Series and Transforms PDF Lecture Notes

Series and Transforms February 9, 2025 Contents 1 Introduction 3 1.1 Series............................... 3 1.2 Why is it important?....................... 3 1....

Series and Transforms February 9, 2025 Contents 1 Introduction 3 1.1 Series............................... 3 1.2 Why is it important?....................... 3 1.3 Unifying topics of this course.................. 7 1.4 How is the course taught?.................... 7 2 Fourier Series 8 2.1 Interesting and maybe useful facts about Fourier series.... 12 2.1.1 Complex Fourier Series................. 12 2.1.2 Fourier Series and square integrable functions..... 13 2.2 Convergence............................ 17 2.3 Different Periods......................... 20 2.4 Differentiation/integration within a series........... 21 3 Legendre polynomials 21 3.1 Other orthogonal systems.................... 27 3.2 Bessel functions.......................... 28 4 Integral transforms 31 4.1 Laplace Transform........................ 31 4.2 Fourier Transform........................ 37 5 Series and ODEs 44 5.1 Background material on series.................. 44 5.2 Ordinary Differential Equations................. 49 5.3 The Frobenius method for second order linear ODEs..... 53 1 Preface The lecture is based on the series of lectures given by Prof. Nicolas Dirr between the years 2019 and 2023. The course covers four main topics: Fourier Series, the Fourier transform, Legendre polynomials and the Frobe- nius method. In addition to the lecture notes, a ‘cheat sheet’ for each topic will be made available, which summarizes the main content of the section. The lecture notes contain some non-examinable material which is marked in purple. This applies mostly to particularly difficult proofs. In this case, you should be familiar with the statement of the Theorem and its application, but are not required to know the details of the proof. The course notes will be updated throughout the semester. Raphael Winter January 2025 Cardiff. 2 1 Introduction 1.1 Series In Foundations you have learned about series - ”infinite sums” that are actually limits in disguise, i.e. ∞ N X 1 X 1 2 = lim , n N →∞ n2 n=1 n=1 | {z } partial sum or the geometric series ∞ 1 X = xn. 1−x n=0 In the second example we see that a function can be expressed as a series: There is a function on the left hand side and a series on the right hand side. In this case the series on Pthe right is a series of powers of the variable x, so all partial sums SN (x) = N n=0 x n are polynomials. If the function is periodic, it is more natural to use sin and cos in the series, so that the partial sums are so-called trigonometric polynomials: ∞ X f (x) = a0 + (an cos(nx) + bn sin(nx)) n=1 N X = a0 + lim (an cos(nx) + bn sin(nx)). N →∞ n=1 Such a series is called Fourier series, this will be our first topic in this course. Note that if you consider a series as a limit of its partial sums, then any power series is a special case of approximating a function by polynomi- als. We will encounter, however, other ways of approximating functions by polynomials in this course. 1.2 Why is it important? ˆ Historically, Fourier and power series were convenient because they allow to transform differential equations in algebraic equations. This offered a way to solve differential equations in a pre-computer age. 3  Indeed, Fourier found the Fourier series when studying heat distribu- 2 u(t, x, y) + ∂ 2 u(t, x, y). For tion, i.e. the equation ∂t u(t, x, y) = ∂xx yy example, consider the differential equation u′′ (x) = −u(x) with periodic boundary conditions, i.e. we want u(−π) = u(π). Let us use a Fourier series: Making the assumption ∞ X u(x) = (an cos(nx) + bn sin(nx)), n=0 we get (ignoring issues with differentiating series) ∞ X ∞ X 2 2 (−n an cos(nx) − n bn sin(nx)) = − (an cos(nx) + bn sin(nx)), n=1 n=1 which implies n2 an = an and n2 bn = bn , i.e. an = bn = 0 for n = ̸ 1, leaving us with linear combinations of sin(x) and cos(x) as the only possible solutions. ˆ Today Fourier series (and the Fourier transform, its variant for non- periodic functions defined on all of R) are extremely important as a means to store, compress and analyze data. CDs and even the record- ing of these lectures would be difficult, slow and expensive without it. A piece of music is a changing air pressure over time. In prin- ciple, one would need to store the pressure at any point in time for the duration of the concert. In old times this was done by engraving a pattern of waves on a vinyl disc. It worked, but it was difficult to store on a smartphone! We could instead use a Fourier series as above to represent the function p(t) ”air pressure at time t.” ∞ X p(t) = (an cos(nt) + bn sin(nt)) n=0 Two additional insights make this extremely useful: First, our ears can hear only signals with not too low and not too high frequencies (between 20Hz and 20kHz). (Elephants can hear very low frequencies, bats very high ones, instead.) So we need only finitely many of the coefficients, i.e. for a human ear N2 X p(t) = (an cos(nt) + bn sin(nt)). n=N1 4 This is now a couple of numbers that can easily be stored on comput- ers, phones, sent by email etc. Second, there is a clever way to compute such a ”finite” Fourier series on computers very quickly, called Fast Fourier Transform (FFT). (For n data points you need n log n multi- plications, which is almost optimal for large n, as you should touch each point at least once.) Even though this will not be a topic of this course, we can find Fourier series for a function of two or more variables. A black-and white image, for simplicity, can be described by a function u(x, y) for the grey value in the point (x, y). The Fourier coefficients can then be stored digitally and proecessed. For example, when storing a movie, we can store only the change of the image instead of the entire image for each time. This would be impossible with old-fashioned video tapes. Another example, this time from Paul Lutus’ webpage arachnoid.com: Tidal information, i.e. the time and water height of low and high tide in a place at the seaside, was collected by observations in vast volumes of tide tables. Tides are cause by three periodic motions: The rotation of the earth around itself, the rotation of the earth around the sun, and the rotation of the moon around the earth. This makes it an ideal application for Fourier techniques. ”First, tidal observations for a particular location are collected and placed in a data set that correlates tidal heights with time. Then a Fourier Transform is applied to create a compact representation of the original data. Just as in the earlier examples, the compact representation contains within it the essence of the original wave pattern, but in a much smaller size, and with the important property that the original waves can be modeled for any time in the past or future. Finally, an inverse Fourier Transform is applied to the compact form to provide a tidal prediction for a particular time and place.” An app on your smartphone can replace thousands of printed pages for decades to come! 5 ˆ This course is also an important foundation for pure mathematics, more specifically for the theory of linear operators: You may have seen or will see soon that linear maps between finite dimensional vector spaces can be represented by matrices, once we have fixed a basis for the vector space. For 2  sure  you  have seen the standard basis of R , 1 0 the unit vectors and. Any vector is a linear combination of 0 1 these, and in addition they are orthogonal , there scalar product (dot product) is zero. A set of functions, like