n = 2 ( 2πnx f(x)cos 0 b n = 2 ( 2πnx f(x)sin 0 then the exonential Fourier series + ( 2πnx f 0 (x) = a n cos n=0 equals f(x) almost everywhere; i.e., ) dx, n = 1,2,... ) dx, n = 1,2,... ) + + n=1 ( 2πnx ) b n sin 0 f(x) f 0(x) 2 dx = 0. Proof: for n 0, by using a n = c n +c n, b n = i(c n c n ) and Euler s relations. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 13/50

Trigonometric and Exonential Series Fourier series aroximation What are the exressions of the Fourier coefficients if the eriodic function is known over another interval than [0, ]? Decomosition theorem Let f(x) be a -eriodic function that is square integrable over [a,b] (with b a = ). Then the trigonometric and comlex-exonential Fourier-series coefficients are given by a 0 = 1 a n = 2 b n = 2 c n = 1 b a f(x)dx b ( 2πnx f(x)cos a ( 2πnx f(x)sin a b b a 2πnx f(x)e i ) dx, n = 1,2,... ) dx, n = 1,2,... dx, n Z Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 14/50

Trigonometric and Exonential Series Fourier series aroximation Proof: for a -eriodic function g(x), we have that a+ a g(x)dx = = = = 0 a+ g(x)dx+ g(x)dx+ g(x) dx a 0 }{{} change variables x = x + a g(x)dx+ g(x)dx+ a 0 0 g(x +)dx }{{} eriodicity: g(x +) = g(x ) a g(x)dx+ g(x)dx+ a 0 0 g(x )dx 0 0 0 g(x)dx Then, it suffices to aly this result either to f(x), f(x)cos ( ) 2πnx, f(x)sin ( ) 2πnx, or f(x)e i 2πnx to obtain the result. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 15/50

Trigonometric and Exonential Series Fourier series aroximation Examle: Fourier series of the 1-eriodic square wave { 1, 0 x < 1/2 Function to reresent: f(x) =, with eriodicity = 1. 1, 1/2 x < 1 1 Comutation of the exonential Fourier coefficients 1 1 c n = 0 f(x)e 2inπx 2 1 dx = 0 e 2inπx dx e 2inπx dx 1 2 = 1 e inπ e inπ e 2inπ { 0 if n is even = 2inπ 2inπ 2i nπ if n is odd 2 Comutation of the trigonometric coefficients: a 0 = c 0 = 0 a n = c n +c n = 0, { for n 1 0 if n is even b n = i(c n c n) = 4 for n 1 and n odd nπ + 4 3 Final result: f(x) = nπ sin(2πnx) n = 1 n odd Reminder: e inπ = ( 1) n and e 2inπ = 1. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 16/50

Trigonometric and Exonential Series Fourier series aroximation Examle: Fourier series of the 1-eriodic sawtooth wave Function to reresent: f(x) = x, where 1/2 x < 1/2, with eriodicity = 1. 1 Comutation of the trigonometric coefficients : a 0 = 2 1/2 1/2 xdx = 0 1/2 for n 1, a n = 2 xcos(2nπx)dx 1/2 = 2 xsin(2nπx) 1/2 2nπ 2 1/2 = 0 1/2 for n 1, b n = 2 xsin(2nπx)dx 1/2 = 2 xcos(2nπx) 1/2 2nπ +2 1/2 1/2 = ( 1)n+1 nπ + ( 1) n+1 2 Final result: f(x) = sin(2πnx) nπ n=1 1/2 1/2 Reminder: cos(nπ) = ( 1) n and sin(nπ) = 0. 1/2 sin(2nπx) dx 2nπ cos(2nπx) dx 2nπ (by arts) (by arts) Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 17/50

Trigonometric and Exonential Series Fourier series aroximation 1-eriodic square-wave Fourier series aroximation 1.5 Fourier series truncated at n = 99 1 0.5 0 0.5 1 1.5 1 0.5 0 0.5 1 1.5 2 Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 18/50

