B. Differential Equations A differential equation is an equation of the form

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1 B Differential Equations A differential equation is an equation of the form ( n) F[ t; x(, xʹ (, x ʹ ʹ (, x ( ; α] = 0 dx d x ( n) d x where x ʹ ( =, x ʹ ʹ ( =,, x ( = n A differential equation describes the behaviour of a state variable over continuous time which explains the use of derivatives (the time derivative measure the marginal changes at each instant of time) In general the above equation can be written explicitly in a more commonly used form ( n) ( n ) x ( = f ( t; x(, xʹ (, x ʹ ʹ (,, x ( ; α) The order of a differential equation is the order of the highest derivative that appears in the equation A differential equation is said to be linear if f is linear function of the state variable and its derivatives A differential equation is said to be autonomous if time does not enter directly in the function f as an argument We shall be dealing only with differential equations of first order It can be shown that any differential equation of higher order can be transformed into a system of first order differential equations by introducing new variables Example: Consider the second-order linear differential equation x ʹ ʹ ( = axʹ ( + bx( (0) Define a new variable y( = xʹ ( Then yʹ ( = x ʹ ʹ ( Putting these values in (0), we get the system of first order differential equations involving two variables x and y: xʹ ( = y( (0ʹ ) yʹ ( = ay( + bx( The theory of differential equations is in many ways similar to theory of difference equations Therefore in some cases we shall just state the results, leaving the derivations as exercises B Solving a Linear First Order Differential Equation (a) AUTONOMOUS EQUATIONS: Consider a linear autonomous equation of the following form: dx( = ax( + b () As in the case of difference equation, the generation solution to () is given by a sum of two terms x and x, where x solves the homogeneous part of () (complementary function), and c p p c n x is any particular solution to () (called a particular integral)first let us solve for the homogeneous component: dx = ax (ʹ ) Equation (ʹ ) can be written as 3

2 dx = a x Integrating both sides and simplifying, we can write the solution to equation (ʹ ) as at x( = c exp where c is an arbitrary constant at Thus the complementary function is given by xc = cexp For the particular integral we again proceed with a trial solution that x ( = k (a dx constan Then = 0 Putting these values in () we get ak + b = 0 b k = a b Hence the particular integral is given by x p = a Thus the general solution to the autonomous differential equation () is at ( b) x( = c exp + a where c is an arbitrary constant whose value is to be determined from the given initial or any other boundary condition (b) NON-AUTONOMOUS EQUATIONS: Let us consider a non-autonomous first order differential equation of the form: dx( = ax( + b( () at The complementary function is once again given by x ( = c exp So we have to find only a particular solution Suppose b ( takes a particular functional form given by b t b( = B exp Then the differential equation given in () takes the form: dx ( b t = ax( + B exp (ʹ ) Let us now proceed with a trial solution, b t dx x( = k exp Then b t = kb exp B Substituting these values to the above equation and simplifying, k = Thus we get b a B a particular solution to (ʹ ) as b t xp( = exp b a Hence the general solution to (ʹ ) is at B b t x( = cexp + exp, b a ; c an arbitrary constant b a c When b = a, the differential equation becomes 4

3 dx ( a t = a x( + B exp (ʺ ) So the complementary function now is a t xc ( = cexp In order to find a particular solution, let us proceed with a trial solution a t x( = kt exp dx Then a t a t = k exp + a kt exp Putting these values back into the above equation and a t simplifying, k = B Thus the particular integral is given by xp( = Bt exp Therefore the general solution to (ʺ ) is a t x( = ( c + B exp, c an arbitrary constant However it may not always be possible to find a suitable trial solution for a nonautonomous difference equation even when the functional form of b ( is known There is a more general method of solving any non-autonomous first order differential equation which requires knowledge of two concepts: (i) exact differential equation, (ii) integrating factor Let us consider a first order non-autonomous differential equation of the form M ( dx + N( = 0 (3) dx N( ie, = M ( This differential equation is said to be an exact differential equation if there exists a F F function F ( such that = M ( and = N( The function F ( is called the primitive function F F Remark Since we know that for any function F (, =, an easy criterion for checking whether a given differential equation of the general form M N M ( dx + N( = 0 is exact or not is to verify if = If this holds then the given differential equation is exact, otherwise not It is easy to see that if a differential equation is exact and if we can identify the primitive function F (, then such an exact differential equation can be readily solved in the following way: M ( dx + N( = 0 F F dx + = 0 df ( = 0 Integrating, we get the general solution to this equation as F ( = c where c is an arbitrary constant (constant of integration) 5