for example the continuous functions on the interval [−π, π] which are 2π periodic are a vector space: You can add them and multiply them by a scalar, and for these operations the commutative and distributive law hold. These function spaces are infinite dimensional, so we cannot hope to express any function as a sum of scalar multiples of finitely many ”basis functions.” But if we take countably many and allow limits, we can, by using a Fourier series: X∞ f (x) = (an cos(nx) + bn sin(nx)) n=0 The sin and cos are a basis for this function space and they are or- thogonal with respect to a scalar product defined by integrals: Z π ⟨f, g⟩ := f (x)g(x)dx −π Plug in some of cos(nx), sin(mx) and you will get zero unless they are both the same function. (We will do this in detail later.) The good thing about this infinite basis is that it is made up of eigen- vectors (here ”eigenfunctions”) of the second derivative, which is a linear map sending functions to functions. In other words, the matrix representation of the map u 7→ u′′ in this basis is a diagonal matrix. This is why differential operators look simpler if we use Fourier series, and why these series are fundamental for their study. Fourier series have applications on other areas of pure mathematics, even in number theory. A lot of work on Fourier analysis was done in the 1920s by the pure mathematicians Hardy and Littlewood. (Read Hardy’s ”A mathematician’s Apology” for his self-description as pure mathematician.) 6 Other topics of this course Similar principles apply to the Fourier and Laplace transforms, which are similar to Fourier series, but for function which are not necessarily periodic. We will also see how functions can be approximated by certain polyno- mials. Again these are very useful both in applied and in pure mathematics. In addition we will see how power series can help to get information on ODEs. We will review a bit what you have learned in year one, and then focus on ODEs with a singularity, i.e. where the coefficients are not defined in a certain point, because you would have to divide by zero, e.g. 1 y ′′ = y, x2 which is not defined for x = 0. They are important in physics (gravitational or electrostatic potential!). We would like to know how many solutions there are and how they behave near this singular point. (Of course the solution of any initial value problem is perfectly well defined if we start away from x = 0, but what happens if we let x tend to zero?) 1.3 Unifying topics of this course There are two ideas that run through all the course : ˆ Series: How can we use series of functions to solve mathematical prob- lems, both pure and applied? ˆ Orthogonality: The series often are a generalization of the standard basis of a vector space, i.e. we express a function as an ”infinite linear combination” of functions which are orthogonal to each other in some suitable scalar product. 1.4 How is the course taught? This course is aimed at all students, so there will not be many detailed proofs, rather explanations. If you are interested in making the material of this course rigorous then there is a year 3 course for you: Fourier and Functional Analysis. We will have fortnightly exercise sheets with longer ex- ercises on the material, and regular multiple choice tests on learning central for extra practice. Sometimes I will put extra some material and examples, depending on how the course is progressing. 7 2 Fourier Series Here we consider functions f : R → R which are 2π periodic, i.e. f (x+2π) = f (x) for all x ∈ R. Note that this is exactly the same as picking an interval of length 2π (we usually choose [−π, π]) and requiring that the values at the endpoints are the same (i.e. here f (−π) = f (π).) Given such a function on [−π, π], we can always extend it periodically to all of R by repeating it on any adjacent interval of length 2π. Our aim is to find numbers an , bn , n ∈ {0, 1, 2,...} called Fourier coef- ficients such that ∞ X f (x) = a0 + (an cos(nx) + bn sin(nx)), (2.1) n=1 which means that for all (or at least very many) x ∈ R the limit on the r.h.s. (series is a limit in disguise!) exists and equals the function value at that point x. The right hand side is called a Fourier Series. We will discuss ˆ how to find the coefficients given f ; ˆ criteria for f and x such that the equality (2.1) holds; ˆ examples. But first we need to recall some more properties of the trigonometric func- tions and introduce some short hand notation. Definition 2.1 (T -periodic functions). 1. Let T > 0 be a real number. A function f : R → R is called T-periodic, if and only if f (x + T ) = f (x) for all x ∈ R. 2.We call C2π the set of all continuous functions f : R → R such that f is 2π-periodic, i.e. f (x + 2π) = f (x) for all x ∈ R. sin and cos are 2π-periodic. x 7→ sin(25x) is (2π)/25-periodic etc. Recall that for trig functions the addition theorems hold sin(a + b) = sin(a) cos(b) + cos(a) sin(b), cos(a + b) = cos(a) cos(b) − sin(a) sin(b). As a consequence, for n, m ∈ Z, 1 sin(nx) sin(mx) = (cos((n − m)x) − cos((n + m)x)) , (2.2) 2 1 cos(nx) cos(mx) = (cos((n − m)x) + cos((n + m)x)) , (2.3) 2 1 cos(nx) sin(mx) = (sin((n + m)x) − sin((n − m)x)). (2.4) 2 8 The proof of (2.2) and (2.3) follows by induction from the addition theorems. This implies the following lemma simply using the relations above for the integrand and then doing the integrals: Lemma 2.1 (Orthogonality relations). For natural numbers n, m we have Z π sin(nx) cos(mx)dx = 0, (2.5) −π Z π Z π (sin(nx))2 = (cos(nx))2 = π. (2.6) −π −π Furthermore, if n ̸= m Z π Z π sin(nx) sin(mx)dx = 0 = cos(nx) cos(mx)dx. (2.7) −π −π Proof. For the proof of (2.5), we use that f (x) = sin(nx) cos(mx) is an odd function, hence the integral vanishes. For the remaining identities, we first observe that for any k ∈ N, k ̸= 0 we have Z π Z π cos(kx)dx = sin(kx)dx = 0. (2.8) −π −π We then use (2.2) and (2.3) to rewrite Z π Z π 1
sin(nx) sin(mx)dx = cos((n − m)x) − cos((n + m)x) dx, 2 Z −π π Z−π π 1
cos(nx) cos(mx)dx = cos((n − m)x) + cos((n + m)x) dx. −π −π 2 For n ̸= m, we infer (2.7) from (2.8). For n = m we obtain using (2.8) Z π Z π 1
sin(nx) sin(nx)dx = cos((n − n)x) − cos((n + n)x) dx −π 2 Z−π π 1 = dx = π. −π 2 Repeating the argument for cos gives Z π Z π 1
cos(nx) cos(nx)dx = cos((n − n)x) + cos((n + n)x) dx −π 2 Z−ππ 1 = dx = π, −π 2 as claimed. 9  Why orthogonality relation? In a vector space we can define a scalar product, which takes two vectors as input and gives a number as output and is linear in each comonent and (conjugate)-symmetric. An example is the dot product of two vectors in Rn :         a c a c , = · = ac + bd. b d b d If the scalar product between two vectors is zero, then we call these vectors orthogonal. We can do the same in a vector space of suitable (see year 3) 2π-periodic functions: Z π ⟨f, g⟩ = f (x)g(x)dx. −π With this scalar product (called L2 scalar product) the functions sin(nx), cos(mx) are always orthogonal to each other, unless they are actually the same func- tion. Further, any vector  in R2  can be written as a linear combination of the 1 0 two orthogonal vectors , : 0 1       a 1 0 =a +b. b 0 1 Now we want to take a function f ∈ C2π and find coefficients an , bn so that we can write this function as ”linear combination” of the orthogonal ”basis vectors” sin(nx), cos(nx). This function space is so big that finite sums will not do, we will need ”infinite sums”, i.e. a series in mathematical terms, the Fourier series: Definition 2.2 (Fourier coefficients). Let f ∈ C2π. Then its Fourier Coef- ficients are defined as Z π 1 a0 := f (x)dx (2.9) 2π −π 1 π Z an = f (x) cos(nx)dx for n ∈ N, n ≥ 1, (2.10) π −π Z π 1 bn = f (x) sin(nx)dx for n ∈ N, n ≥ 1, (2.11) π −π 10 The series ∞ X S(f )(x) := a0 + (an cos(nx) + bn sin(nx)), n=1 is called the Fourier Series of f. (Note that there is no coefficient b0 as sin(0x) = 0 for all x!). This is the unique choice for the coefficients that has a chance of making the Fourier series converge to the function, i.e. it is a necessary condition fo equality in (2.1). We can see this by simply multiplying (2.1) by sin(nx) and integrating from −π to π: By the orthogonality relations, on the r.h.s. everything drops out except πbn. By multiplying by cos(nx) we get the formula for bn. As cos(0x) = 1, we get a0 by simply integrating (2.1) over [−π, π]. Remark 2.1. Definition 2.2 works not only for continuous functions, but for all functions for which the integrals are defined and finite, in particular for functions which are continuous except for finitely many points at which they jump. (Recall the Riemann integral is defined for such functions!) It remains to investigate for which x and under which conditions (2.1) really holds, i.e. S(f )(x) = f (x). But before going further in that direction, we will compute the Fourier coefficients of some functions as examples. Example 2.1. (Triangle function) if x ∈  2k − 21
π, 2k + 12
π   

 x − 2kπ, f (x) = k∈Z −x + (2k + 1)π, if x ∈ 2k + 12 π, 2k + 32 π This is in C2π. It is odd, i.e. f (−x) = −f (x). As cos is even, the area under the graph of cos(nx)f (x) from −π to 0 is the same as the one form 0 to π, but with opposite sign, so they cancel out and a0 = 0, an = 0, n ∈ N. We find ∞ 4 X (−1)k+1 S(f )(x) = sin((2k − 1)x). π (2k − 1)2 k=1 Important Fact: If a function f is odd (f (−x) = −f (x)), all the an are zero, if it is even, (f (−x) = f (x)) all the bn are zero. Example 2.2. 3 1 1 f (x) = sin4 (x) = − cos(2x) + cos(4x). (2.12) 8 2 8 11 This is an example of a finite Fourier Series, and can be verified using the addition theorems. By the identity (2.12) the Fourier coefficients are bn = 0 for all n ∈ N, 3 1 1 a0 = , a1 = 0, a2 = − , a3 = 0, a4 = , 8 2 8 ak = 0 for all k ≥ 5. End of lecture January 30 Example 2.3. Consider the function f : R → R given by f (x) = x in [−π, π) extended periodically. This is continuously differentiable everywhere except at integer multiples of π. Then ∞ X (−1)n+1 S(f )(x) = 2 sin(nx). n n=1 (The function is odd, so only the bn appear!) 2.1 Interesting and maybe useful facts about Fourier series 2.1.1 Complex Fourier Series Using eix = cos(x) + i sin(x) we can define Fourier Series for complex valued functions. Here we will only use real valued functions, but even for these the notation is much simpler and computations get more transparent, because we get rid of the addition theorems for trig functions and need just rules for exponentials! Lemma 2.2. (Complex Fourier Series) For a 2π-periodic function f its complex Fourier series X N X inx cn e = lim cn einx N →∞ n∈Z n=−N has coefficients Z π 1 cn = f (x)e−inx dx. 2π −π They are related to the coefficients in (2.2) by c0 = a0 1 cn = (an − ibn ) , n > 0 2 1 cn = (a−n + ib−n ) = c−n , n < 0. 2 12 2.1.2 Fourier Series and square integrable functions A 2π-periodic function f is called square integrable or f ∈ L2 (−π, π) if the integral Z π (f (x))2 dx −π exists and is a finite number. This is in particular the case if it is Riemann integrable or if the integral is an improper Riemann integral with a finite limit value. The resulting class of funtions will be studied more in year 3. It suffices to say that they are far more than continuous or piecewise continous functions. This class of functions is particularly well suited for the Fourier Series: Lemma 2.3. If f is 2π-periodic and square integrable and ∞ X S(f )(x) = a0 + ((an cos(nx) + bn sin(nx)), n=1 then for all N ∈ N 1. N X Z π 2πa20 + π (a2n + b2n ) ≤ (f (x))2 dx n=1 −π 2. limn→∞ an = limn→∞ bn = 0 3. ∞ X Z π 2πa20 + π (a2n + b2n ) = (f (x))2 dx n=1 −π Proof. The proof of 1. follows from ∞ !2 Z π X 0≤ f (x) − ((an cos(nx) + bn sin(nx)). −π n=0 Do the square and use the definition of the Fourier coefficients and the orthogonality P∞ relations. 2. follows from 1., as the partial sums of the series (a 2 +b2 ) are monotone and bounded, so the series is convergent, and its n=0 n n terms have to converge to zero. (2. is not sufficient to conclude convergence of the Fourier series itself!) The proof of 3. proof is beyond this course, see year 3. 13  Note 2 R π that we2use cos(nx), sin(nx), which are not normalised w.r.t L , i.e. −π (cos(nx)) dx ̸= 1 etc., so we need the factors of π above. The interesting part about 3. is that it pushes further the analogy be- tween sin(nx) and cos(nx) as basis functions and the standard basis vectors of Rn : If       a 1 0 =a +b , b 0 1 √ then the euclidean length of this vector is a2 + b2. This makes it natural to define the so-called L2 -norm of a function as sZ π ∥f ∥2 = (f (x))2 dx, −π so that v u ∞ X u ∥f ∥2 = t2πa20 + π (a2n + b2n ). n=1 You will see in year 3 that Fourier series always converge to f in the sense of this L2 -distance: v uZ N !2 u π X lim t f (x) − ((an cos(nx) + bn sin(nx)) dx = 0 N →∞ −π n=0 There is yet another way the L2 norm goes well with the Fourier trans- form. As PNyou know, a polynomial of degree N is a function of the form n p(x) = n=0 an x. In analogy, we call a trigonometric polynomial of degree N a function of the form N X p(x) = α0 + ((αn cos(nx) + βn sin(nx)) n=0 for real numbers αn , βn and a finite natural number N. The N -the partial sum of a Fourier series is a trigonometric polynomial, and it is the best approximation to f in the sense of L2 : Definition 2.3. For any 2π-periodic f we define N X SN (f )(x) = a0 + ((an cos(nx) + bn sin(nx)) n=1 the N-th partial sum of the Fourier series. 14 Lemma 2.4. Let f be square integrable, then for any trigonometric polyno- mial p(x) of degree N we have Z π Z π 2 (f (x) − SN (f )(x)) dx ≤ (f (x) − p(x))2 dx. −π −π In words: The partial sums of the Fourier series are the best approximation by trigonometric polynomials to a given function, if ”best” is understood in the L2 sense or equivalently as minimizing the mean square error of the approximation. Proof. Let N ∈ N be a natural number. Then for an arbitrary trigonometric polynomial p of degree N we compute Z π Z π N
2 X
2 f (x) − p(x) dx = f (x) − (α0 + (αn cos(nx) + βn sin(nx))) −π −π n=1 N X
= ⟨f, f ⟩ − 2 2πα0 a0 + π (αn an + βn bn ) n=1 N X + 2πα02 + π (αn2 + βn2 ). n=1 Adding and subtracting πa2n + πb2n we obtain Z π
2 f (x) − p(x) dx −π N X =⟨f, f ⟩ − 2πa20 −π (a2n + b2n ) n=1 XN +2π(a0 − α0 )2 + π (an − αn )2 + (bn − βn )2. n=1 The first line is independent of the coefficients αn , βn. The second line becomes minimal for βn = bn , αn = an. Hence the expression becomes minimal if p is the exactly N − th partial Fourier sum of f , which was to be shown. The following Lemma shows that, if you have computed the coefficients an (f ), bn (f ) for a function f and an (g), bn (g) for another function g, then you can find the coefficients for λf (x)+κg(x) without doing a single integration. 15 Lemma 2.5 (Linearity). Let f, g be square integrable 2π periodic functions, and let an (f ), bn (f ) and an (g), bn (g) be their respective Fourier coefficients. Then for any λ, κ ∈ R the Fourier coefficients of λf + κg are given by an (λf + κg) = λan (f ) + κan (g), bn (λf + κg) = λbn (f ) + κbn (g). Proof. Follows from the linearity of the Riemann integral. We have 1 π Z
