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- 1. DIFFERENTIAL EQUATIONS Boundary Value Problems & Fourier Series Paul Dawkins
- 2. Differential Equations Table of Contents Preface ...................................................................................................................................... 2 Boundary Value Problems & Fourier Series ........................................................................... 4 Introduction ......................................................................................................................................... 4 Boundary Value Problems ................................................................................................................... 5 Eigenvalues and Eigenfunctions ......................................................................................................... 11 Periodic Functions, Even/Odd Functions and Orthogonal Functions .................................................... 28 Fourier Sine Series ............................................................................................................................. 36 Fourier Cosine Series ......................................................................................................................... 47 Fourier Series ..................................................................................................................................... 56 Convergence of Fourier Series ............................................................................................................ 64 © 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx
- 3. Differential Equations Preface Here are my online notes for my differential equations course that I teach here at Lamar University. Despite the fact that these are my “class notes” they should be accessible to anyone wanting to learn how to solve differential equations or needing a refresher on differential equations. I’ve tried to make these notes as self contained as possible and so all the information needed to read through them is either from a Calculus or Algebra class or contained in other sections of the notes. A couple of warnings to my students who may be here to get a copy of what happened on a day that you missed. 1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learn differential equations I have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of your fellow class mates to see if there is something in these notes that wasn’t covered in class. 2. In general I try to work problems in class that are different from my notes. However, with Differential Equation many of the problems are difficult to make up on the spur of the moment and so in this class my class work will follow these notes fairly close as far as worked problems go. With that being said I will, on occasion, work problems off the top of my head when I can to provide more examples than just those in my notes. Also, I often don’t have time in class to work all of the problems in the notes and so you will find that some sections contain problems that weren’t worked in class due to time restrictions. 3. Sometimes questions in class will lead down paths that are not covered here. I try to anticipate as many of the questions as possible in writing these up, but the reality is that I can’t anticipate all the questions. Sometimes a very good question gets asked in class that leads to insights that I’ve not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are. 4. This is somewhat related to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get you in trouble. As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class. © 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx
- 4. Differential Equations © 2007 Paul Dawkins iii http://tutorial.math.lamar.edu/terms.aspx
- 5. Differential Equations Boundary Value Problems & Fourier Series Introduction In this chapter we’ll be taking a quick and very brief look at a couple of topics. The two main topics in this chapter are Boundary Value Problems and Fourier Series. We’ll also take a look at a couple of other topics in this chapter. The main point of this chapter is to get some of the basics out of the way that we’ll need in the next chapter where we’ll take a look at one of the more common solution methods for partial differential equations. It should be pointed out that both of these topics are far more in depth than what we’ll be covering here. In fact you can do whole courses on each of these topics. What we’ll be covering here are simply the basics of these topics that well need in order to do the work in the next chapter. There are whole areas of both of these topics that we’ll not be even touching on. Here is a brief listing of the topics in this chapter. Boundary Value Problems – In this section we’ll define the boundary value problems as well as work some basic examples. Eigenvalues and Eigenfunctions – Here we’ll take a look at the eigenvalues and eigenfunctions for boundary value problems. Periodic Functions and Orthogonal Functions – We’ll take a look at periodic functions and orthogonal functions in section. Fourier Sine Series – In this section we’ll start looking at Fourier Series by looking at a special case : Fourier Sine Series. Fourier Cosine Series – We’ll continue looking at Fourier Series by taking a look at another special case : Fourier Cosine Series. Fourier Series – Here we will look at the full Fourier series. Convergence of Fourier Series – Here we’ll take a look at some ideas involved in the just what functions the Fourier series converge to as well as differentiation and integration of a Fourier series. © 2007 Paul Dawkins 4 http://tutorial.math.lamar.edu/terms.aspx
- 6. Differential Equations Boundary Value Problems Before we start off this section we need to make it very clear that we are only going to scratch the surface of the topic of boundary value problems. There is enough material in the topic of boundary value problems that we could devote a whole class to it. The intent of this section is to give a brief (and we mean very brief) look at the idea of boundary value problems and to give enough information to allow us to do some basic partial differential equations in the next chapter. Now, with that out of the way, the first thing that we need to do is to define just what we mean by a boundary value problem (BVP for short). With initial value problems we had a differential equation and we specified the value of the solution and an appropriate number of derivatives at the same point (collectively called initial conditions). For instance for a second order differential equation the initial conditions are, y ( t0 ) = y0 y¢ ( t0 ) = y0 ¢ With boundary value problems we will have a differential equation and we will specify the function and/or derivatives at different points, which we’ll call boundary values. For second order differential equations, which will be looking at pretty much exclusively here, any of the following can, and will, be used for boundary conditions. y ( x0 ) = y0 y ( x1 ) = y1 (1) y ¢ ( x0 ) = y0 y ¢ ( x1 ) = y1 (2) y ¢ ( x0 ) = y0 y ( x1 ) = y1 (3) y ( x0 ) = y0 y¢ ( x1 ) = y1 (4) As mentioned above we’ll be looking pretty much exclusively at second order differential equations. We will also be restricting ourselves down to linear differential equations. So, for the purposes of our discussion here we’ll be looking almost exclusively at differential equations in the form, y ¢¢ + p ( x ) y ¢ + q ( x ) y = g ( x ) (5) along with one of the sets of boundary conditions given in (1) – (4). We will, on occasion, look at some different boundary conditions but the differential equation will always be on that can be written in this form. As we’ll soon see much of what we know about initial value problems will not hold here. We can, of course, solve (5) provided the coefficients are constant and for a few cases in which they aren’t. None of that will change. The changes (and perhaps the problems) arise when we move from initial conditions to boundary conditions. One of the first changes is a definition that we saw all the time in the earlier chapters. In the earlier chapters we said that a differential equation was homogeneous if g ( x ) = 0 for all x. Here we will say that a boundary value problem is homogeneous if in addition to g ( x ) = 0 we also have y0 = 0 and y1 = 0 (regardless of the boundary conditions we use). If any of these are not zero we will call the BVP nonhomogeneous. © 2007 Paul Dawkins 5 http://tutorial.math.lamar.edu/terms.aspx