4 Thus if we are given an exact difference equation to solve, all that we have to do is to find the corresponding primitive function F ( F How to find the primitive function: we know that = M ( Hence the function F ( must contain a partial integral of M ( Let us start with a trial primitive function of the form F( = M( + φ ( (4) [Note that in (4), M ( is integrated partially with respect to ie, in integrating the function we treat t as given] If (4) is indeed the primitive function corresponding to (3), then its partial derivative with respect to t should be equal to N ( Using this relation we can identify the φ ( function and consequently identify the F ( function Once F ( is found we can equate it with an arbitrary constant to get the general solution to (3) Example: Consider a differential equation of the form: (3x + 4x dx + (x + = 0 (3ʹ ) Here M ( = (3x + 4x and N ( = (x + The first step is to check whether (3ʹ ) is an exact differential equation or not M N M N From the equation we find, = 4 x and = 4 x Thus = and the differential equation is exact So in order to solve (3ʹ ) we have to find the corresponding primitive function F ( Let us proceed with the trial function: F( = M( + φ ( ie, F( = (3x + 4x + φ ( 3 F( = ( x + x + φ( From above, F dφ = x + If the above trail function is indeed the primitive function corresponding to (3ʹ ), then d x + φ = x + t dφ ie, = t Integrating, φ ( t ) = t + c, where c is an arbitrary constant Thus we have derived the primitive function as 3 F ( = x + x t + t + c The general solution to (3ʹ ) is then given by, F ( = c, or 3 x + x t + t + c = c Combining the two arbitrary constants c and c we may write this solution as 3 x + x t + t = c 6

5 Remark In order to find the primitive function of the exact differential equation given in F (3), we can proceed in an alternative way Since = N(, the function F ( must contain a partial integral of N( as well So we can write the trial solution as F( = N( + ψ ( x) (4ʹ ) Note once again that N ( is to integrated partially with respect to t, ie, in integrating the function we have to treat x as given If this is indeed the primitive function corresponding to (3), then its partial derivative with respect to x should be equal to M ( Using this relation we can identify the ψ (x) function and consequently identify the F ( function Both this procedures should give us the same primitive function (Verify that for the above example, the alternative procedure described here will generate the same functional form of the primitive function) If the given differential equation is not exact, then it is possible to transform it into an exact differential equation by multiplying both sides with a common factor This common factor that transforms a non-exact differential equation to an exact one is called an integrating factor We shall only discuss the procedure for solving a linear non-autonomous first order differential equation which is not exact Consider a first order linear differential equation of the form dx = ax + b( (5) Writing this in the M ( dx + N( = 0 format, we get dx + { ax b( } = 0 (5ʹ ) Thus M ( = and N ( = { ax + b( } Obviously, this differential equation is not exact as long as a 0 Suppose I is the integrating factor (which is yet unknown) Then multiplying both sides of (5ʹ ) by I should convert it to an exact differential equation Hence the following is an exact differential equation: I dx + I{ ax b( } = 0 (5ʺ ) The new M and N function corresponding to (5ʺ ) are given by M ( = I and N ( = I{ ax + b( } Since (5ʺ ) is an exact differential equation (by construction), M N di = That is, = ai Integrating, we can identify the integrating factor as, log I = + K ( K being the constant of integration) K or, I = Aexp ( A = exp ) Any function of the above form will act as the integrating factor for (5ʹ ), whatever be the value of A Hence for simplicity we set A = Thus using the integrating factor I = exp, we can write (5) as an exact differential equation in the following way: exp dx exp { ax + b( } = 0 (5ʹ ʺ ) To find the primitive function corresponding to (5ʹ ʺ ), we proceed with the trial function 7

