Last Update: April 4, 2011

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1 Math ES Winter Last Update: April 4, Introduction to Partial Differential Equations Disclaimer: his lecture note tries to provide an alternative approach to the material in Sections..6 in the textbook. It is not a replacement of the textbook. able of contents Introduction to Partial Differential Equations Introduction Separation of variables Separation of variables the idea Eigenvalue problem Separation of variables the method Fourier series Introduction Representation of general f Convergence of Fourier series Complex form of the Fourier series Fourier Cosine and Sine series he heat equation Heat equation with Neumann boundary condition Heat equation with nonhomogeneous boundary condition Heat equation with forcing Heat equation in higher dimensions Uniqueness of solution Wave Equations Wave equations with x over a finite interval Wave equations with <x < Introduction. Unlike ordinary differential equations which describe the trajectory of one object as it moves, partial differential equations arise when people model mathematically the collective evolution of many objects. In the latter case, both the evolution in time and the interaction between different objects have to be taken into account, as a consequence, a typical partial differential equation involves both time derivative and spatial derivatives, for example the heat equation: t x, t = β u xx, t, < x < L, t > u, t=ul, t=, t >, ux, = fx, < x < L. 3 his equation models the temperature change of a rod of length L. Here ux, t represents the temperature at x at time t. he initial condition ux, = fx models the fact that at time t =, the temperature at x is fx. For a partial differential equation, initial conditions alone is not enough. In the above heat equation, we also have boundary conditions u, t = ul, t = which models the fact that the temperature at the two ends of the rod is fixed at. When instead of a rod we consider a higher dimensional object, then the heat equation reads we omit the boundary and initial conditions here = β u 4 t where u is called the Laplacian of u, defined in two and three dimensions as u= u x + u y and u = u x + u y + u z 5

2 respectively. Remark. he Laplacian is the most important and most useful object in partial differential equations, both in theory and in applications. It appears whenever our equation models the evolution of a large number of objects so large that their motion have to be assumed to be random. For example, suppose a million ants are moving randomly inside a fixed region, then the density ρx, t the number of ants per say square meter is governed by the two dimensional heat equation ρ = β ρ. 6 t Of course other terms can be added. Say there are aardvarks wondering around the region eating ants, then a forcing term should be added on the right hand side: ρ = β ρ + fx, t 7 t where fx, t describes how many ants are being eaten at location x at time t.. Separation of variables... Separation of variables the idea. How to solve PDEs? One approach is to reduce a PDE into ODEs. Let s look at the one dimensional heat equation t = β u x; u, t=ul, t=; ux, = fx. 8 How do we reduce it to ODEs? Remember that the difference between an ODE and a PDE is that, the unknown function in an ODE has only one variable, while that in a PDE has more than one. So naturally we try to separate the variables and hope that X, can be obtained by solving ODEs. Substitute u=x into the equation we obtain ux, t=xx t 9 Xx t= β X x t. he two new unknown functions X, are still coupled together. Fortunately, we can obtain a single equation for each through the following technique: We divide both sides by X to obtain t β t = X x Xx. Now the crucial observation is that, the left hand side depends only on t, while the right hand side is independent of t the only possibility is then both sides are constants. hus there is a constant λ such that or equivalently t β t =λ; Viola! We have successfully decoupled X and! But wait a minute. here are some problems here: X x Xx =λ, λβ = ; X λx =. 3. X is subject to a second order ODE, and we know that second order ODEs need either two initial values or a pair of boundary values to make the solution unique. Which one and what are the conditions?. is subject to a first order ODE, we need one initial value to solve it. In other words =?

3 3. What is λ after all? Although we will spend most of the remaining lectures answering the above questions, we will try to summarize the answers there.. Recall that we have set u = Xx t, and the boundary condition for the PDE is u, t = ul, t=. Substitute u by X, we reach for all t. hus it is reasonable to ask hus the full ODE for X is. We try the same thing, the initial value for u suggests Xt=XLt= 4 X =XL=. 5 X λx = ; X=XL=. 6 Xx = fx. 7 But this is not gonna work he restriction on X is too severe! Put it another way, from this we reach which means fx has to satisfy Xx = fx/ 8 f λf = 9 which is definitely not a reasonable requirement. he fix of this situation will be revealed in the section about separation of variable together with the section about Fourier series. Note that in the following we will see that when solving the equation, the exact value of does not need to be specified. 3. Interestingly, the requirement that there are nonzero solutions it is easily seen that X = is always a solution unforunately a useless one to X λx = ; X=XL=, excludes most values of λ. In other words, for most values of λ, the above ODE has only solution. Now we start to explore in full details the answers... Eigenvalue problem. Recall that to solve the heat equation we tried to find solutions of the form t = β u x; u, t=ul, t=; ux, = fx. and finally obtained the equation for X to be u=xxt X λx = ; X=XL=, 3 Our task is then to find nonzero aka nontrivial solutions. As we have mentioned, this is only possible for some values of λ, such λ s are called eigenvalues for the system, and the corresponding nonzero solutions are called eigenfunctions. Now we try to find out the eigenvalues/eigenfunctions. Recalling our theory of constant coefficient ODEs, we write down the auxiliary equation r λ=. 4

4 here are three cases:. λ >. hen the general solution is X = C e λ x + C e λ x. 5 Setting X =XL= we have C + C =, 6 L C + e λ L C =. 7 e λ Solving this we reach C = C =. herefore for any λ >, the only solution is. In other words, none of λ > is an eigenvalue.. λ =. he general solution is Setting X =XL= we reach X = C +C x. 8 C = 9 C + C L =. 3 Clearly the only possibility is again C = C =. herefore is not an eigenvalue either. 3. λ <. he general solution is X =C cos λ he boundary conditions then give hus we have x + C sin λ x. 3 C = 3 C cos λ L + C sin λ L =. 33 λ L=n λ = n. 34 L for any integer n. Note that ± n yield the same λ, and n = λ =. So we should restrict the range of n to positive integers. Summarizing, the eigenvalues to the problem X λx = ; X=XL=, 35 are n, λ n = L,, 36 he corresponding eigenfunctions are n x X n =C sin, L n =,, 37 where C is any nonzero constant. Remark. Note that if X is an eigenfunction corresponding to an eigenvalue λ, then C X is also an eigenfunction corresponding to the same eigenvalue, for any C. We summarize: When solving t = β u x; u, t=ul, t=; ux, = fx. 38 by assuming u = Xxt, the only possible nontrivial X s are n x CX n with X n = sin, n =,, 39 L

