Differential Equations

(1) General Differential Equations

Economists use differential equations largely in the context of dynamical systems, i.e. in systems where time, t, is one of the variables. However, differential equations are defined more generally than this. In this section, we provide general definitions and revert only to including time as the explicit variable in the next section.

Ordinary Differential Equation: an ordinary differential equation of nth order is the following implicit relationship:

(x, y, y , .., y(n)) = 0

where x is a variable, y is an unknown function of x and y , y , .., y(n) are the n derivatives of y.

Intuitively, a differential equation is an equation involving derivatives of an unknown function y. The problem is one of finding this function - thus a solution to a differential equation is a function y = j (x) which satisfies:

(x, j (x), j (x), .., j (n)(x)) = 0

For conditions establishing the existence of a solution, j (x), the Cauchy-Peano theorem, we refer to any text on this matter and shall thus pass it over in silence here.

We shall concern ourselves throughout with first order differential equations (FODE), so that we have:

(x, y, y )

As x is included explicitly, then this is a "non-autonomous" system; if x were excluded, then we have an autonomous system. We can convert from non-autonomous to autonomous systems via a change of variable technique which we shall not pursue here. We shall also focus the bulk of our attention on linear FODE. This is defined as follows:

Linear FODE: a differential equation is a linear first order differential equation if it can be written in the form:

a(x)y + b(x)y = c(x)

where a, b, c are functions of x, where a(x), b(x) are referred to as "coefficients" and c(x) is referred to as the "second member".

Theorem: The general solution of a linear FODE is the sum of the particular solution of the complete equation, a(x)y +b(x)y = c(x) and the solution of the equation without second member, a(x)y +b(x)y = 0.

Proof: let y0 be a particular solution to the complete equation. Let y be the general solution such that y = y0 + z. We must prove that z is a solution to the equation without the second member. As y solves a(x)y + b(x)y = c(x), then a(x)[y0 + z ] + b(x)[y0 + z] = c(x), or:

a(x)yo + a(x)z + b(x)y0 + b(x)z = c(x)

Since y0 is a particular solution to the complete equation, the a(x)y0 + b(x)y0 - c(x) = 0, thus the previous equation reduces to:

a(x)z + b(x)z = 0

thus z is a solution to the equation without the second member.

Bernoulli Equation: A differential equation is called a Bernoulli equation if it can be written in the form:

y + a(x)y = b(x)ym

where a, b are functions of x and m is a constant (m 0, m 1).

Resolution: if y 0, rewrite the Bernoulli equation as:

y /ym + a(x)/ym-1 = b(x)

and let z = 1/ym-1 and z = (1-m)y /ym so y /ym = z /(1-m). Thus, rearraning:

z /(1-m) + a(x)z = b(x)

which is a linear FODE we can solve.

Ricatti Equation: A differential equation is called a Riccati equation if it can be written in the form:

y = a(x)y2 + b(x)y + c(x)

where a, b, c are functions of x.

Resolution: Let y1 be a particular solution of the Riccati equation. Then, setting y = y1 + z, then this becomes:

y1 + z = a(x)(y1+z)2 + b(x)(y1 + z) + c(x)

since y1 is a particular solution then we obtain y1 - a(x)y12 - b(x)y1 - c(x) = 0, so, after some algebra, the previous equation becomes:

z = a(x)z2 + 2[a(x)y1 + b(x)]z

which is a Bernoulli equation for m = 2, which we can solve.

(2) Dynamical Systems of Differential Equations

In our previous section, we defined a differential equation as a general function. Now, we shall consider time explicitly and thus consider differential equations (t, x, x , .., x(n)) where, note, time, t R+, is now the variable and x(t) is a function of time (and x , x , etc. are its first and higher order derivatives). We shall in this section focus our attention exclusively on systems of linear first order differential equations. This translates effectively to a system of n differential equations of the following form:

dx1(t)/dt = a11x1(t) + a12x2(t) + ..... a1nxn(t) + b1(t)

dx2(t)/dt = a21x1(t) + a22x2(t) + ..... a2nxn(t) + b2(t)

............................................................................

dxn(t)/dt = an1x1(t) + an2x2(t) + ..... annxn(t) + bn(t)

or, letting x (t) = [dx1(t)/dt, dx2(t)/dt, ... dxn(t)/dt] , x(t) = [x1(t), x2(t), .... xn(t)] , b(t) = [b1(t), b2(t), ..., bn(t)] , and letting:

 a11 a12 .... a1m A = a21 a22 .... .... .... .... .... an1 an2 .... anm

be a matrix of (constant) coefficients, then the system can be rewritten as:

x (t) = Ax(t) + b(t)

Throughout the following, the term t will be dropped as an argument of x (t) and x(t) if no confusion is risked.