6 F( = M + φ ( = exp + φ( = xexp + φ( d Hence F ax φ at = exp + Again from (5ʹ ʺ ) N( = exp { ax + b( } Equating the two, dφ = exp b( φ = b t exp ( ) Thus the primitive function is given by at F( = xexp exp b( Hence the general solution to a linear non-autonomous difference equation of the form given in (5) is xexp exp b( = c Rearranging, we can write the general solution as at x( = exp c + exp b( (6) ( ) where c is an arbitrary constant whose value is to be determined from the given boundary condition B Autonomous First Order Differential Equation: Steady States and Stability Consider an autonomous first order differential equation of the form x ʹ ( = f ( x( ; α) (7) The steady state and stability of this autonomous differential equation is defines as follows Steady state: A point x X is a steady state of the dynamic system represented by the differential equation (7) if it satisfies the following condition: f ( x; α) = 0 In other words, the steady state can be found by setting x ʹ ( = 0 Stability: A dynamical system with a steady state x X is said to be asymptotically stable if lim x( = x t Example: Let us consider a linear difference equation of the form dx( = ax( + b at ( b) We have seen that this function has a general solution x( = c exp + a dx b Setting x ʹ ( = 0, ie, = 0, we can derive the steady state as x =, which is a nothing but the particular integral as derived earlier Hence the general solution to the above equation can be written as x( = cexp at + x Thus no matter what the value of c is (ie, no matter what the initial condition is), the system will approach its steady state as t tends to infinity if and only if a < 0 Thus the stability condition here is given by a < 0 8

7 Remark By comparing this example with corresponding difference equation, one can readily see the similarity between the two cases However, one should note that in the case of difference equation stability depends on whether the parameter a is greater than unity or less than unity On the other hand, in the case of differential equation stability depends on whether the parameter a is positive or negative B3 Solving a System of First Order Differential Equations (Linear and Autonomous) (a) n dimensional system: Consider a system of linear and autonomous differential equations of the form x ʹ ( t ) = Ax( + b (8) where A is an n n matrix of constant coefficients As in the case of system of difference equations, we first identify the particular solution (integral) as p x ( = A b = x (the steady state vector) Next, to find the complementary function, let us consider the homogenous system x ʹ ( = Ax( (8ʹ ) If the n dimensional square matrix A has n distinct roots then we can use the corresponding matrix of the eigenvectors M to diagonalize A and the transform the homogeneous system given by (8ʹ ) to a simpler system of the form: yʹ ( = Λy( (8ʺ ) where Λ is a diagonal matrix with all the distinct eigenvalues of A as its diagonal elements Equation (8ʺ ) represents a system of n independent linear and autonomous differential equations of the following kind: dy = λ y dy = λ y dyn = λ n yn λt λt The solution to this system is given by y( = c exp, y( = c exp,, λ y nt n ( = c n exp, where c, c,, c n are all arbitrary constants We can then find the general solution to the original system given by (8) as x( t ) = My( A b The system will be asymptotically stable if all the eigenvalues are negative; will be unstable if all the eigenvalues are positive; and will be saddle point stable if some of the eigenvalues are positive and others are negative If A matrix is not digonizable, then we can transform it to its Jordan canonical form by using a similarity transformation and solve the resulting simplified system to arrive at the general solution to (8) We shall however discuss this procedure only for the dimensional case 9