5 where C is any nonzero constant. Now we turn to solving eigenvalue problems. he general procedure is the following: Given an ODE with a parameter λ,. Find the general solution.. Subject the general solution to the boundary values and determine the eigenvalues. Example Find the eigenvalues and eigenfunctions of Solution. he auxiliary equation is here are three cases.. λ >. We have Subject to the boundary conditions: which reduces to y + λ y =, <x <; y =, y =. 4 r +λ=. 4 y = C cos λ x + C sin λ x. 4 C = 43 λ C sin λ + λ C cos λ = 44 We can take C if and only if cos λ = or equivalently his leads to C cos λ =. 45 λ = n +, n =,,,,, 46 λ = n +, n =,,, 47 So λ =n +/, n =,, are among the eigenvalues, and the corresponding eigenfunctions are y n x=c n sin n + x, n =,,, 48. λ =. he general solution is Subject to the boundary conditions: hus is not an eigenvalue. 3. λ <. he general solution is Subject to boundary conditions, λ y =C + C x. 49 C = 5 C =. 5 y =C e λ x + C e λ x. 5 e λ C + λ C + C = 53 e λ x he only solution is C = C =. hus any λ < is not an eigenvalue. C =. 54

6 Summarizing, we have λ= n + /, n =,, are the eigenvalues, and the corresponding eigenfunctions are y n x=c n sin n + x, n =,,, 55 Example 4... Find the eigenvalues and eigenfunctions for y + λ y =, <x <; y= y, y = y. 56 Solution. Again there are three cases.. λ >. he general solution is y = C cos λ x + C sin λ x. 57 Subject to the boundary conditions we have C = C cos λ + C sin λ 58 λ C = λ C sin λ + λ C cos λ. 59 Re-arrange we have cos sin λ λ C + C sin cos λ λ C = 6 C =. 6 or equivalently cos sin λ λ sin λ cos λ C C =. 6 o be able to have nonzero solution C C, the necessary and sufficient condition is which gives det cos sin λ λ sin cos λ λ = 63 cos λ + sin λ = 64 which reduces to cos λ = cos λ = λ = n λ = n, n =,,, 65 hus from the case λ > we obtain eigenvalues n, n =,, 3, and eigenfunctions a n cosn x + b n sinn x.. λ =. he general solution is Subject to the boundary conditions: y =C + C x. 66 C = C + C 67 C = C. 68 his leads to C = and puts no restriction on the value of C. hus is an eigenvalue and the corresponding eigenfunctions are y = a for arbitrary a. 3. λ <. he general solution is y =C e λ x + C e λ x. 69

7 Subject to boundary conditions: C +C = C e λ λ + C e 7 λ C λ C = λ C e λ λ λ C e. 7 Canceling λ from the second equation, we obtain C C =C e λ λ C e. 7 Now adding this to the first equation and subtract this from the first equation, we obtain C = C e λ 73 λ C = C e 74 If C C, then either or =e λ 75 λ =e. 76 Either case, we have λ = which is outside the case under current discussion. In other words, there is no negative eigenvalue. Summarizing, the eigenvalues are, n,,, 3, with corresponding eigenfunctions where a, a n, b n are arbitrary constants. y =a, y n = a n cosn x +b n sinn x, n =,, 3, Separation of variables the method. Recall our main result from last section: When solving t = β u x; u, t=ul, t=; ux, = fx. 78 by assuming u = Xxt, the only possible nontrivial X s are n x CX n with X n = sin, n =,, 79 L where C is any nonzero constant. here is a negative way of looking at this result: If fx C X n for some n and some C, there is no way that the solution to the PDE can be written as Xx t. In other words, few problems can be solved by setting u = X. What saves the day is the following idea: Given a PDE, although it is not possible for u = X, is it possible that u is a sum of solutions of the type X? Somewhat suprisingly, this idea works for a large class of PDEs: Linear PDEs. A PDE is linear if linear combinations of solutions are still solutions. For example, consider our heat equation t = β u x; u, t=ul, t=; ux, = fx. 8 If u, u,, u k solves the above problem with initial values f x,, f k x, then it is easy to verify that c u + + c k u k solves the above problem with initial value c f + + c k f k. We now can state the following method of separation of variables. he method consists of several steps: Given a linear PDE,. Separate the variables: ux, t =Xxt.. Solve the eigenvalue problem to obtain all eigenfunctions X n. 3. Solve the corresponding n.

8 4. Write 5. Determine a n through ux, t= a n X n n. 8 fx=ux, = a n X n n. 8 Example Consider the heat flow problem t = β u x; u, t=ul, t=; ux, = fx. 83 with β =3, L =, fx=sin x 6 sin 4 x. Solve it using separation of variables. Solution.. Separate the variables. Write u =Xxt. We reach he equations for X and are then X = 3X he value of does not need to be specified.. Solve the eigenvalue problem. We discuss the three cases: i. λ >. We have 3 = X = λ. 84 X X λx = ; X=X =; 85 3λ = ; 86 X λx = ; X=X =; 87 Subjecting to boundary conditions, we have We quickly see that no solution except C = C = exists. ii. λ=. We have which subjecting to boundary conditions gives It is clear that C = C =. iii. λ <. he general solution is X =C cos λ Subjecting to boundary conditions, X = C e λ x + C e λ x. 88 C +C = 89 C e λ + C e λ =. 9 X = C + C x 9 C =, 9 C + C =. 93 x + C sin λ x. 94 C =, 95 C cos λ + C sin λ =. 96 which gives C =, C sin λ =. 97

9 herefore nonzero solution exists if and only if sin λ Summarizing, the eigenvalues are with corresponding eigenfunctions 3. With λ n obtained, the equation for n becomes or he solution is given by 4. Now we write =, or λ = λ n = n,,, 3,. 98 λ n = n, n =,, 3, 99 X n = c n sinn x, n =,, 3, ux, t= n 3λ n = n + 3n =. n t= n e 3nt. 3 n c n e 3nt sinn x. 4 From this it is clear that n c n can be treated as one single arbitrary constant the exact value of n never needs to be explicitly found out! hus we write ux, t= and will use the initial value to find out c n s. 5. Equate One reasonably conclude that sin x 6 sin 4x= fx=ux, = Finally we see that the solution to the problem is given by c n e 3nt sinn x. 5 c n sinn x. 6 c =, c 4 = 6, c = c 3 = c 5 = =. 8 ux, t =e 3t sin x 6e 48t sin 4 x. 9 Example Find the formal solution to the heat flow problem Solution. t = u x ; u, t=u, t=; ux, = fx= sin n x. n. Separate the variables. Write ux, t=xxt, we reach. Solve the eigenvalue problem X = X = X = λ. X X λx =, X=X =.. Clearly some theory is needed here. For example, why should the representation be unique? sinx 6 sin4 x= c n sinn x 7