bn (λf + κg) = λf (x) + κg(x) sin(nx)dx π −π 1 π 1 π Z Z =λ f (x) sin(nx)dx + κ g(x) sin(nx)dx π −π π −π = λbn (f ) + κbn (g). The proof for the coefficients an is analogous. End of lecture February 2 16 2.2 Convergence QUESTION: Is the resulting Fourier series S(f )(x) equal to the original function? ˆ The series may be convergent or divergent at a given x. ˆ Even if it is convergent, its sum may not be equal to f. ˆ The convergence may be uniform for all x, or slower at certain x. A complete answer is far beyond the scope of this course, we will only give some partial answers. The convergence issue is tricky, for example there are continuous 2π periodic functions for which the Fourier series does not converge to the function for certain x. On the other hand, for differentiable functions the Fourier series converges everywhere. First let us recall some material from Foundation II: Definition 2.4. (i) Given f : R and x0 ∈ R we say that the real number L is the right- handed limit (or limit from the right) of f at x0 , if for every ε > 0 there exists a δ = δ(ε) > 0 such that |f (x) − L| < ε 0 < x − x0 < δ. In this case we write limx→x+ f (x) = L. 0 (ii) Given f : R and x0 ∈ R we say that the real number L is the left- handed limit (or limit from the left) of f at x0 , if for every ε > 0 there exists a δ = δ(ε) > 0 such that |f (x) − L| < ε − δ < x − x0 < 0. In this case we write limx→x− f (x) = L. 0 In this course we will use the short-hand notation f (x0 +) := lim f (x), f (x0 −) := lim f (x). x→x+ 0 x→x− 0 We apply the same idea to the difference quotient: Even if a function is not differentiable at x0 , we can determine the slope from the left and from the right. 17 Definition 2.5. (i) Given f : R → R and x0 ∈ R such that f (x0 +) exists we say that the real number L is the right-handed derivative (or limit from the right) of f at x0 , if for every ε > 0 there exists a δ = δ(ε) > 0 such that f (x) − f (x0 +) − L < ε 0 < x − x0 < δ. x − x0 In this case we write f+′ (x0 ) = L. (ii) Given f : R → R and x0 ∈ R such that f (x0 −) exists we say that the real number L is the left-handed derivative (or limit from the left) of f at x0 , if for every ε > 0 there exists a δ = δ(ε) > 0 such that f (x) − f (x0 −) −L 0 and K(p, x) = 0 else. But for which f and p does this integral exist? First, the integral up tp +∞ is an improper Riemann integral, i.e. a limit of Riemann integrals: Z ∞ Z b f (x)e−px dx = lim f (x)e−px dx. 0 b→∞ 0 31 Second, f (x) should not grow too fast or the limit would be ∞: Say p = 1 and f (x) = ex , then Z b Z b Z b −px x −x f (x)e dx = e e dx = 1dx = b → ∞. 0 0 0 For p > 1, however, the limit exists. This will happen often: The Laplace transform of f is defined for p > p0 for some p0 depending on f, but not necessarily for p ≤ p0. Here we do not aim for the largest possible class of functions for which the Laplace transform is defined, but for one we can work with in year 2. This leads to the following definition: Definition 4.1. Let f : [0, ∞) → R be piecewise continuous and of exponen- tial growth, i.e. suppose there exists M > 0 and p0 ∈ R s.t. |f (x)| ≤ M ep0 x for all x ≥ 0. Then the Laplace transform of f, L[f ] : (p0 , ∞) → R : p 7→ L[f, p] is defined as Z ∞ Z b −px L[f, p] := f (x)e dx := lim f (x)e−px dx, p > p0. 0 b→∞ 0 Lemma 4.1. The integral in the previous definition is defined and finite. Proof: f is piecewise continuous and e−px is continuous. Moreover |f (x)e−px | ≤ M e(p0 −p)x ≤ M, so the integrand is bounded on [0, b]. Hence Rb −px dx is defined as Riemann integral for all b > 0. Moreover for 0 f (x)e c>b Z b Z c Z c Z d −px −px −px f (x)e dx − f (x)e dx ≤ |f (x)|e dx ≤ M e(p0 −p)x dx 0 0 b c M = e−|p−p0 |b (1 − e−|p0 −p|(c−b) ), |p0 − p| so the limit exists and is a real number by a Cauchy criterion similar to the one for sequences. (Replace b and c by m and n.) Remark 4.1. In order to simplify notation, we often write F (p) instead of L[f, p] for the Laplace transform 32 Example 4.1. The following list of Laplace transforms will be used fre- quently: f (x) F (p) defined for 1 1 p p>0 n! xn pn+1 p > 0, n ∈ N0 e−ax 1 p+a p+a>0 p cos(ax) p2 +a2 p > 0, a ∈ R a sin(ax) p2 +a2 p > 0, a ∈ R e−ap H(x − a) p p > 0, a > 0 Here H is the Heaviside function,  0, x 0 h x i L f , p = aL[f, pa] a 3. Multiplying by an exponential becomes a shift (attention, minus be- comes plus!) : Let a ∈ R be a number. Then L[e−ax f (x), p] = L[f, p + a]. 33  4. A shift and multiplying by the Heaviside function turns into multiplying by an exponential: For a > 0 L[f (x − a)H(x − a), p] = e−pa L[f, p] 5. For p > p0 ≥ 0 as in Def. 4.1 the Laplace transform L[f, p] has derivatives of all order (even if f has not) which are given by the following formula: (Multiplying by x becomes a derivative) L[xn f (x), p] = (−1)n (F (p))(n). The proof is given in class and usually a straightforward application of integration rules. The ”inverse” of the last property is what makes the Laplace transform useful for solving ODEs: Lemma 4.3. Suppose f is n times differentiable and the derivatives are continuous on [0, ∞). Moreover, suppose that for some p lim f (k) (x)e−px = 0 for all k ≤ n − 1. x→∞ Then for this p the Laplace transform of f (n) exists and n−1 X (n) L[f , p] = − pk f (n−k−1) (0) + pn L[f, p]. k=0 End of lecture March 4 The following Theorem is optional. You will learn more about that type of integral in your complex analysis courses. We will not use it actively. However, the important consequence for us is that it shows that the Laplace transform is injective: Different functions have different Laplace transforms. The Laplace transform defines a function uniquely. Lemma 4.4. (Mellin’s inverse formula) Suppose f is continuous and has growth as in as in 4.1. Z γ+iT 1 f (x) = L−1 [F, x] = lim esx F (s) ds 2πi T →∞ γ−iT Z T 1 = lim e(γ+is)x F ((γ + is)) ds 2π T →∞ −T Here γ and T are real, and i is the imaginary unit. γ is chosen such that the integral exists, the result does not depend on it ( see complex analysis). 34  Note that the r.h.s. of Mellin’s inverse formula is for γ = 0 (i.e. case p0 = 0) Z T Z ∞ Z T Z ∞ 1 isx −yis 1 lim e e f (y)dy ds = lim eis(x−y) f (y)dy ds 2π T →∞ −T 0 2π T →∞ −T 0 ∞ Z ∞ eiT (x−y) − e−iT (x−y) sin(T (x − y)) Z 1 1 = lim f (y)dy = lim f (y)dy π T →∞ 0 2i(x − y) π T →∞ 0 (x − y) Z ∞ sin(T (y − x)) 1 ∞ sin(z) Z 1 1 = lim f (y)dy = lim f (x + zT −1 )dz π T →∞ 0 (y − x) π T →∞ T −T x z/T Z ∞ 1 sin(z) = lim f (x + zT −1 )dz π T →∞ −T x z the integrand converges to f (x) and the domain of integration to (−∞, ∞). The claim follows then from Z ∞ sin(z) dz = π. −∞ z (This integrand is continuous everwhere, its value at z = 0 is 1.) The integrals are improper Riemann integrals. Attention: The problem with this proof is that the convergence f (x + zT −1 ) → f (x) is not uniform on R, only pointwise: For big z the convergence is much slower! For large z one has to use that the integral of the sine over one period is zero, while for big z and T the function 1/zf (x + z/T ) is almost constant on a period of the sine. Here further conditions on f are needed. If f has a jump, the convergence would be similar to the case of the Fourier series. Remark 4.2. The inversion theorems for integral transforms play the role the convergence theorems played for series. Both types of theorems state we we can get back the original function from something we computed by inte- gration, be it the Fourier or Legendre coefficients or the Laplace transfrom. In practice in this class we will not use the inversion formula. Instead, we will use the properties of Lapalce transform to express a given F (p) by scalings, linear combinations, derivatives etc. of functions in our list of examples and read off the inverse Laplace transfrom from the list, as in the following example: 35 Example 4.2. Find the inverse Laplace transform of p F (p) = 2. p + 3p + 2 We use partial fractions, a useful technique for many problems: We write the polynomial in the denominator as product (p − p1 )(p − p2 ), where p1 and p2 are its roots: p2 + 3p + 2 = (p + 1)(p + 2). Now we look for A and B such that p A B = +. p2 + 3p + 2 p+1 p+2 We find p −1 2 = +. p2 + 3p + 2 p+1 p+2 According to our list, 1/(p + 1) is the transform of e−1x = e−x , (a = 1), while 1/(p + 2) is the transform of e−2x , (a = 2), so by linearity we get p L[2e−2x − e−x , p] =. p2 + 3p + 2 Laplace transform and ordinary differential equations Consider an initial value problem for a linear ODE with a right hand side g(t) and constant coefficients of order n, i.e. c0 f (t) + c1 f ′ (t) +... cn f (n) (t) = g(t), f (0) = a0 , f ′ (0) = a1 ,... , f (n−1) (0) = an−1 for real numbers c1 ,... , cn , a0 ,... an−1. (Note that the variable is called t, not x, because t stands for time. This makes of course no difference.) By taking the Laplace transform, using linearity and Lemma 4.3 (assuming the decay conditions on the function are satisfied), we get for L[f, p] = F (p), L[G, p] = G(p), n X j X n G(p) = c0 F (p) + c1 pF (p) +... cn p F (p) − cj pk−1 f (j−k) (0) j=1 k=1 n X n X j X = F (p) cj pj − cj pk−1 aj−k. j=0 j=1 k=1 36 By solving this algebraic equation for F (p) we have found the Laplace trans- form of our solution: G(p) + nj=1 cj jk=1 pk−1 aj−k P P F (p) = Pn j j=0 cj p Now all we have to do is to invert the Laplace transform, and the initial value problem for the ODE is solved. The following example shows these ideas in practice: Example 4.3. Find a solution f to the initial value problem f ′′ + f = cos(t), f (0) = f ′ (0) = 0. Here according to our list G(p) = L[cos, p] = p/(p2 + 1), so, because f (0) = f ′ (0) = 0, we get p G(p) = F (p)(1 + p2 ) − 0 ⇒ F (p) =. (p2 + 1)2 p In order to find the inverse Laplace transform of (p2 +1)2 , note that 1 d 1 p − = 2. 2 dp p2 + 1 (p + 1)2 According to our list, 1/(p2 + 1) is the Laplace transform of sin(t). By property 5. we get d 1 L[t sin(t), p] = − 2 , dp p + 1 so   1 p L t sin(t), p = 2 , 2 (p + 1)2 t and our solution is f (t) = 2 sin(t). (Note that we cannot use partial fractions for the inversion, because the denominator has no real numbers as roots!) 4.2 Fourier Transform The Fourier transform is both in pure and applied mathematics the most important integral transform. It can be seen as the extension of the Fourier series to function f : R → R which are not periodic. Here we will deal with it at a more informal level, more details will be given in year 3. 37 Definition 4.2. We say a function f : R → R is integrable or in L1 (R) if the following integral exists and is finite: Z ∞ |f (x)| dx < ∞. −∞ The proper notion of integration is the Lebesgue integral to be introduced in year 3, we will interpret it here as improper Riemann integral. Definition 4.3. (Fourier Transfrom) For f : R → R (or f : R → C) integrable and piecewise continuous, its Fourier transfrom F[f, p] denoted by fˆ : R → C, is defined as the following improper Riemann integral: Z ∞ fˆ(p) = f (x) e−2πixp dx. −∞ Remarks: ˆ This formula is similar to the formula for the complex Fourier coeff- cients. ˆ Even if the function is real valued, the Fourier transform is complex valued, but we can express it by sin and cos as we did for the Fourier series. This makes however all formulas very awkward, so we will not do it. ˆ The assumption of piecewise continuity ensures that it is Riemann integrable on bounded intervals, it can be dropped when working with Lebesgue integrals. ˆ The existence of the improper Riemann integral in the definition can be shown following the steps we did for the Laplace transform, observing |f (x) e−2πixp | = |f (x)| |e−2πixp | = |f (x)| 1 = |f (x)|. ˆ The kernel of this integral transform is K(p, y) = e−2πixp. ˆ Constant functions or polynomials are NOT integrable on R. (The inte- gral of the modulus over R would be infinity!) Their Fourier transform does not exist in the sense of the previous definition. There is, however, a way to do it nevertheless (not examinable in year 2!): The Fourier transform is a bijection on so-called Schwartz functions: These are in- finitely often differentiable functions on R such that the function and 38  all its derivatives decay at ±∞ faster than any polynomial. An exam- 2 ple is f (x) = e−x. By techniques from functional analysis the Fourier transform can be extended as bijection from the Schwartz functions to a very large class of generalised functions called tempered distri- butions, which contain for example constants, polynomials, integrable and square integrable functions. For example, the generalised Fourier transform of the constant function f (x) = 1 is the so-called Dirac dis- tribution. In this way one can even define the Fourier transform of periodic functions (which are not integrable) and thus see the Fourier series as special case of the Fourier transform. ˆ For square integrable R∞ functions (called functions in L2 (R)), i.e func- tions such that −∞ (f (x))2 dx exists and is finite, the following defini- tion of the Fourier transform works: Z ˆ f (p) = lim f (x)e−2πix·p dx R→∞ |x|≤R (The limit is in the sense of convergence in L2 , (mean square conver- gence) neither pointwise nor uniform, see year 3. ) ˆ Sometimes the Fourier transform is defined as Z ∞ fˆ(p) = f (x) e−ixp dx. −∞ Translation between both versions is simply by replacing p̃ = 2πp. Properties: The following properties can be shown in a way very similar to the corresponding properties for the Laplace transfrom. We always assume that the Fourier transforms of all functions involved exist. ˆ Linearity: Linearity For any complex numbers a and b, if h(x) = af (x) + bg(x), then ĥ(p) = afˆ(p) + bĝ(p). ˆ Translation and multiplication by number on unit circle: For any real number z, if h(x) = f (x − z), then ĥ(p) = e−2πizp fˆ(p). For any real number z, if h(x) = e2πizx f (x), then ĥ(p) = fˆ(p − z). ˆ Scaling: For a non-zero real number a if h(x) = f (ax), then 1 ˆ p  ĥ(p) = f. |a| a 39  ˆ Reocvering the integral: Z ∞ fˆ(0) = f (x) dx. −∞ ˆ Derivatives become multiplications: Suppose f is continuously differ- entiable function, and both f and f ′ are integrable. Then the Fourier transform of the derivative is given by fc′ (p) = 2πi pfˆ(p). More generally, (n) (p) = (2πip)n fˆ(p). fd The last property allows us to transform differential equations into algebraic equations in a similar way as we did with the Laplace transfrom. Lemma 4.5. The inverse of the Fourier transform is given by the following integral Z ∞ f (x) = fˆ(p) e2πixp dp −∞ (Note the difference is in the exponent: No negative sign.) 2 Example 4.4. Let f (x) = e−ax. Then r ˆ π − (πp)2 f (p) = e a. a 40 Convolution Definition 4.4. Let f : R → R and g : R → R. (or f, g : C → C) The convolution of f and g is the function f ∗ g : R → R (or C → C) defined as Z ∞ (f ∗ g)(x) := f (y)g(x − y)dy. −∞ (This can be extended to many other classes of functions, see year 3/4). Example 4.5. We saw that we can write the N -th partial sum of the Fourier series SN (f ) as f ∗ DN , where sin((n + 12 )x) Dn (x) =. sin(x/2) (This is not integrable, but the convolution can be defined nevertheless.) Example 4.6. Solving a PDE by Fourier transform: The heat equation ∂ ∂2 u(t, x) = u(t, x) ∈ (0, T) × R ∂t ∂x2 u(0, x) = f (x) is solved by the convolution with the so-called heat kernel Z ∞ 1 − |x−y|2 u(t, x) = f (y) √ e 4t dy. −∞ 4πt Solving a PDE by Fourier transform: The last property can be proved by Fourier transform: Transforming (in x) both sides of the heat equation gives for any p the ODE ∂ û(t, p) = (2πip)2 û(t, p) = −4π 2 p2 û(t, p), ∂t