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.
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 0 0 1 1 - 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, ) 0.8 0.6 0.4 0. 0 0 1 3 4 5 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
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
3 Forcing Function 1 -cos(t) 0 1-cos(t) -1 - -3 0 1 3 4 5 6 7 8 9 10
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
-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 1 1 1 -s e 6 - + 3 s(s + )(s + 3) s s + s + 3 1 1 1 -s e 6 - + 3 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)
Now let s inspect this solution as t gets large. Graphing y(t) we get -t -3t e - e On [0, 1) 0.35 0.3 0.5 0. 0.15 Note that as t gets large y(t) converges to 1/6. 0.1 0.05 0 0 1 3 4 5 6 7 8 9 10 -t -3t e - e 1 1 -(t-1) 1-3(t-1) + - e + e On [1, ) 6 3
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.
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) 5 5 5 t - 5 t - 10 1 5 10 5 10 5 5 5 1,[10, ) = u (t) + u (t) = u (t) t - 5 - 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/ 4 1 1 1 Ys () 5 5 4 8 s s 4 s s 4 Used #13. Details omitted.
For ease of computing the inverse Laplace transform we break things up a bit. Let 1 1 1 1 1 1 h( t) L t sin t 4 8 s s 4 4 8 5s 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) 1 1 1 y t 5 sin( t 5),[5,10) 5 4 8 1 1 1 1 t 5 sin( t 5) t 10 sin( t 10), [10, ) 4 8 4 8
Sinusoid wrapped around a straight 0, [0,5) line segment. 1 1 1 y t 5 sin( t 5),[5,10) 5 4 8 1 1 1 1 t 5 sin( t 5) t 10 sin( t 10), [10, ) 4 8 4 8 Next we simplify the expression that is on [10, ) and graph it.
Simplify 0, [0,5) 1 1 1 y t 5 sin( t 5),[5,10) 5 4 8 1 1 1 1 t 5 sin( t 5) t 10 sin( t 10), [10, ) 4 8 4 8 t 5 1 sin ( 5) 4 4 8 t 10 1 sin ( 10) 4 4 8 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 1 5 1 1 1 1 + - sin(t - 5) + sin(t - 10) = + sinusoid 5 4 5 8 8 4 Next we adjoin the graph over [10, ).
The graph of the complete solution is 0.5 Look at the behavior as t 0, [0,5) gets large. Just as we 1 1 1 predicted earlier a y t 5 sin( t 5),[5,10) 5 4 8 sinusoid wraps around 1 1 1 1 the line y = 1/4. t 5 sin( t 5) t 10 sin( t 10), [10, ) 4 8 4 8
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.