1 Systems of Differential Equations

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1 March, Systems of Differential Equations Let U e an open suset of R n, I e an open interval in R and : I R n R n e a function from I R n to R n The equation ẋ = ft, x is called a first order ordinary differential equation in R n We emphasize here that x is an n dimensional vector in R n We also consider the initial value prolem ẋ = ft, x, xt 0 = x 0 2 where t 0 I and x 0 U A solution to the IVP 2 is a differentiale function xt from an open suinterval J I containing t 0 such that ẋt = ft, xt for t J The general solution to is an expression xt, c where c is an n dimensional constant vector in R n such that every solution of can e written in the form for some choice of c If we write out the DE in coordinates, we get a system of first order differential equations as follows ẋ = f t, x,, x n ẋ n = f n t, x,, x n Fact: The n th order scalar DE is equivalent to a simple n dimensional system Consider 4 y n = ft, y, y,, y n 5 Letting x = y, x 2 = y,, x n = y n, we get

2 March, ẋ = x 2 ẋ 2 = x ẋ n = ft, x,, x n If we have a solution yt to 5, and set x = yt, x 2 t = y t,, x n t = y n t, then xt = x t,, x n t is a solution to the system 6 Conversely, if we have a solution xt = x t,, x n t to the system 6, then putting yt = x t gives a solution to 5 The following existence and uniqueness theorem is proved in more advanced courses TheoremExistence-Uniqueness Theorem for systems U e an open set in R n, and let I e an open interval in R Let ft, x e a C function of the variales t, x defined in I U with values in R n Let t 0, x 0 I 0 U Then, there is a unique solution xt to the initial value prolem ẋ = ft, x, xt 0 = x 0 6 If the right side of the system ft, x does not depend on time, then one calls the system autonomous or time-independent Otherwise, one calls the system non-autonomous or time-dependent There is a simple geometric description of autonomous systems in R n In that case, we consider ẋ = fx 7 where f is a C function defined in an open suset U in R n We think of f as a vector field in U and solutions xt of 7 as curves in U which are everywhere tangent to f Linear Systems of Differential Equations: General Properties The system ẋ = Atx + gt 8

3 March, in which At is a continuous n n matrix valued function of t and gt is a continuous n vector valued function of t is called a linear system of differential equations or a linear differential equation in R n As in the case of scalar equations, one gets the general solution to 8 in two steps First, one finds the general solution x h t to the associated homogeneous system ẋ = Atx 9 Then, one finds a particular solution x p t to 8 and gets the general solution to 8 as a sum xt = x h t + x p t Accordingly, we will examine ways of doing oth tasks Let y i t e a collection of R n valued functions for i k We say that they form a linearly independent set of functions if whenever c,, c k are k scalars such that c y t + c 2 y 2 t + + c k y k t = 0 for all t, we have that c = = c k = 0 An n n matrix Φt of linearly independent solutions to the homogeneous linear system 9 is called a fundamental matrix for 9 A necessary and sufficient condition for the matrix of solutions Φt to e a fundamental matrix is that detφt 0 for some or any t If y t,, y n t are n solutions to 9, and Φt is the matrix whose columns are the functions y i t, then the function W t = W y,, y n t = detφt is called the Wronskian of the collection {y t,, y n t} of solutions It is then a fact that W t vanishes at some point t 0 if and only if it vanishes at all point t The general solution to 9 has the form xt = Φtc where Φt is any fundamental matrix for 9 and c is a constant vector

4 March, Thus, we have to find fundamental matrices and particular solutions We will do this explicitly elow for n = 2, and constant matrices A To close this section, we oserve an analogy etween systems x = Ax and the scalar equation x = ax One can define the matrix expa = e A y the power series e A = I + A + A2 2! + A! + It can e shown that the matrix series on the right side of this equation converges for any A The series represents a matrix with many properties analogous to the usual exponential function of a real variale In particular, for a real numer t, the matrix function t e ta is a differentiale matrix valued funtion and its derivative, computed y differentiating the series term y term satisfies I + ta + t2 A 2 2! + t A! d dt eta = Ae ta + It follows that, for each vector x 0, the vector function is the unique solution to the IVP xt = e ta x 0 x = Ax, x0 = x 0 Hence, the matrix e ta is a fundamental matrix for the system x = Ax This oservation is useful in certain circumstances, ut, in general, it is hard to compute e ta directly In practice the methods involving eigenvalues descried in the next secion are easier to use to find the general solution to x = Ax

5 March, Linear Homogeneous Systems with Constant Coefficients Consider the system ẋ = Ax 0 where A is a constant n n matrix and x is an n vector in R n In the one-dimensional scalar case, we found solutions using exponential functions, so it seems reasonale to try to find a solution of the form xt = e rt ξ where r is a real constant and ξ is a non-zero constant vector Plugging in, we get ẋt = re rt ξ = Ae rt ξ for all t Since e rt is never zero, we can cancel it and get or rξ = Aξ ri Aξ = 0 Thus, r is a scalar such that there is a non-zero vector ξ such that ξ is a solution of the system of linear equations Thus, r must e an eigenvalue of A with associated eigenvector ξ see section 6 for the definitions and properties Let us first consider the case in which A has a real eigenvalue with associated eigenvector ξ Then, reversing the aove steps shows that the function xt = e rt ξ is a solution to 0 Now, suppose that the characteristic polynomial zr = detri A has n distinct real roots That is, we suppose that zr = r r r r 2 r r n