10 he discussion is almost identical to that of the above problem so we omit it here. he eigenvalues are λ n = n, n =,, 3, with eigenfunctions sinn x. 3. Solve the equation for n : which gives 4. he solution is then represented as 5. he initial value suggests which gives he solution is then given by n + n n = 3 n t= n e nt. 4 ux, t= c n e nt sinn x. 5 sin n x= c n n sinn x 6 c n = n. 7 ux, t= n e nt sinn x. 8 Example Solve the vibrating string problem with u t = u α x, <x<l, t >; u, t=ul, t=, ux, = fx, x, = gx 9 t Solution. α = 3, L =, fx=3sin x+ sin 3 x, gx=.. Separate the variables. Write ux, t=xxt. Substitute into the equation, we have X = 9 X 9 = X X. As the LHS is dependent only on t and the RHS is independent of t, both ratios must be equal to a same constant λ. hus the equations for X and are X λx =, X=X=; 9λ =. 3 Similar to the heat equation case, there is no need to specify the initial values,.. Solve the eigenvalue problem X λx =, X=X=; 4 his is the same problem as appeared in our heat equation example. he eigenvalues and eigenfunctions are 3. Solve the equation Substitute λ= n, we have which solution is λ= n, X n = c n sinn x,,, 5 n 9 λ n n =. 6 n + 9n n = 7 n t =A n cos3nt +B n sin3 n t. 8

11 4. Write down ux, t: ux, t= A n cos3nt+b n sin3 n t c n sinn x. 9 It is clear that the constant c n can be absorbed by A n and B n. hus we have ux, t= 5. o apply the initial values, we compute and Now compare with the initial values: We reasonably conclude that Finally the solution is given by A n cos3 n t+b n sin3nt sinn x. 3 ux, = A n sinn x 3 t x, = 3 n B n sinn x. 3 3 sin x+ sin 3 x = = A n sinn x, 33 3 nb n sinn x. 34 A =3, A 3 =, A n = n, 3; B n = for all n. 35 ux, t=3cos6 t sinx+ cos39t sin3x. 36 Now what if the initial value f is not given as an explicit sum of eigenfunctions? Can we still carry out separation of variables? he answer is yes, guaranteed by the theory of Fourier series and more generally by the theory of Sturm-Liouville problems. 3. Fourier series. 3.. Introduction. We have seen the execution of the method of separation of variables.. Separate the variables: ux, t =Xxt.. Solve the eigenvalue problem to obtain all eigenfunctions X n. 3. Solve the corresponding n. 4. Write 5. Determine a n through ux, t = a n X n n. 37 fx=ux, = a n X n n. 38 In the final step, we need to determine the coefficients a n by expanding the initial value fx into a possibly infinite linear combination of the eigenfunctions X n. In our previous examples, we have only dealt with f already written into such combinations. Naturally one would like to ask the following questions:. How about general f? Can we always write f = a n X n, where X n are eigenfunctions?. As the summation may be infinite, what do we mean by a n X n? he answers to the two questions form the core of the theory of Sturm-Liouville problems, to which the theory of Fourier series is a special case.

12 Definition 8. A Fourier series is the infinite sum We notice the following: his can be written as If a + { a n cos n x where X, X n, Y n are all the eigenfunctions of the problem + b n sin n x }. 39 a X + {a n X n + b n Y n } 4 X + λx =, X=X, X =X. 4 f = a + { a n cos n x + b n sin n x } 4 in any reasonable interpretation, then fx = fx +. hat is f has to be a periodic function of period. In other words, the most we can expect from a Fourier series as presented here is that it can represent most periodic functions. As we will see soon, this expectation is fulfilled. In the following we will answer the two questions one by one:. How about general f? Can we always write f = a n X n, where X n are eigenfunctions?. As the summation may be infinite, what do we mean by a n X n? 3.. Representation of general f. Given an arbitrary function f with the only restriction that it is periodic with period, for what a, a n, b n is there any hope to say fx= a + { a n cos n x + b n sin n x }? 43 At first look this is a totally nonsensical question: We are given one single equation with infinitely many unknowns and are asked to solve it! However, looking at it a bit longer, one realizes that we are given not one single equation, but infinitely many fix the value of x and we obtain one equation, and the equality has to hold for all x! Although there are infinitely many equations, but they are quite hard to analyze. o simplify the situation, we try to get rid of the functions and replace them by numbers. his is fulfilled through weighted integration. hat is, we multiply the equation by appropriate functions and then integrate. Knowledge from linear algebra tells us that we need the same number of equations as the unknowns to make the system uniquely solvable. hus we need to find the same number of appropriate functions as the unknowns. Notice that each unknown is paired with an eigenfunction: So naturally we try these guys. a with, a n with cos n x, b n with sin n x. 44 First we multiply the equation by and integrate from to. a fxdx = + { a n cos n x + b n sin n x } dx a = dx + { a n cos n x } dx + b n sin n x dx = a. 45 So we have a = fxdx. 46

13 Next we multiply the equation by cos m x for some m< and integrate from to. fx cos m x dx = a + { a n cos n x + b n sin n x } cos m x dx a = x cosm dx + a n cos n x x cosm dx We evaluate them one by one: First term. a + Second term. a n cos n x cosm x dx = = + b n sin n x cosm x dx. 47 cosm x dx = a x sinm m N = ; 48 a n n +mx cos a n n + mx cos a n cos n m x n mx + cos dx dx + dx. 49 Now the first term is evaluated similar to the a case: a n n + mx cos dx = a + nx sinm N =. 5 m + n For the second term, there are two cases.. n=m. In this case a n n mx cos dx = a m dx=a m. 5. n m. In this case a n n mx cos Summarizing, we have Last term. b n sin n x cosm x dx= a n x sinm N =. 5 m n a n cos n x { x cosm dx = am n = m n m. 53 dx= b n n + mx sin + sin n mx dx 54 Similar to the previous case we finally reach b n sin n x x cosm dx=. 55 Putting things together, we have fx cos m x dx = a m a m = fx cos m x dx. 56