which has the solution 2 p2 t 2 p2 t û(t, p) = e−4π û(0, p) = e−4π fˆ(p). \x2 2 We recognise from the example above that √ 1 e− 4t = e−4(πp) t. The claim 4πt now follows from the following lemma: 41 Lemma 4.6. (Fourier transform turns convolution into multiplication) If Z ∞ h(x) = (f ∗ g)(x) = f (y)g(x − y) dy, −∞ then ĥ(p) = fˆ(p) · ĝ(p). An important property of convolutions is that a convolution f ∗ g is already differentiable if only one of the two functions, say g, is differentiable. (I.e. at least one of the two must to be differentiable, the other need not be.) This follows from differentiating under the integral: Z ∞ Z ∞ ∂ ∂ ∂ (f ∗ g)(x) = f (y)g(x − y)dy = (f (y)g(x − y))dy ∂x ∂x −∞ −∞ ∂x Z ∞ ∂ ∂ = f (y) (g(x − y))dy = (f ∗ g)(x). −∞ ∂x ∂x This allows us to approximate integrable (L1 ) or square integrable (L2 ) functions by differentiable functions: Let 2 2p fˆn (p) := e−4π n fˆ(p) then Z ∞ 1 |x−y|2 fn (x) = f (y) q e−n 4 dy −∞ 4π n which is differentiable, as the second function in the convolution is. We see immediately that fˆn (p) → fˆ(p) pointwise. One can show that fn converges to f, but not pointwise or uniform. It converges in the L1 or L2 sense, depending on whether f is integrable (L1 ) or square integrable (L2 ): Z ∞ Z ∞ 1 |f (x)−fn (x)|dx → 0 if f ∈ L or |f (x)−fn (x)|2 dx → 0 if f ∈ L2 −∞ −∞ This fact will be very important in further analysis modules. End of lecture March 18 RSquare ∞ integrable functions and Fourier transform If −∞ |f (x)|2 dx < ∞, then the Fourier transform exists not as improper Riemann integral, but as L2 -limit, see above. We have, however, particularly nice formulas. The general proof will be given in year 3. Lemma 4.7. Suppose all transforms and integrals are defined. Then 42  ˆ Parseval: Z ∞ Z ∞ f (x)g(x) dx = fˆ(p)ĝ(p) dp. −∞ −∞ here the bar denotes complex conjugation, it can be omitted for real valued functions. ˆ Plancherel: Z ∞ Z ∞ 2 2 |f (x)| dx = fˆ(p) dp. −∞ −∞ R∞ In other words: The L2 scalar product ⟨f, g⟩ = −∞ f (x)g(x)dx is in- variant P 2 under the R Fourier transform. Note the similarity with Fourier series: (a + b2 ) = π (f (x))2 dx. n n n −π 43 5 Series and ODEs 5.1 Background material on series Definition 5.1. The series generated by the sequence (an )n∈N of real or complex numbers is the sequence (sn )n∈N where n X sn := ai , i=1 is called called the n-th partial sum of the series. If limn→∞ sn exists, we say the series is convergent and call its limit the sum of the series and write X∞ an := lim sn. n→∞ n=1 Pn If the sequence of partial sums i=1 |ai | is convergent, then we call the series absolutely convergent. Lemma 5.1. A series which is absolutely convergent is always convergent, butPthe converse is not always true. Pm If P exists M > 0 such that n=1 |an | < M for |an | is bounded (i.e. there all m ∈ N), then the series an converges absolutely. Lemma P P convergence test) If |an | ≤ bn P 5.2. (Dominated for all nP∈ N and bn converges, then an is absolutely convergent with an ≤ bn. P Lemma 5.3. (Zero test for DIVERGENCE) If an is convergent, then limn→∞ an = 0. This yields the following test: If it is not true that limn→∞ an = 0, then you know that the series is not convergent. If, however, limn→inf ty an = 0, no conclusion can be drawn. Lemma 5.4. (Ratio test) Let (an )n∈N be a sequence of nonnegative real numbers. Consider the sequence defined by an+1 ℓn := , an where terms with an = 0 are omitted. ˆ If limn→∞ ℓn < 1, then the series P n an is convergent. ˆ If limn→∞ ℓn > 1, then the series P n an is not convergent. 44  ˆ In any other case no conclusion can be drawn. By taking the modulus, we can apply this test for any real or complex series. It becomes in this case a test for absolute convergence. Lemma 5.5. (Leibniz’ test or alternating series test) Let (an )n∈N be a decreasing sequence of nonnegative real numbers with lim(an ) = 0. Then n a is convergent. P n (−1) n Attention: In general n (−1)n an need not be absolutely convergent. P Definition 5.2. (Power series) Apower series is a series of the form ∞ X an (z − z0 )n , n=0 where z0 is a real or complex number, (an )n∈N is a real or complex sequence and z is a real or complex number. an is caled the n-th coefficient. Note that all partial sums are polynomials. Typically, a powerPseries will represent a function, where z above is the ∞ variable, i.e. f (z) = n=0 an (z −z0 )n. The name ”power series” comes from the powers of the variable over which we sum. Example 5.1. ∞ X zn ez = n! n=0 ∞ X (−1)n 2n+1 z3 z5 sin z = z =z− + − ··· (2n + 1)! 3! 5! n=0 ∞ X (−1)n z2 z4 cos z = z 2n = 1 − + − ··· (2n)! 2 4! n=0 Definition 5.3. P∞(Radius of Convergence) The radius of convergence R of a n power series n=0 an (z − z0 ) is a nonnegative real number, possibly zero, or infinity, such that the series converges if |z − z0 | < R and diverges if |z − z0 | > R. This is equivalent to ∞ ( ) X n r = sup |z − z0 | an (z − z0 ) converges n=0 On the boundary, where |z − z0 | = R, no conclusion regarding convergence is possible, see next example. 45  1 Example 5.2. For z0 = 0 and an = n we get the power series ∞ X 1 n z. n n=1 For |z| < 1 it converges absolutely and for |z| > 1 it diverges by the ratio test. For z = 1 it diverges (Foundations I, Cauchy test), while for z = −1 it converges, but not absolutely (Leibniz or alternating series test.) So the radius of convergence is R = 1,and for |z| = R no conclusion is possible. Remark 5.1. If a power series ∞ n P n=0 an (z − z0 ) converges for some z, it converges absolutely for any w with |w − z0 | < |z − z0 |. Indeed, convergence implies by the zero test that limn→∞ an (z − z0 )n = 0, so there exists some N ∈ N such that 1 > |an (z − z0 )n | = |an ||z − z0 |n for n ≥ N. (Choose ε = 1 in the definiton of convergence.) Let 0 < δ = |w − z0 ||z − z0 |−1 < 1. Then ∞ ∞ ∞ X X X 1 |an ||w − z0 |n = |an ||z − z0 |n δ n < δn ≤ < ∞, 1−δ n=N n=N n=N P∞ so n=1 |an ||w − z0 |n < ∞. Therefore we may replace ”convergent” by ”ab- solutely convergent” in the defintion of the radius of convergence. Lemma 5.6. (Formula for the radius of convergence) The ratio test implies that |an | R = lim , n→∞ |an+1 | if this limit exists. (Note that the quotient is the other way round compared to the one in the ratio test, because we need R limn→∞ |a|an+1 n| | ≤ 1, so R is the limit of the inverse!) P∞ z n |an | Example 5.3. For n=0 n! , we get |an+1 | = n → ∞, so R = +∞, i.e. |an | convergence for all real or complex numbers. For ∞ 1 n P n=1 n z we get |an+1 | = → 1, so R = 1. For an = n! we get |a|an+1 n| 1 P∞ n 1 − n+1 | = 1/n → 0, so n=1 n!z converges only for z = 0! P∞The z in thenpower series can be seen as a variable, so the power series n=0 an (z − z0 ) is a function with domain a disc (or an interval, if we con- sider only real numbers) around z0 with radius R, its radius of convergence. A function which is in such a way represented by a power series is called analytic. 