(A) The Homogeneous System

If b(t) = 0, then x (t) = Ax(t) is homogeneous. The solution to a homogenous system can be expressed as follows:

Theorem: Let x = Ax is a homogeneous linear first-order system. If x = velt is a solution to this system (where v = [v1, v2, ..., vn] ], then l is an eigenvalue of A and v is the corresponding eigenvector.

Proof: If x = velt, then x = l velt and thus substituting for x and x , the homogeneous system can be rewritten as l velt = Avelt, which, dividing through by elt, yields us the eigenvalue system l v = Av or (A - l I)v = 0. In other words, for a non-trivial solution, it must be that |A-l I| = 0, which is the characteristic equation of matrix A. Thus, l is an eigenvalue of A and v is its associated eigenvector.

As the matrix A has n eigenvalues, l 1, .., l n and n associated eigenvectors, v1, v2, .., vn, then each term vielit is a solution to the homogeneous system x = Ax. The following theorem establishes that any linear combination of these terms are also solutions to x = Ax:

Theorem: if A is a real n n matrix with n distinct eigenvalues, l 1, .., l n and associated eigenvectors, v1, v2, ..., vn, then z(t) = ・/font> i=1ncivielit is also a solution to the homogeneous system x = Ax where c1, .., cn are arbitrary, possibly complex, constants.

Proof: We wish to prove that as v1el1t, v2el2t, .., vnelnt are all independent solutions to the system x = Ax, then so is their linear combination z(t) = ・/font> i=1n civielit. This is easily noticed as, taking first derivatives of z(t), we obtain z (t) = ・/font> icil ivi elit which as l ivi = Avi, then z (t) = ・/font> iciAvielit = Az(t) by the definition of z(t). Thus, z(t) is a solution to the system x = Ax.

The matrix F (t) = [v1el1t, v2el2t, .., vnelnt] is sometimes referred to as the "fundmental matrix" as vielit are linearly independent of each other (a result of l 1, l 2, .., l n being distinct eigenvalues). This implies that any solution x(t) to the system x = Ax can be expressed as a unique combination of the vectors in the fundamental matrix. (we omit the proof). Consequently, what is commonly referred to as the general solution to the system x = Ax is given as:

x(t) = ・/font> i=1n civielit

where, as noted earlier, c1, .., cn are arbitrary, possibly complex, constants. If the eigenvalues are not dinstinct, things get a bit complicated but nonetheless, as repeated roots are not robust, or "structurally unstable" (i.e. do not survive small changes in the coefficients of A), then these can be generally ignored for practical purposes (cf. Murata, 1977).

Let us now turn to another interesting issue. Recall that a matrix A is "diagonalizable" if there is a matrix, P, such that P-1AP is a diagonal matrix. We now turn to the following:

Theorem: An n-square matrix is diagonalizable if and only if it has n independent eigenvectors.

Proof: Define the modal matrix P = [v1, v2, .., vn], thus P is a (n n) matrix whose n columns are n eigenvectors of A. Thus, as Avi = l ivi for i = 1, .. n, then A[v1, v2, .., vn] = [l 1v1, l 2v2, .., l nvn], or simply AP = PL where L is a diagonal matrix with the eigenvalues l 1, l 2, .., l n of A arrayed along the diagonal, i.e.

 l 1 0 .... 0 L = 0 l 2 .... .... .... .... .... 0 0 .... l n

As AP = PL , then obviously P-1AP = L , thus the matrix P diagonalizes A. For P-1 to exist, the columns of P, i.e. the eigenvectors vi, must be linearly independent. Conversely, if P is non-singular, P-1 exists and P-1AP = L , i.e. P diagonalizes A.

For the next set of theorems, it is worth noting that Taylor's expansion of the function (t) = eat around t = 0 is:

(t) = eat = 1 + at/1! + a2t2/2! + a3t3/3! + ....

As a consequence, the following theorem can be stated:

Theorem: The solution of x (t) = Ax(t), x(0) = x0 is x(t) = eAtx0.

Proof: Taylor's expansion of x(t) around t = 0 yields:

x(t) = x(0) + x (0)t/1! + x (0)t2/2! + x (0)t3/3! + ....