Trigonometric and Exonential Series Fourier series aroximation Exercises Comute the exonential and trigonometric Fourier series of 1 f(x) = cos(3x+1) over [0,2π] 2 f(x) = cos 2 (x) over [0,2π] 3 f(x) = cos(3x+1) over [0,π] 4 f(x) = e x 1 over [0,2] 5 f(x) = +e x+2 +e x+1 +e x +e x 1 + over [0,1] Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 19/50

Link with ower series Trigonometric and Exonential Series Fourier series aroximation Consider a comlex function, g(z), analytic in the oen disk of center 0 and radius 1. It is known to be equal to its Taylor series g(z) = g (n) (0) z n n! n 0 Thanks to the convergence on the unit circle, we also have that g(e i 2πx g (n) (0) ) = e i2πnx n! n 0 which shows that the exonential Fourier series decomosition of the -eriodic (comlex) function f(x) = g(e i2πx ) is given by { 0, n 1 c n = g (n) (0) n!, n 0. NOTE: coefficients with negative indices can be obtained from the comlex conjugate of analytic functions. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 20/50

Link with ower series Trigonometric and Exonential Series Fourier series aroximation Exercises Comute the exonential and trigonometric Fourier series of 1 f(x) = ln(1 e ix /2) over [0,2π] 2 f(x) = (1 e ix /2) 1/2 over [0,2π] 3 f(x) = e eix over [0,2π] Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 21/50

Truncated Fourier series Trigonometric and Exonential Series Fourier series aroximation A truncated Fourier series is the artial sum f N (x) = c n e 2iπnx, with c n = 1 Proerties n N f(x)e i2πnx 0 dx. The mean-square difference between f(x) and f N (x) is given by 1 f(x) f N(x) 2 dx = c n 2 0 n N+1 f N (x) is the orthogonal rojection of f(x) onto the exonential basis e 2iπnx/, for n N f N (x) minimizes mean-square difference between f(x) and any Fourier series truncated at n N Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 22/50

Gibbs henomenon Trigonometric and Exonential Series Fourier series aroximation The convergence of the truncated Fourier series near a discontinuity exhibits a constant overshoot of aroximately 9% of the jum, known as Gibbs henomenon (see slide 18). 1.5 1 0.5 Fourier series truncated at n = 99 overshoot 0 0.5 1 1.5 1 0.5 0 0.5 1 1.5 2 Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 23/50

Trigonometric and Exonential Series Fourier series aroximation Pointwise convergence of the Fourier series Dirichlet conditions Under technical conditions (finite number of discontinuities and extrema, boundedness), the truncated Fourier series converges to the function at each oint where it is continuous to the mean of the left and right limits of the function at each oint where it is discontinuous Examle: The 1-eriodic square-wave is discontinuous at 0, 1/2 and 1. At these oints, the Fourier series + 4 nπ sin(2πnx) n=1,n odd converges to 0, and at the other oints, it converges to the square wave. NOTE: In the general case, convergence for almost every value of x was roven only recently (Carleson, 1966). Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 24/50

Parity Fourier Series Trigonometric and Exonential Series Fourier series aroximation even/odd functions A function f(x) is said to be even whenever f(x) = f( x). A function f(x) is said to be odd whenever f(x) = f( x). Theorem If f(x) is an even -eriodic function, then its trigonometric Fourier series does not contain any sin terms + ( ) 2πnx f(x) = a 0 + a n cos n=1 If f(x) is an odd -eriodic function, then its trigonometric Fourier series contains only sin terms + ( ) 2πnx f(x) = b n sin n=1 Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 25/50

Parity Fourier Series Trigonometric and Exonential Series Fourier series aroximation Let us define c n = 1 Fourier coefficients /2 0 f(x)e 2iπnx/ dx. The comlex Fourier coefficients of an even function are real-valued and result from /2 c n = 2 f(x)cos ( 2πnx/ ) dx 0 = c n +c n odd function are ure imaginary numbers and result from /2 c n = i 2 f(x)sin ( 2πnx/ ) dx 0 = c n c n Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 26/50

