4 KIYOSHI IGUSA0. Differential and difference equationsWe have two days to go over the basics of linear differential equations.Differential equations is a one semester course and we don’t have timeto cover it in detail. However, to do Markov chains you just need tounderstand how first order linear differential equations work.0.1. Linear differential equations in one variable. In the first lec-ture, I discussed linear differential equations (Diffeq’s) in one variableof arbitrary order. I presented the problem and the complete solutionbut without proof. These missing proofs I appended at the end sothat these notes will faithfully represent the style and content of thelectures.The problem in degree d = 2 is to find a function y = f(t) so that:(0.1) y!!+ ay!+ by + c = 0where a, b, c are constants. (The degree, or order is the number of timesthat the variables are differentiated. In this case the degree is 2 sincewe have y!!.)0.1.1. particular solution. A (one) solution y = f0(t) of this equationis called a particular solution. It is really easy to find:y = f0(t) =−cb.This is a constant function. It’s derivative (and higher derivatives) arezero: y!= y!!= 0. So, when you plug it into Equation (0.1) you get0 + 0 + by + c = 0 ⇒ y = −c/b.If b = 0 then the answer isy = f0(t) =−cat.This is also easy to see: y!= −c/a and y!!= 0. So,y!!+ ay!+ by + c = 0 + a(−c/a) + 0 + c = −c + c = 0.Now, suppose you have another solution y = f(t).0.1.2. homogeneous equation.Lemma 0.1. If f0(t), f(t) are two solutions of the differential equationthen the differencey = f(t) − f0(t)is a solution of the homogeneous equationy!!+ ay!+ by = 0.(This is the original equation minus the constant term c.)MATH 56A SPRING 2008 STOCHASTIC PROCESSES 5This lemma implies that the solutions of the Diffeq are given by:f(t) = f0(t) + (all solutions of the homogeneous equation)The solutions of homogeneous equation have good theoretical prop-erties:Lemma 0.2. If f1(t) is a solution of the homogeneous equation thenso is αf1(t) for any (constant) scalar α.Lemma 0.3. If f1, f2are two solutions of the homogeneous equationthen so is y = f1+ f2.These two lemmas imply that we have a vector space:Theorem 0.4. The set of solutions of the homogeneous equation forma vector space. The dimension of this space is equal to the degree d ofthe differential equation.As you remember from linear algebra, every vector space has a basis:f1(t), f2(t), ··· , fd(t). So, every solution of the homogeneous equationis a linear combination of the basis elements:a1f1+ a2f2+ ··· + adfd.This means that the solutions of the original Diffeq are given by:f(t) = f0(t) + a1f1(t) + a2f2(t) + ··· + adfd(t)where a1, ··· , adare arbitrary scalars.Thus, we need d linearly independent solutions of the homogeneousdifferential equation.0.1.3. solutions for homogeneous equation of degree one. I first did thecase of degree one. This is an equation of the form:y!+ ay = 0or:y!= −ay.I.e., y is decreasing at a rate proportional to its size. This is theequation of exponential decay which you learn in calculus. The solutionisy = y0e−at.6 KIYOSHI IGUSA0.1.4. solutions in higher order. To explain the idea, I gave a specificexample:y!!− 5y!+ 6y + 1 = 0.The particular solution is f0(t) = −1/6 and the homogeneous equationis:y!!− 5y!+ 6y = 0.I rewrote this using the differential operator D =ddt:D2y − 5Dy + 6y = 0or(D2− 5D + 6)y = 0.Now we can factor the operator (the thing that is operating on y):(D − 2) (D − 3)y!"# $=0= 0.To solve this, we look at the part (D − 3)y. If this is zero then thewhole thing is zero. But the equation(D − 3)y = 0is the first order equationy!− 3y = 0.The solution is y = C1e3tfor some constant C1. This gives one basiselement f1(t) = e3t. To get the other one, we go back and rewrite thedifferential equation as:(D − 3)(D − 2)y = 0.Then we get the solution y = C2e2tand f2(t) = e2t. So, the generalsolution of the Diffeq isf(t) = −16+ C1e3t+ C2e2t.Theorem 0.5. If we have a linear differential equation in one variableof order d given by a polynomial in D of degree d with d distinct rootsλ1, λ2, ··· , λdthen the functions eλ1t, eλ2t, ··· , eλdtform a basis for thevector space of all solutions of the associated homogeneous equation.In the example, d = 2, the polynomial is D2− 5D + 6 which hasroots λ1= 3, λ2= 2. The analysis of the general case is very similar.There are two points which I expanded on in order to give a completedescription of the answer:• complex roots,• multiple roots.MATH 56A SPRING 2008 STOCHASTIC PROCESSES 70.1.5. complex roots. What happens in the case when λ is a complexnumber? I gave an example to start:y!!+ 2y!+ 5y = 0.In terms of differential operators this is:(D2+ 2D + 5)y = 0.The roots of this polynomial are:λ±= −1 ± 2i, i =√−1.So, a basis for the solution space is eλ+t, eλ−t. But what are thesefunctions? Here I switched to letters: λ±= a ±bi where a = −1, b = 2.eλ+t= eatebit= eat(cos bt + i sin bt)eλ−t= eate−bit= eat(cos bt − i sin bt)If we add these and divide by 2 or subtract and divide by 2i we gettwo other basic solutions of the homogeneous equation:f1(t) = eatcos btf2(t) = eatsin btIn our particular example, we havef1(t) = e−tcos 2tf2(t) = e−tsin 2t.These two functions form the real basis for the 2-dimensional vectorspace of all solutions of the second order homogeneous diffeq.0.1.6. multiple roots. Suppose that the polynomial has multiple roots.For example, suppose the equation is:y!!+ 4y!+ 4y = 0(D2+ 4D + 4)y = 0.This factors as:(D + 2)2y = 0.The roots are λ = −2, −2. I.e., −2 is a double root.We know that one of the solutions is f1(t) = e−2t. There must be onemore. We cannot take the same function twice. The general answer isf2(t) = teλt= e−2t.If λ is a triple root we get f3= t2eλtand so on.The derivation, which I did not give in class, is easy:D(teλt) = eλt+ tλeλt.8 KIYOSHI IGUSASo,(D − λ)teλt= eλt(D − λ)(D − λ)teλt= (D − λ)eλt= 0.This gives the complete description of the solution of a linear differ-ential equation in one variable.0.1.7. proofs. Students should feel free to skip this subsection and goon to the section on epidemic modeling. This is only for those whowant to see all the details.Going back to our original equationy!!+ ay!+ by + c = 0we rewrite this as(D2+ aD + b)y + c = 0ϕ(y) = −cwhereϕ = D2+ aD + b.The point is that ϕ is a linear operator, i.e.,ϕ(αf1+ βf2) =