Section 6.4 DEs with Discontinuous Forcing Functions

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1 Section 6.4 DEs with Discontinuous Forcing Functions Key terms/ideas: Discontinuous forcing function in nd order linear IVPs Application of Laplace transforms Comparison to viewing the problem s solution as composite of separate solutions to separate IVPs over non-overlapping intervals. Since we are considering nonhomogeneous DEs the solution consists of the solution to the associated homogeneous DE plus a particular solution of the nonhomogeneous DE.

2 We focus on examples of nonhomogeneous initial value problems in which the forcing function is discontinuous. The general equations appear like We will proceed by example. ay +by +cy = g(t), y 0 = y, y 0 = y heaviside(t - ) Example: On-Off Forcing Term The IVP is 1, [0,π) y'' + y = f(t) = = 1- u π(t), y(0) = 0, y'(0) = 0 0, [π, ) t This IVP can be interpreted as describing the motion of a mass-spring system with no damping which is initially at rest, to which a constant external force is applied for π units of time. Possibly, the force is caused by an electromagnet attracting the mass, or even an additional weight was added to the original weight but taken off at t = π. This IVP can also be interpreted as describing the charge on a condenser in a simple circuit containing a capacitor, an inductor, and an impressed voltage source. The initial charge on the capacitor is zero farads (y(0) = 0), the initial current is zero ampere (y'(0) = 0), and the driving force is 1 volt for π seconds and zero there after. Ref: Farlow

3 1, [0,π) y'' + y = f(t) = = 1- u π(t), y(0) = 0, y'(0) = 0 0, [π, ) Take the Laplace transform of both sides and simplify to obtain an expression for Y(s) = L{y} for which we can compute the inverse Laplace transform. -πs 1 e L{y'' + y} = L{1- u π(t)} s Y(s) + Y(s) = - s s -πs Solve for Y(s): 1 e - -πs 1- e -πs 1 Y(s) = s s = 1- e s +1 s(s +1) s(s +1) We can use partial fractions to show that We used the first and second derivative properties and the initial conditions, as well as #1 and #1 from the table 1 = 1 s s s(s +1) s +1 -πs Rewriting Y(s) we get 1 s 1 s -πs 1 s Y(s) = 1 e = e Taking inverse transforms we have So we have y(t) = 1- cos(t) - u (t) +u (t)cos(t - π) π π = 1- cos(t) - u (t) 1+ cos(t) π 1- cos(t), [0,π) y(t) = -cos(t), [π, s s +1 s s +1 s s +1 Used trig. Identity. -πs -1 e L = u π (t) s ) Just add terms on [π, ] to get -cos(t). 1 cos(t) -πs #1 #13-1 se L = u π(t)cos(t - π) s +1

4 3 Forcing Function 1 -cos(t) 0 1-cos(t)

5 Example: Solve IVP 0, [0,1) y'' + 5y' + 6y = u 1(t) = y(0) = 0,y'(0) = 1, [1, ) What to expect when we take Laplace transforms: (1) the characteristic equation is quadratic with distinct real roots so we will have exponential functions in the interval [0, 1]. The exponential functions will converge to zero. (Why?) () on [1, ) we can expect to add on shifted exponential functions because of the unit step function. (3) Since the characteristic equation is a quadratic which will nicely factor we will need to apply partial fractions. Steps: (1) Take the Laplace transform. () Solve for Y(s). (3) Use partial fractions to split Y(s) into simple terms. (4) Carefully find inverse transforms. Taking Laplace Transforms: We will use the linearity, the first and second derivative properties, and the initial conditions. L{y'' + 5y' + 6y} = L{y''} + 5L{y'} + 6L{y} = L{u (t)} e [s Y(s) - sy(0) - y'(0)] + 5[sY(s) - y(0)] + 6Y(s) = s 1 -s The DE is damped. 0 0

6 -s e [s Y(s) - 0s - ]+5[sY(s) - 0]+6Y(s) = s -s e s + 5s + 6Y(s) - = s -s -s e e + + Y(s) = s = s = s + 5s + 6 (s + )(s + 3) Solve for Y(s): e -s s e + = + (s + )(s + 3) (s + )(s + 3) (s + )(s + 3) s(s + )(s + 3) We wrote it this way to make it simpler to do partial fractions. Use partial fractions: Omitting the details. (s + )(s + 3) (s + ) (s + 3) See # Carefully taking inverse transforms: = - (s + ) (s + 3) (s - (-)) (s - (-3)) y(t) = -t -3t e - e + -s -s e s e s(s + )(s + 3) s s + s s e s s - (-) s - (-3) 1 1 -(t-1) 1 u 1(t) - u 1(t)e + u 1(t)e 6 3 See #1 & #13-3(t-1)