Signal extensions Fourier Series Trigonometric and Exonential Series Fourier series aroximation When a signal is known only over half of its eriod [0,/2], it is customary to extend it 1 either by assuming even arity 2 or by assuming odd arity Then, its Fourier coefficients are given by the even/odd formulas of revious slide. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 27/50

Signal extensions Fourier Series Trigonometric and Exonential Series Fourier series aroximation Examle: extensions of f(x) = e x /2 Comutation of c n = 1 e x e 2iπnx/ dx 0 = 1 [ ] e (1 2iπn/)x /2 1 2iπn/ 0 = ( 1)n e /2 1 = ( 1)n e /2 1 2iπn 2 +4π 2 n 2 (+2iπn) Even case: c n = c n +c n = 2( 1)n e /2 1 2 +4π 2 n 2 Odd case: c n = c n c n = i4πn( 1)n e /2 1 2 +4π 2 n 2 Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 28/50

Partial differential equations General considerations on PDE s The Wave Equation The Heat Equation A artial differential equation (PDE) is an equation involving a function of several variables and its artial derivatives. A PDE is characterized by its linearity: the function/its derivatives are only involved with ower 1; order: the maximum differentiation order; Origin of PDE s Conservation laws: e.g., light intensity of moving objects df(x,y,z,t) = 0 f dt t +u f x x +u f y y +u f z z = 0 if u x = dx/dt, u y = dy/dt and u z = dz/dt. Physical fields (electromagnetism, gravitation): least-action rincile Geometry: e.g., iso-surfaces, caustics. Image rocessing: otimization rocesses. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 29/50

General considerations on PDE s The Wave Equation The Heat Equation Examle of PDE s Partial differential equations in 3 satial dimensions (x, y, z) with, ossibly, the time dimension, t: 1 Heat equation: f t = k ( f xx +f yy +f zz ) where f(x,y,z,t) is the temerature measured in Kelvins; 2 Lalace equation: f xx +f yy +f zz = 0 where f(x,y,z) is, e.g., a static electric otential without sources; 3 Poisson equation: f xx +f yy +f zz = ρ/ε where f(x,y,z) is, e.g., a static electric otential generated by a charge density ρ(x, y, z); 4 Wave equation: c 2( f xx +f yy +f zz ) ftt = 0 where f(x,y,z,t) is, e.g., a ressure wave roagating at seed c in a homogenous liquid. NOTE: f xx +f yy +f zz def = 2 f is the 3D Lalacian; 1D and 2D versions of these PDEs can be obtained by relacing it with f xx +f yy or f xx. In this course, however, we will limit the dimensions to two. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 30/50

Homogenous linear PDE s General considerations on PDE s The Wave Equation The Heat Equation A linear sace of solutions Homogenous linear PDE s have the roerty that if f and g are solutions, then af +bg is also a solution of the PDE (a,b scalars). Examle of roof with Lalace equation: from f xx +f yy +f zz = 0 g xx +g yy +g zz = 0 we get a(f xx +f yy +f zz )+b(g xx +g yy +g zz ) = 0 (af xx +bg xx )+(af yy +bg yy )+(af zz +bg zz ) = (af +bg) xx +(af +bg) yy +(af +bg) zz = Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 31/50

General considerations on PDE s The Wave Equation The Heat Equation Boundary conditions Functional conditions PDE s of order N require N initial conditions in order to be solved comletely. Tyically, these initial conditions consist of differential equations satisfied by the solution on a hyersurface; e.g., at a oint x 0 for PDE s in dimension 1 (i.e., ODE s); on a curve/line for PDE s in dimension 2; on a surface for PDE s in dimension 3. within a volume for PDE s in dimension 4 (e.g., time+sace). NOTE: Without initial conditions, the integration constants, usual in ODE s, become integration functions. Order N PDE s require N integration functions, in general. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 32/50

