1 7 Laplace ransform. Solving linear ODE wih piecewise coninuous righ hand sides In his lecure I will show how o apply he Laplace ransform o he ODE Ly = f wih piecewise coninuous f. Definiion. A funcion f() is piecewise coninuous on he inerval I = [a,b] if i is defined and coninuous on his inerval excep, probably, a finie number of poins,,,..., k, a each of which he lef and righ limis of his funcion exis (i.e., all he disconinuiies of he firs ype, or jump disconinuiies). Example. Consider he funcion he graph of which is given in he figure..5. f() =, 4,, > 4, f() Figure : Piecewise coninuous funcion f() from Example You can see ha here is one poin of disconinuiy, namely = 4, a which f() has he disconinuiy of he firs ype, or jump disconinuiy. We have lim 4 =, lim =, f() +f() hey boh exis and are no equal o each oher. The value of he jump is. Wha is he Laplace ransform of f()? We can find i using he definiion: L f} = = 4 4 f()e s d ( )e s d+ = s e s 4 s e s 4 = s e 4s s + s e 4s = s (e 4s ). 4 e s d MATH66: Inro o ODE by Arem Novozhilov, Fall 3
2 Example 3 (Heaviside funcion). The mos imporan example for us will be he Heaviside funcion, defined as, <, u() =, >. Noe ha L u()} = s, he same as for L }, because we are only ineresed in funcions on [, ). This yields ha for any funcion f() defined on [, ) i is rue ha L f()} = L u()f()}. Addiionally o he Heaviside funcion, we consider shifed Heaviside funcion u( a) for a nonnegaive a (see he figure). u() u( a) a Figure : Heaviside funcion u() (lef) and shifed Heaviside funcion u( a) (righ) I is a simple exercise o check, using he definiion, ha L u( a)} = e as. Now I can sae he final propery of he Laplace ransform ha we will use (here are many more acually): 6 Time shifing. Le L f()} = F(s), hen To prove i, consider L u( a)f( a)} = L u( a)f( a)} = e as F(s). u( a)f( a)e s d = a f( a)e s d, because u( a) = when < a. Now make he subsiuion ξ = a, hen d = dξ and if = a hen ξ =. We obain f(ξ)e s(ξ+a) dξ = e sa L f}.
3 Le us find he Laplace ransform of he funcion in Example. Example 4 (Coninue Example ). Noe ha using he shifed Heaviside funcion we can consruc for any a < b he funcion u( a) u( b), such ha his funcion is equal o when (a,b) and zero oherwise (hink his ou!) This means u( a) u( b) a b Figure 3: u( a) u( b) for b > a ha for any f() he funcion f() ( u( a) u( b) ) equals f(), when (a,b), and oherwise. Now consider again, 4, f() =, > 4. This, using he previous, can be represened as Now, using Propery 6, f() = ( u() u( 4) ) + u( 4) = u()+u( 4). exacly as we already found in Example. L f} = L u()+u( 4)} = s +e 4s s Example 5. Find he Laplace ransform for, <, f() =, < <,, >. Again, using he properies of he Heaviside funcion, we can wrie f() = ( u( ) u( ) ) = u( ) u( ). However, we canno apply Propery 6 direcly o his expression since we need he expressions of he form f( a)u( a). To deal wih i, consider = ( +) = ( ) +( )+., 3
4 Similarly, Hence we have = ( +) = ( ) +4( )+4. which implies ha f() = ( ( ) +( )+ ) u( ) ( ( ) +4( )+4 ) u( ), L f} = ( s 3 + s + ) ( e s + s s s + 4 ) e s. s Now we are ready o sar solving ODE wih piecewise coninuous righ hand side. Example 6. Solve where We have y +y +y = f(), y() =, y () =, f() =, <,, >. f() = u() u( ). Applying he Laplace ransform o boh sides, we have (s +s+)y s = s ( e s ), or, afer some rearrangemen, Since, using he parial fracions, Y = s s(s+) e s. hen we find s(s+) = s s+ (s+), y() = L Y} = ( e ( ) e ( ) ( ) ) u( ), using Propery 6. Noe ha we can rewrie our soluion as, <, y() = e ( ), >. The graph of he soluion in given in he figure below. Example 7. Solve where f() as in he figure below. y +y +y = f(), y() =, y () =, 4
5 y() Figure 4: Soluion y() in Example 6 f() π Figure 5: f() in Example 7 π We find ha f() = u( π) u( π). Applying he Laplace ransform o he ODE, we find from where s Y s+sy +Y = s (e πs e πs ), Y = s+ s +s+ + s(s +s+) (e πs e πs ). Noe ha using parial fracion decomposiion, we have herefore, finally we have s(s +s+) = s s+ s +s+, Y(s) = s (e πs e πs )+ s+ s +s+ ( e πs +e πs ). Now our ask is o find he inverse Laplace ransform for Y. We have L s} = u(), hence } L s (e πs e πs ) = u( π) u( π). 5
6 For he second erm in Y(s) consider firs herefore s+ s +s+ = s+/+/ (s+/) +( 3/) = s+/ (s+/) +( 3/) + 3/ 3 (s+/) +( 3/), } ( s+ 3 L s = e / cos +s+ + sin 3 3. Finally, using Propery 6, we have y() = L Y} = L s (e πs e πs )+ s+ } s +s+ ( e πs +e πs ) = 3 = u( π) u( π)+e (cos / + 3 sin u() 3 3( π) e (cos ( π)/ + 3( π) sin u( π)+ 3 3( π) +e (cos ( π)/ + 3( π) sin u( π). 3 Below is he graph of he soluion. y() π π Figure 6: Soluion y() depending on ime in Example 7 Example 8. Solve y +y = f() = cos, < π/,, > π/, y() = 3, y () =. We can rewrie he righ hand side as f() = cos ( u() u( π/) ). Noe ha o invoke Propery 6 we need he funcion wih he same argumen as he argumen of he Heaviside funcion. f() = u()cos cos( π/+π/)u( π/) = u()cos+sin( π/)u( π/), 6
7 where I used or, cos(a+b) = cosacosb sinasinb. Therefore, afer applying he Laplace ransform, I will find (s +)Y = 3s + s s + + s + e π/, Y(s) = 3s s + s s + + s (s +) + (s +) e π/. To find he inverse Laplace ransform, we will need } } L s (s +), L (s +). For his, using he propery of he differeniaion of he frequency, noe ha L sin} = s (s +), L cos} = s (s +). Hence For he second one hence Finally we find } L s (s +) = sin. (s +) = ( ) s + s (s +), } L (s +) = (sin cos). y() = L Y} = 3cos sin+ sin+ ( sin( π/) ( π/)cos( π/) ) u( π/). By noing ha sin( π/) = cos and cos( π/) = sin, we can simplify his expression as 3cos sin+ y() = sin, <, 5 π 4 cos+ 4 cos, >. Insead of conclusion I said ha he examples in his lecure are he main reason we need he Laplace ransform. This is indeed rue, bu only parially. There are wo more cases when Laplace ransform becomes indispensable heoreical and compuaional ool: Firs, when f() is an arbirary periodic funcion differen from simple sine and cosine funcions. Second, when we consider soluions of he equaions wih impulsive righ hand side (hink of an insananeous impac). Unforunaely, due o ime limiaion, we do no cover hese exciing opics, and I refer you o he exbook. 7