Variation of Parameters

Transcription

1 Variation of Parameters James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University April 13, 218 Outline Variation of Parameters Example One

2 We eventually want to look at a simple linear partial differential equation called the cable equation which arises in information transmission such as in a model of a neuron. there are many places you can get the requisite background for this kind of model, but here we will concentrate on the use of analysis in physical modeling. First, we need to talk about a technique in differential equations called Variation of Parameters. We can use the ideas of the linear independence of functions in the solution of nonhomogeneous differential equations. Here the vector space is C 2 a, b] for some interval a, b] and we use the linear independence of the two solutions of the homogeneous linear second order model to build a solution to the nonhomogeneous model. Rather than doing this in general, we will focus on a specific model: 2 d 2 x dt 2 x(t) = f (t) x() = 1, x(5) = 4. where is a nonzero number. The homogeneous solution xh solves 2 d 2 x x(t) = dt 2 and has the form xh(t) = Ae t t + Be We want to find a particular solution, called xp, to the model. Hence, we want xp to satisfy 2 x p (t) xp(t) = f (t) Since we don t know the explicit function f we wish to use in the nonhomogeneous equation, a common technique to find the particular solution is the one called Variation of Parameters or VoP. In this technique, we take the homogeneous solution and replace the constants A and B by unknown functions u1(t) and u2(t). Then we see if we can derive conditions that the unknown functions u1 and u2 must satisfy in order to work. So we start by assuming xp(t) = u1(t)e t + u2(t)e t x p(t) = u 1(t)e t 1 u1(t) t e + u 2 (t)e t 1 + u2(t) e t ( ) ( = u 1(t)e t + u 2 (t)e t + u1(t) 1 t 1 ) e + u2(t) e t

3 We know there are two solutions to the model, e t t and e which are linearly independent functions. Letting x1(t) = e t and x2(t) = e t, consider the system ] ] ] x1(t) x2(t) φ(t) x 1 (t) x 2 (t) = ψ(t) f (t) For each fixed t, the determinant of this system is not zero and so is a unique solution for the value of φ(t) and ψ(t) for each t. Hence, the first row gives us a condition we must impose on our unknown functions u 1 (t) and u 2 (t). We must have u 1(t)e t + u 2 (t)e t = This simplifies the first derivative of xp to be x p(t) = u1(t) 1 e t + u2(t) 1 e t Now take the second derivative to get x p (t) = u 1(t) 1 e t + u 2 (t) 1 e t + u1(t) 1 2 e t + u2(t) 1 2 e t Now plug these derivative expressions into the nonhomogeneous equation to find f (t) = (u 2 2(t) 1 e t u 1 (t) 1 ) ( t e + 2 u2(t) 1 2 e t 1 ) + u1(t) t 2 e ( ) u2(t)e t + u1(t)e t Now factor out the common u1(t) and u2(t) terms to find after a bit of simplifying that f (t) = (u 2 2(t) 1 e t u 1 (t) 1 ) t e +u2(t)(e t t e ) + u1(t)(e t e t ) We see the functions u1 and u2 must satisfy f (t) 2 = u 2(t) 1 e t u 1 (t) 1 t e

4 This gives us a second condition on the unknown functions u1 and u2. Combining we have u 1(t)e t + u 2 (t)e t = u 1(t) 1 e t + u 2 (t) 1 e t = f (t) 2 This can be rewritten in a matrix form: e t e t 1 e t 1 e t ] ] u 1 (t) u 2 (t) = f (t) 2 ] We then use Cramer s Rule to solve for the unknown functions u 1 and u 2. Let W denote the matrix ] e t e t W = 1 t e 1 e t Then the determinant of W is det(w ) = 2 and by Cramer s Rule ] e t u 1(t) = f (t) 1 2 e t det(w ) = 1 2 f (t)e t and ] e t u 2(t) = 1 t e f (t) 2 det(w ) Thus, integrating, we have u1(t) = 1 t 2 u2(t) = 1 t 2 = 1 2 f t (t)e f (u) e u du f (u) e u du where is a convenient starting point for our integration.