Boundary conditions Fourier Series General considerations on PDE s The Wave Equation The Heat Equation Examles: solve the following equations 1 f x (x,y) = 0 f(x,y) = C 1 (y) 2 f x (x,y,z) = 0 f(x,y,z) = C 1 (y,z) 3 f xx (x,y) = 0 f x (x,y) = C 1 (y) f(x,y) = xc 1 (y)+c 2 (y) { f(x,0) = x 4 f tt (x,t)+f(x,t) = 0 under the conditions f t (x,0) = x 2 General solution of y +y = 0 by denoting y(t) = f(x,t): y(t) = C 1 cost+c 2 sint, { hence f(x,t) = C 1 (x)cost+c 2 (x)sint. f(x,0) = C1 (x) = x Boundary conditions: f t (x,0) = C 2 (x) = x 2 Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 33/50

General considerations on PDE s The Wave Equation The Heat Equation Non-homogenous linear equations Non-homogenous linear equations can be ut under the form D{f} = g where D{f} is a linear differential oerator a linear combination of artial derivatives of f. General solution The solutions of D{f} = g are of the form f = f 0 +f 1 where f 0 is a articular solution of the PDE; i.e., D{f 0 } = g f 1 is any solution of the homogenous PDE; i.e., D{f 1 } = 0 Proof: If f 0 is such that D{f 0 } = g, then by linearity D{f f 0 } = 0 which means that f f 0 is a solution of the homogenous equation. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 34/50

General considerations on PDE s The Wave Equation The Heat Equation Aroaches for solving homogenous linear PDE s Solving PDE s in general cases is often very difficult even much more than ODE s. General techniques often consist in reverting to ODE s: 1 directly, when derivatives w.r.t. only one variable are involved 2 after a change of variables to revert to case 1 3 by looking for searable solutions (solutions of ODE s) able to reresent (by linear combination) all the solutions of the PDE For an examle of case 1, see slide 33. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 35/50

General considerations on PDE s The Wave Equation The Heat Equation Aroaches for solving homogenous linear PDE s Examle for case 2: f x (x,y)+f y (x,y) = 0. { x = u Solution: change of variables g(u,v) = f(u,u v). y = u v Then g u (u,v) = f x(u,u v)+f y (u,u v) = 0 g(u,v) = C 1 (v). f(x,y) = C 1 (x y). f Verification: x = C 1 (x y) f hence f x +f y = 0. y = C 1 (x y) Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 36/50

General considerations on PDE s The Wave Equation The Heat Equation Aroaches for solving homogenous linear PDE s Examle for case 3: f x (x,y)+f y (x,y) = 0. Solution: look for searable solutions f(x,y) = f 1 (x)f 2 (y). Then f x (x,y) = f 1 (x)f 2(y) and f y (x,y) = f 1 (x)f 2 (y). Hence f x +f y = 0 f 1 (x) f 1 (x) + f 2 (y) f 2 (y) = 0 f 1 (x) f 1 (x) = C 1 = f 2 (y) f 2 (y) in such a way that the searable solution has the form where C 1 is an arbitrary scalar. f(x,y) = e C1x e C1y = e C1(x y) Finally, a more general solution is obtained by linear combinations of such searable solutions f(x,y) = C 1 α C1 e C1(x y) = f 1 (x y). Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 37/50

General considerations on PDE s The Wave Equation The Heat Equation General solution for the 1D wave equation The 1D wave equation also known as the vibrating string equation is of the form f tt c 2 f xx = 0 Solution without initial conditions The solution of the 1D wave equation is given by f(x,t) = f 0 (ct x)+f 1 (ct+x) where f 0 and f 1 are arbitrary 1D functions. This solution is a suerosition of a wave roagating in the ositive x direction, and of a wave roagating in the negative x direction. at time t = 0 at time t = 5 at time t = 10 full wave forward roagating wave backward roagating wave Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 38/50

General considerations on PDE s The Wave Equation The Heat Equation General solution for the 1D wave equation { x = u+v u v Proof: By changing variables f(u+v, ct = u v c ) = g(u,v). ( ) Then g u (u,v) = f x u+v, u v c + 1 c f ( ) t u+v, u v c, and the second order artial derivative ( g uv = f xx 1 ) 1 c f xt +( c f tx 1 ) c 2f tt = f xx 1 c 2f tt = 0 Hence, g(u,v) = C 1 (u)+c 2 (v) which roves that f(x,y) is the sum of a function of ct x and a function of ct+x. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 39/50

