Differential equations Piecewise Continuous Functions, guaranteed and verified 100% Pass

1


Piecewise Continuous Functions


Recall if:

𝑢(𝑡) = 0 if 𝑡 < 0
1 𝑦 = 𝑢𝑎 (𝑡)
= 1 if 𝑡 ≥ 0
𝑢𝑎 (𝑡) = 𝑢(𝑡 − 𝑎) = 0 if 𝑡 < 𝑎
= 1 if 𝑡 ≥ 𝑎
Then:
1 𝑎
ℒ(𝑢(𝑡)) =
𝑠
𝑒−𝑎𝑠
ℒ(𝑢(𝑡 − 𝑎)) = .
𝑠




Theorem: Translation on the 𝑡-axis.

If ℒ(𝑓 (𝑡)) exists for 𝑠 > 𝑐, then ℒ(𝑢(𝑡 − 𝑎)𝑓 (𝑡 − 𝑎)) = 𝑒 −𝑎𝑠 𝐹(𝑠)
and ℒ −1 (𝑒 −𝑎𝑠 𝐹 (𝑠)) = 𝑢 (𝑡 − 𝑎)𝑓(𝑡 − 𝑎).



Notice that:

𝑢(𝑡 − 𝑎)𝑓(𝑡 − 𝑎) = 0 if 𝑡 < 𝑎
= 𝑓(𝑡 − 𝑎) if 𝑡 ≥ 𝑎.

, 2


Proof:
𝑤=∞
𝑒 −𝑎𝑠 𝐹 (𝑠) = 𝑒 −𝑎𝑠 ∫𝑤=0 𝑒 −𝑠𝑤 𝑓(𝑤) 𝑑𝑤
𝑤=∞
= ∫𝑤=0 𝑒 −𝑠(𝑤+𝑎) 𝑓(𝑤) 𝑑𝑤
Let 𝑡 = 𝑤 + 𝑎

𝑑𝑡 = 𝑑𝑤

𝑡=∞
𝑒 −𝑎𝑠 𝐹 (𝑠) = ∫𝑡=𝑎 𝑒 −𝑠𝑡 𝑓 (𝑡 − 𝑎) 𝑑𝑡
𝑡=∞
= ∫𝑡=0 𝑒 −𝑠𝑡 𝑢(𝑡 − 𝑎)𝑓(𝑡 − 𝑎) 𝑑𝑡

= ℒ(𝑢(𝑡 − 𝑎)𝑓(𝑡 − 𝑎)).


𝑒−𝑠
Ex. Let 𝐹 (𝑠) = . Find ℒ −1 (𝐹 (𝑠)).
𝑠+2



ℒ −1 (𝑒 −𝑎𝑠 (𝐺 (𝑠))) = 𝑢(𝑡 − 𝑎)𝑔(𝑡 − 𝑎)

where ℒ(𝑔(𝑡)) = 𝐺 (𝑠).


𝑒−𝑠 1
So for ℒ −1 ( ), 𝑎 = 1 and 𝐺 (𝑠) = .
𝑠+2 𝑠+2


1
From a Laplace transform table, if 𝑔(𝑡) = 𝑒 −2𝑡 , then 𝐺 (𝑠) = .
𝑠+2


𝑒−𝑠
ℒ −1 ( ) = 𝑢(𝑡 − 1)𝑔(𝑡 − 1) = 𝑢(𝑡 − 1)𝑒 −2(𝑡−1) .
𝑠+2