25.4. Solution Using Fourier Series. Introduction. Prerequisites. Learning Outcomes

Transcription

1 Soution Using Fourier Series 25.4 Introduction In this Section we continue to use the separation of variabes method for soving PDEs but you wi find that, to be abe to fit certain boundary conditions, Fourier series methods have to be used eading to the fina soution being in the (rather compicated form of an infinite series. The techniques wi be iustrated using the two-dimensiona Lapace equation but simiar situations often arise in connection with other important PDEs. Prerequisites Before starting this Section you shoud... Learning Outcomes On competion you shoud be abe to... HELM (2008: Section 25.4: Soution Using Fourier Series be famiiar with the separation of variabes method be famiiar with trigonometric Fourier series sove the 2-D Lapace equation for given boundary conditions and utiize Fourier series in the soution when necessary 35

2 1. Soutions invoving infinite Fourier series We sha iustrate this situation using Lapace s equation but infinite Fourier series can aso be necessary for the heat conduction and wave equations. We reca from the previous Section that using a product soution u(x, t = X(xY (y in Lapace s equation gives rise to the ODEs: X X = K Y Y = K To determine the sign of K and hence the appropriate soutions for X(x and Y (y we must impose appropriate boundary conditions. We wi investigate soving Lapace s equation in the square 0 x 0 y for the boundary conditions u(x, 0 = 0 u(0, y = 0 u(, y = 0 u(x, = U 0, a constant. See Figure 7. y u = U 0 u =0 u =0 u =0 x Figure 7 (a We must first deduce the sign of the separation constant K: if K is chosen to be positive say K = λ 2, then the X equation is X = λ 2 X with genera soution X = Ae λx + Be λx whie the Y equation becomes Y = λ 2 Y with genera soution Y = C cos λy + D sin λy If the sign of K is negative K = λ 2 the soutions wi change to trigonometric in x and exponentia in y. These are the ony two possibiities when we sove Lapace s equation using separation of variabes and we must ook at the boundary conditions of the probem to decide which is appropriate. Here the boundary conditions are periodic in x (since u(0, y = u(, y and non-periodic in y which suggests we need a soution that is periodic in x and non-periodic in y. 36 HELM (2008: Workbook 25: Partia Differentia Equations

3 Thus we choose K = λ 2 to give X(x = (A cos λx + B sin λx Y (y = (Ce λy + De λy (Note that had we chosen the incorrect sign for K at this stage we woud ater have found it impossibe to satisfy a the given boundary conditions. You might ike to verify this statement. The appropriate genera soution of Lapace s equation for the given probem is u(x, y = (A cos λx + B sin λx(ce λy + De λy. (b Inserting the boundary conditions produces the foowing consequences: u(0, y = 0 gives A = 0 u(, y = 0 gives sin λ = 0 i.e. λ = nπ where n is a positive integer 1, 2, 3,.... Whie n = 0 aso satisfies the equation it eads to the trivia soution u = 0 ony. u(x, 0 = 0 gives C + D = 0 i.e. D = C At this point the soution can be written ( enπy u(x, y = BC sin This can be convenienty written as ( nπy u(x, y = E sin sinh where E = 2BC. nπy e At this stage we have just one fina boundary condition to insert to obtain information about the constant E and the integer n. Our soution (1 gives u(x, = E sin sinh(nπ and ceary this is not compatibe, as it stands, with the given boundary condition u(x, = U 0 = constant. The way to proceed is again to superpose soutions of the form (1 for a positive integer vaues of n to give ( nπy u(x, y = E n sin sinh (2 n=1 from which the fina boundary condition gives U 0 = E n sin sinh(nπ 0 < x <. (3 n=1 = b n sin where b n = E n sinh(nπ. n=1 (1 HELM (2008: Section 25.4: Soution Using Fourier Series 37