6 March, where r i r j for i j Then, letting v i e an eigenvector for r i for each i, we get that each function x i t = e r it v i is a solution, and these are linearly independent functions since their Wronskian W t at t = 0 equals detv, v 2,, v n which is the determinant of the matrix whose columns are the v is The set of vectors {v, v 2,, v n } is linearly independent since these are eigenvectors for distinct eigenvectors Thus, the general solution to 0 has the form xt = c x t + c 2 x 2 t + + c n x n t In the cases of multiple roots or complex roots, one has to do some extra work and we will consider those situations later in the high dimensional case For now, let us specialize to the case of two dimensional systems 2 Two dimensional homogeneous systems of linear differential equations with constant coefficients Consider the system ẋ = a x + y ẏ = c x + d y where a A = c d is a constant 2 2 real matrix We compute the eigenvalues r, r 2 These are the roots of the characteristic polynomial r 2 trar + deta Case : Both roots are real and distinct Say these are r > r 2 Step Compute the eigenvectors v for r and v 2 for r 2, repectively Then, we get solutions of the form

7 March, x = e r t v, x 2 = e r 2t v 2 These turn out to e linearly independent, so the general solution is xt = c e r t v + c 2 e r 2t v 2 where c and c 2 are constants In this case, just as we did for second order scalar equations, we call x t the first fundamental solution to 2 and x 2 t the second fundamental solution to 2 Case 2: Both roots are real and equal Say the common root is r We get one solution x t of the form x t = e r t v where v is an eigenvector for r Next, we have two sucases Sucase 2a: There are two linearly independent eigenvectors, say ξ, η for the eigenvalue r In this case the general solution is An example of this is the system xt = e r t c ξ + c 2 η ẋ = 2x ẏ = 2y with general solution xt = e 2t c 0 + c 2 0 Sucase 2: All eigenvectors for r are multiples of v In this case we proceed as follows Let us try to find another linearly independent solution of the form

8 March, We get x 2 t = e r t v 0 + te r t v or ẋ 2 t = r e rt v 0 + tr e rt v + e rt v = Ae rt v 0 + te rt v, r v 0 + tr v + v = Av 0 + tv Setting the constant and terms with t equal we get that v is an eigenvector, and v 0 satisfies the linear system A r Iv 0 = v 2 Finding the solution v 0, we can in fact otain a second linearly independent solution of the aove form Thus, the general solution has the form xt = c e r t v + c 2 e r t v 0 + tv where v is an eigenvector associated to r and v 0 satisfies 2 Note that this involves solving the two systems of equations Example Consider the system A r Iv = 0, A r Iv 0 = v ẋ = 4x y ẏ = x + 6y The matrix is A = 4 6

9 March, with characteristic equation r 2 0r + 25 = r 5 2 Hence, r = 5 is a root of multiplicity two Since the upper right corner of the matrix is not zero, we can otain an eigenvector v of the form v = r a = 5 4 Thus, we get one non-zero solution as x t = e t = For the second independent solution we have x t = e t v 0 + te t where or, A 5Iv 0 = 4 5 v = v In scalar form, with v 0 =, this is v 2 4 v v 2 = v + v 2 = Note that these two equations are multiples of each other that will always e the case with a root of multiplicity two and 0 So, we can use the first equation Setting v =, we get v 2 = 2 The general solution is

10 March, xt = c e 5t + c 2 e 5t [ 2 + t The solutions to this last equation are all vectors of the form with ξ 2 ξ 2 aritrary, so we can pick v 0 = and get a second linearly independent 0 solution as The general solution is x 2 t = e t 0 + te t 0 xt = c x t + c 2 x 2 t Note that we could also choose v 0 = independent solution as x 2 t = e t + te t 0 ] and get a second linearly This last choice is consistent with the method using first and second fundamental solutions we will define later Remark Note that we used the method aove when there are not two linearly independent eigenvectors for the eigenvalue We did not check whether this is the case, so why does this work? The answer is that if there were indeed two linearly independent eigenvectors for the eigenvalue, then the system would not have had any solutions, so the fact that we could solve the system justifies the approach The proof of this requires more linear algera and will have to e deferred to a more advanced course This method generalizes to n dimensional systems with eigenvalues of multiplicity greater than one although the linear algera required is more complicated Case : The roots are α ± iβ where β > 0 Here we use complex variales We have a complex solution of the form