8 (b) dimensional system: Consider a two dimensional system of the form dx = ax + ax + b dx = a + + x ax b The co-efficient matrix is given by a a A = a a p x x The particular integral can be found as x () t =, where is obtained by solving x x a x + a x + b = 0 This particular solution is also the steady state + + = 0 the equations: ax ax b solution of the dynamic system In order to find the complementary function, we derive the characteristic roots and the corresponding characteristic vectors of A Case (i): the roots are real and distinct e Let the two roots be λ and λ with the associated characteristic vectors e = e e and e = respectively Then the transformation matrix that diagonalizes A is given e by M = ( e, e), which has the two eigenvectors e and e as its columns As in the difference equation case, the general solution to the system is then given by: xi ( = ec exp + ec exp + x (0) x = + + ( ec exp ec exp x where x and x are the two steady state values The arbitrary constraints c and c will be determined by the given boundary conditions The system will be stable if both λ and λ are negative; will be unstable if λ and λ both positive; and will be saddle point stable if one of the eigenvalues is negative and the other is not As in the case of difference equation, we can find the equation of the saddle path in the following way Let λ 0 and λ 0 Choosing the initial values such that c (the > < (9) 0

9 coefficient associated with the unstable root λ ) vanishes, we can derive corresponding stable time paths of x and x as: x ( = ec exp + x x( = ec exp + x t Eliminating c expλ, we can write the equation of the saddle path as: e x ( = ( xi ( x ) + x e Case (ii): the roots are real and repeated Let the repeated root be denoted by λ In this case the matrix A is not diagonaizable However, as in the case of difference equations, by a similarity transformation, we can λ still convert A to its canonical form C = 0 λ e If the eigenvector associated with λ be e =, then the invertible transformation e e v v e matrix that converts A to C is given by M =, where ( A λ I) = e v v e As in the difference equation case, we can then define a new set of variables y ( t ) M = x ( t ) and transform the homogeneous system x ʹ ( = Ax( to a simplified system given by: y ʹ ( = Cy( yʹ λ y ie, = yʹ 0 λ y The solution for the second equation is given by λt y ( = c exp Putting this solution back in the first equation, we get a non-autonomous differential equation of the form λt y ʹ = λy + c exp, which has a solution y( = ( c + c exp Given the solutions to y, once again the general solutions for as x t = M yt + x x t and x t can be derived Stability of the system depends on the value of λ The system is stable if λ < 0 and is unstable if λ > 0 Case (iii): the roots are complex conjugate When the eigenvalues of A are comple (0) would still be a solution to the system However, as in the difference equation case, we can simplify the system to get a solution in terms of real numbers so that the stability property of the system becomes visible

10 If the Matrix A has a specific form such that a = a = a(say) and a = a = b(say) a b so that A looks like this:, then as in the difference equation case, we can b a derive the eigen values and eigen vectors of A as a ± ib and, respectively i i Then directly using (0), we can obtain the general solution for this special case as: ( a+ ib) t ( a ib) t xt = c exp + c exp + x ( a+ ib) t ( a ib) t xt = ic exp + ic exp + x (0ʹ ) Again applying the second part of De Moivre s Theorem, the solution can be simplified to at xt = exp ( c3cos θ t+ c4sin θ + x at xt = exp ( c3sin θ t c4cos θ + x where c 3 = ( c + c) and c 4 = ( c + c) i are two arbitrary constants (not necessarily real) To determine the stability property, observe that as in the difference equation case, the two terms cos θ t and sin θ t move in a periodic manner, taking values between and +, and returning to the same value after every π θ period Hence x and x will also move in cyclically fashion However, whether they approach the steady state over time or not depends crucially on the term exp at If a < 0, the system will approach the steady state over time, albeit cyclically If, a > 0 the system will move away from the steady state If a = 0, the system will exhibit limit cycles, moving in the same orbit period after period, neither approaching the steady state, nor moving away from it Remark Note that though the methods of solving difference and differential equations are very similar, and often the stability properties are also somewhat similar, in the complex root case the stability condition for a system of differential equations is substantially different from that of difference equation Recall that in a system of difference equations with complex roots a ± ib, stability requires that r = a + b < But in a system of differential equations with complex roots a ± ib, stability requires that a < 0 Thus in the latter case, only the real part of the complex root is important for stability, the imaginary part does not play any role a b If the matrix A does not follow the specific form given by, we can construct a b a matrix M and transform the x-system into a system of new variables y such that the new co-efficient matrix of the y-system has the specific form Once we solve the y-system, we can derive the general solution for the x-system as: xt = Myt + x The stability property of the x-system will be the same as the y-system