Our goal is to solve a general constant coecient linear second order. this way but that will not always happen). Once we have y 1, it will always

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1 October 5 Relevant reading: Section 2.1, 2.2, 2.3 and 2.4 Our goal is to solve a general constant coecient linear second order ODE a d2 y dt + bdy + cy = g (t) 2 dt where a, b, c are constants and a 0. We will nd that, taking the initial guess y 1 (t) = e rt, we will always be able to nd a solution y 1 of the corresponding homogeneous equation (sometimes you may nd two solutions this way but that will not always happen). Once we have y 1, it will always be possible to nd another solution y 2 for the homogeneous equation using reduction of order, and hence a particular solution y p of the inhomogeneous equation using variation of constants. However our guess e rt will only work in general after we have discussed Euler's identity, and we will need to study a little bit of complex analysis to fully appreciate this identity.. Complex numbers Remember that the imaginary unit i = 1 is a hypothetical solution of the equation x = 0. Obviously this equation has no real solutions, but it turns out that one way to solve a polynomial that we cannot solve is to simply adjoin a formal symbol, representing a theoretical solution to that 1

2 equation, into our number system. If our new, bigger number system is to follow the usual rules of arithmetic, then numbers including an imaginary part will have to add, subtract, multiply and divide in a very specic way: e.g. if a 1, b 1, a 2, b 2 are real we must have (a 1 + ib 1 ) + (a 2 + ib 2 ) = (a 1 + a 2 ) + i (b 1 + b 2 ) (a 1 + ib 1 ) (a 2 + ib 2 ) = (a 1 + a 2 ) i (b 1 + b 2 ) (a 1 + ib 1 ) (a 2 + ib 2 ) = (a 1 a 2 b 1 b 2 ) + i (a 1 b 2 + b 1 a 2 ) a 1 + ib 1 = a 1a 2 b 1 b 2 + i a 2b 1 a 1 b 2 a 2 + ib 2 a b 2 2 a b 2 2 (a 2, b 2 ) (0, 0) For the rst three we just expand out and apply the rule i 2 = 1; for the last one we rationalize by multiplying the numerator and denominator by a 2 ib 2 and proceeding as in the third line. In particular there is only one way to develop arithmetic consistently, with an adjoined imaginary unit, if the usual rules will apply. On the other hand, applying these rules exactly as listed will never (as far as I am aware) lead to inconsistencies. remark. If you are still uncomfortable with complex numbers, you can actually realize them as real matrices by replacing the number a + ib 2

3 by the matrix a b b a (a, b R) Two matrices of this form will always commute under the usual matrix multiplication rule, so we are motivated to view these new matrices as a type of number. If we work out the rules of arithmetic for these matrixnumbers (identifying multiplication with matrix multiplication and A with B B 1 A when det B 0) then we will obtain the same rules which prevailed when we just treated i = 1 as a regular number. You can check all these claims yourself as an exercise; in particular, if the usual systems of real numbers and algebra of real matrices are self-consistent, then a larger system including i = 1 must be self-consistent as well. remark. Apart from dening terms today, we will not use the theories of complex analytic or holomorphic functions in this class; it will be enough to recognize that complex numbers extend the real numbers and obey specic rules listed above. However note that more advanced studies of dierential equations are greatly facilitated by understanding complex analytic functions (even when you are only interested in real-valued solutions!) so I highly recommended taking a class on Complex Analysis if you are able to do so. Already in this class there are many concepts that we can only appreciate in a heuristic way because the true reasons for certain results cannot even be written easily without complex analysis! (for example: the function 1 1+x 2 is 3

4 globally smooth and real-valued for all x R but its power series expansion around x = 0 has radius of convergence equal to 1, even though the power series for e x converges at every x R; it is very hard to explain these result about a real valued function without complex analysis.) Note that the modulus of a complex number z = a + ib C with a, b R is dened using the Pythagorean theorem z 2 = a 2 + b 2 Also the complex conjugate z is dened z = a ib Also we have the following identity: zz = z 2 Also it is a fact that z 1 z 2 = z 1 z 2. Not all smooth functions have convergent power series 4

5 A real-valued function f in one real variable x is said to be real analytic if, near any point x 0, it can be expanded as a convergent power series, f (x) = k=0 f (k) (x 0 ) k! (x x 0 ) k which converges to f for all x close enough to x 0. A complex-valued function in one complex variable is similarly called analytic if it can be expanded as a convergent power series near any point. The reason we introduce a new word for such functions is that not all smooth functions are analytic. This is a fascinating topic by itself but we will be satised with a simple example to illustrate the point. Consider the following function: e 1/(1 x2 ) x < 1 f (x) = 0 x 1 You can show by direct computation that lim f (k) (x) = lim f (k) (x) = 0 x 1 x 1 This is because dierentiation will pull down inverse powers of 1 ± x, but those inverse powers cannot compete with the exponential! It is actually possible to show that f has continuous derivatives of all orders everywhere, 5

