of Dicontinuou Function Week November 7, 06 Week
of Dicontinuou Function Example: If f (t) = u (t) + t (u 3(t) u 6(t)), then what i f (5)? Recall, that by definition: u (5) =, u 3(5) = and u 6(5) = 0. Then f (5) = + 5 ( 0) = 6 If f (t) = ( + t)u (t) + in t/u π(t) e t ( u π/ (t) ), then what i f (π)? Recall, that by definition: u (π) =, u π(π) = 0 and u π/ (π) =. Then f (π) = + π + 0 0 = + π Week
of Dicontinuou Function (cont.) Example: Find Lf (t)}, where f (t) = t +, t < 3 e t + t, t 3 Write f (t) = (t + )( u 3 (t)) + (e t + t )u 3 (t), and re-group the term a follow: then f (t) = (t + ) + u 3 (t)(t t ) + e t u 3 (t), Lf (t)} = + + Lu 3(t)(t t )} + Le t u 3 (t)} () To apply # of Table, we need to hift both f (t) = t t and f (t) = e t : f (t) = t t 5t 5t 9 + 9 = (t 3) 5t 0 = (t 3) 5(t + 3 + 3) = (t 3) 5(t 3) 5, f (t) = e t 3+3 u 3 (t) = e 3 u 3 (t)e t 3 Then continuing () we have Lf (t)} = + + L u 3 (t) [ (t 3) 5(t 3) 5 ]} + e 3 Lu 3 (t)e t 3 } Finally, applying # of Table and the linearity to the lat two LT, we obtain: Lf (t)} = + + e 3 3 5 e 3 5 e 3 + e3 3 Week
of Dicontinuou Function (cont.) Example: te 4t, t < Find Lf (t)}, where f (t) = 0, t < π/ e t in t, t π/ Write f (t) = te 4t ( u (t))+u π (t)e t in t = te 4t u (t)te 4t +u π (t)e t in t, then Lf (t)} = ( + 4) Lu(t)te 4t } + Lu π (t)e t in t} () To apply # of Table, we need to hift both f (t) = te 4t and f (t) = e t in t a follow: f (t) f (t) [ = (t + )e 4(t +) = e 4 (t )e 4(t ) + e 4(t )] = e (t π + π ) in [(t π + π ] ) = e π e (t π ) in [(t π ] ) + π = e π e (t π ) in (t π ) uing that in(x + π) = in x. Then Week
of Dicontinuou Function (cont.) [ ] Lu (t)te 4t } = e 4 Lu (t)(t )e 4(t ) } + Lu (t)e 4(t ) } = e 4 e ( + 4) + e 4 e + 4, Lu π (t)e t in t} = e π L u π (t)e (t π ) in (t π ) } Finihing (), we have Lf (t)} = = e π e π ( + ) + 4 π ( + 4) e 4 ( + 4) e 4 + 4 e π ( + ) + 4 Week
of Dicontinuou Function (cont.) Example: Find L G()}, where G() = Conider G() = Let firt find: e + : e + = e F (), where F () = +. f (t) = L F ()} = 3 L Then applying # of Table we obtain: 3 ( + ) 9 4 } = 3 e t/ inh 3 t L G()} = u (t)f (t ) = 3 u(t)e (t )/ inh 3 (t ) Week
of Dicontinuou Function (cont.) Alternatively, uing PFD, we write: F () = ( )( + ) = A + B + then A( + ) + B( ) =, and etting = : B( 3) = B = /3 = : A 3 = A = /3 A( + ) + B( ) =, ( )( + ) Then, we obtain f (t) = L F ()} = } 3 L } 3 L + = 3 et 3 e t Finally, from the above, we have L G()} = u (t)f (t ) = 3 u(t) [ e (t ) e (t )] Week