- 7. Differential Equations It is important to now remember that when we say homogeneous (or nonhomogeneous) we are saying something not only about the differential equation itself but also about the boundary conditions as well. The biggest change that we’re going to see here comes when we go to solve the boundary value problem. When solving linear initial value problems a unique solution will be guaranteed under very mild conditions. We only looked at this idea for first order IVP’s but the idea does extend to higher order IVP’s. In that section we saw that all we needed to guarantee a unique solution was some basic continuity conditions. With boundary value problems we will often have no solution or infinitely many solutions even for very nice differential equations that would yield a unique solution if we had initial conditions instead of boundary conditions. Before we get into solving some of these let’s next address the question of why we’re even talking about these in the first place. As we’ll see in the next chapter in the process of solving some partial differential equations we will run into boundary value problems that will need to be solved as well. In fact, a large part of the solution process there will be in dealing with the solution to the BVP. In these cases the boundary conditions will represent things like the temperature at either end of a bar, or the heat flow into/out of either end of a bar. Or maybe they will represent the location of ends of a vibrating string. So, the boundary conditions there will really be conditions on the boundary of some process. So, with some of basic stuff out of the way let’s find some solutions to a few boundary value problems. Note as well that there really isn’t anything new here yet. We know how to solve the differential equation and we know how to find the constants by applying the conditions. The only difference is that here we’ll be applying boundary conditions instead of initial conditions. Example 1 Solve the following BVP. æp ö y ¢¢ + 4 y = 0 y ( 0 ) = -2 y ç ÷ = 10 è4ø Solution Okay, this is a simple differential equation so solve and so we’ll leave it to you to verify that the general solution to this is, y ( x ) = c1 cos ( 2 x ) + c2 sin ( 2 x ) Now all that we need to do is apply the boundary conditions. -2 = y ( 0 ) = c1 10 = y ( p ) = c2 4 The solution is then, y ( x ) = -2 cos ( 2 x ) + 10sin ( 2 x ) We mentioned above that some boundary value problems can have no solutions or infinite solutions we had better do a couple of examples of those as well here. This next set of examples will also show just how small of a change to the BVP it takes to move into these other possibilities. © 2007 Paul Dawkins 6 http://tutorial.math.lamar.edu/terms.aspx
- 8. Differential Equations Example 2 Solve the following BVP. y ¢¢ + 4 y = 0 y ( 0 ) = -2 y ( 2p ) = -2 Solution We’re working with the same differential equation as the first example so we still have, y ( x ) = c1 cos ( 2 x ) + c2 sin ( 2 x ) Upon applying the boundary conditions we get, -2 = y ( 0 ) = c1 -2 = y ( 2p ) = c1 So in this case, unlike previous example, both boundary conditions tell us that we have to have c1 = -2 and neither one of them tell us anything about c2 . Remember however that all we’re asking for is a solution to the differential equation that satisfies the two given boundary conditions and the following function will do that, y ( x ) = -2cos ( 2 x ) + c2 sin ( 2 x ) In other words, regardless of the value of c2 we get a solution and so, in this case we get infinitely many solutions to the boundary value problem. Example 3 Solve the following BVP. y ¢¢ + 4 y = 0 y ( 0 ) = -2 y ( 2p ) = 3 Solution Again, we have the following general solution, y ( x ) = c1 cos ( 2 x ) + c2 sin ( 2 x ) This time the boundary conditions give us, -2 = y ( 0 ) = c1 3 = y ( 2p ) = c1 In this case we have a set of boundary conditions each of which requires a different value of c1 in order to be satisfied. This, however, is not possible and so in this case have no solution. So, with Examples 2 and 3 we can see that only a small change to the boundary conditions, in relation to each other and to Example 1, can completely change the nature of the solution. All three of these examples used the same differential equation and yet a different set of initial conditions yielded, no solutions, one solution, or infinitely many solutions. Note that this kind of behavior is not always unpredictable however. If we use the conditions y ( 0 ) and y ( 2p ) the only way we’ll ever get a solution to the boundary value problem is if we have, y (0) = a y ( 2p ) = a © 2007 Paul Dawkins 7 http://tutorial.math.lamar.edu/terms.aspx
- 9. Differential Equations for any value of a. Also, note that if we do have these boundary conditions we’ll in fact get infinitely many solutions. All the examples we’ve worked to this point involved the same differential equation and the same type of boundary conditions so let’s work a couple more just to make sure that we’ve got some more examples here. Also, note that with each of these we could tweak the boundary conditions a little to get any of the possible solution behaviors to show up (i.e. zero, one or infinitely many solutions). Example 4 Solve the following BVP. y ¢¢ + 3 y = 0 y (0) = 7 y ( 2p ) = 0 Solution The general solution for this differential equation is, y ( x ) = c1 cos ( ) 3 x + c2 sin ( 3x ) Applying the boundary conditions gives, 7 = y ( 0 ) = c1 ( ) 0 = y ( 2p ) = c1 cos 2 3 p + c2 sin 2 3 p ( ) Þ ( c2 = -7 cot 2 3 p ) In this case we get a single solution, y ( x ) = 7 cos ( ) ( 3 x - 7 cot 2 3 p sin ) ( 3x ) Example 5 Solve the following BVP. y ¢¢ + 25 y = 0 y¢ ( 0 ) = 6 y ¢ ( p ) = -9 Solution Here the general solution is, y ( x ) = c1 cos ( 5 x ) + c2 sin ( 5 x ) and we’ll need the derivative to apply the boundary conditions, y ¢ ( x ) = -5c1 sin ( 5 x ) + 5c2 cos ( 5 x ) Applying the boundary conditions gives, 6 = y¢ ( 0 ) = 5c2 Þ c2 = 6 5 -9 = y¢ (p ) = -5c2 Þ c2 = 9 5 This is not possible and so in this case have no solution. All of the examples worked to this point have been nonhomogeneous because at least one of the boundary conditions have been non-zero. Let’s work one nonhomogeneous example where the differential equation is also nonhomogeneous before we work a couple of homogeneous examples. Example 6 Solve the following BVP. © 2007 Paul Dawkins 8 http://tutorial.math.lamar.edu/terms.aspx
- 10. Differential Equations æp ö 5 y ¢¢ + 9 y = cos x y¢ ( 0 ) = 5 yç ÷ = - è2ø 3 Solution The complimentary solution for this differential equation is, yc ( x ) = c1 cos ( 3 x ) + c2 sin ( 3 x ) Using Undetermined Coefficients or Variation of Parameters it is easy to show (we’ll leave the details to you to verify) that a particular solution is, YP ( x ) = 1 cos x 8 The general solution and its derivative (since we’ll need that for the boundary conditions) are, y ( x ) = c1 cos ( 3 x ) + c2 sin ( 3 x ) + 1 cos x 8 y ¢ ( x ) = -3c1 sin ( 3 x ) + 3c2 cos ( 3 x ) - 1 sin x 8 Applying the boundary conditions gives, 5 = y¢ ( 0 ) = 3c2 Þ c2 = 5 3 - 5 = y ( p ) = -c2 3 2 Þ c2 = 5 3 The boundary conditions then tell us that we must have c2 = 5 3 and they don’t tell us anything about c1 and so it is can be arbitrarily chosen. The solution is then, y ( x ) = c1 cos ( 3 x ) + 5 sin ( 3 x ) + 1 cos x 3 8 and there will be infinitely many solutions to the BVP. Let’s now work a couple of homogeneous examples that will also be helpful to have worked once we get to the next section. Example 7 Solve the following BVP. y ¢¢ + 4 y = 0 y (0) = 0 y ( 2p ) = 0 Solution Here the general solution is, y ( x ) = c1 cos ( 2 x ) + c2 sin ( 2 x ) Applying the boundary conditions gives, 0 = y ( 0 ) = c1 0 = y ( 2p ) = c1 So c2 is arbitrary and the solution is, y ( x ) = c2 sin ( 2 x ) and in this case we’ll get infinitely many solutions. Example 8 Solve the following BVP. © 2007 Paul Dawkins 9 http://tutorial.math.lamar.edu/terms.aspx