6 and in fact f (k) (±1) = 0 k N However f is not analytic at ±1 even though it is smooth everywhere. Indeed the Taylor expansion of f near x 0 = 1 is identically zero, yet f (1 ε) > 0 for any 0 < ε < 1. This is not some isolated example either; it is possible to construct a function which is smooth everywhere and analytic nowhere. There are so many functions like this that it would be reasonable to think that a typical smooth function is nowhere analytic. Usually if we have some smooth real-valued function of one real variable, we won't know whether it is analytic unless we analyze very carefully the behavior of f (k) (x) jointly in k and x; in particular, we need information about very high derivatives of f. Complex analysis is studied by mathematicians largely due to its beauty and its utility in areas such as number theory. From a practitioner's point of view, one great reason to study complex analysis is that it gives us a shortcut for showing that a function is analytic. denition. A complex valued function f (z)of a complex variable z = x + iy is said to be holomorphic if the following limit exists at every point z 0 : f (z 0 ) = lim z z0 f (z) f (z 0 ) z z 0 Note that the derivative f (z) of a function of a complex variable must take the same value no matter how we take the limit. 6

7 example. Let f (z) = z 2, then f (z 0 ) = lim z z0 z 2 z 2 0 z z 0 = lim z z0 (z + z 0 ) (z z 0 ) z z 0 = lim z z0 (z + z 0 ) = 2z 0 hence f (z) = 2z. theorem. If f is holomorphic then it is analytic. Observe the power of the theorem: simply by checking the existence of a single (complex) derivative, we automatically nd that there are innitely many derivatives and the innite Taylor series of f near any point converges to f. Therefore one way to show that a real valued function of a real variable is real analytic is to extend it to some complex function and show that the extension is holomorphic; then, the extension will be complex analytic, hence its restriction to the real axis is real analytic. Conversely, a smooth function which is not analytic at a point x 0 cannot be extended to a function which is holomorphic on an disk centered at x 0. The above theorem is sucient reason to seek holomorphic functions. Our goal for the remainder of the lecture is to nd a holomorphic extension of the exponential function. Complex exponential function Remember that the function e x is dened by the following convergent power series: f (x) = k=0 x k k! 7

8 Let us dene a function g (z) by the following formula: g (z) = k=0 z k k! Note that the monomials z k are holomorphic, just as we saw that z 2 is holomorphic. It turns out that the innite sum for g converges for all z C, and we can check this using the ratio test (which is still valid for complex series). Also g (x) = f (x) for all x R. Before we fully accept g as the new exponential function, we should check that it satises the following key property of exponentials: e x+y = e x e y Namely we need to show that if z, w C then g (z + w) = g (z) g (w) Indeed we have g (z + w) = Expand using the binomial theorem: g (z + w) = (z + w) n n=0 n n=0 k=0 1 n! n! ( ) n z k w n k k The double sum is absolutely convergent (consider a corresponding series for 8

9 e z + w ) and using this fact we can conclude that we can re-arrange the terms however we like. In particular let us introduce a new variable, j = n k, and re-write the sum this way: g (z + w) = j=0 k=0 1 (k + j)! ( k + j k ) z k w j = j=0 k=0 1 k!j! zk w j therefore g (z + w) = ( j=0 ) ( w j ) z k = g (w) g (z) j! k! k=0 which was what we wanted. We can also check by termwize dierentiation g (z) = g (z) so g has another key property of the exponential function, namely that it is its own derivative. At this point we realize that the function g, a holomorphic function, has so many properties in common with the usual exponential function that it is only reasonable to identify the two. Hence we write: e z = k=0 z k k! z C 9

10 We want a way to evaluate e z for given z = x + iy. We have e z = e x e iy so it suces to evaluate e iy for real y. Using the power series representation and gathering real and imaginary terms together, we obtain e iy = ) ) (1 y2 2! + y4 4! y6 6! i (y y3 3! + y5 5! y7 7! +... which we recognize as the power series for cosine and sine functions. Hence e iy = cos y + i sin y and therefore e z = e x (cos y + i sin y) You will also see Euler's identity written e iθ = cos θ + i sin θ which shows that e iθ parametrizes the unit circle when 0 θ 2π. In particular it shows that any complex number can be represented in polar form as z = re iθ 10

11 where r 0 and θ R. Another Model Problem For Quiz 3 you need to know problems 1-21 already covered and the following additional problem: 22. The equation d 2 y dt 2 + tdy dt + y = 0 has solutions y 1 = e t2 /2 and y 2 = e t2 /2 t 0 e s2 /2 ds which have non-vanishing Wronskian (you do not have to prove any of that for this problem). Find a solution to the following initial value problem: d 2 y + t dy dt + y = 0 2 dt y (0) = 1 y (0) = 1 Solution. Any solution of this homogeneous ODE can be written as a 11

12 linear combination: y (t) = c 1 y 1 (t) + c 2 y 2 (t) Plugging in the initial conditions gives 1 = c 1 y 1 (0) + c 2 y 2 (0) 1 = c 1 y 1 (0) + c 2 y 2 (0) Since y 1 (0) = 1 and y 2 (0) = 0 we get 1 = c 1 On the other hand since y 1 (0) = 0 and y 2 (0) = 1 we get 1 = c 2 Therefore y (t) = e t2 /2 e t2 /2 t 0 e s2 /2 ds 12

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