Lecture 31. Systems of Differential Equations1. Systems of Equations2. Linear AlgebraDifferential EquationsLECTURE 31Systems of Differential EquationsTo this point, we’ve only discussed individual differential equations. But it’s quite rare that asituation in the real world is modeled using only a single function: quite often, there are severalinterplaying factors at work in the evolution of something.A good example is population dynamics. It would be possible to model the size of a singlepopulation using a single differential equation, making certain assumptions about death and birthrates (namely, that they are constant). But in general, this won’t be the case: the death rateof a prey species is dependant on the size of a predator population and the size of the predatorpopulation will depend on the number of prey. To be able to write down a model for the size ofthe prey population, we need to know the predator population, and vice versa. This would thengive us a system of two interlocked differential equations.We’ll turn our attention to this area next. An example of a system of first order linear equationsisx01= 3x1+ x2x02= 2x1− 4x2.We call a system like this coupled because we need to know what x1is to know what x2is and viceversa.It’s important to note that there will be a lot of similarities between our discussion here and ourearlier discussion of second and higher order linear equations. There’s a very good reason for this:any higher order linear equation can be written as a system of first order differential equations.Let’s see how this is done.Example 31.1. Write the following second order differential equation as a system of first orderlinear differential equations.y00+ 4y0− y = 0 y(0) = 2 y0(0) = −2All that’s required to rewrite this equation as a first order system is a very simple change ofvariables. In fact, this is always the change of variables to use for a problem like this. We setx1(t) = y(t)x2(t) = y0(t).Then we havex01= y0= x2x02= y00= y − 4y0= x1− 4x2.Notice how we used the original differential equation to obtain the second equation. The firstequation, x01= x2, is always something you should expect to see when doing this, just by virtue ofthe change of variables we use.1Differential Equations Lecture 31: Systems of Differential EquationsAll we have left to do is to convert the initial conditions.x1(0) = y(0) = 2x2(0) = y0(0) = −2Thus our original initial value problem has been transformed into the systemx01= x2x1(2) = 0x02= 2x1−23x2x2(2) = −2 Let’s do an example to see how this works for higher order linear equations.Example 31.2. Writey(4)+ ty000− 2y00− 3y0− y = 0as a system of first order differential equations.We want to start by making an analogous change of variables as in Example 31.1. The onlydifference is that, since our equation in this example is fourth order, we will need four new variablesinstead of just two.x1= yx2= y0x3= y00x4= y000Then we havex01= y0= x2x02= y00= x3x03= y000= x4x04= y(4)= y + 3y0+ 2y00− ty000= x1+ 3x2+ 2x3− tx4as our system of equations. To be able to solve these, we need to review some facts about systems of equations and linearalgebra.1. Systems of EquationsIn this section, we will restrict our attention only to the linear algebra that might come upwhen studying systems of differential equations. This is far from a complete treatment, so if you’recurious, taking a linear algebra course would be a good idea.Suppose we start with a system of n equations with n unknowns, x1, x2, . . . , xn.a11x1+ a12x2+ . . . + a1nxn= b1(31.1)a21x1+ a22x2+ . . . + a2nxn= b2...an1x1+ an2x2+ . . . + annxn= bn2Differential Equations Lecture 31: Systems of Differential EquationsHere’s the basic fact about systems of equations with the same number of unknowns as equa-tions, such as (31.1).Theorem 31.1. Given a system of n equations with n unknowns, there are three possibilities forthe number of solutions:(1) no solutions;(2) exactly one solution;(3) infinitely many solutions.We have one more definition to give: a system of equations such as (31.1) is called nonhomo-geneous if at least one bi6= 0. If every bi= 0, the system is called homogeneous. A homogeneoussystem has the following form.a11x1+ a12x2+ . . . + a1nxn= 0 (31.2)a21x1+ a22x2+ . . . + a2nxn= 0...an1x1+ an2x2+ . . . + annxn= 0Notice that there is always at least one solution, given byx1= x2= . . . = xn= 0.This solution is called the trivial solution. This means that it is impossible for a homogeneoussystem to have zero solutions, and Theorem 31.1 can be modified as follows.Theorem 31.2. Given a homogeneous system of n equations with n unknowns, there are twopossibilities for the number of solutions:(1) exactly one solution, the trivial solution;(2) infinitely many non-zero solutions in addition to the trivial solution.2. Linear AlgebraWhile we could, in principle, solve the systems of equations (31.1) and (31.2) directly, we havesome very powerful tools available to us. This is why linear algebra was invented. The main”objects” of study in linear algebra are matrices and vectors.An ntimesn matrix (sometimes referred to as an n-dimensional matrix) is an array of numberswith n rows and n columns. It’s possible to consider matrices with different numbers of rows andcolumns, but this is more general than we will need. An n × n matrix has the formA =a11a12. . . a1na21a22. . . a2n............an1an2. . . ann.There’s one special matrix we will need to be familiar with; this is the n-dimensional identitymatrixIn=1 0 . . . 00 1 . . . 0............0 0 . . . 1.3Differential Equations Lecture 31: Systems of Differential EquationsWe will be focusing on 2 × 2 matrices in this class; the principles of everything we will discussextend to higher dimensional matrices, but the computations are much simpler in 2 dimensions.Matrix addition and subtraction are fairly straightforward: everything is done componentwise.The same goes for multiplying a matrix by a constant, called scalar multiplication: we just multiplyevery component of the matrix by that constant. This will be illustrated in the following example.Example 31.3. Given the matricesA =3 1−2 5and B =−2 01 4,compute A − 2B.The first thing to do is to compute 2B.2B = 2−2 01 4=−4 02 8Then we haveA − 2B =3 1−2 5−−4 02 8=7 1−4 −3. Notice that these operations require the
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