14 Finally multiply both sides by sin m x and integrate, we reach b m = fx sin m x dx. 57 So our plan was successful! Definition 9. Let f be a piecewise continuous function on the interval,. he Fourier series of f is the trignometric series fx a + { a n cos n x + b n sin n x } 58 with a n, b n given by a n = fx cos n x dx. n =,,, 59 b n = fx sin n x dx. n =,, 6 Remark. Note that we applied many times termwise integration fk f k 6 in the above calculation. In calculus we know that this is not automatically true. In our situation, this fact puts restriction on what we mean by f = a + { a n cos n x + b n sin n x } in a reasonable way his reasonable way has to allow termwise integration! he symbol is used, instead of =, also emphasizes this fact. Remark. In the above calculation, we have used the following relations: cos n x cosm x { n=m dx= n m, 63 sin n x cosm x dx=, 64 sin n x x sinm dx = 6 { n = m n m, 65 hese are called orthogonality conditions. In general, if X n are eigenfunctions of an eigenvalue problem defined by a second order ODE over a, b, we have b X m X n w dx= 66 whenever n m. Here w is a particular weight function. a Remark. he interval, can be replaced by any one period. In other words, instead of we can use,, etc. Example Compute the Fourier series for fx=x, < x <. 67. Of course, in the exercises or exams, the function f is always only explicitly given over one period, so we have to use that particular period. For example, if the problem says, f = x for < x < and is periodic with period and blah blah, then we should replace by in all the formulas.

15 Solution. Note that what s implicit here is that f is periodic with period =. We compute each of the coefficients. First Next Finally Summarizing, we have a = xdx=; 68 a n = x cosn x dx = x dsinn x n = x sinnx N sinn x dx n =. 69 b n = x sinn x dx = x dcosn x n = x cosn x N cosn xdx n = n n n = n n = n+. 7 n fx n+ sinn x. 7 n Remark 4. he above calculation can be significantly simplified if we consider the even/oddness of the functions. A function fx is even if fx = f x for all x, while it is odd if fx = f x for all x. An important property is that fx gx= 7 if one of f, g is even and the other odd. Furthermore, for a periodic function f with period, if fx = f x for x then f is even, and similar result holds for f odd. Application of this property to the calculation of Fourier series is immediate after we notice that cos n x n x is even while sin is odd. hus in the above example, as fx = x is odd, we immediately reach a n = x cosn xdx=. 73 Example Compute the Fourier series for { <x< fx =. 74 x <x < Solution. his time the period is and n x first a = fx dx = becomes n x. We compute dx+ x dx = a = ; 75

16 next and finally Summarizing, we have fx + a n = n x fx cos dx = n x cos dx+ = n x n sin N + n = + n x x sin N n = n x n n cos N = = n n n n x x cos n x xdsin n x sin dx dx n n ; 76 b n = n x fx sin dx = n x sin dx + n x x sin dx = n x n cos N n x xdcos n = n x cos n x cos N n n n = n n x cos dx n x n sin N n n n cos n x n n + n x sin. 78 Remark 6. More generally, any set of nonzero functions {f n } satisfying, for some w b f m f n w dx = whenever n m 79 a is said to be an orthogonal system with respect to weight w on the interval a, b. If furthermore we have b f n w dx=, n =,, 3, 8 then {f n } is called an orthonormal system. he main property of an orthogonal system is that if a fx c f +c f + 8 then the coefficients can be determined through fx fmx wx dx c m =. 8 fm x dx Example ,.3.7 Show that the set of functions { cos x, sin n n x,, cos x, sin x, } 83

17 is an orthonormal system on, with respect to the weight function wx. hen find the orthogonal expansion for { <x< fx= <x< in terms of this orthonormal system. Solution. Verify orthonormality. 84. Integrating product of different functions gives. We compute for n m, n, m =,, note that w = n x m x cos cos dx = cosn + m xdx + cosn mx dx 85 As neither n + m nor n m is zero, we have cosn + m x dx= n +m sinn +m x N =, 86 cosn mxdx = sinn m xdx =. 87 n m Similarly we compute for n m. Finally we can compute Note that this time n=m is OK. n x m x sin sin dx= 88 n x m x cos sin dx = 89. Integrating the square of any function in the list gives. We compute n x cos dx = +cos n x dx=. 9 Similarly we have hus the set of functions is an orthonormal system. Orthogonal expansion for Recall that if n x sin dx =. 9 { <x< fx= <x< 9 fx c f +c f + 93 then the coefficients can be determined through fx fmx wx dx c m =. fm x dx In case of our system we write 94 fx= { } n x n x a n cos +b n sin 95

18 and compute and hus finally we have fx 3.3. Convergence of Fourier series. a n = b n = = = = = n x fx cos dx n x cos dx n sinn / n sin n x N = n+ n. 96 n x fx sin dx n x sin = n = dx cosn x N n. 97 n+ n x n x cos + sin. 98 n heorem 8. Pointwise convergence If f and f are piecewise continuous on,, and let f has Fourier expansion fx a + { a n cos n x + b n sin n x } 99 then we have a + { a n cos n x In particular, at x=±, we have a + { a n cos n x + b n sin n x } = f x + + f x. + b n sin n x } = f + + f. Remark 9. Recall that a function is piecewise continuous on a, b if it is continuous at every point in a, b except possibly for a finite number of points at which f has a jump discontinuity, that is f x +, f x both exist but are not equal. Recall that f x + = lim fx+h, f x = lim h>,h h<,h fx + h= lim h>,h fx h. Remark. In particular, for functions specified in the theorem, we have whenever f is continuous at x. fx= a + { a n cos n x + b n sin n x } 3

19 Example..3.7 Consider the Fourier series for fx with period and fx = x for < x <. f is continuous for < x < but f = = f +. As a consequence its Fourier series will converge to f with period and { x < x < f x= 4 x=± Example..3.9 Consider the Fourier series for fx with period 4 and { <x< fx =. 5 x <x < We see that f is continuous for < x < and < x <, while has jump discontinuity at,,. As a consequence its Fourier series converges to f with period 4 and <x< / x= f x=. 6 x <x< 3/ x=± Remark 3. Gibbs phenomenon,.3.39 When jump discontinuities are present, the convergence around them suffers from the so-called Gibbs phenomenon, that is, the partial sum N a { + a n cos n x + b n sin n x } always overshoot f by approximately 9% of the size of the jump. As a consequence the convergence is not uniform which means it may be misleading to use the graph of the partial sum as an indicator of what the actually function looks like. Gibbs phenomenon disappears as soon as the function is continuous everywhere. In fact, in this case the convergence is uniform: heorem 4. Uniform convergence of Fourier series Let f be a continuous functions on, and periodic of period. If f is piecewise continuous on,, then the Fourier series for f converges uniformly to f on, and hence on any interval. hat is, for each ε >, there exists an integer N that depends on ε such that fx for all N N and all x,. a + N { a n cos n x 7 + b n sin n x } < ε 8 For f with appropriate continuity/smoothness, we can differentiate and integrate term by term. heorem 5. Differentiation and Integration Let fx be continuous on, and -periodic. Let f x and f x be piecewise continuous on,. hen the Fourier series of f x can be obtained from that of f through termwise differentiation. Let fx be piecewise continuous on,. hen we can integrate its Fourier series termwise. Remark 6. Note that the integration of a Fourier series is not a Fourier series anymore. Example b Find the functions represented by the series obtained by the termwise integration of the given series from to x. 4 n= sin n +x n+ fx 9