- 11. Differential Equations y ¢¢ + 3 y = 0 y (0) = 0 y ( 2p ) = 0 Solution The general solution in this case is, y ( x ) = c1 cos ( ) 3 x + c2 sin ( 3x ) Applying the boundary conditions gives, 0 = y ( 0 ) = c1 ( 0 = y ( 2p ) = c2 sin 2 3 p ) Þ c2 = 0 In this case we found both constants to be zero and so the solution is, y ( x) = 0 In the previous example the solution was y ( x ) = 0 . Notice however, that this will always be a solution to any homogenous system given by (5) and any of the (homogeneous) boundary conditions given by (1) – (4). Because of this we usually call this solution the trivial solution. Sometimes, as in the case of the last example the trivial solution is the only solution however we generally prefer solutions to be non-trivial. This will be a major idea in the next section. Before we leave this section an important point needs to be made. In each of the examples, with one exception, the differential equation that we solved was in the form, y ¢¢ + l y = 0 The one exception to this still solved this differential equation except it was not a homogeneous differential equation and so we were still solving this basic differential equation in some manner. So, there are probably several natural questions that can arise at this point. Do all BVP’s involve this differential equation and if not why did we spend so much time solving this one to the exclusion of all the other possible differential equations? The answers to these questions are fairly simple. First, this differential equation is most definitely not the only one used in boundary value problems. It does however exhibit all of the behavior that we wanted to talk about here and has the added bonus of being very easy to solve. So, by using this differential equation almost exclusively we can see and discuss the important behavior that we need to discuss and frees us up from lots of potentially messy solution details and or messy solutions. We will, on occasion, look at other differential equations in the rest of this chapter, but we will still be working almost exclusively with this one. There is another important reason for looking at this differential equation. When we get to the next chapter and take a brief look at solving partial differential equations we will see that almost every one of the examples that we’ll work there come down to exactly this differential equation. Also, in those problems we will be working some “real” problems that are actually solved in places and so are not just “made up” problems for the purposes of examples. Admittedly they will have some simplifications in them, but they do come close to realistic problem in some cases. © 2007 Paul Dawkins 10 http://tutorial.math.lamar.edu/terms.aspx
- 12. Differential Equations Eigenvalues and Eigenfunctions As we did in the previous section we need to again note that we are only going to give a brief look at the topic of eigenvalues and eigenfunctions for boundary value problems. There are quite a few ideas that we’ll not be looking at here. The intent of this section is simply to give you an idea of the subject and to do enough work to allow us to solve some basic partial differential equations in the next chapter. Now, before we start talking about the actual subject of this section let’s recall a topic from Linear Algebra that we briefly discussed previously in these notes. For a given square matrix, A, r r if we could find values of l for which we could find nonzero solutions, i.e. x ¹ 0 , to, r r Ax = l x r then we called l an eigenvalue of A and x was its corresponding eigenvector. It’s important to recall here that in order for l to be an eigenvalue then we had to be able to find nonzero solutions to the equation. So, just what does this have to do with boundary value problems? Well go back to the previous section and take a look at Example 7 and Example 8. In those two examples we solved homogeneous (and that’s important!) BVP’s in the form, y ¢¢ + l y = 0 y (0) = 0 y ( 2p ) = 0 (1) In Example 7 we had l = 4 and we found nontrivial (i.e. nonzero) solutions to the BVP. In Example 8 we used l = 3 and the only solution was the trivial solution (i.e. y ( t ) = 0 ). So, this homogeneous BVP (recall this also means the boundary conditions are zero) seems to exhibit similar behavior to the behavior in the matrix equation above. There are values of l that will give nontrivial solutions to this BVP and values of l that will only admit the trivial solution. So, for those values of l that give nontrivial solutions we’ll call l an eigenvalue for the BVP and the nontrivial solutions will be called eigenfunctions for the BVP corresponding to the given eigenvalue. We now know that for the homogeneous BVP given in (1) l = 4 is an eigenvalue (with eigenfunctions y ( x ) = c2 sin ( 2 x ) ) and that l = 3 is not an eigenvalue. Eventually we’ll try to determine if there are any other eigenvalues for (1), however before we do that let’s comment briefly on why it is so important for the BVP to homogeneous in this discussion. In Example 2 and Example 3 of the previous section we solved the homogeneous differential equation y ¢¢ + 4 y = 0 with two different nonhomogeneous boundary conditions in the form, y (0) = a y ( 2p ) = b In these two examples we saw that by simply changing the value of a and/or b we were able to get either nontrivial solutions or to force no solution at all. In the discussion of eigenvalues/eigenfunctions we need solutions to exist and the only way to assure this behavior is © 2007 Paul Dawkins 11 http://tutorial.math.lamar.edu/terms.aspx
- 13. Differential Equations to require that the boundary conditions also be homogeneous. In other words, we need for the BVP to be homogeneous. There is one final topic that we need to discuss before we move into the topic of eigenvalues and eigenfunctions and this is more of a notational issue that will help us with some of the work that we’ll need to do. Let’s suppose that we have a second order differential equations and its characteristic polynomial has two real, distinct roots and that they are in the form r1 = a r2 = - a Then we know that the solution is, y ( x ) = c1e 1 + c2e 2 = c1ea x + c2e - a x r x r x While there is nothing wrong with this solution let’s do a little rewriting of this. We’ll start by splitting up the terms as follows, y ( x ) = c1ea x + c2 e - a x c1 a x c1 a x c2 - a x c2 - a x = e + e + e + e 2 2 2 2 Now we’ll add/subtract the following terms (note we’re “mixing” the ci and ± a up in the new terms) to get, c1 a x c1 a x c2 - a x c2 - a x æ c1 - a x c1 - a x ö æ c2 a x c2 a x ö y ( x) = e + e + e + e +ç e - e ÷+ç e - e ÷ 2 2 2 2 è2 2 ø è2 2 ø Next, rearrange terms around a little, 1 y ( x) = 2 ( c1ea x + c1e- a x + c2ea x + c2e- a x ) + 1 ( c1ea x - c1e- a x - c2ea x + c2e- a x ) 2 Finally, the quantities in parenthesis factor and we’ll move the location of the fraction as well. Doing this, as well as renaming the new constants we get, ea x + e - a x ea x - e - a x y ( x ) = ( c1 + c2 ) + ( c1 - c2 ) 2 2 ax -a x ax -a x e +e e -e = c1 + c2 2 2 All this work probably seems very mysterious and unnecessary. However there really was a reason for it. In fact you may have already seen the reason, at least in part. The two “new” functions that we have in our solution are in fact two of the hyperbolic functions. In particular, e x + e- x e x - e- x cosh ( x ) = sinh ( x ) = 2 2 So, another way to write the solution to a second order differential equation whose characteristic polynomial has two real, distinct roots in the form r1 = a , r2 = - a is, © 2007 Paul Dawkins 12 http://tutorial.math.lamar.edu/terms.aspx
- 14. Differential Equations y ( x ) = c1 cosh (a x ) + c2 sinh (a x ) Having the solution in this form for some (actually most) of the problems we’ll be looking will make our life a lot easier. The hyperbolic functions have some very nice properties that we can (and will) take advantage of. First, since we’ll be needing them later on, the derivatives are, d d dx (cosh ( x ) ) = sinh ( x ) dx (sinh ( x ) ) = cosh ( x ) Next let’s take a quick look at the graphs of these functions. Note that cosh ( 0 ) = 1 and sinh ( 0 ) = 0 . Because we’ll often be working with boundary conditions at x = 0 these will be useful evaluations. Next, and possibly more importantly, let’s notice that cosh ( x ) > 0 for all x and so the hyperbolic cosine will never be zero. Likewise, we can see that sinh ( x ) = 0 only if x = 0 . We will be using both of these facts in some of our work so we shouldn’t forget them. Okay, now that we’ve got all that out of the way let’s work an example to see how we go about finding eigenvalues/eigenfunctions for a BVP. Example 1 Find all the eigenvalues and eigenfunctions for the following BVP. y ¢¢ + l y = 0 y (0) = 0 y ( 2p ) = 0 Solution We started off this section looking at this BVP and we already know one eigenvalue ( l = 4 ) and we know one value of l that is not an eigenvalue ( l = 3 ). As we go through the work here we need to remember that we will get an eigenvalue for a particular value of l if we get non-trivial solutions of the BVP for that particular value of l . In order to know that we’ve found all the eigenvalues we can’t just start randomly trying values of l to see if we get non-trivial solutions or not. Luckily there is a way to do this that’s not too bad and will give us all the eigenvalues/eigenfunctions. We are going to have to do some cases however. The three cases that we will need to look at are : l > 0 , l = 0 , and l < 0 . Each of © 2007 Paul Dawkins 13 http://tutorial.math.lamar.edu/terms.aspx