As x (t) = Ax(t), then x (t) = Ax (t) = AAx(t) = A2x(t). Similarly, x (t) = A3x(t) and so on. Thus, at t = 0, we have x (0) = Ax(0) = Ax0, x (0) = A2x(0) = A2x0, x (0) = A3x(0) = A3x0, etc. from the initial condition x(0) = x0. Thus, replacing these in the Taylor's expansion:

x(t) = x0 + Ax0t/1! + A2x0t2/2! + A3x0t3/3! + ....

or, factoring out x0:

x(t) = [I + At/1! + A2t2/2! + A3t3/3! + ....]x0

where I is the identity matrix. But, as established earlier, we know that eAt = [I + At/1! + A2t2/2! + A3t3/3! + ....], so x(t) = eAtx0.

We can now turn to the following:

Theorem: The solution of x (t) = Ax(t), x(0) = x0, A diagonalizable is:

x(t) = eAtx0 = PeLtP-1x0

where P = [v1, v2, ..., vn] is the modal matrix whose columns are eigenvectors of A and L is a diagonal matrix whose diagonal elements are distinct eigenvalues of A.

Proof: Distinct eigenvalues ensure linearly independent eigenvectors and hence non-singularity of P and, by our previous theorem, the diagonalizability of A. Thus, P-1AP = L or A = PL P-1. Thus, A2 = AA = (PLP-1)(PLP-1) = PLIL P-1 = PL 2P-1. Similarly, A3 = PL3P-1 and so on. Now, recall that:

eAt = [I + At/1! + A2t2/2! + A3t3/3! + ....]

so, substituting in for A, A2, etc. and recalling that I = PP-1, then:

eAt = [PP-1 + (PL P-1)t/1! + (PL 2P-1)t2/2! + (PL 3P-1) t3/3! + ....]

or factoring out P to the left and P-1 to the right:

eAt = P[I + Lt/1! + L2t2/2! + L 3t3/3! + ....]P-1

but, as we know by definition, eLt = [I + L t/1! + L 2t2/2! + L 3t3/3! + ....], thus this reduces to:

eAt = PeLtP-1

hence:

x(t) = eAtx0 = PeLtP-1x0

as was to be shown.

Now, recall that the fundamental matrix was defined as F(t) = [v1el1t, v2el2t, .., vnelnt] where each column is an independent solution of the homogeneous system, x (t) = Ax(t). Also, recall that the general solution was:

x(t) = ・/font> icivielit

or, letting c = [c1, .., cn]:

x(t) = F (t)c

It is elementary to note, then, that F (t) = PeLt by the definition of P and L . Thus:

x(t) = PeLtc

But we also know that x(t) = PeLtP-1x0, thus it must be that c = P-1x0.

Thus, in short, a solution to the homogeneous system x = Ax can be obtained by trying a solution x(t) = c1v1el1t + c2v2el2t +....+ cnvnelnt where l 1, l 2, ..., l n are the eigenvalues of A, v1, v2, .., vn are its eigenvectors and c1, c2, .., cn the constants to be determined by the initial conditions.

(B) The Typical System

Let us now turn to a typical, non-homogeneous system of linear first order differential equations. Thus, turning away from the homogeneous case, we are now considering the system:

x (t) = Ax(t) + b

where b 0 and, note, b is not a function of time. Consider now the following:

Theorem: The solution to x = Ax + b with initial condition x(0) = x0 is the following: x(t) = eAtk - A-1b where k = x0 + A-1b or, in alternative form, provided A is diagonalizable, x(t) = PeL tP-1k - A-1b.

Proof: Let y = x + A-1b. Then, as b is independent of time, taking the time derivative, y = x . Thus, substituting, y = Ax + b = Ax + AA-1b = A(x + A-1b) = Ay, i.e. we obtain a homogenous system y = Ay. We know that the solution to a homogeneous system is y(t) = eAty0 = PeL tP-1y0. For the first, note that y = eAty0 implies x(t) + A-1b = eAt[x(0) + A-1b] or simply x(t) = eAt[x(0) + A-1b] - A-1b or, by the definition of k, x(t) = eAtk - A-1b. For the second, y(t) = PeL tP-1y0 implies x(t) + A-1b = PeL tP-1[x(0) + A-1b] or x(t) = PeL tP-1[x(0) + A-1b] - A-1b, or, once again, by definition of k, x(t) = PeL tP-1k - A-1b.

It can be noticed that the latter term x(t) = PeL tP-1k - A-1b can be expressed as:

x(t) = c1v1el 1t + c2v2el 2t +....+ cnvnel nt + xp

or:

x(t) = F (t)c + xp

where l 1, l 2, ..., l n are the eigenvalues of A and v1, v2, .., vn are its associated eigenvectors, so the fundamental matrix F (t) = [v1el 1t, v2el 2t, .., vnel nt] = PeL t; the constants c1, c2, .., cn are determined by the initial conditions, i.e. c = P-1k = P-1[x(0) + A-1b]; and xp is the particular integral (xp = A-1b).

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