46 Definition 5.4. A function f : R → R P is called real analytic at x0 ∈ R if there exists δ > 0 and a power seriesP ∞ n n=0 an (x − x0 ) with radius of ∞ n convergence R ≥ δ > 0 such that f (x) = n=0 an (x − x0 ) for |x − x0 | < δ. A function f : C → C is called P complex analytic at z0 ∈ C if there exists δ > 0 and a power series P ∞ a n=0 n (z − z 0 )n with radius of convergence R ≥ δ > 0 such that f (z) = ∞ n n=0 an (z − z0 ) for |z − z0 | < δ. (Complex analytic and complex differentiable is equivalent, but this will not play a role in this class.) Example 5.4. ez , sin(z), cos(z) are complex (and real) analytic everywhere. Instead, P f (z) = z, i.e. f (x + iy) = x − iy is not complex analytic. Suppose z= ∞ n=0 na (z − z 0 ) n , then by plugging in z = x, i.e. a real number, we get a1 = 1 but an = 0 for n ̸= 1. This give however a wrong result for z = iy, so no such power series can exists. For |z −z0 | < R, the functions of partial sums fN (z) := N n P n=0 an (z −z0 ) converge uniformly, so we may differentiate inside the series and obtain the following lemma. We formulate it for real numbers only, as the complex derivative is not part of this class. Lemma 5.7. If ∞ X f (x) = an (x − x0 )n on |x − x0 | < R, n=0 then ∞ X f ′ (x) = nan (x − x0 )n−1 , n=1 and, more generally, ∞ X n! f k (x) = an (x − x0 )n−k , (n − k)! n=k In particular, a real analytic function is infinitely often differentiable and 1 (n) an = f (x0 ). n! Remark 5.2. (Optional) Any real analytic function is infinitely often differ- entiable, but not any infinitely often differentiable function is real analytic. A counterexample is ( − 1 2 f (x) = e 1−|x| , |x| < 1 0 else. 47  By taking limits from the left and right, one can show that all deriva- P∞ in x = 1 exist tives and are equal zero. So all coefficients in the power series n should be zero, which clearly gives a wrong result for x < 1. a n=0 n (x−1) The coefficients of a power series are uniquely determined by the function 1 (n) it represents, because an = n! f (x0 ). This holds in the complex case as well. Lemma 5.8. Suppose for some R > 0 and for all |z − z0 | < R ∞ X ∞ X n an (z − z0 ) = bn (z − z0 )n , n=0 n=0 then an = bn for all n ∈ N. Lemma 5.9. (Sums and products of power series) Suppose that ∞ X ∞ X an (z − z0 )n , bn (z − z0 )n n=0 n=0 have both radius of convergence at least R. Then for |z −z0 | < R and a, b ∈ C ∞ X ∞ X ∞ X n n a an (z − z0 ) + b bn (z − z0 ) = (a an + b bn )(z − z0 )n (linearity) n=0 n=0 n=0 ∞ ∞ ∞ n ! ! ! X X X X an (z − z0 )n bn (z − z0 )n = ak bn−k (z − z0 )n (Cauchy product) n=0 n=0 n=0 k=0 48 5.2 Ordinary Differential Equations We consider here scalar equations only, i.e. the solutions of the ODE will take value sin R, not in Rn. the general case is left to year 3. Let us define a geenral ODE: Definition 5.5. D ⊆ R × Rm+1 and G : D → R be a real valued function. For some interval I ⊆ R a function y : I → R : x 7→ y(x) is called a solution of the m-th order (implicit) ordinary differential equation G(x, y, y ′ ,... , y (m) ) = 0 if ˆ y is m times continuously differentiable on I, ˆ (x, y(x), y ′ (x),... , y (m) (x)) ∈ D for all x ∈ I, ˆ G(x, y(x), y ′ (x),... , y (m) (x)) = 0 for all x ∈ I. This implicit form appears naturally in many applications, but is awk- ward for formulating and proving general theorems. However, an implicit equation can often be turned locally in an explicit one, as in the following example: Example 5.5. Consider n = 1, m = 1 and G(x, y, y ′ ) = y 2 − (y ′ )2. By solving for y ′ we can consider instead the two explicit ODEs y ′ = y and y ′ = −y. This motivates the restriction to thew following more narrow class of equations, the so-called m − th order explicit equations Definition 5.6. Let D ⊆ R × Rm and G : D → R be a real valued function. For some interval I ⊆ R a function y : I → R : x 7→ y(x) is called a solution of the m-th order (explicit) ordinary differential equation, y (m) = G(x, y, y ′ ,... , y (m−1) ) if ˆ y is m times continuously differentiable on I, ˆ (x, y(x), y ′ (x),... , y (m−1) (x)) ∈ D for all x ∈ I, ˆ y (m) (x) = G(x, y(x), y ′ (x),... , y (m−1) (x)) for all x ∈ I. 49  Note that this is a special case of the previous definition for e y, y ′ ,... , y (m) ) = G(x, y, y ′ ,... , y (m−1) ) − y (m). G(x, Definition 5.7. A function y is called a local solution of an ODE near a point x0 ∈ R if there exists δ > 0 s.t. y : (x0 − δ, x0 + δ) → R is a solution. We are mainly interested in local solutions. they can be extended to so-called maximal solutions, see year 3. Remark 5.3. If G(x, z,... , zm+1 ) as in Def. 5.5 has the property that ∂ G(x, z,... , zm+1 ) ̸= 0 at (x, z,... , zm+1 ), ∂zm+1 then the implicit ODE can be re-written as explicit ODE near (x, z,... , zm+1 ) as in the example above. A local solution of the resulting explicit ODE will be a local solution of the original implicit ODE. We will consider some ex- amples where this is not possible and use some power-series based methods. Example 5.1. I = R, m = 1, G(x, y) = 0 Any constant function is a solution. This example shows that in order to have only one solution (uniqueness), we have to complement the ODE by additional conditions, typically prescribing the value of y and the first m − 1 derivatives at the left boundary point of I (initial condition). The following example shows that this is not always sufficient: Example 5.2. y(0) = 0 (5.16) ′ √ y = 2 y (5.17) We can check that for any r ≥ 0 the following are continuously differen- tiable solutions:  0, 0≤x≤r yr (x) = 2 (x − r) , x > r We may also encounter situations where no solution exists: Example 5.3.  ′ 1 if x ≤ 0 y = (5.18) −1 else y(−1) = 0 (5.19) 50  We see immediately that the derivative cannot be continuous, so this equation cannot have a continuously differentiable solution. In order to show existence of exactly one solution, we need in addition to prescribing initial conditions also to restrict the class of equations we will be considering. In order to avoid such pathologies, we will always assume that our coefficients are at least continuous in x. Lemma 5.10. (Existence and uniqueness for the initial value problem) Sup- pose G in the explicit ODE y (m) = G(x, y, y ′ ,... , y (m−1) ) is at least continuous in x and at least continuously differentiable in all other variables. Suppose (x0 , y0,0 ,... y0,m−1 ) is a point in its domain. Then there is exactly one local solution near x0 of the so-called initial value problem y (m) = G(x, y, y ′ ,... , y (m−1) ) y(x0 ) = y0,0 ,... (m−1) y (x0 ) = y0,m−1 (This can be generalised slightly, see year 3.) Definition 5.8. An implicit linear differential equation of order m is a differential equation which can be written as c0 (x)y(x) + c1 (x)y ′ (x) +... + cm−1 y (m−1) (x) + cm (x)y m (x) = r(x), i.e. m X ck (x)y (k) (x) = r(x) k=0 with the so-called coefficients ck (x) and r(x) continuous functions in x. It is called a homogeneous linear differential equation if r(x) = 0 (constant zero function.) Remark 5.4. If y1 (x) and y2 (x) are two solutions of a homogeneous linear differential equation and a and b are real numbers, then ay1 (x) + by2 (x) (i.e. a linear combination of the solutions) is again a solution. Definition 5.9. An explicit linear differential equation of order m is a dif- ferential equation which can be written as m X m ′ (m−1) y (x) = c0 (x)y(x)+c1 (x)y (x)+...