- 15. Differential Equations these cases gives a specific form of the solution to the BVP which we can then apply the boundary conditions to see if we’ll get non-trivial solutions or not. So, let’s get started on the cases. l >0 In this case the characteristic polynomial we get from the differential equation is, r2 + l = 0 Þ r1,2 = ± -l In this case since we know that l > 0 these roots are complex and we can write them instead as, r1,2 = ± l i The general solution to the differential equation is then, y ( x ) = c1 cos ( ) l x + c2 sin ( lx ) Applying the first boundary condition gives us, 0 = y ( 0 ) = c1 So, taking this into account and applying the second boundary condition we get, ( 0 = y ( 2p ) = c2 sin 2p l ) This means that we have to have one of the following, c2 = 0 or ( sin 2p l = 0 ) However, recall that we want non-trivial solutions and if we have the first possibility we will get the trivial solution for all values of l > 0 . Therefore let’s assume that c2 ¹ 0 . This means that we have, ( sin 2p l = 0 ) Þ 2p l = np n = 1, 2,3,K In other words, taking advantage of the fact that we know where sine is zero we can arrive at the second equation. Also note that because we are assuming that l > 0 we know that 2p l > 0 and so n can only be a positive integer for this case. Now all we have to do is solve this for l and we’ll have all the positive eigenvalues for this BVP. The positive eigenvalues are then, 22 ænö n ln = ç ÷ = n = 1, 2,3,K è 2ø 4 and the eigenfunctions that correspond to these eigenvalues are, ænxö yn ( x ) = sin ç ÷ n = 1, 2,3,K è 2 ø © 2007 Paul Dawkins 14 http://tutorial.math.lamar.edu/terms.aspx
- 16. Differential Equations Note that we subscripted an n on the eigenvalues and eigenfunctions to denote the fact that there is one for each of the given values of n. Also note that we dropped the c2 on the eigenfunctions. For eigenfunctions we are only interested in the function itself and not the constant in front of it and so we generally drop that. Let’s now move into the second case. l =0 In this case the BVP becomes, y ¢¢ = 0 y (0) = 0 y ( 2p ) = 0 and integrating the differential equation a couple of times gives us the general solution, y ( x ) = c1 + c2 x Applying the first boundary condition gives, 0 = y ( 0 ) = c1 Applying the second boundary condition as well as the results of the first boundary condition gives, 0 = y ( 2p ) = 2c2p Here, unlike the first case, we don’t have a choice on how to make this zero. This will only be zero if c2 = 0 . Therefore, for this BVP (and that’s important), if we have l = 0 the only solution is the trivial solution and so l = 0 cannot be an eigenvalue for this BVP. Now let’s look at the final case. l<0 In this case the characteristic equation and its roots are the same as in the first case. So, we know that, r1,2 = ± -l However, because we are assuming l < 0 here these are now two real distinct roots and so using our work above for these kinds of real, distinct roots we know that the general solution will be, y ( x ) = c1 cosh ( ) -l x + c2 sinh ( -l x ) Note that we could have used the exponential form of the solution here, but our work will be significantly easier if we use the hyperbolic form of the solution here. Now, applying the first boundary condition gives, 0 = y ( 0 ) = c1 cosh ( 0 ) + c2 sinh ( 0 ) = c1 (1) + c2 ( 0 ) = c1 Þ c1 = 0 Applying the second boundary condition gives, © 2007 Paul Dawkins 15 http://tutorial.math.lamar.edu/terms.aspx
- 17. Differential Equations ( 0 = y ( 2p ) = c2 sinh 2p -l ) Because we are assuming l < 0 we know that 2p -l ¹ 0 and so we also know that ( ) sinh 2p -l ¹ 0 . Therefore, much like the second case, we must have c2 = 0 . So, for this BVP (again that’s important), if we have l < 0 we only get the trivial solution and so there are no negative eigenvalues. In summary then we will have the following eigenvalues/eigenfunctions for this BVP. n2 ænxö ln = yn ( x ) = sin ç ÷ n = 1, 2,3,K 4 è 2 ø Let’s take a look at another example with slightly different boundary conditions. Example 2 Find all the eigenvalues and eigenfunctions for the following BVP. y ¢¢ + l y = 0 y¢ ( 0 ) = 0 y¢ ( 2p ) = 0 Solution Here we are going to work with derivative boundary conditions. The work is pretty much identical to the previous example however so we won’t put in quite as much detail here. We’ll need to go through all three cases just as the previous example so let’s get started on that. l >0 The general solution to the differential equation is identical to the previous example and so we have, y ( x ) = c1 cos ( ) l x + c2 sin ( lx ) Applying the first boundary condition gives us, 0 = y ¢ ( 0 ) = l c2 Þ c2 = 0 Recall that we are assuming that l > 0 here and so this will only be zero if c2 = 0 . Now, the second boundary condition gives us, 0 = y ¢ ( 2p ) = - l c1 sin 2p l ( ) Recall that we don’t want trivial solutions and that l > 0 so we will only get non-trivial solution if we require that, ( sin 2p l = 0 ) Þ 2p l = np n = 1, 2,3,K Solving for l and we see that we get exactly the same positive eigenvalues for this BVP that we got in the previous example. 2 2 ænö n ln = ç ÷ = n = 1, 2,3,K è 2ø 4 © 2007 Paul Dawkins 16 http://tutorial.math.lamar.edu/terms.aspx
- 18. Differential Equations The eigenfunctions that correspond to these eigenvalues however are, ænxö yn ( x ) = cos ç ÷ n = 1, 2,3,K è 2 ø So, for this BVP we get cosines for eigenfunctions corresponding to positive eigenvalues. Now the second case. l =0 The general solution is, y ( x ) = c1 + c2 x Applying the first boundary condition gives, 0 = y ¢ ( 0 ) = c2 Using this the general solution is then, y ( x ) = c1 and note that this will trivially satisfy the second boundary condition, 0 = y ¢ ( 2p ) = 0 Therefore, unlike the first example, l = 0 is an eigenvalue for this BVP and the eigenfunctions corresponding to this eigenvalue is, y ( x) = 1 Again, note that we dropped the arbitrary constant for the eigenfunctions. Finally let’s take care of the third case. l<0 The general solution here is, y ( x ) = c1 cosh ( ) -l x + c2 sinh ( -l x ) Applying the first boundary condition gives, 0 = y ¢ ( 0 ) = -l c1 sinh ( 0 ) + -l c2 cosh ( 0 ) = -l c2 Þ c2 = 0 Applying the second boundary condition gives, 0 = y¢ ( 2p ) = -l c1 sinh 2p -l ( ) As with the previous example we again know that 2p -l ¹ 0 and so sinh 2p -l ¹ 0 . ( ) Therefore we must have c1 = 0 . © 2007 Paul Dawkins 17 http://tutorial.math.lamar.edu/terms.aspx
- 19. Differential Equations So, for this BVP we again have no negative eigenvalues. In summary then we will have the following eigenvalues/eigenfunctions for this BVP. n2 ænxö ln = yn ( x ) = cos ç ÷ n = 1, 2,3,K 4 è 2 ø l0 = 0 y0 ( x ) = 1 Notice as well that we can actually combine these if we allow the list of n’s for the first one to start at zero instead of one. This will often not happen, but when it does we’ll take advantage of it. So the “official” list of eigenvalues/eigenfunctions for this BVP is, n2 ænxö ln = yn ( x ) = cos ç ÷ n = 0,1, 2, 3, K 4 è 2 ø So, in the previous two examples we saw that we generally need to consider cases for l as different values will often lead to different general solutions. Do not get too locked into the cases we did here. We will mostly be solving this particular differential equation and so it will be tempting to assume that these are always the cases that we’ll be looking at, but there are BVP’s that will require other/different cases. Also, as we saw in the two examples sometimes one or more of the cases will not yield any eigenvalues. This will often happen, but again we shouldn’t read anything into the fact that we didn’t have negative eigenvalues for either of these two BVP’s. There are BVP’s that will have negative eigenvalues. Let’s take a look at another example with a very different set of boundary conditions. These are not the traditional boundary conditions that we’ve been looking at to this point, but we’ll see in the next chapter how these can arise from certain physical problems. Example 3 Find all the eigenvalues and eigenfunctions for the following BVP. y ¢¢ + l y = 0 y ( -p ) = y (p ) y¢ ( -p ) = y ¢ (p ) Solution So, in this example we aren’t actually going to specify the solution or its derivative at the boundaries. Instead we’ll simply specify that the solution must be the same at the two boundaries and the derivative of the solution must also be the same at the two boundaries. Also, this type of boundary condition will typically be on an interval of the form [-L,L] instead of [0,L] as we’ve been working on to this point. As mentioned above this kind of boundary conditions arise very naturally in certain physical problems and we’ll see that in the next chapter. As with the previous two examples we still have the standard three cases to look at. l >0 The general solution for this case is, y ( x ) = c1 cos ( ) l x + c2 sin ( lx ) © 2007 Paul Dawkins 18 http://tutorial.math.lamar.edu/terms.aspx