General considerations on PDE s The Wave Equation The Heat Equation 1D wave equation with boundary+initial conditions Problem: Solve the 1D wave equation under the boundary conditions f(0,t) = 0 and { f(l,t) = 0, for all t 0 f(x,0) = f0 (x) and the initial value conditions f t (x,0) = f 1 (x) Solving method 1 Find all the solutions to the PDE that are searable, and that satisfy the boundary conditions 2 Exress arbitrary solutions to the PDE as linear combination of the searable solutions, and ensure that it satisfies the initial value conditions. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 40/50

General considerations on PDE s The Wave Equation The Heat Equation 1D wave equation: 1. Searable solutions Problem: Find the searable functions f(x, t) = u(x)v(t) such that f tt c 2 f xx = 0, and f(0,t) = f(l,t) = 0 for all t 0. Solution 0 = 2{ u(x)v(t) } { } u(x)v(t) t 2 c 2 2 x 2 = u(x)v (t) c 2 u (x)v(t) Assuming that there exist t 0 such that v(t 0 ) 0, then we obtain two second order ODE s u (x) v (t 0 ) c 2 v(t 0 ) u(x) = 0 and v (t) λc 2 v(t) = 0 }{{} λ General solution, without boundary conditions (yet) u(x) = u 1 e λx +u 2 e λx and v(t) = v 1 e c λt +v 2 e c λt, λ 0 If λ = 0, then u(x) = u 1 +u 2 x, and v(t) = v 1 +v 2 t. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 41/50

General considerations on PDE s The Wave Equation The Heat Equation 1D wave equation: 1. Searable solutions Enforcing the boundary conditions f(0,t) = f(l,t) = 0 for all t is equivalent to u(0) = u(l) = 0. Hence { { 0 = u1 +u 2 0 = u1 +u if λ 0, 0 = u 1 e L λ +u 2 e L λ or if λ = 0, 2 0 = u 1 +Lu 2 This has a non-zero solution only for λ 0 and e L λ = e L λ. This means that 2L λ = 2niπ, n 0 integer λ = n2 π 2 L. 2 Final solution, with boundary conditions ( ) u(x) = u 1 e i πnx L e i πnx L ( πnx ) and v(t) = v 1 e iπnct L +v 2 e iπnct L = Csin L where n 0. Basis of wave solutions satisfying boundary conditions ( f n (x,t) = e i πnct πnx ) L sin, n Z\{0} L NOTE: Emhasis on exonential Fourier series, contrary to Kreyszig 10e. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 42/50

General considerations on PDE s The Wave Equation The Heat Equation 1D wave equation: 2. Comlete resolution Problem: Find the comlete solution of the 1D wave equation with the boundary+initial conditions { f(x,0) = f0 (x) and f(0,t) = f(l,t) = 0 for all t 0. f t (x,0) = f 1 (x) Solution Exress the solution as a linear combination of searable solutions ( πnx ) sin L f(x,t) = b n e i πnct L n 0 This exression automatically satisfies both the wave equation, and f(0,t) = f(l,t) = 0. The initial conditions at t = 0 f 0 (x) = n 0b ( πnx ) n sin and f 1 (x) = i πnc ( πnx ) L L b nsin L n 0 show that f 0 and f 1 have to be decomosed as an odd arity trigonometric Fourier series of eriod = 2L. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 43/50

General considerations on PDE s The Wave Equation The Heat Equation 1D wave equation: 2. Comlete resolution Using the antisymmetry of the sinus f 0 (x) = n 1(b ( πnx ) n b n )sin L f 1 (x) = i πnc ( πnx ) L (b n +b n )sin L n 1 2 L ( πnx ) b n b n = L f 0(x)sin dx 0 L Hence b n +b n = 2i L ( πnx ) πnc f 1(x)sin dx 0 L General solution of the wave equation where b n = 2 L L 0 f(x,t) = n 0 ( f 0 (x) il ) πnc f 1(x) sin b n e i πnct L ( πnx ) sin L ( πnx ) dx. L Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 44/50

