Lecture Notes for Math 251: ODE and PDE. Lecture 22: 6.5 Dirac Delta and Laplace Transforms

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1 Lecture Notes for Math : ODE and PDE. Lecture : 6. Dirac Delta and Laplace Transforms Shawn D. Ryan Spring 0 Dirac Delta and the Laplace Transform Last Time: We studied differential equations with discontinuous forcing functions. Now we want to look at when that forcing function is a Dirac Delta. When we considered step functions we considered these to be like switches which get turned on after a given amount of time. In applications this represented a new external force applied at a certain time. What if instead we wanted to apply a large force over a short period of time? In applications this represents a hammer striking an object once or a force applied only at one instant. We want to introduce the Dirac Delta function to accomplish this goal.. The Dirac Delta Paul Dirac (90-98) was a British Physicist who studied quantum Mechanics. Famous Quote on Poetry: In Science one tries to tell people, in such a way as to be understood by everyone, something that no one ever knew before. But in poetry, it s exactly the opposite. There are several ways to define the Dirac delta, but we will define it so it satisfies the following properties: Definition. (Dirac Delta) The Dirac delta at t = c, denoted δ(t c), satisfies the following: () δ(t c) = 0, when t c () c+ɛ δ(t c)dt =, for any ɛ > 0 c ɛ (3) c+ɛ f(t)δ(t c)dt = f(c), for any ɛ > 0. c ɛ We can think of δ(t c) as having an infinite value at t = c, so that its total energy is, all concentrated at a single point. So think of the Dirac delta as an instantaneous impulse at time t = c. The second and third properties will not work when the limits are the endpoints of any interval including t = c. This function is zero everywhere but one point, and yet the integral is. This is why it is a function, the Dirac delta is not really a function, but in higher mathematics it is referred to as a generalized function or a distribution.

2 . Laplace Transform of the Dirac Delta Before we try solving IVPs with Dirac Deltas, we will need to know its Laplace Transform. By definition L{δ(t c)} = 0 e st δ(t c)dt = e cs () by the third property of the Dirac delta. Notice that this requires c > 0 or the integral would just vanish. Now let s try solving some IVPs involving Dirac deltas Example. Solve the following initial value problem y + 3y 0y = δ(t ), y(0) =, y (0) = 3 () We begin by taking the Laplace Transform of both sides of the equation. So s Y (s) sy(0) y (0) + 3(sY (s) y(0)) 0 e s (3) By partial fractions decomposition we have (s + 3s 0)Y (s) s 3 = e s () e s (s + )(s ) + s + 3 () (s + )(s ) = Y (s)e s + Y (s) (6) Y (s) = Y (s) = (s + )(s ) = 7 s 7 s + s + 3 (s + )(s ) = s + s + (7) (8) Take inverse Laplace Transforms to get y (t) = 7 et 7 e t (9) y (t) = e t + e t (0) and the solution is then y(t) = y (t )u (t) + y (t) () ( = u (t) 7 e(t ) ) 7 e (t ) + e t + e t () ( = u (t) 7 et ) 7 e t 0 + e t + e t (3)

3 Notice that even though the exponential in the transform Y (s) came originally from the delta, once we inverse the transform the corresponding term becomes a step function. This is generally the case, because there is a relationship between the step function u c (t) and the delta δ(t c). We begin with the integral { t 0, t < c δ(u c)du = (), t > c = u c (t) () The Fundamental Theorem of Calculus says u c(t) = d ( t ) δ(u c)du = δ(t c) (6) dt Thus the Dirac delta at t = c is the derivative of the step function at t = c, which we can think of geometrically by remembering that the graph of u c (t) is horizonatal at every t c, hence at those points u c (t) = 0 and it has a jump of one at t = c. Example 3. Solve the following initial value problem. y + y + 9y = δ(t ) + e t, y(0) = 0, y (0) = (7) First we Laplace Transform both sides and solve for Y (s). Thus s Y (s) sy(0) y (0) + (sy (s) y(0)) + 9 e s + s (s + s + 9)Y (s) + = e s + s (8) (9) e s s + s (s )(s + s + 9) (0) s + s + 9 = Y (s)e s + Y (s) = Y 3 (s) () Next, we have to prepare Y (s) for the inverse transform. This will require completing the square for Y (s) and Y 3 (s), while we will need to first use partial fractions decomposition on Y (s). I leave the details to you to verify, so we now have everything in the correct form for the 3

4 inverse transform Y (s) = So their inverse transforms are s + s + 9 = (s + ) + = (s + ) + Y = (s )(s + s + 9) = ( ) s s + (s + ) + = ( ) (s + ) + s (s + ) + = ( ) s s + (s + ) + 3 (s + ) + = ( s s + (s + ) + 3 ) (s + ) + Y 3 (s) = (s + s + 9) = (s + ) + = (s + ) + y (t) = 3 t sin( t) () y (t) = (et e t cos( t) 3 e t sin( t)) (3) y 3 (t) = e t sin( t). () Thus, since our original transformed function was we obtain Y (s)e s + Y (s) Y 3 (s) () y(t) = u (t)y (t ) + y (t) y 3 (t) (6) ( = u (t) e t+ sin( t ) ) (7) + ( e t e t cos( t) 3 e t sin( ) t) (8) e t sin( t). (9)

5 Example. Solve the following initial value problem. y + 6y = u 3 (t) + δ(t ), y(0) =, y (0) = (30) Again, we begin by taking the Laplace Transform of the entire equation and applying our initial conditions, we get Solving for Y (s) s Y (s) sy(0) y (0) + 6 e 3s s (s + 6)Y (s) s = e 3s s + e s (3) + e s (3) e 3s s(s + 6) + e s s s + (33) s + 6 = Y (s)e 3s + Y (s)e s + Y 3 (s) (3) The only one of these we need to use partial fractions on is the first one. The rest can be dealt with directly, all they need is a little modification. and so the associated inverse transforms are Our solution is the inverse transform of and this will be Y (s) = s(s + 6) = 8 s s 8 s + 6 Y (s) = s + 6 = s + 6 Y 3 (s) = s + s + 6 = s s s + 6 (3) (36) (37) y (t) = 8 cos(t) (38) 8 y (t) = sin(t) (39) y 3 (t) = cos(t) + sin(t). (0) Y (s)e 3s + Y (s)e s + Y 3 (s) () y(t) = u 3 (t)y (t 3) + u (t)y (t ) + y 3 (t) () ( = u 3 (t) 8 ) cos((t 3)) + 8 u (t) sin((t )) + cos(t) + sin(t) (3) ( = u 3 (t) 8 ) cos(t ) + 8 u (t) sin(t ) + cos(t) + sin(t) () HW 6. #,,6,8,