5 Hence, the particular solution to the nonhomogeneous time independent equation is xp(t) = u1(t) e t + u2(t) e t ( = 1 t ) ( f (u) e u du e t 1 t + f (u) e u du )e t. 2 The general solution is thus x(t) = xh(t) + xp(t) = A1e t + A2e t 1 t t 2 e f (u) e u 1 t du + 2 e t f (u) e u du for any real constants A1 and A2. Finally, note we can rewrite these equations as x(t) = A1e t + A2e t 1 t f (u) e u t 1 t du f (u) e u t du or x(t) = A1e t + A2e t 1 t f (u) ( e u t e u t ) du 2 In applied modeling work, the function ew e w 2 arises frequently enough to be given a name. It is called the hyperbolic sine function and is denoted by the symbol sinh(w). Hence, we can rewrite once more to see x(t) = A1e t + A2e t 1 t ( ) u t f (u) sinh du Finally, sinh is an odd function, so we can pull the minus side inside by reversing the argument into the sinh function. This gives x(t) = A1e t + A2e t 1 t ( ) t u + f (u) sinh du

6 Since sinh(t/) and cosh(t/) are just linear combinations of e t/ and e t/, we can also rewrite as x(t) = A1 cosh( t ) + A2 sinh( t ) + 1 t ( ) t u f (u) sinh du Now let s use the initial conditions x() = 1 and x(5) = 4. Hence, 1 = x() = A1 4 = x(5) = cosh( 5 ) + A2 sinh( 5 ) + 1 which is solvable for A1 and A2. We find A1 = 1 ( 1 A2 = sinh( 5 ) 4 cosh( 5 ) ( ) 5 u f (u) sinh du ( 5 u f (u) sinh We can use this technique on lots of models and generate similar solutions. Here is another example. ) ) du Consider u (t) + 9u(t) = 2t u() = 1 u () = 1 The characteristic equation is r = which has the complex roots ±3i. Hence, the two linearly independent solutions to the homogeneous equation are u1(t) = cos(3t) and u2(t) = sin(3t). We set the homogeneous solution to be uh(t) = Acos(3t) + B sin(3t) where A and B are arbitrary constants. The nonhomogeneous solution is of the form up(t) = φ(t) cos(3t) + ψ(t) sin(3t). We know the functions φ and ψ must then satisfy ] ] ] cos(3t) sin(3t) φ (t) 3 sin(3t) 3 cos(3t) ψ = (t) 2t

7 Applying Cramer s rule, we have φ (t) = 1 ] sin(3t) = 2 3 2t 3 cos(3t) 3 t sin(3t) and ψ (t) = 1 3 ] cos(3t) = 2 3 sin(3t) 2t 3 t cos(3t) Thus, integrating, we have φ(t) = 2 t u sin(3u) du ψ(t) = 2 t u cos(3u) du. The general solution is therefore u(t) = Acos(3t) + B sin(3t) ( 2 t ) ( 2 t ) u sin(3u) du cos(3t) + u cos(3u) du sin(3t). This can be simplified to u(t) = A cos(3t) + B sin(3t) + 2 t u{sin(3t) cos(3u) sin(3u) cos(3t)} du = A cos(3t) + B sin(3t) + 2 t u sin(3t 3u) du. Applying Leibnitz s rule, we find u (t) = 3A sin(3t) + 3B cos(3t) + 2 t sin(3t 3t) t 3u cos(3t 3u) du = 3A sin(3t) + 3B cos(3t) + 2 t 3u cos(3t 3u) du Applying the boundary conditions and using Leibnitz s rule, we find 1 = u() = A and 1 = u () = 3B. Thus, A = 1 and B = 1 3 giving the solution u(t) = cos(3t) sin(3t) + 2 t u sin(3t 3u) du.

8 Consider u (t) + 9u(t) = 2t u() = 1 u(4) = 1 The model is the same as the previous example except for the boundary conditions. We have 1 = u() = A 1 = u(4) = A cos(12) + B sin(12) Thus, since A = 1, we have and so B sin(12) = 1 cos(12) u sin(12 3u) du. 4 u sin(12 3u) du. B = 1 + cos(12) u sin(12 3u) du. sin(12) We can then assemble the solution using these constants. Homework Use VoP to solve Consider u (t) 4u(t) = f (t) u() = 1 u(8) = 1 where f is a continuous function Consider u (t) + 4u(t) = f (t) u() = 1 u(6) = 1 where f is a continuous function.