General considerations on PDE s The Wave Equation The Heat Equation 1D wave equation with initial conditions Problem: Solve the 2D wave equation under the initial value conditions f(x,0) = f 0 (x) and f t (x,0) = f 1 (x) D Alembert solution The full solution of the wave equation with initial value conditions is f(x,t) = f 0(x ct)+f 0 (x+ct) 2 Proof: by direct check (do it!). + 1 2c x+ct x ct f 1(u)du NOTE: No boundary conditions here. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 45/50

General considerations on PDE s The Wave Equation The Heat Equation The heat equation The heat equation is at the genesis of Fourier reresentations: Joseh Fourier invented Fourier series to solve the roblem of heat roagation (1822). 1D heat roagation roblem The 1D heat equation takes the form f t = c 2 f xx with boundary conditions f(0,t) = f(l,t) = 0, for all t 0 and initial value condition f(x,0) = f 0 (x). Same solving method as for the wave equation. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 46/50

General considerations on PDE s The Wave Equation The Heat Equation 1D heat equation: 1. Searable solutions Problem: Find the searable functions f(x, t) = u(x)v(t) such that f t c 2 f xx = 0, and f(0,t) = f(l,t) = 0 for all t 0. Solution 0 = { u(x)v(t) } c 2 2{ u(x)v(t) } t x 2 = u(x)v (t) c 2 u (x)v(t) Assuming that there exist t 0 such that v(t 0 ) 0, then we obtain two ODE s u (x) v (t 0 ) c 2 v(t 0 ) u(x) = 0 and v (t) λc 2 v(t) = 0 }{{} λ General solution, without boundary conditions (yet) u(x) = u 1 e λx +u 2 e λx and v(t) = v 1 e λc2t, λ 0 If λ = 0, then u(x) = u 1 +u 2 x, and v(t) = v 1. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 47/50

General considerations on PDE s The Wave Equation The Heat Equation 1D heat equation: 1. Searable solutions Enforcing the boundary conditions f(0,t) = f(l,t) = 0 for all t is equivalent to u(0) = u(l) = 0. Hence { { 0 = u1 +u 2 0 = u1 +u if λ 0, 0 = u 1 e L λ +u 2 e L λ or if λ = 0, 2 0 = u 1 +Lu 2 This has a non-zero solution only for λ 0 and e L λ = e L λ. This means that 2L λ = 2niπ, n 0 integer λ = n2 π 2 L. 2 Final solution, with boundary conditions ( ) u(x) = u 1 e i πnx L e i πnx L ( πnx ) and v(t) = v 1 e n 2 π 2 c 2 t L 2 = Csin L where n 0. Basis of solutions satisfying boundary conditions ( f n (x,t) = e n2 π 2 c 2 t πnx ) L 2 sin, n Z + \{0} L Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 48/50

General considerations on PDE s The Wave Equation The Heat Equation 1D heat equation: 2. Comlete resolution Problem: Find the comlete solution of the 1D heat equation with the boundary+initial conditions f(x,0) = f 0 (x) and f(0,t) = f(l,t) = 0 for all t. Solution Exress the solution as a linear combination of searable solutions f(x,t) = ( b n e n2 π 2 c 2 t πnx ) L 2 sin L n 1 This exression automatically satisfies both the heat equation, and f(0,t) = f(l,t) = 0. The initial condition at t = 0 f 0 (x) = n 1b ( πnx ) n sin L show that f 0 and f 1 have to be decomosed as an odd arity trigonometric Fourier series of eriod = 2L. Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 49/50

General considerations on PDE s The Wave Equation The Heat Equation 1D heat equation: 2. Comlete resolution General solution of the heat equation where b n = 2 L f(x,t) = n 1 ( πnx f 0(x)sin 0 L L b n e n2 π 2 c 2 t L 2 ) dx. ( πnx ) sin L Prof. Thierry Blu ENGG 2420: Fourier series and artial differential equations 50/50