- 20. Differential Equations Applying the first boundary condition and using the fact that cosine is an even function (i.e. cos ( - x ) = cos ( x ) ) and that sine is an odd function (i.e. sin ( - x ) = - sin ( x ) ). gives us, ( ) ) ( ( ) c1 cos -p l + c2 sin -p l = c1 cos p l + c2 sin p l ( ) c cos (p l ) - c sin (p l ) = c cos (p l ) + c sin (p l ) 1 2 1 2 -c sin (p l ) = c sin (p l ) 2 2 0 = 2c sin (p l ) 2 This time, unlike the previous two examples this doesn’t really tell us anything. We could have ( ) sin p l = 0 but it is also completely possible, at this point in the problem anyway, for us to have c2 = 0 as well. So, let’s go ahead and apply the second boundary condition and see if we get anything out of that. ( ) ) ( - l c1 sin -p l + l c2 cos -p l = - l c1 sin p l + l c2 cos p l ( ) ( ) l c sin (p l ) + l c cos (p l ) = - 1 2 l c sin (p l ) + 1 l c cos (p l ) 2 l c sin (p l ) = -1 l c sin (p l ) 1 2 l c sin (p l ) = 01 So, we get something very similar to what we got after applying the first boundary condition. Since we are assuming that l > 0 this tells us that either sin p l = 0 or c1 = 0 . ( ) ( ) Note however that if sin p l ¹ 0 then we will have to have c1 = c2 = 0 and we’ll get the ( ) trivial solution. We therefore need to require that sin p l = 0 and so just as we’ve done for the previous two examples we can now get the eigenvalues, p l = np Þ l = n 2 n = 1, 2,3,K Recalling that l > 0 and we can see that we do need to start the list of possible n’s at one instead of zero. So, we now know the eigenvalues for this case, but what about the eigenfunctions. The solution for a given eigenvalue is, y ( x ) = c1 cos ( n x ) + c2 sin ( n x ) and we’ve got no reason to believe that either of the two constants are zero or non-zero for that matter. In cases like these we get two sets of eigenfunctions, one corresponding to each constant. The two sets of eigenfunctions for this case are, yn ( x ) = cos ( n x ) yn ( x ) = sin ( n x ) n = 1, 2,3,K © 2007 Paul Dawkins 19 http://tutorial.math.lamar.edu/terms.aspx
- 21. Differential Equations Now the second case. l =0 The general solution is, y ( x ) = c1 + c2 x Applying the first boundary condition gives, c1 + c2 ( -p ) = c1 + c2 (p ) 2p c2 = 0 Þ c2 = 0 Using this the general solution is then, y ( x ) = c1 and note that this will trivially satisfy the second boundary condition just as we saw in the second example above. Therefore we again have l = 0 as an eigenvalue for this BVP and the eigenfunctions corresponding to this eigenvalue is, y ( x) = 1 Finally let’s take care of the third case. l<0 The general solution here is, y ( x ) = c1 cosh ( ) -l x + c2 sinh ( -l x ) Applying the first boundary condition and using the fact that hyperbolic cosine is even and hyperbolic sine is odd gives, ( ) ( ) c1 cosh -p - l + c2 sinh -p -l = c1 cosh p -l + c2 sinh p - l ( ) ( ) -c sinh ( -p 2 -l ) = c sinh (p 2 -l ) 2c sinh (p 2 -l ) = 0 Now, in this case we are assuming that l < 0 and so we know that p -l ¹ 0 which in turn ( ) tells us that sinh p -l ¹ 0 . We therefore must have c2 = 0 . Let’s now apply the second boundary condition to get, ( ) - l c1 sinh -p -l = -l c1 sinh p -l ( ) 2 -l c sinh (p -l ) = 0 1 Þ c1 = 0 By our assumption on l we again have no choice here but to have c1 = 0 . Therefore, in this case the only solution is the trivial solution and so, for this BVP we again have © 2007 Paul Dawkins 20 http://tutorial.math.lamar.edu/terms.aspx
- 22. Differential Equations no negative eigenvalues. In summary then we will have the following eigenvalues/eigenfunctions for this BVP. l n = n2 yn ( x ) = sin ( n x ) n = 1, 2,3,K l n = n2 yn ( x ) = cos ( n x ) n = 1, 2,3,K l0 = 0 y0 ( x ) = 1 Note that we’ve acknowledged that for l > 0 we had two sets of eigenfunctions by listing them each separately. Also, we can again combine the last two into one set of eigenvalues and eigenfunctions. Doing so gives the following set of eigenvalues and eigenfunctions. l n = n2 yn ( x ) = sin ( n x ) n = 1, 2,3,K l n = n2 yn ( x ) = cos ( n x ) n = 0,1, 2, 3, K Once again we’ve got an example with no negative eigenvalues. We can’t stress enough that this is more a function of the differential equation we’re working with than anything and there will be examples in which we may get negative eigenvalues. Now, to this point we’ve only worked with one differential equation so let’s work an example with a different differential equation just to make sure that we don’t get too locked into this one differential equation. Before working this example let’s note that we will still be working the vast majority of our examples with the one differential equation we’ve been using to this point. We’re working with this other differential equation just to make sure that we don’t get too locked into using one single differential equation. Example 4 Find all the eigenvalues and eigenfunctions for the following BVP. x 2 y ¢¢ + 3 xy¢ + l y = 0 y (1) = 0 y (2) = 0 Solution This is an Euler differential equation and so we know that we’ll need to find the roots of the following quadratic. r ( r - 1) + 3r + l = r 2 + 2r + l = 0 The roots to this quadratic are, -2 ± 4 - 4l r1,2 = = -1 ± 1 - l 2 Now, we are going to again have some cases to work with here, however they won’t be the same as the previous examples. The solution will depend on whether or not the roots are real distinct, double or complex and these cases will depend upon the sign/value of 1 - l . So, let’s go through the cases. 1 - l < 0, l > 1 In this case the roots will be complex and we’ll need to write them as follows in order to write © 2007 Paul Dawkins 21 http://tutorial.math.lamar.edu/terms.aspx
- 23. Differential Equations down the solution. r1,2 = -1 ± 1 - l = -1 ± - ( l - 1) = -1 ± i l - 1 By writing the roots in this fashion we know that l - 1 > 0 and so l - 1 is now a real number, which we need in order to write the following solution, ( ) ( y ( x ) = c1 x -1 cos ln ( x ) l - 1 + c2 x -1 sin ln ( x ) l - 1 ) Applying the first boundary condition gives us, 0 = y (1) = c1 cos ( 0 ) + c2 sin ( 0 ) = c1 Þ c1 = 0 The second boundary condition gives us, ( 0 = y ( 2 ) = 1 c2 sin ln ( 2 ) l - 1 2 ) In order to avoid the trivial solution for this case we’ll require, ( ) sin ln ( 2 ) l - 1 = 0 Þ ln ( 2 ) l - 1 = np n = 1, 2,3,K This is much more complicated of a condition than we’ve seen to this point, but other than that we do the same thing. So, solving for l gives us the following set of eigenvalues for this case. 2 æ np ö ln = 1+ ç ÷ n = 1, 2,3,K è ln 2 ø Note that we need to start the list of n’s off at one and not zero to make sure that we have l > 1 as we’re assuming for this case. The eigenfunctions that correspond to these eigenvalues are, æ np ö yn ( x ) = sin ç ln ( x ) ÷ n = 1, 2,3,K è ln 2 ø Now the second case. 1 - l = 0, l = 1 In this case we get a double root of r1,2 = -1 and so the solution is, y ( x ) = c1 x -1 + c2 x -1 ln ( x ) Applying the first boundary condition gives, 0 = y (1) = c1 The second boundary condition gives, 0 = y ( 2 ) = 1 c2 ln ( 2 ) 2 Þ c2 = 0 © 2007 Paul Dawkins 22 http://tutorial.math.lamar.edu/terms.aspx