20 with { < x < fx =. <x< Solution. We know that the integrated series represents x x dx < x < { x + < x < fxdx = x = = x. + x <x < dx+ dx < x < 3.4. Complex form of the Fourier series Recall that e iθ = cosθ + sinθ which leads to hus we can write { a + If we set c = a, c n = Now we compute and similarly c n = a n ib n a n cos n x cosθ = eiθ + e iθ + b n sin n x }, sin θ = eiθ e iθ. i = a + bn + i = a + an e i nx e i nx an ib n i +e nx i e nx an i bn an + i bn, c n = for n =,, 3, then we have a + { a n cos n x = fx cos n x Summarizing, we have the complex Fourier series: fx 4. Fourier Cosine and Sine series. e i nx nx a + n + ib n i e. 3 +b n sin n x } = c n e i nx. 4 i sinn x dx= nx i fxe dx, 5 c n = fxe i nx. 6 c n e i nx, cn = nx i fxe dx. 7 Definition 8. Let fx be piecewise continuous on the interval,. he Fourier cosine series of f is a + a n cos n x, 8 where a n = n x fx cos dx, n =,, 9 while the Fourier sine series of f is where b n = n x fx sin b n sin n x. dx, n =,, 3,

21 From the looks we are tempted to say that a Fourier cosine series is half of a Fourier series, and a Fourier sine series is the other half. In some sense this is indeed true, as we will see now. Consider a function f defined on the interval,. We now extend it to a periodic function with period in two ways.. We first extend f evenly, { fx <x< f e x= f x < x < and then extend f e x to a -periodic function f ex. Now consider the Fourier series of f e. We compute a = f e xdx= fx dx + f xdx= fxdx. 3 a n = = = b n = f e x cos n x fx cos n x fx cos n x dx dx + f e x sin n x f x cos n x dx. 4 dx dx = 5 as f e is even, sin n x is odd, and the integration is on an interval symmetric with respect to. hus the Fourier series for f e is f e a + a n cos n x, a n = n x fx cos dx, n =,, 6 which is exactly the Fourier cosine series of f.. We first extend f oddly, { fx <x< f o x = f x < x < 7 and then extend f o x to a -periodic function f ox. Similar to the even extension case, we conclude that the Fourier series for f o is exactly the Fourier sine series for f. Remark 9. An alternative way to understand Fourier cosine and sine series is the following. We notice the following facts: he eigenfunctions for the problem are exactly he eigenfunctions for the problem are exactly X λx =, X=X= 8 sin n x,,, 3, 9 X λx =, X =X = 3 cos n x, n =,,, 3, 3 he general theory of eigenvalue problems guarantees that the set of eigenfunctions form an orthogonal system. hus { sin n x } { n x} is an orthogonal system over, with weight wx and so is cos. Now recall the property of orthogonal systems:

22 If {f n } is an orthogonal system over a, b with weight wx, then any function f defined over a, b can be expanded into with When we apply this theory to { sin n x } we obtain with as b n = fx c n f n 3 b fx f nxwxdx c n = a b. 33 f n x wxdx a fx Similarly, when we apply the theory to { cos n x b n sin n x, 34 fx sin n x dx sin n x = n x fx sin dx dx 35 sin n x dx=. 36 } we obtain the Fourier cosine formulas. Remark 3. Each of the two approaches have their advantages and disadvantages. he first approach, treating Fourier cosine and sine series as special cases of Fourier series, has the advantage that the convergence theorems of Fourier cosine/sine series can be obtained immediately by adapting corresponding theorems for the Fourier series. he second approach, treating all three Fourier series, Fourier cosine series, and Fourier sine series as examples of the general theory of eigenfunctions, has the advantage that it reveals deeper understanding, and makes moving on to general separation of variables, like those will be treated in Math337, easier. Example Compute the Fourier sine series for Solution. We have =. Compute b n = x sinn x dx = x dcosn x n = x cosn x N n = n x dsinn x n n = n x sinn x N n n = n n n n cosn x N = n n n n fx =x, <x<. 37 = n n+ + 4 n 3 cosn xxdx sinn xdx n. 38

23 herefore the Fourier sine series is fx n n+ + 4 n n 3 sinnx. 39 Example Compute the Fourier cosine series for Solution. We have =. First next herefore a n = = = fx=e x, <x <. 4 a = e x cosn xdx cosn xde x cosn xe x N + n e x dx=e. 4 e x sinn x dx = e n +n sinn xde x = e n +n e x sinn x N n = e n n So the Fourier cosine series is given by e x cosn xdx e x cosn xdx = e n n a n. 4 fx e + a n = e n +n. 43 e n Example Find the solutio nto the heat flow problem Solution. We use separation of variables. +n cosn x. 44 = 5 u t x, <x <, t > 45 u, t =u, t =, t > 46. Separate the variables. Write u =Xxt, the equation becomes he equations for X and are then. Solve the eigenvalue problem ux, = fx = cos x, <x<. 47 X = 5X 5 = X = λ. 48 X X λx =, X=X=; 49 5λ =. 5 X λx =, X=X=; 5

24 As we have solved it before, we omit the details. he eigenvalues are n, n =,, 3,, and the corresponding eigenfunctions are b n sinn x, n =,, 3, 5 3. Solve the equation for. For each eigenvalue n, the corresponding n satisfies 4. Write down the expansion for u: n + 5n n = n =c n e 5nt. 53 ux, t= Clearly the c n s can be absorbed into the b n s, so we simply write ux, t = c n e 5nt b n sinn x. 54 b n e 5nt sinn x Determine b n using the initial condition. We have cos x=ux, = b n sinn x. 56 hus all we need to do is to find the Fourier sine series for cos x. As the interval is, we have =. We compute for n =,, 3, b n = cosx sinn xdx = sinn xdx sinn x cosxdx = n cosn x N sinn + x+sinn x dx = n n sinn +xdx sinn xdx. 57 We evaluate For the last term, there are two cases. sinn + xdx= n + cosn + xn = n+. 58 n + If n =, then sinn x= and sinn xdx=. 59 If n, we compute sinn xdx= n cosn xn = n. 6 n Putting everything together, we have n n n+ b n = n n n n + n =. 6 n Noticing that, when n is even, we have n =. hus the above formula can be simplified by setting n = k to b k = k k 3, k =,, 3, 6 k +