ODE with Dicontinuou Right Hand Side We now conider an IVP for the econd order ODE with a dicontinuou RHS: Solve: y + y + y = h(t), y(0) = 0, y () =, where h(t) = in t, t < π/, t π/ We apply the LT to both ide of the given ODE, then ue propertie of linearity and LT of derivative and apply given IC to have the following: Ly } + Ly } + Ly} = Lh(t)} Ly} y(0) y (0) + (Ly} y(0)) + Ly} = Lh(t)} Now, we find: Lh(t)} ( + + )Ly} = + Lh(t)} where we have ued that in t = co(t π = Lin t[ u π (t)] u π (t)} = Lin t} Lu π (t) in t} Lu π (t)} = + L u π (t) co(t π } π ) e = π + e π + e ). Hence, Week
ODE with Dicontinuou Right Hand Side (cont.) ( + + )Ly} = + + e π + e π Ly} = + + + ( + )( + + ) e π ( + )( + + ) e π ( + + ) Then where y(t) = I + II + III + IV, } I = L + + } II = L ( + )( + + ) } III = L e π ( + )( + + ) } IV = L e π ( + + ) We compute each ILT eparately: } I = L ( + ) = e t in t (by #8 of Table) + Week
ODE with Dicontinuou Right Hand Side (cont.) For II, we firt apply PFD: ( + )( + + ) = A + B + + C + D + +, then (A + B)( + + ) + (C + D)( + ) =, from which we obtain: A = 5, B = 5, C = 5, D = 3 5, hence II = [ } }] + 3 L 5 + L + ( + ) = + 5 } } +L + L + + ( + ) + L + [ L ( + ) + } + }] = 5 co t + 5 in t + 5 e t co t + 5 e t in t (by #4,5,8,9 of Table) } Next, we write III = L e π F (), where F () = and firt decompoe F () uing PFD: ( + )( + + ) = A + B + + ( + )( + + ), C + D + +, then (A + B)( + + ) + (C + D)( + ) =, from which we obtain: Week
ODE with Dicontinuou Right Hand Side (cont.) A = 5, B = 5, C = 5, D = 4 5, hence, if f (t) = L F ()} f (t) = [ } }] + L 5 + L + 4 + ( + ) = [ } L + 5 + } } }] L + L + + ( + ) + 3L + ( + ) + = 5 co t 5 in t + 5 e t co t + 3 5 e t in t (by #4,5,8,9 of Table) Then by # of Table: III = u π (t)f (t π ) and III = 5 u π (t) [ co(t π ) in(t π ) +e (t π ) co(t π ) + 3e (t π ) in(t π ) ] = ] [ 5 u π (t) in t + co t + e t+ π in t 3e t+ π co t uing co(t π ) = in t and in(t π ) = co t Week
ODE with Dicontinuou Right Hand Side (cont.) } Latly, we write IV = L e π F (), where F () = and decompoe F () uing PFD: ( + + ) = A + B + C + +, ( + + ), then A( + + ) + (B + C) =, from which we obtain: A =, B =, C =. Hence, if f (t) = L F ()} then } } } f (t) = L + + = L + L + + } + ( + ) + +L ( + ) = + e t co t + e t in t (by #,8,9 of Table) + Then by # of Table: IV = u π (t)f (t π ) and IV = u π (t) [ + e (t π ) co(t π ) + e (t π ) in(t π ] ) [ ] = u π (t) + e t+ π in t e t+ π co t uing co(x π ) = in x and in(x π ) = co x Week
ODE with Dicontinuou Right Hand Side (cont.) Collecting what we have obtained above: y(t) = e t in t 5 co t + 5 in t + 5 e t co t + 5 e t in t + [ ] 5 u π (t) in t + co t + e t+ π in t 3e t+ π co t ] +u π (t) [ + e t+ π in t e t+ π co t = 5 co t + 5 in t + 5 e t co t 4 5 e t in t [ +u π (t) 5 in t + 5 co t + 6 ] π 8 π 5 e t+ in t 5 e t+ co t Week