- 24. Differential Equations We therefore have only the trivial solution for this case and so l = 1 is not an eigenvalue. Let’s now take care of the third (and final) case. 1 - l > 0, l < 1 This case will have two real distinct roots and the solution is, y ( x ) = c1 x -1+ 1- l + c2 x -1- 1- l Applying the first boundary condition gives, 0 = y (1) = c1 + c2 Þ c2 = -c1 Using this our solution becomes, y ( x ) = c1 x -1+ 1- l - c1 x -1- 1- l Applying the second boundary condition gives, 0 = y ( 2 ) = c1 2-1+ 1- l - c1 2-1- 1- l ( = c1 2-1+ 1- l - 2-1- 1- l ) Now, because we know that l ¹ 1 for this case the exponents on the two terms in the parenthesis are not the same and so the term in the parenthesis is not the zero. This means that we can only have, c1 = c2 = 0 and so in this case we only have the trivial solution and there are no eigenvalues for which l < 1 . The only eigenvalues for this BVP then come from the first case. So, we’ve now worked an example using a differential equation other than the “standard” one we’ve been using to this point. As we saw in the work however, the basic process was pretty much the same. We determined that there were a number of cases (three here, but it won’t always be three) that gave different solutions. We examined each case to determine if non-trivial solutions were possible and if so found the eigenvalues and eigenfunctions corresponding to that case. We need to work one last example in this section before we leave this section for some new topics. The four examples that we’ve worked to this point we’re all fairly simple (with simple being relative of course…), however we don’t want to leave without acknowledging that many eigenvalue/eigenfunctions problems are so easy. In many examples it is not even possible to get a complete list of all possible eigenvalues for a BVP. Often the equations that we need to solve to get the eigenvalues are difficult if not impossible to solve exactly. So, let’s take a look at one example like this to see what kinds of things can be done to at least get an idea of what the eigenvalues look like in these kinds of cases. Example 5 Find all the eigenvalues and eigenfunctions for the following BVP. y ¢¢ + l y = 0 y (0) = 0 y ¢ (1) + y (1) = 0 © 2007 Paul Dawkins 23 http://tutorial.math.lamar.edu/terms.aspx
- 25. Differential Equations Solution The boundary conditions for this BVP are fairly different from those that we’ve worked with to this point. However, the basic process is the same. So let’s start off with the first case. l >0 The general solution to the differential equation is identical to the first few examples and so we have, y ( x ) = c1 cos ( ) l x + c2 sin ( lx ) Applying the first boundary condition gives us, 0 = y ( 0 ) = c1 Þ c1 = 0 The second boundary condition gives us, 0 = y (1) + y¢ (1) = c2 sin ( l )+ l c2 cos( l) = c ( sin ( l ) + 2 l cos ( l ) ) So, if we let c2 = 0 we’ll get the trivial solution and so in order to satisfy this boundary condition we’ll need to require instead that, 0 = sin ( l ) + l cos ( l ) sin ( l)=- l cos ( l ) tan ( l ) = - l Now, this equation has solutions but we’ll need to use some numerical techniques in order to get them. In order to see what’s going on here let’s graph tan ( l ) and - l on the same graph. Here is that graph and note that the horizontal axis really is values of l as that will make things a little easier to see and relate to values that we’re familiar with. © 2007 Paul Dawkins 24 http://tutorial.math.lamar.edu/terms.aspx
- 26. Differential Equations So, eigenvalues for this case will occur where the two curves intersect. We’ve shown the first five on the graph and again what is showing on the graph is really the square root of the actual eigenvalue as we’ve noted. The interesting thing to note here is that the farther out on the graph the closer the eigenvalues come to the asymptotes of tangent and so we’ll take advantage of that and say that for large enough n we can approximate the eigenvalues with the (very well known) locations of the asymptotes of tangent. How large the value of n is before we start using the approximation will depend on how much accuracy we want, but since we know the location of the asymptotes and as n increases the accuracy of the approximation will increase so it will be easy enough to check for a given accuracy. For the purposes of this example we found the first five numerically and then we’ll use the approximation of the remaining eigenvalues. Here are those values/approximations. l1 = 2.0288 l1 = 4.1160 ( 2.4674 ) l 2 = 4.9132 l 2 = 24.1395 ( 22.2066 ) l 3 = 7.9787 l 3 = 63.6597 ( 61.6850 ) l 4 = 11.0855 l 4 = 122.8883 (120.9027 ) l 5 = 14.2074 l 5 = 201.8502 (199.8595 ) ln » ( 2 n -1) 2 ln » 2 n -1 2 p 4 p2 n = 6, 7,8,K The number in parenthesis after the first five is the approximate value of the asymptote. As we can see they are a little off, but by the time we get to n = 5 the error in the approximation is 0.9862%. So less than 1% error by the time we get to n = 5 and it will only get better for larger value of n. The eigenfunctions for this case are, yn ( x ) = sin ( ln x ) n = 1, 2,3,K where the values of l n are given above. So, now that all that work is out of the way let’s take a look at the second case. l =0 The general solution is, y ( x ) = c1 + c2 x Applying the first boundary condition gives, 0 = y ( 0 ) = c1 © 2007 Paul Dawkins 25 http://tutorial.math.lamar.edu/terms.aspx
- 27. Differential Equations Using this the general solution is then, y ( x ) = c2 x Applying the second boundary condition to this gives, 0 = y ¢ (1) + y (1) = c2 + c2 = 2c2 Þ c2 = 0 Therefore for this case we get only the trivial solution and so l = 0 is not an eigenvalue. Note however that had the second boundary condition been y ¢ (1) - y (1) = 0 then l = 0 would have been an eigenvalue (with eigenfunctions y ( x ) = x ) and so again we need to be careful about reading too much into our work here. Finally let’s take care of the third case. l<0 The general solution here is, y ( x ) = c1 cosh ( ) -l x + c2 sinh ( -l x ) Applying the first boundary condition gives, 0 = y ( 0 ) = c1 cosh ( 0 ) + c2 sinh ( 0 ) = c1 Þ c1 = 0 Using this the general solution becomes, y ( x ) = c2 sinh ( -l x ) Applying the second boundary condition to this gives, 0 = y¢ (1) + y (1) = - l c2 cosh ( ) -l + c2 sinh ( -l ) = c2 ( -l cosh ( ) -l + sinh ( -l )) Now, by assumption we know that l < 0 and so -l > 0 . This in turn tells us that sinh ( ) -l > 0 and we know that cosh ( x ) > 0 for all x. Therefore, - l cosh ( ) -l + sinh ( -l ¹ 0 ) and so we must have c2 = 0 and once again in this third case we get the trivial solution and so this BVP will have no negative eigenvalues. In summary the only eigenvalues for this BVP come from assuming that l > 0 and they are given above. So, we’ve worked several eigenvalue/eigenfunctions examples in this section. Before leaving this section we do need to note once again that there are a vast variety of different problems that we can work here and we’ve really only shown a bare handful of examples and so please do now walk away from this section believing that we’ve shown you everything. © 2007 Paul Dawkins 26 http://tutorial.math.lamar.edu/terms.aspx
- 28. Differential Equations The whole purpose of this section is to prepare us for the types of problems that we’ll be seeing in the next chapter. Also, in the next chapter we will again be restricting ourselves down to some pretty basic and simple problems in order to illustrate one of the more common methods for solving partial differential equations. © 2007 Paul Dawkins 27 http://tutorial.math.lamar.edu/terms.aspx
- 29. Differential Equations Periodic Functions, Even/Odd Functions and Orthogonal Functions This is going to be a short section. We just need to have a brief discussion about a couple of ideas that we’ll be dealing with on occasion as we move into the next topic of this chapter. Periodic Function The first topic we need to discuss is that of a periodic function. A function is said to be periodic with period T if the following is true, f ( x + T ) = f ( x) for all x The following is a nice little fact about periodic functions. Fact 1 If f and g are both periodic functions with period T then so is f + g and fg. This is easy enough to prove so let’s do that. ( f + g )( x + T ) = f ( x + T ) + g ( x + T ) = f ( x ) + g ( x ) = ( f + g )( x ) ( fg )( x + T ) = f ( x + T ) g ( x + T ) = f ( x ) g ( x ) = ( fg )( x ) The two periodic functions that most of us are familiar are sine and cosine and in fact we’ll be using these two functions regularly in the remaining sections of this chapter. So, having said that let’s close off this discussion of periodic functions with the following fact, Fact 2 2p sin (w x ) and cos (w x ) are periodic functions with period T = . w Even and Odd Functions The next quick idea that we need to discuss is that of even and odd functions. Recall that a function is said to be even if, f (-x) = f ( x) and a function is said to be odd if, f (-x) = - f ( x) The standard examples of even functions are f ( x ) = x 2 and g ( x ) = cos ( x ) while the standard examples of odd functions are f ( x ) = x 3 and g ( x ) = sin ( x ) . The following fact about certain integrals of even/odd functions will be useful in some of our work. © 2007 Paul Dawkins 28 http://tutorial.math.lamar.edu/terms.aspx