4 What we have here is a Fourier (sine series for the function f(x = U 0 0 < x <. Recaing the work on haf-range Fourier series ( an odd function with period 2. Hence we define 23.5 we must extend this definition to produce f(x = { U0 0 < x < U 0 < x < 0 f(x + 2 = f(x iustrated in Figure 8. f(x U 0 2 x U 0 Figure 8 (c We can now appy standard Fourier series theory to evauate the Fourier coefficients b n in (3. We obtain b n = E n sinh nπ = 4U ( 0 nπx sin dx 2 0 (Reca that, in genera, b n = 2 the mean vaue of f(x sin over a period. Here, because ( nπx f(x is odd, and hence f(x sin is even, we may take haf the period for our averaging process. Carrying out the integration E n sinh nπ = 2U 0 nπ (1 cos nπ i.e. E n = 4U 0 nπ sinh nπ n = 1, 3, 5,... 0 n = 2, 4, 6,... (Since f(x is a square wave with haf-period symmetry we are not surprised that ony odd harmonics arise in the Fourier series. Finay substituting these resuts for E n into (2 we obtain the soution to the given probem as the infinite series: ( nπy u(x, y = 4U sin sinh 0 π n sinh nπ n=1 (n odd 38 HELM (2008: Workbook 25: Partia Differentia Equations

5 Task Sove Lapace s equation to determine the steady state temperature u(x, y in the semi-infinite pate 0 x 1, y 0. Assume that the eft and right sides are insuated and assume that the soution is bounded. The temperature aong the bottom side is a known function f(x. First write this probem as a mathematica boundary vaue probem paying particuar attention to the mathematica representation of the boundary conditions: Your soution Since the sides x = 0 and x = 1 are insuated, the temperature gradient across these sides is zero i.e. u u = 0 for x = 0, 0 < y < and = 0 for x = 1, 0 < y <. x x The third boundary condition is u(x, 0 = f(x. The fourth boundary condition is ess obvious: since the soution shoud be bounded (ie not grow and grow we must demand that u(x, y 0 as y. (See figure beow. y u x =0 u x =0 0 u = f(x 1 x HELM (2008: Section 25.4: Soution Using Fourier Series 39

6 Now use the separation of variabes method, putting u(x, y = X(xY (y, to find the differentia equations satisfied by X(x, Y (y and decide on the sign of the separation constant K: Your soution We have boundary conditions which, ike the worked exampe above, are periodic in x. Hence the differentia equations are, again, X = λ 2 X Y = +λ 2 Y putting the separation constant K as λ 2. Write down the soutions for X, for Y and hence the product soution u(x, y = X(xY (y: Your soution X = A cos λx + B sin λx so Y = Ce λy + De λy u = (A cos λx + B sin λx(ce λy + De λy (4 Impose the derivative boundary conditions on this soution: Your soution 40 HELM (2008: Workbook 25: Partia Differentia Equations

7 u x = ( λa sin λx + λb cos λx(ceλy + De λy Hence u x (0, y = 0 gives λb(ceλy + De λy = 0 for a y. The possibiity λ = 0 can be excuded this woud give a trivia constant soution in (4. Hence we must choose B = 0. The condition u (1, y = 0 gives x λa sin λ(ce λy + De λy = 0 Choosing A = 0 woud make u 0 so we must force sin λ to be zero i.e. choose λ = nπ where n is a positive integer. Thus, at this stage (4 becomes u = A cos nπx(ce nπy + De nπy = cos nπx(ee nπy + F e nπy (5 Now impose the condition that this soution shoud be bounded: Your soution The region over which we are soving Lapace s equation is semi-infinite i.e. the y coordinate increases without imit. The soution for u(x, y in (5 wi increase without imit as y due to the term e nπy (n being a positive integer. This can be avoided i.e. the soution wi be bounded if the constant E is chosen as zero. Finay, use Fourier series techniques to dea with the fina boundary condition u(x, 0 = f(x: Your soution HELM (2008: Section 25.4: Soution Using Fourier Series 41

8 Your soution Superposing soutions of the form (5 (with E = 0 gives u(x, y = F n cos(nπx e nπy (6 n=0 so the boundary condition gives f(x = F n cos nπx n=0 We have here a haf-range Fourier cosine series representation of a function f(x defined over 0 < x < 1. Extending f(x as an even periodic function with period 2 and using standard Fourier series theory gives with F n = 2 F 0 2 = f(x cos nπx dx n = 1, 2,... f(x dx. Hence (6 is the soution of this given boundary vaue probem, the integras giving us in principe the Fourier coefficients F n for a given function f(x. 42 HELM (2008: Workbook 25: Partia Differentia Equations