7 Now let s inspect this solution as t gets large. Graphing y(t) we get -t -3t e - e On [0, 1) Note that as t gets large y(t) converges to 1/ t -3t e - e 1 1 -(t-1) 1-3(t-1) + - e + e On [1, ) 6 3

8 0, [0,5) Example: Solve IVP y'' + 4y = g(t), y(0) = 0,y'0) = 0, g(t) = (t - 5) / 5,[5,10) 1,[10, ) Let s discuss the forcing function g(t). Initially it is zero on [0, 5), then rises on a straight line of slope 1/5 on [5, 10), and finally becomes a constant there after. This type of forcing function is called ramp loading and has the graph. Undamped. Next let s discuss the form of the solution. Since the characteristic equation is r + 4 = 0, the roots are complex so we expect a sinusoid. However, on [0, 5) the trivial solution y = 0 satisfies the DE y'' + 4y = 0. For t > 10 the DE becomes y'' + 4y = 1. By observation y = ¼ is a particular solution of the nonhomogeneous equation so we expect the solution to be a sinusoid + the particular solution; y = C 1 sin(t) + C cos(t) + ¼. (This portion of the solution oscillates about the line y = ¼.) The solution on [5, 10) has a forcing function which is a polynomial of degree 1. In this case we expect that the solution here will oscillate about a linear function. All these observations are based on the forms of the trial solution for DEs of the form ay'' + by' + cy = polynomial.

9 In order to construct the solution we first express g(t) in terms of unit step functions and the apply Laplace transforms. 0, [0,5) g(t) = (t - 5) / 5,[5,10) t - 5 t - 5 t - 5 g(t) = u 5 (t) + (jump)u 10(t) = u 5(t) + 1- u 10(t) t - 5 t ,[10, ) = u (t) + u (t) = u (t) t u (t) t - 10 Taking the Laplace transform of the DE, applying the first and second derivative properties, and the initial conditions gives us 5s 10s s Y ( s) e e 5s 10s e e 4 5s Y( s) Applying partial fractions to So we have 1 s s 4 we get 5s 10s 5s 10s e e e e 5s s 4 1/ 4 1/ 4 s s 4 1/ 4 1/ Ys () s s 4 s s 4 Used #13. Details omitted.

10 For ease of computing the inverse Laplace transform we break things up a bit. Let h( t) L t sin t 4 8 s s s 10s Recall that we have things multiplied by the term e e 5 Then using #13 and making the shifts required for the inverse transform we have solution given by 1 y ( t) u 5( t) h( t 5) u10( t) h( t 10) 5 It helps to express the solution as a piecewise function. 0, [0,5) y t 5 sin( t 5),[5,10) t 5 sin( t 5) t 10 sin( t 10), [10, )

11 Sinusoid wrapped around a straight 0, [0,5) line segment y t 5 sin( t 5),[5,10) t 5 sin( t 5) t 10 sin( t 10), [10, ) Next we simplify the expression that is on [10, ) and graph it.

12 Simplify 0, [0,5) y t 5 sin( t 5),[5,10) t 5 sin( t 5) t 10 sin( t 10), [10, ) t 5 1 sin ( 5) t 10 1 sin ( 10) t t If we subtract these expressions and rearrange terms. The terms involving t/4 add to zero and the two constants add to 5/4. The expression on [10, ) is sin(t - 5) + sin(t - 10) = + sinusoid Next we adjoin the graph over [10, ).

13 The graph of the complete solution is 0.5 Look at the behavior as t 0, [0,5) gets large. Just as we predicted earlier a y t 5 sin( t 5),[5,10) sinusoid wraps around the line y = 1/4. t 5 sin( t 5) t 10 sin( t 10), [10, )

14 1/4 Zero Sinusoid about linear function. Sinusoid about horizontal line y = ¼. Note that in this example the forcing function g is continuous but g' is discontinuous at t = 5 and t = 10. It follows that the solution y(t) and its first two derivatives are continuous everywhere, but y" has discontinuities at t = 5 and at t = 10 that correspond to the discontinuities in g' at those points.

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