25 Putting everything together, the solution is given by ux, t= k= k k 3 k + t e 5k sin k x he heat equation. In this final section, we focus on the heat equation. Recall that we have already solved the follow version of heat equation t = β u x, <x <L, t > 64 u, t=ul, t=, t > 65 ux, = fx, <x<l. 66 which models the evolution of temperature of a rod with initial temperature distribution fx and with both ends kept at temperature. In the following we will discuss other cases: Instead of keeping both ends at temperature, we insulate them. In this case the boundary conditions become x, t = L, t=, t >. 67 x Instead of keeping both ends at temperature, we keep them at temperatures U, U. he boundary conditions are then u, t=u, ul, t =U. 68 We heat up or cool down the rod during the process. he equation then becomes t = β u + Px, <x<l, t >. 69 x Instead of a rod, we consider say a rectangular plate occupying, L, W. he equation then becomes two dimensional, and we have more freedom specifying boundary conditions. 5.. Heat equation with Neumann boundary condition. Example Solve, t= x x t Solution. We use separation of variables.. Separate the variables. Write ux, t=xxt. hen the equation leads to = 3 u x, <x<, t > 7, t =, t >, 7 ux, = x, <x<. 7 which gives X =3X 3 = X X = λ 73 X λx =, X =X =, 74. Solve the eigenvalue problem 3λ =. 75 X λx =, X =X =, 76

26 We discuss the three cases. a. λ <. he general solution is X =C cos λ We compute X = λ C sin λ he boundary conditions then lead to hus C =, and λ C sin λ with corresponding eigenfunctions b. λ=. he general solution is x + C sin x λ + λ C cos + λ C cos x. 77 λ λ = n. Consequently the eigenvalues are x. 78 λ C =, 79 λ =. 8 λ n = n, n =,, 3, 8 X n = a n cosn x. 8 X = C + C x 83 the boundary conditions then gives C = which means is an eigenvalue and the corresponding eigenfunctions are X =a for a constant a. c. λ >. he general solution is X =C e λ x + C e λ x. 84 he boundary conditions leads to λ C λ C = 85 λ e λ C λ e λ C =. 86 Solving it gives C = C =. herefore there is no positive eigenvalues. Summarizing, the eigenvalues are with eigenfunctions 3. Solve the equation for n : 4. Write down the solution 5. Find out the coefficients. We have λ n = n, n =,,, 3, 87 X n =a n cosn x, n =,,, 3, n +3n n = n = n e 3nt. 88 ux, t= n= ux, = n= a n e 3nt cosn x. 89 a n cosn x. 9 All we need to do is to find the cosine series for x: < x <. We compute 3 a = xdx =, 9 a n = x cosn xdx = x sinn x N sinn xdx = n n n. 93

27 Note that n = for all n even. hus we have Summarizing, we have ux, t= k= a k =, a k+ = 5.. Heat equation with nonhomogeneous boundary condition. Example Solve 4 k t k + e 3k+ cosk + x. 95 = u t x, <x<, t > 96 u, t =5, u, t=, t > 97 ux, = sin 3 x sin 5x, <x<. 98 Solution. his time we have to get rid of the nonhomogeneous boundary c onditions first. Since otherwise, say we solve X n subject to X n = 5, X n =, then clearly there is no guarantee that the infinite sum an X n n 99 will satisfy the same boundary conditions. herefore, before proceeding with the method of separation of variables, we need to find a function vx, t such that v t = v x, v, t=5, v, t =. 3 If we can find such v, then ũ = u v would satisfy = u t x, < x <, t > 3 u, t=, u, t =, t > 3 ux, = sin 3x sin 5 x vx,, <x <. 33 which can then be solved using separation of variables. Such v is easy to find. he simplest one is the linear function which takes 5 at and at : Now ũ = u v satisfies We use separation of variables. vx =5+ 5 x. 34 = u t x, <x<, t > 35 u, t =, u, t=, t > 36. Separate the variables. We reach ux, = sin 3 x sin 5x 5 5 x, <x< Note that the Fourier cosine series is for = We see that our a here corresponds to a / in the formula. X λx =, X=X=, 38 λ =. 39 f a + a n cosn x 9

28 . Solve the eigenvalue problem X λx =, X=X=, 3 he eigenvalues are λ n = n, and the eigenfunctions are X n = b n sinn x. 3. Solve n : 4. Write down the solution 5. Determine the coefficients. We have n +n n = n = n e nt. 3 ũx, t= sin 3 x sin 5x 5 5 x = b n e nt sinn x. 3 b n sinn x. 33 Clearly we only need to find out the Fourier sine series for the function + x. We compute + x sinn xdx = + x dcosn x n = + x cosn x N cosn xdx n = n. 34 n Putting things together, we have Finally ũx, t=e 8t sin3x e 5t sin5 x + n e nt sinn x. 35 n ux, t =5+ 5 x+e 8t sin3 x e 5t sin5x+ n e nt sinnx. 36 n or equivalently ux, t = x 3 e t sin x + 5 e 8t sin x + 6 e 5t sin5x+ n n= Heat equation with forcing. e 8t sin3 x + 5 e 3t sin4 x + n e nt sinn x. 37 Example Find a formal solution to the initial-boundary value problem = u + 4x, <x<, t > ; 38 t x u, t = u, t=, t >, 39 ux, = sin x, <x <. 3 Solution. he idea is as follows. Find a function wx such that hen ũ =u w would satisfy w + 4x=, w=w =. 3 ũ = ũ t x, < x <, t > ; 3 ũ, t = ũ, t =, t >, 33 ũx, = sin x wx, < x <. 34