- 30. Differential Equations Fact 3 1. If f ( x ) is an even function then, L L ò f ( x ) dx = 2ò f ( x ) dx -L 0 2. If f ( x ) is an odd function then, L ò f ( x ) dx = 0 -L Note that this fact is only valid on a “symmetric” interval, i.e. an interval in the form [ - L, L ] . If we aren’t integrating on a “symmetric” interval then the fact may or may not be true. Orthogonal Functions The final topic that we need to discuss here is that of orthogonal functions. This idea will be integral to what we’ll be doing in the remainder of this chapter and in the next chapter as we discuss one of the basic solution methods for partial differential equations. Let’s first get the definition of orthogonal functions out of the way. Definition 1. Two non-zero functions, f ( x ) and g ( x ) , are said to be orthogonal on a £ x £ b if, b ò a f ( x ) g ( x ) dx = 0 2. A set of non-zero functions, { f ( x )} , is said to be mutually orthogonal on a £ x £ b (or just i an orthogonal set if we’re being lazy) if fi ( x ) and f j ( x ) are orthogonal for every i ¹ j . In other words, b ì0 i¹ j òa f i ( x ) f j ( x ) dx = í îc > 0 i = j Note that in the case of i = j for the second definition we know that we’ll get a positive value from the integral because, b b ò a f ( x ) f ( x ) dx = ò a é f ( x ) ù 2 i i ë û i dx > 0 Recall that when we integrate a positive function we know the result will be positive as well. Also note that the non-zero requirement is important because otherwise the integral would be trivially zero regardless of the other function we were using. Before we work some examples there are a nice set of trig formulas that we’ll need to help us with some of the integrals. © 2007 Paul Dawkins 29 http://tutorial.math.lamar.edu/terms.aspx
- 31. Differential Equations 1 sin a cos b = ésin (a - b ) + sin (a + b )ù 2ë û 1 sin a sin b = écos (a - b ) - cos (a + b ) ù 2ë û 1 cos a cos b = écos (a - b ) + cos (a + b ) ù 2ë û Now let’s work some examples that we’ll need over the course of the next couple of sections. ¥ ì æ np x ö ü Example 1 Show that ícos ç ÷ý is mutually orthogonal on - L £ x £ L . î è L ø þn = 0 Solution This is not too difficult to do. All we really need to do is evaluate the following integral. L ó æ np x ö æ mp x ö ô cos ç ÷ cos ç ÷ dx õ- L è L ø è L ø Before we start evaluating this integral let’s notice that the integrand is the product of two even functions and so must also be even. This means that we can use Fact 3 above to write the integral as, L L ó æ np x ö æ mp x ö ó æ np x ö æ mp x ö ô cos ç ÷ cos ç ÷ dx = 2ô cos ç ÷ cos ç ÷ dx õ- L è L ø è L ø õ0 è L ø è L ø There are two reasons for doing this. First having a limit of zero will often make the evaluation step a little easier and that will be the case here. We’ll discuss the second reason after we’re done with the example. Now, in order to do this integral we’ll actually need to consider three cases. n=m=0 In this case the integral is very easy and is, L L ò -L dx = 2ò dx = 2 L 0 n=m¹0 This integral is a little harder than the first case, but not by much (provided we recall a simple trig formula). The integral for this case is, L L L ó 2 æ np x ö ó 2 æ np x ö ó æ 2np x ö ô cos ç ÷ dx = 2ô cos ç ÷ dx = ô 1 + cos ç ÷ dx õ- L è L ø õ0 è L ø õ0 è L ø L æ L æ 2np x ö ö L =çx+ sin ç ÷÷ = L + sin ( 2np ) è 2np è L øø 0 2np © 2007 Paul Dawkins 30 http://tutorial.math.lamar.edu/terms.aspx
- 32. Differential Equations Now, at the point we need to recall that n is an integer and so sin ( 2np ) = 0 and our final value for the is, L L ó 2 æ np x ö ó 2 æ np x ö ô cos ç ÷ dx = 2ô cos ç ÷ dx = L õ- L è L ø õ0 è L ø The first two cases are really just showing that if n = m the integral is not zero (as it shouldn’t be) and depending upon the value of n (and hence m) we get different values of the integral. Now we need to do the third case and this, in some ways, is the important case since we must get zero out of this integral in order to know that the set is an orthogonal set. So, let’s take care of the final case. n¹m This integral is the “messiest” of the three that we’ve had to do here. Let’s just start off by writing the integral down. L L ó æ np x ö æ mp x ö ó æ np x ö æ mp x ö ô cos ç ÷ cos ç ÷ dx = 2ô cos ç ÷ cos ç ÷ dx õ- L è L ø è L ø õ0 è L ø è L ø In this case we can’t combine/simplify as we did in the previous two cases. We can however, acknowledge that we’ve got a product of two cosines with different arguments and so we can use one of the trig formulas above to break up the product as follows, L L ó æ np x ö æ mp x ö ó æ np x ö æ mp x ö ô cos ç ÷ cos ç L ÷ dx = 2ô cos ç L ÷ cos ç L ÷ dx õ- L è L ø è ø õ0 è ø è ø æ (n - m)p x ö æ ( n + m )p x ö L ó = ô cos ç ÷ + cos ç ÷ dx õ0 è L ø è L ø L é L æ ( n - m)p x ö L æ ( n + m )p x öù =ê sin ç ÷+ sin ç ÷ú ê ( n - m )p ë è L ø ( n + m)p è L ø ú0 û L L = sin ( ( n - m ) p ) + sin ( ( n + m ) p ) ( n - m)p (n + m)p Now, we know that n and m are both integers and so n - m and n + m are also integers and so both of the sines above must be zero and all together we get, L L ó æ np x ö æ mp x ö ó æ np x ö æ mp x ö ô cos ç ÷ cos ç ÷ dx = 2ô cos ç ÷ cos ç ÷ dx = 0 õ- L è L ø è L ø õ0 è L ø è L ø So, we’ve shown that if n ¹ m the integral is zero and if n = m the value of the integral is a positive constant and so the set is mutually orthogonal. In all of the work above we kept both forms of the integral at every step. Let’s discuss why we did this a little bit. By keeping both forms of the integral around we were able to show that not ¥ ì æ np x ö ü only is ícos ç ÷ý mutually orthogonal on - L £ x £ L but it is also mutually orthogonal î è L ø þn = 0 © 2007 Paul Dawkins 31 http://tutorial.math.lamar.edu/terms.aspx
- 33. Differential Equations on 0 £ x £ L . The only difference is the value of the integral when n = m and we can get those values from the work above. Let’s take a look at another example. ¥ ì æ np x ö ü Example 2 Show that ísin ç ÷ý is mutually orthogonal on - L £ x £ L and on î è L ø þn = 1 0£ x£ L. Solution First we’ll acknowledge from the start this time that we’ll be showing orthogonality on both of the intervals. Second, we need to start this set at n = 1 because if we used n = 0 the first function would be zero and we don’t want the zero function to show up on our list. As with the first example all we really need to do is evaluate the following integral. L ó æ np x ö æ mp x ö ô sin ç ÷ sin ç ÷ dx õ- L è L ø è L ø In this case integrand is the product of two odd functions and so must be even. This means that we can again use Fact 3 above to write the integral as, L L ó æ np x ö æ mp x ö ó æ np x ö æ mp x ö ô sin ç ÷ sin ç ÷ dx = 2ô sin ç ÷ sin ç ÷ dx õ- L è L ø è L ø õ0 è L ø è L ø We only have two cases to do for the integral here. n=m Not much to this integral. It’s pretty similar to the previous examples second case. L L L ó 2 æ np x ö ó 2 æ np x ö ó æ 2np x ö ô sin ç ÷ dx = 2ô sin ç ÷ dx = ô 1 - cos ç ÷ dx õ- L è L ø õ0 è L ø õ0 è L ø L æ L æ 2np x ö ö L =çx- sin ç ÷÷ = L - sin ( 2np ) = L è 2np è L øø 0 2np Summarizing up we get, L L ó 2 æ np x ö ó 2 æ np x ö ô sin ç ÷ dx = 2ô sin ç ÷ dx = L õ- L è L ø õ0 è L ø n¹m As with the previous example this can be a little messier but it is also nearly identical to the third case from the previous example so we’ll not show a lot of the work. © 2007 Paul Dawkins 32 http://tutorial.math.lamar.edu/terms.aspx
- 34. Differential Equations L L ó æ np x ö æ mp x ö ó æ np x ö æ mp x ö ô sin ç ÷ sin ç ÷ dx = 2ô sin ç ÷ sin ç ÷ dx õ- L è L ø è L ø õ0 è L ø è L ø æ ( n - m )p x ö æ ( n + m )p x ö L ó = ô cos ç ÷ - cos ç ÷ dx õ0 è L ø è L ø L é L æ (n - m)p x ö L æ ( n + m )p x öù =ê sin ç ÷- sin ç ÷ú ê ( n - m)p ë è L ø ( n + m )p è L øú0 û L L = sin ( ( n - m ) p ) - sin ( ( n + m ) p ) ( n - m)p (n + m)p As with the previous example we know that n and m are both integers a and so both of the sines above must be zero and all together we get, L L ó æ np x ö æ mp x ö ó æ np x ö æ mp x ö ô sin ç ÷ sin ç ÷ dx = 2ô sin ç ÷ sin ç ÷ dx = 0 õ- L è L ø è L ø õ0 è L ø è L ø So, we’ve shown that if n ¹ m the integral is zero and if n = m the value of the integral is a positive constant and so the set is mutually orthogonal. ¥ ì æ np x ö ü We’ve now shown that ísin ç ÷ý mutually orthogonal on - L £ x £ L and on î è L ø þn = 1 0£ x£ L. We need to work one more example in this section. ¥ ¥ ì æ np x ö ü ì æ np x ö ü Example 3 Show that ísin ç ÷ý and ícos ç ÷ý are mutually orthogonal on î è L ø þn = 1 î è L ø þn = 0 -L £ x £ L . Solution This example is a little different from the previous two examples. Here we want to show that together both sets are mutually orthogonal on - L £ x £ L . To show this we need to show three things. First (and second actually) we need to show that individually each set is mutually orthogonal and we’ve already done that in the previous two examples. The third (and only) thing we need to show here is that if we take one function from one set and another function from the other set and we integrate them we’ll get zero. Also, note that this time we really do only want to do the one interval as the two sets, taken together, are not mutually orthogonal on 0 £ x £ L . You might want to do the integral on this interval to verify that it won’t always be zero. So, let’s take care of the one integral that we need to do here and there isn’t a lot to do. Here is the integral. © 2007 Paul Dawkins 33 http://tutorial.math.lamar.edu/terms.aspx