11 March, x c t = e α+iβt ξ where ξ is the complex eigenvector [ α+iβ a ] associated to the eigenvalue α+iβ Note that here we used the condition 0 It is possile to show that, for 2 2 real matrices with non-real complex conjugate eigenvalues, the off diagonal terms and c are, in fact, non-zero To find two independent solutions to 0, in this case, one need only take the real and imaginary parts of the complex solution x c t Thus, if x t is this real part and x 2 t is this complex part, then the general solution will have the form xt = c x t + c 2 x 2 t If we write express the complex eigenvector ξ as u + iv, then and Let us do an example Consider the system x t = e αt [cosβtu sinβtv] x 2 t = e αt [cosβtv + sinβtu] ẋ = 2 x y ẏ = 2 x + 4 y The matrix A is given y The characterstic polynomial is

12 March, with roots r 2 6r + 4 r = 6 ± = ± i 5 We seek a complex eigenvector ξ = ξ, ξ 2 for the eigenvalue r = +i 5 We get the equation ri Aξ = 0 or r 2 2 r 4 ξ ξ 2 = 0 0 Because the matrix is singular, we need only consider the first row equation Setting ξ =, we get r 2ξ + ξ 2 = 0 ξ 2 = 2 r = 2 + i 5 = i 5 Thus, we get the first fundamental solution of the form x = e +i 5t i 5 = e +i 5t 0 + i 5

13 March, and the second fundamental solution of the form x 2 = e i 5t +i 5 = e +i 5t 0 i 5 The real and imaginary parts of x t are R def = e t cos 5t sin 0 5t 5 I def = e t sin 5t The general real solution is + cos 0 5t 5 xt = c R + c 2 I The first and second fundamental solutions and the complex solution Consider the system a ẋ = c d x with characteristic polynomial zr = r 2 a + dr + ad c For convenience in finding the general solution to this system, we define the first and second fundamental solutions The general solution can e expressed as a linear comination of these two solutions The definitions depend on the nature of the roots We assume that 0 Corresponding solutions can e defined if = 0 and c 0, ut, for the sake of simplicity, we avoid that here If fact, one could simply change coordinates, interchanging x and y and reduce to the present case

14 March, Case : real distinct roots r > r 2, 0 First Fundamental Solution: x t = e r t r a Second Fundamental Solution: x 2 t = e r 2t r 2 a Case 2: real equal roots r = r 2, 0 First Fundamental Solution: x t = e r t r a Second Fundamental Solution: x 2 t = te r t r a +e r t r a+

15 March, Case : complex roots r = α + βi with β > 0 Complex Solution: x c t = e tα+βi α+iβ a = e tα+βi [ α a + i 0 β ] First Fundamental Solution: x t = e [cosβt αt α a sinβt 0 β ] Second Fundamental Solution: x 2 t = e [sinβt αt α a + cosβt 0 β ]

16 March, An alternate method for 2 dimensional systems: Elimination and reduction to scalar equations First, we give some examples to descrie the elimination method to solve two dimensional systems Example : Consider the system We can write ẋ = x + y ẏ = x y y = ẋ x 5 from the first equation and sustitute into the second equation getting ẏ = ẍ ẋ = x ẋ x This gives the second order scalar equation for x ẍ 2x = 0 We know how to solve this The characteristic equation is r 2 2 with roots r = ± 2 and general solution Then, we get y from 5 as xt = c e 2t + c 2 e 2t yt = ẋ x = c 2e 2t c 2 2e 2t c e 2t c 2 e 2t,

17 March, so, the general solution to the system is xt yt c = e 2t + c 2 e 2t c 2e 2t c 2 2e 2t c e 2t c 2 e 2t = c e 2t + c 2 2 e 2t 2 This method can most often e used for two dimensional homogeneous systems Which method is est? In my opinion, this method is est when the eigenvalue is real of multiplicity two and the matrix method is est in the other cases Example Let us apply the method of elimination to the system we considered aove We get ẋ = x ẏ = x + y x = ẏ y ẋ = ÿ ẏ = x = ẏ y or ÿ 2ẏ + y = 0 The general solution is yt = c e t + c 2 te t This gives xt = ẏ y = c e t + c 2 e t + c 2 te t c e t + c 2 te t = c 2 e t

18 March, and the general solution xt yt = c 2 e t c e t + c 2 te t 0 = c e t + c 2 e t t Finally, we descrie some general aspects of the method of elimination We consider the system x = ax + y y = cx + dy We assume, that either or c is not 0 Otherwise, the system is diagonal and easily solvale Assuming 0, we use the elimination method to find a second order equation for xt We will see that the characteristic polynomial for this second order equation is the same as the characteristic a polynomial of c d The latter characteristic polynomial is Now, r 2 a + dr + ad c or x = ax + y = ax + cx + dy = ax + dx c adx x a + dx + ad c = 0 Now, we can find the general solution xt of this second order equation, and then get yt in the original system from

19 March, y = x ax If, = 0 ut c 0, we use the elimination method to find a second order equation for yt We get the general solution for yt, and then get xt from x = y dy c