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Fourier Series Chapter 1 Fourier Series 1.1 Motivation The motivation behind this topic is as follows, Joseph-Louis Fourier, (1768- 1830), a French engineer (and mathematician) discussed heat flow through a bar which gives rise to the so-called Heat Diffusion Problem, ∂2u 1 ∂u = (1.1) ∂x2 K ∂t where u = u(x, t),K>0 is a constant depending on the thermal properties of the bar, u(0,t)=0=u(L, t), and u(x, 0) = f(x), where f is given at the outset. Think of f as being the initial state of the bar at time t =0,andu(x, t)as being the temperature distribution along the bar at the point x in time t.The boundary conditions or conditions at the end-points are given in such a way that the bar’s “ends” are kept at a fixed temperature, say 0 degrees (whatever) and we can assume that most of the bar is at a temperature close to room temperature, for simplicity. We apply the method of Separation of Variables first. Like Daniel Bernoulli before him, Fourier assumed that the solution he was looking for had the form, u(x, t)=f(x)g(t), (1.2) where we need to find these two functions f,g of one variable. Substituting this expression into the diffusion equation (1.1) we find, ∂2u 1 ∂u 0= − ∂x2 K ∂t 1 = f (x)g(t) − f(x)g(t). K Now, it must be the case that f(x) =0and g(t) =0,otherwise u(x, t)=0for such x and t. This, however is not a sustainable conclusion on physical grounds (since the rod can’t have its temperature equal to 0 except at the end-points). So, dividing both sides by the product f(x)g(t) and separating out terms in x 1 2 The ABC’s of Calculus from terms in t we get f (x) 1 g(t) = . (1.3) f(x) K g(t) Now this equality, (1.3), is valid for any value of x where 0 <x<Land any value of t where t>0. So we can let x = x0 where this number x0 is any fixed number in the interval [0,L]. The left side of (1.3) is a constant while the right hand side is a function of t. On the other hand, we can do the same thing with t, that is, we can set t = t0 in the right of (1.3) and leave the x alone. Then the right side becomes a constant and the left side is a function of x.Thepoint is that these two constants must be equal because of the equality (1.3). We’ll denote the common value of these constants by the symbol −λ so that (1.3) can be rewritten as f (x) 1 g(t) = = −λ. f(x) K g(t) These equalities can be recast as two equations, namely, f (x) = −λ ⇒ f (x)+λf(x)=0, (1.4) f(x) and 1 g(t) = −λ ⇒ g(t)+λK g(t)=0. (1.5) K g(t) At this point we still don’t know the value of this mysterious number λ but this will become clearer later. Now (1.5) is a constant coefficient first order linear differential equation. We also know that its general solution is given by −Kλt g(t)=c1 e . Arguing on physical grounds, the bar should reach a steady state as t →∞ (i.e., the whole bar should ultimately be at a temperature of 0 as t →∞.) This means that λ>0, or else g(t) is exponentially large (since K>0too).Okay, now that we know that λ>0 we can write down the general solution of the constant coefficient second order linear differential equation, (1.4), as √ √ f(x)=c2 sin( λx)+c3 cos( λx). (1.6) where c2, c3 are constants. Combining these expressions for f and g we get, √ √ −Kλt u(x, t)=(c2 sin λx + c3 cos λx) c1 e . (1.7) At this point in the analysis all we know is that if u(x, t) looks like (1.2) then it must be expressible as (1.7). But we still don’t know λ or the cs!So,Fourier figures the solution looks like (1.7) and in order to get some values for the constants therein he must use the boundary conditions given at the ends of the rod, u(0,t)=0=u(L, t), “b.c.”, for short. We note that these b.c. are really saying that f(0)g(t)=0=f(L)g(t) for all t>0. Since g(t) =0wemusthave, f(0) = 0 and f(L)=0, Functions and their properties 3 But these two conditions on f now determine λ, but the λ is not