+cm−1 y (x)+r(x) = r(x)+ ck (x)y (k) (x) k=0 51 i.e. m−1 X m y (x) = r(x) + ck (x)y (k) (x) k=0 with the so-called coefficients ck (x) and r(x) continuous functions in x. It is called a homogeneous explicit linear differential equation if r(x) = 0 (con- stant zero function.) Remark 5.5. If cm (x0 ) ̸= 0, then an implicit linear equation can be written locally near x0 as an explicit linear equation by dividing by cm (x). The following lemma will be proved in year 3. Lemma 5.11. A homogeneous explicit linear ODE of order m m X m y (x) = ck (x)y (k) (x) k=0 has always m linearly independent solutions y1 (x),... ym (x). Example 5.6. y ′′ (x) = y(x) has the two linearly independent solutions y1 (x) = ex and y2 (x) = e−x. Remark 5.6. By linearly independent solutions we mean: If there are num- bers c1 ,... cm such that for some x0 and δ > 0 m X ck yk (x) = 0 for all x ∈ (x0 − δ, x0 + δ) k=1 then c1 = c2... = cm = 0. (Actually we can choose the interval to be the domain of the coefficient functions.) Example 5.7. aex + be−x = 0 is true for a = −b and x = 0, but it is never simultaneously true for all x in an open interval unless a = b = 0 : Suppose b ̸= 0. Then a 1 = − e2x , b which is false for all except possibly one x ∈ R. 52 5.3 The Frobenius method for second order linear ODEs In this chapter we will see how we can use power series to find a basis of solutions to a linear second order ODE even if the coefficient multiplying the second derivative is zero at some points. Consider a(x)y ′′ + p(x)y ′ + q(x)y = 0 (5.20) and suppose that a, p and q are analytic. Formally, we can divide by a(x), obtaining p(x) q(x) y ′′ + p̃(x)y ′ + q̃(x)y = 0, p̃(x) = , q̃(x) =. (5.21) a(x) a(x) Suppose a(x0 ) = 0. In this case p̃ and q̃ may still be analytic, as for example if x0 = 0, a(x) = x2 , p(x) = 2x2 , q(x) = x4 : We can extend 2x2 x4 p̃(x) = , q̃(x) = x2 x2 analytically to x0 = 0 resulting in p̃(x) = 2, q̃(x) = x2. If this is not possible, then the new coefficients are indeed singular at x0. As you may know from complex analysis, there are however good and bad singularities. A ”good” singularity is one such that the function near x0 looks like a sum of negative powers of x − x0. Suppose a(x) = x3 in our example above, then q̃(x) would still be analytic near 0 but p̃(x) = 2x−1 diverges as x → 0 like a negative power of x. Such singularities are called ”poles:” Definition 5.10. Suppose f : (x0 − δ, x0 + δ) \ {x0 } → R. (f may not be defined at x0 , but x0 is a cluster point of its domain.) We say f has a pole of order n at x0 if and only if (x − x0 )n f (x) can be extended to a function defined in x0 and analytic near x0 , but limx→x0 (x − x0 )n f (x) ̸= 0. Example 5.8. 1 1 f (x) = 2 + + x3 (x − 1) x−1 has a pole of order 2 at x0 = 1 and is analytic near any other point. Note it does not have a pole of order 3 (or higher) at 1: While (x − 1)3 f (x) = x − 1 + (x − 1)2 + (x − 1)3 x3 is anlaytic near 1, limx→1 (x − 1)3 f (x) = 0. (A similar definition holds for functions C → C.) Note that if f has a pole of order n at x0 then near x0 ∞ X f (x) = ak (x − x0 )k + g(x), k=−n 53 where g(x) is analytic near x0. We have seen that by dividing by the leading coefficient a(x), we can reduce (5.20) to (5.21), so from now on, we will deal with the form (5.21) only. Definition 5.11. Consider y ′′ (x) + p(x)y ′ (x) + q(x)y(x) = 0 on some interval I ⊆ R. We call a point x0 ∈ I ˆ an ordinary point when both p and q are analytic x = x0. ˆ a regular singular point if p has a pole up to order 1 at x = x0 and q has a pole of order up to 2 at x = x0. ˆ an irregular singular point in all other cases. Remarks: ˆ The so-called Frobenius method outlined below works for ordinary and regular singular points. ˆ Without loss of generality (w.l.o.g.) we will often assume x0 = 0. We can easily recover the general case by considering ỹ(x) = y(x − x0 ) instead. ˆ If, for example, one of p or q is analytic, and the other one has a pole of at most the order stated in the definition, then the point is a regular singular point. (This is meant by the words ”up to” in the definition.) ˆ If x0 is a regular singular point, then there are analytic functions p̃ and q̃ such that p̃(x) q̃(x) p(x) = , q(x) =. x − x0 (x − x0 )2 Example 5.9. x0 = 0. 1 a)y ′′ (x) + xy ′ (x) + x−2 y(x) = 0, b)y ′′ (x) + y ′ (x) + x−2 y(x) = 0 sin(x) c)y ′′ (x) + cos(x)y ′ (x) + x sin(x)y(x) = 0 d)y ′′ (x) + x−2 y ′ (x) + x−1 y(x) = 0 54 Frobenius’ method consists in solving the ODE p(x) ′ q(x) y ′′ (x) + y (x) + 2 y(x) = 0 x x near x0 = 0 by a power series of the form ∞ X r y(x) = x am xm m=0 where r need not be an integer P∞ and a0 ̸= 0.P(the last condition fixes r, or we m −1 ∞ m −2 P∞ m etc.). could for example write m=0 x = x m=1 x =x m=2 x p and q are assumed to be analytic. ∞ X y ′ (x) = (m + r)am xm+r−1 m=0 ∞ X y ′′ (x) = (m + r − 1)(m + r)am xm+r−2 m=0 Substitute this into our ODE after multiplying through by x2 : ∞ X ∞ X ∞ X 2 m+r−2 m+r−1 x (m + r − 1)(m + r)am x + xp(x) (m + r)am x + q(x) am xm+r m=0 m=0 m=0 X∞ ∞ X ∞ X = (m + r − 1)(m + r)am xm+r + p(x) (m + r)am xm+r + q(x) am xm+r m=0 m=0 m=0 X∞ = [(m + r − 1)(m + r)am xm+r + p(x)(m + r)am xm+r + q(x)am xm+r ] m=0 X∞ = [(m + r − 1)(m + r) + p(x)(m + r) + q(x)] am xm+r m=0 = [r(r − 1) + p(x)r + q(x)] a0 xr X∞ + [(m + r − 1)(m + r) + p(x)(m + r) + q(x)] am xm+r = 0 m=1 The expression r (r − 1) + p (0) r + q (0) = I(r) 55 is known as the indicial polynomial, which is quadratic in r. I(r) = 0 is known as indicial equation. r must solve the indicial equation in order for a solution with a0 ̸= 0 to exists: For each power of the variable x the coefficient must be zero in order for a power series P to be constant zero. ∞ m Recall P∞ that p and q are analytic, let us say p(x) = m=0 pm x , q(x) = m r m=0 qm x. Then the lowest order term in the last line above is I(r)x , which must be zero, i.e. I(r) = 0. As I(r) is a polynomial of degree 2, it has at most two roots, r1 ≥ r2. a0 ̸= 0 is arbitrary, we may choose it as 1 w.l.o.g. This is necessary, because the equation is linear, so with y1 also cy1 is a solution for c ∈ R. The other coefficients can then be determined recursively as k−1 X (j + r)p(k−j) (0) + q (k−j) (0) 1 aj = ak −I(k + r) (k − j)! j=0 (Do not memorise this formula!) We see that there is a problem if the two roots differ only by an integer: If r2 = r1 + k, then the formula leads to a division by zero! For a linear ODE we expect however two linearly independent solutions. This means we have to find the solution for the smaller root in a different way: 1. If I(r) = 0 has two real solutions r1 and r2 and r1 − r2 ̸∈ Z, then we have two linearly independent power series solutions ∞ X ∞ X y1 (x) = xr1 am xm , y2 (x) = xr2 bm xm , m=0 m=0 where the coefficients can be found recursively as above. 2. If I(r) = 0 has two real solutions r1 and r2 and r1 − r2 ∈ Z, (this includes the case of one root with multiplicity 2), then we have two linearly independent solutions, but in general only one is a power series as above: Suppose r1 ≥ r2. ∞ X ∞ X y1 (x) = xr1 am xm , y2 (x) = C ln(x)y1 (x) + xr2 bm xm , m=0 m=0 here C is to be determined and may be zero. The coefficient br1 −r2 is not determined and can be set arbitrarily. 3. No real roots: Not considered in this module. 56 Remark: The roots of the indicial equation are the only exponents r such that 0 < | lim x−r y(x)| < ∞, x→0 for a solution y(x), where we assume that 0 is the regular singular point. y(x) behaves like a0 xr as x → 0. Let us conclude this section by some examples. Examples 1. 4xy ′′ + 2y ′ + y = 0 0 is a regular singular point. We re-write 1 x y ′′ + 2 y′ + y = 0, x 4x2 to see p(0) = 12 , q(0) = 0, so I(r) = r (r − 1) + p (0) r + q (0) = r (r − 1) + 12 r, with roots r1 = 0 and r2 = 21. The difference between these roots in NOT an integer, so we have two (linearly independent) solutions ∞ ∞ 1 X X y1 (x) = x 2 ak m m , y2 (x) = bm xm , m=0 m=0 and by solving the recursion we find ∞ ∞ 1 X (−1)m m X (−1)m m y1 (x) = x 2 x , y2 (x) = x. (2m + 1)! (2m)! m=0 m=0 2. 4x2 y ′′ − 4x2 y ′ + (1 − 2x)y = 0. We get (1−2x) −x ′ ′′ 4 y + y + y = 0, x x2 so p(0) = 0 and q(0) = 41 , i.e. I(r) = r(r − 1) + 14 = (r − 12 )2 , so a repeated zero r = 21. We expect a logaritmic solution and find 1 1 X y1 (x) = x 2 ex , y2 (x) = ln(x)x 2 ex + bm xm. m=0 57 Attention: xy ′′ + 2y ′ + xy = 0 gives I(r) = r(r − 1) + 2r with roots 0 and −1 which differ by an integer. Nevertheless, the solutions do not contain a logarithm. They are ∞ ∞ ! ! sin(x) X cos(x) X y1 (x) = = x0 1 + am xm , y2 (x) = = x−1 1 + bm xm. x x m=1 m=1 Here the coefficient in front of the logarithm happens to be zero. You find this only by plugging the ansatz in the ODE. 58