- 35. Differential Equations L ó æ np x ö æ mp x ö ô sin ç ÷ cos ç ÷ dx õ- L è L ø è L ø The integrand in this case is the product of an odd function (the sine) and an even function (the cosine) and so the integrand is an odd function. Therefore, since the integral is on a symmetric interval, i.e. - L £ x £ L , and so by Fact 3 above we know the integral must be zero or, L ó æ np x ö æ mp x ö ô sin ç ÷ cos ç ÷ dx = 0 õ- L è L ø è L ø So, in previous examples we’ve shown that on the interval - L £ x £ L the two sets are mutually orthogonal individually and here we’ve shown that integrating a product of a sine and a cosine gives zero. Therefore, as a combined set they are also mutually orthogonal. We’ve now worked three examples here dealing with orthogonality and we should note that these were not just pulled out of the air as random examples to work. In the following sections (and following chapter) we’ll need the results from these examples. So, let’s summarize those results up here. ¥ ¥ ì æ np x ö ü ì æ np x ö ü 1. ícos ç ÷ý and ísin ç ÷ý are mutually orthogonal on - L £ x £ L as î è L ø þn = 0 î è L ø þn = 1 individual sets and as a combined set. ¥ ì æ np x ö ü 2. ícos ç ÷ý is mutually orthogonal on 0 £ x £ L . î è L ø þn = 0 ¥ ì æ np x ö ü 3. ísin ç ÷ý is mutually orthogonal on 0 £ x £ L . î è L ø þn = 1 We will also be needing the results of the integrals themselves, both on - L £ x £ L and on 0 £ x £ L so let’s also summarize those up here as well so we can refer to them when we need to. © 2007 Paul Dawkins 34 http://tutorial.math.lamar.edu/terms.aspx
- 36. Differential Equations L ì2 L if n = m = 0 ó æ np x ö æ mp x ö ï 1. ô cos ç ÷ cos ç ÷ dx = í L if n = m ¹ 0 õ- L è L ø è L ø ï0 î if n ¹ m ìL if n = m = 0 L ïL ó æ np x ö æ mp x ö ï 2. ô cos ç ÷ cos ç ÷ dx = í if n = m ¹ 0 õ0 è L ø è L ø ï2 ï0 î if n ¹ m L ó æ np x ö æ mp x ö ì L if n = m 3. ô sin ç ÷ sin ç ÷ dx = í0 if n ¹ m õ- L è L ø è L ø î L ìL ó æ np x ö æ mp x ö ï if n = m 4. ô sin ç ÷ sin ç ÷ dx = í 2 õ0 è L ø è L ø ï0 if n ¹ m î L ó æ np x ö æ mp x ö 5. ô sin ç ÷ cos ç ÷ dx = 0 õ- L è L ø è L ø With this summary we’ll leave this section and move off into the second major topic of this chapter : Fourier Series. © 2007 Paul Dawkins 35 http://tutorial.math.lamar.edu/terms.aspx
- 37. Differential Equations Fourier Sine Series In this section we are going to start taking a look at Fourier series. We should point out that this is a subject that can span a whole class and what we’ll be doing in this section (as well as the next couple of sections) is intended to be nothing more than a very brief look at the subject. The point here is to do just enough to allow us to do some basic solutions to partial differential equations in the next chapter. There are many topics in the study of Fourier series that we’ll not even touch upon here. So, with that out of the way let’s get started, although we’re not going to start off with Fourier series. Let’s instead think back to our Calculus class where we looked at Taylor Series. With Taylor Series we wrote a series representation of a function, f ( x ) , as a series whose terms were powers of x - a for some x = a . With some conditions we were able to show that, f ( n) ( a ) ¥ f ( x) = å ( x - a) n n =0 n! and that the series will converge to f ( x ) on x - a < R for some R that will be dependent upon the function itself. There is nothing wrong with this, but it does require that derivatives of all orders exist at x = a . Or in other words f ( ) ( a ) exists for n = 0,1, 2,3, K Also for some functions the value of R n may end up being quite small. These two issues (along with a couple of others) mean that this is not always the best way or writing a series representation for a function. In many cases it works fine and there will be no reason to need a different kind of series. There are times however where another type of series is either preferable or required. We’re going to build up an alternative series representation for a function over the course of the next couple of sections. The ultimate goal for the rest of this chapter will be to write down a series representation for a function in terms of sines and cosines. We’ll start things off by assuming that the function, f ( x ) , we want to write a series representation for is an odd function (i.e. f ( - x ) = - f ( x ) ). Because f ( x ) is odd it makes some sense that should be able to write a series representation for this in terms of sines only (since they are also odd functions). What we’ll try to do here is write f ( x ) as the following series representation, called a Fourier sine series, on - L £ x £ L . ¥ æ np x ö åB n =1 n sin ç è L ø ÷ There are a couple of issues to note here. First, at this point, we are going to assume that the series representation will converge to f ( x ) on - L £ x £ L . We will be looking into whether or not it will actually converge in a later section. However, assuming that the series does converge © 2007 Paul Dawkins 36 http://tutorial.math.lamar.edu/terms.aspx
- 38. Differential Equations to f ( x ) it is interesting to note that, unlike Taylor Series, this representation will always converge on the same interval and that the interval does not depend upon the function. Second, the series representation will not involve powers of sine (again contrasting this with Taylor Series) but instead will involve sines with different arguments. Finally, the argument of the sines, np x , may seem like an odd choice that was arbitrarily chosen L and in some ways it was. For Fourier sine series the argument doesn’t have to necessarily be this but there are several reasons for the choice here. First, this is the argument that will naturally arise in the next chapter when we use Fourier series (in general and not necessarily Fourier sine series) to help us solve some basic partial differential equations. The next reason for using this argument is that fact that the set of functions that we chose to work { }n ¥ with, sin ( np x ) L in this case, need to be orthogonal on the given interval, - L £ x £ L in this =1 case, and note that in the last section we showed that in fact they are. In other words, the choice of functions we’re going to be working with and the interval we’re working on will be tied together in some way. We can use a different argument, but will need to also choose an interval on which we can prove that the sines (with the different argument) are orthogonal. So, let’s start off by assuming that given an odd function, f ( x ) , we can in fact find a Fourier sine series, of the form given above, to represent the function on - L £ x £ L . This means we will have, ¥ æ np x ö f ( x ) = å Bn sin ç ÷ n =1 è L ø As noted above we’ll discuss whether or not this even can be done and if the series representation does in fact converge to the function in later section. At this point we’re simply going to assume that it can be done. The question now is how to determine the coefficients, Bn , in the series. Let’s start with the series above and multiply both sides by sin ( mp x ) where m is a fixed integer L in the range {1, 2,3,K} . In other words we multiply both sides by any of the sines in the set of sines that we’re working with here. Doing this gives, æ mp x ö ¥ æ np x ö æ mp x ö f ( x ) sin ç ÷ = å Bn sin ç ÷ sin ç ÷ è L ø n=1 è L ø è L ø Now, let’s integrate both sides of this from x = - L to x = L . L L ó æ mp x ö ó ¥ æ np x ö æ mp x ö ô f ( x ) sin ç ÷ dx = ô å Bn sin ç ÷ sin ç ÷ dx õ -L è L ø õ - L n =1 è L ø è L ø At this point we’ve got a small issue to deal with. We know from Calculus that an integral of a finite series (more commonly called a finite sum….) is nothing more than the (finite) sum of the integrals of the pieces. In other words for finite series we can interchange an integral and a series. For infinite series however, we cannot always do this. For some integrals of infinite series we © 2007 Paul Dawkins 37 http://tutorial.math.lamar.edu/terms.aspx

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