unique. Why? The only solutions of (1.4) that satisfy f(0) = f(L) = 0, are those for which the constants c2, c3 satisfy (see (1.6)) √ √ 0=f(0) = c2 sin( λ · 0) + c3 cos( λ · 0) =0+c3 = c3 i.e, c3 = 0 must occur. But if c3 =0thenc2 = 0 (otherwise, c2 =0would imply that u(x, t) = 0 for all x and t, (see (1.7)), an impossible conclusion! Since c3 = 0 it follows from (1.6) that f(x) must now look like √ f(x)=c2 sin λx with c2 = 0. On the other hand, c2 = 0 along with the b.c. f(L)=0gives √ 0=f(L)=c2 sin( λL) √ and this can happen only if λL is a zero of the sine function. Since the zeros of the sine√ function occur at numbers of the form nπ where n is an integer, we see that λL = nπ is necessary, that is, n2π2 λ = = λn, L2 where n is an integer and we show the dependence of λ upon n by the symbol on the right, λn.Sothereareinfinitely many possibilities for λ, as each one of these λ = λn (called eigenvalues) generates a solution fn(x)=c2 sin λnx of the Sturm-Liouville equation f (x)+λn f(x)=0,f(0) = 0 = f(L). These special solutions fn(x) that satisfy both the equation and the boundary conditions are called eigenfunctions of the boundary value problem. Using these fn(x) we can construct u = un(x, t), where (by (1.7) and since c3 =0andλ = λn) 2 2 nπx − Kn π t un(x, t)=c sin e L2 . 2 L We emphasize that, for each integer n, these functions un(x, t) all satisfy (1.1) along with the b.c. u(0,t)=u(L, t)=0. Butdotheyalsosatisfy u(x, 0) = f(x)? Not necessarily, that is, not unless the initial configuration of the rod at time t = 0 resembles that of a sine function and there is no reason why this should be so! So Fourier probably thought ... “ What if one writes . N u(x, t)= bnun(x, t) n=1 N 2 2 nπx − n π Kt = bn sin e L2 , L n=1 4 The ABC’s of Calculus then this new function u(x, t)also satisfies (1.1) along with the b.c. u(0,t)= u(L, t) = 0 (the numbers bn also have to be determined somehow and we will see how this is done below). But, once again, if f(x) is basically arbitrary it is not necessarily true that N nπx f(x)=u(x, 0) = bn sin . L n=1 So, the great insight was the query, “What if f(x) can be represented as an infinite series of such sine functions?”, that is, what if ∞ nπx f(x)= bn sin , (1.8) L n=1 in the sense of convergence of the series on the right to f(x)formostx in [0,L]? It turned out that this could be done and the representation of f given by (1.8) (where the constants bn depend on f) would eventually be an example of a Fourier Series! The solution of the original problem of heat conduction in a bar would then be solved analytically by the infinite series ∞ 2 2 nπx − n π Kt u(x, t)= bn sin e L2 , L n=1 where the bn are called the Fourier coefficients of f on the interval [0,L]. Fourier actually gave a proof of the convergence of the series he developed (in his book on the theory of heat) yet it must be emphasized that D. Bernoulli before him solved the problem of the vibrating string by wrting down the solution in terms of a “Fourier series” too! 1.2 The General Fourier Series Representation If we proceed with the idea of Section 1.1 and instead use a bar of length 2L stretching from −L to L and we assume that the temperature at its ends satisfy the boundary conditions ∂u ∂u u(−L, t)=u(L, t), and (−L, t)= (L, t), ∂x ∂x then one can show that the assumption of separation of variables, (1.2), will lead to the trial form 2 2 nπx nπx − n π Kt un(x, t)= an cos + bn sin e L2 L L from which we obtain the general representation of u as ∞ 2 2 a0 nπx nπx − n π Kt u(x, t)= + an cos + bn sin e L2 , 2 L L n=1 valid for all x, t under consideration. In this case, the initial configuration of the rod given by u(x, 0) = f(x) will force the representation of f in the more Functions and their properties 5 general form ∞ a0 nπx nπx f(x)= + an cos + bn sin .
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