29 and can be solved by separation of variables. Now the equation for w is easy to solve: Integrating twice, we have wx= x3 3 Using the boundary conditions, we reach + Ax+B. 35 wx= x3 3 + x Next we carry out separation of variables:. Separate the variables: X λx =, X=X=; 37. Solve the eigenvalue problem λ =. 38 X λx =, X=X=; 39 he eigenvalues are λ= n, n =,, 3, with corresponding eigenfunctions b n sin n x. 3. Solve the corresponding n : 4. Write down n + n n = n t = n e nt. 33 ũx, t= b n e nt sinn x. 33 hus 5. Determine the coefficients: sin x + x3 3 3 x=ũx, = b n sinn x. 33 We compute the Fourier sine series for x3 x: 3 3 x x sinnxdx = x 3 n 3 3 x dcosn x = x 3 n 3 3 x cosnx N = x cosn x dx cosn xdx n 3 = n x dsinn x = n x sinn x N x sinnxdx = 4 n x sinn xdx = 4 n 3 x dcosn x = 4 n 3 x cosn x N cosn x dx ũx, t =e t sin x + cosn x x dx 3 = 4 n 3 n n 3 n e nt sinn x 334

30 Finally we have or equivalently ux, t= 3 x x3 3 + e t sin x+ 4 n 3 n e nt sinn x 335 ux, t = 3 x x3 3 3e t sin x+ 4 n 3 n e nt sinnx. 336 n= Remark 37. Note that, when the forcing is a function of both x and t, that is t = β t = β u + Px, t, 337 x or when the forcing is independent of t but the dimension is more than, say u x + u y +Px, y 338 the above method does not work anymore. Instead, one needs to integrate the forcing into the method of separation of variables. Such technique will be studied in introductory PDE courses Heat equation in higher dimensions. Example Find a formal solution to the initial-boundary value problem = u t x + u y, < x <, < y <, t > 339, y, t =, y, t=, < y <, t > 34 x x ux,, t = ux,, t=, < x <, t > 34 ux, y, = y, < x <, < y <. 34 Solution. We follow the procedure of separation of variables.. Separate the variables. Write ux, y, t=xxy yt. Substitute into the equation, we have Divide both sides by XY : X Y =X Y + XY. 343 = X X + Y Y. 344 As the left hand side only depends on t while the right hand side is independent of t, both sides have to be constant. Applying the same argument one more time, we conclude that he equations for X, Y, are then X X = λ, Y Y = µ, = λ+ µ. 345 X λx =, X =X =; 346 Y µ Y =, Y =Y =; 347 λ + µ = Solve the eigenvalue problems. Now there are two eigenvalue problems. We solve them one by one. i. Solve X n. X λx =, X =X =; 349 We have eigenvalues λ n = n, n =,,, and eigenfunctions X n = a n cosn x, n =,, 3. ii. Solve Y m. Y µ Y =, Y =Y =; 35

31 We have eigenvalues µ m = m, m =,, 3, and eigenfunctions Y m = b m sinm y, m =,, 3, 3. Solve n,m. We have 4. Write n,m + n + m n,m = n,m = n,m e n +m t. 35 ux, y, t= n= 5. Compute the coefficients. We have ux, y, = n= c nm cosn x sinm ye n +m t. 35 m= c nm cosn x sinm y e n +m t. 353 m= o determine the coefficients, we first need to understand the integrations cosn x sinm y cosn x sinm ydx dy. 354 We compute cosn x sinm y cosn x sinm y dx dy = cosn x cosn xdx sinm y sinm y dy. 355 Recall that for n, n {,,, 3, } we have n =n = cosn x cosn xdx= n =n, 356 n n and for m, m {,, 3, } herefore As a consequence, sinm y sinm y dy = { m =m m m 357 n = n =, m = m cosn x sinm y cosn x sinm ydx dy = n = n, m = m n n or m m Now we compute ux, y, cosn x sinm y dxdy = c m n = 4 c nm n =,, 3, y sinm y dxdy = y sinm ydy = y dcosm y m = y cosm y N m cosm ydy. 359 = m m. 36

32 hus herefore Summarizing, we have y cosn x sinm ydx dy = c m = m m+. 36 cosnxdx y sinm ydy =. 36 c nm =, n =,, 3, ; m =,, 3, 363 ux, y, t= m+ e mt sinm y. 364 m m= 5.5. Uniqueness of solution. We have seen that we can use separation of varaibles to find a solution of the heat equation modula convergence proof. Now as heat equation models a physical process, naturally we ask: Does the solution obtained describe the real physical process? In other words, is it possible that the physical process, the real solution, is some function that cannot be obtained through our method? Now we will show that such things cannot happen, through using the following maximum principle. heorem 39. Let ux, t be a continuously differentiable function that satisfies the heat equation and the boundary conditions t = β u x, <x<l, t >, 365 u, t =ul, t=, t >. 366 hen ux, t attains its maximum value at t =, for some x in, L that is max ux, t= max ux,. 367 t, x L x L Using the maximum principle one easily obtains uniqueness of solution: heorem 4. the initial-boundary value problem = β u t x, <x <L, t >, 368 u, t =ul, t =, t >, 369 ux, = fx, < x < L. 37 has at most one continuously differentiable solution. 6. Wave Equations. he wave equation problem we will mainly discuss reads 4 u t = α u + hx, t, <x < L, t > 37 x u, t =U, ul, t =U, t > 37 ux, = fx, <x <L, 373 x, = gx, t <x <L. 374 Notice that instead of one single initial condition ux, = fx, we have two. he reason is that the wave equation is a second order differential equation with respect to the variable t compare: nd order ODE needs two initial conditions: y and y. 4. When h =, U = U =, Neumann boundary condition = may also be considered. x

33 he separation of variables procedure for solving the above type of wave equation is exactly the same as that for the heat equation: For the general case: Step : Find w to take care of U and U. Set v = u w. Note that the initial values will change to vx, = f w; v x, = g. t Step : Get and solve the eigenvalue problem. Obtain X n, λ n and the range of n. Step : Write Expand v = hx, t= range of n range of n X n n 375 h n tx n 376 Substitute v and h back into the equation to find the equation for n ; Substitute vx, = n X n and v t x, = n X n into the initial conditions to get the initial values for n. Step 3: Solve n. Step 4: Write down the final answer u =v +w. If hx, t = hx is independent of t, the following conceptually-simler approach can be taken: Step : Find w to take care of hx, U and U ; Set v = u w. Note that the initial values will change to vx, = f w; v x, = g. t Step : Get and solve the eigenvalue problem. Obtain X n, λ n and the range of n. Step : Solve n using the obtained λ n. Step 3: Write down v = range of n Absorb all redundant arbitrarn constants. Step 4: Use initial values to determine the coefficients. Step 5: Write down the final answer u =v +w. c n X n n 377 However, the details in solving wave equations are more complicated then that in solving the heat equation. In particular, in the general case, instead of a simple st order linear equation, the equation for n is a nd order nonhomogeneous linear equation whose solution requires undetermined coefficients or even variation of parameters. We will also consider the problem with < x < instead of < x < L, and with h = : u t = α u x, < x <, t > 378 ux, = fx, < x <, 379 x, = gx, < x <. 38 t In this case naïve separation of variables would not yield anything useful: What happens to sin n x L when L becomes infinity? Interestingly, in the case solving the wave equation is in fact much easier, through the following simple formula: ux, t = x+αt fx+αt + fx α t + gy dy. 38 α x αt

34 6.. Wave equations with x over a finite interval. As there is nothing new in the solution procedure, we will just present a few examples here. Example 4. Basic case Find the formal solution of Solution. u t = 9 u x, <x<8, t > 38 u, t=, u8, t=, t > 383 ux, = x, <x<8, 384 x, = 3 + x, t <x< Step : Eigenvalue problems. We separate variables: u = Xx t. Substitute into the equation we get X =9X X t=9 x both must be constants. 386 X So 9 λ = ; X λ X =. 387 Combining the boundary conditions, we get the eigenvalue problem: X λx = ; X=X8=. 388 As this type of problems has been solved in detail in the heat equation section we jump directly to its solution here: n, n λ n = Xn = A n sin 8 8 x, n =,, 3, 389 Step : Solve n. he equation for n is whose characteristic equation is leading to r, = ± 3 n 8 Step 3: Write ux, t = n 9 λ n n = n +9 r +9 i. So 3 n n = D n cos 8 range of n c n n X n = 3 n c n D n cos 8 3 n a n cos t 8 = n n = 39 8 n = n t + E n sin t n t + E n sin 8 3n + b n sin t 8 n t A n sin sin Note that we have absorbed all redundant constants in the last step. Step 4: Determine the coefficients. 8 x n 8 x. 393

35 he above formula for ux, t gives: ux, = n a n sin 8 x, 394 t x, = 3n n b n sin 8 8 x. 395 Comparing against the given initial conditions we have x = n a n sin 8 x, 3 +x= 3n n b n sin 8 8 x. 396 hat is, a n and 3 n b 8 n are Fourier sine coefficients of x and 3 + x, respectively, over < x < 8. We compute a n = 8 n x sin 8 8 x dx = 8 8 n xdcos 4 n 8 x = n x cos n 8 x 8 8 n cos 8 x dx = 8 cosn 8 n n n sin 8 8 = 6 n n = n+ 6 n 6 n odd = n n n even b n = 8 8 n 3+x sin 3 n 8 8 x dx So the final answer is = 8 8 n 3 sin dx+ 3 n 8 8 x 8 8 = n sin 3 n 4 8 x dx +a n = n 8 3 n 4 n cos 8 x + n+ 6 n = n n cosn 6 + n+ n = n n + n+ n = 6 n + 76 n+ 3n 4 = 3n n odd ux, t= n x sin 8 x dx 8 3n n even n 3 n n a n cos t + b n sin t sin x. 399

36 with 6 n odd a n = n 6 ; b n = n n even 4 3n n odd 8 3n n even. 4 Example 4. Basic case with Neumann boundary conditions Solution. u t = 4 u x, < x < 5, t > 4, t=, 5, t=, t > 4 x x ux, = 5, <x <5, 43 x x, = sin, <x <5. 44 t 5 Step. As the steps are similar to the previous problem and examples in the heat equation section, we omit the details. he eigenvalue problem gives λ n = and the equation is Step. he n equation is X λx =, X =X 5=. 45 n, n Xn = A n cos 5 5 x, n =,,, 3, 46 4λ =. 47 n +4 which needs to be solved in two cases: n n : n =D n cos 5 n n = 48 5 n t +E n sin t 5 49 n = : =D +E t. 4 Step 3. We have ux, t = c D +E ta + n= n c n D n cos 5 = a + b t + n a n cos t + b n sin 5 n t + E n sin 5 n 5 t cos n t A n cos 5 x n 5 x. 4 Note that we have recognized that Fourier cosine series is relavent and have written the solution in a form that is convenient for computation of Fourier cosine series. Step 4. Comparing with initial values we have 5 = a + n a n cos 5 x ; 4 x sin = b 5 + n n b n cos 5 5 x. 43 As 5 is one of the terms in a, a cos x, we see that no computation is needed before concluding a =, a n = for all other 5 n.

37 For b n we compute b = 5 5 x sin dx=; 44 5 b n = 5 5 x n x sin cos d x n = 5 n +x n x sin sin dx n 5 5 = 5 n + x 5 n n + cos 5 n x + 5 n cos 5 = 5 n+ 5 + n n n + n = n 5 n n 5 n+ = n 5 4 n n 4 = n = nn 4 n n 4 n odd. 45 n even he above computation is not legitimate when n = dueto the appearance of n in denominators. We have to computer separately b = 5 5 x x sin cos dx= 5 4 x sin dx= So the general formula is still true when n =. So the final answer is ux, t=5+ n n b n sin t cos 5 5 x 47 with b n = nn 4 n odd. 48 n even Example 43. Nonhomogeneous but h = hx Solution. Step. Find wx such that he general solution for w is u t = u +, <x <, t > 49 x u, t =, u, t=3, t > 4 ux, = cosx, <x<, 4 x, = 6, t <x <. 4 w xx + =; w=, w=3. 43 wx=c + C x x. 44 5

38 Using the boundary conditions: So w = C = ; w=3 C +C =3 C =. 45 Set v = u w. hen v satisfies wx=+x x. 46 v tt = v xx 47 v, t=v, t = 48 vx, = cos x +x x 49 v t x, = Step. Eigenvalue problems. We omit details. n, λ n = n Xn = A n sin x, n =,, 3, 43 Step. Solve n. n n + n = n =,, 3, 43 gives n n n =D n cos t + E n sin t. 433 Step 3. Write v. vx, t= n n n a n cos t + b n sin t sin x. 434 Step 4. Find coefficients. Comparing with initial conditions we reach cos x +x x = Now we have a n = = 6 = n a n sin x cos x +x x n sin x dx n cos x sin x dx +x x 435 n n b n sin x. 436 n sin x dx. 437 Note that the two integrals require different techniques the first one trignometric identities, the second one integration by parts. herefore we d better calculate them separately. n cos x sin dx x = n n sin + x + sin x dx = n cos + x n + + cos n x n = cos +n cos +n n + + n = n cos n + + n cos n