Existence And Property Of Laplace Transform

Existence And Property Of Laplace transforms :

Existence Of Laplace Transforms:

     In order to discuss the theorem on existence of transform we need the following definition .

Definition :

        A function f(t) is said to be piecewise continuous in a finite range a≤t≤b , if it is possible to divide the range into a finite number of sub-intervals such that f(t) is continuous inside each sub-interval and approaches finite values as t approaches either end of any interval from the interior.

Theorem :

       Let f(t) be a function which is piecewise continuous on every finite interval in the range t≥0 and satisfies 

        |f(t)|≤Me^αt , for all t≥0        .........(1)
   and for some constants α and M . Then the Laplace transform of f(t) exists for all p>α .

Proof :

          Since f(t) is piecewise continuous ,      e⁻ᵖᵗf(t) is integrable over any finite interval on the t-axis , 

          T

i.e     ∫ e⁻ᵖᵗf(t) dt  exists for all T.

         0

    Hence for all T , 

          T                            T

          ∫ |e⁻ᵖᵗf(t)|dt ≤M∫ e⁻ᵖᵗ e^αt dt 

          0                           0

  = M[e^(α-p)Τ -1]/(α-p)

     ≤ M/(p-α) , for p>α 

                   T

Thus   lim ∫ |e⁻ᵖᵗ f(t)| dt ≤ M/(p-α) 

        T-->∞  0
                                                     ∞
          for   p>α which implies ∫ |e⁻ᵖᵗf(t) | dt
                                ∞                  0
exists and hence ∫ e⁻ᵖᵗ f(t) dt exists for p>α,
                               0

 that is L{f(t)} exists for all p>α .

     From now on we shall assume that the functions under discussion satisfy conditions in above theorem so that the Laplace transform exists on a suitable interval .

Linearity Property Of Laplace Transform :

  Theorem :

                             n

              If   f(t) = Σ Cᵢ fᵢ(t) ,

                            i=1

where Cᵢ are constants then 

                      n

       L{f(t)} = Σ Cᵢ L{fᵢ(t)}  .

                     i=1

Proof :

      By definition , we have 

                     ∞       n

     L{f(t)} = ∫ e⁻ᵖᵗ Σ Cᵢ fᵢ(t) dt 

                    0       i=1

                    ∞

                = ∫ e⁻ᵖᵗ[C₁f₁(t)+ C₂f₂(t)+....+Cₙfₙ(t)]dt

                   0

               ∞                           ∞

        = C₁∫ e⁻ᵖᵗ f₁(t) dt + C₂∫ e⁻ᵖᵗ f₂(t)dt

               0                            0

                                            ∞

                         + ..........+Cₙ∫ e⁻ᵖᵗ fₙ(t) dt

                                            0

          = C₁L{f₁(t)} + C₂L{f₂(t)}+....+CₙL{fₙ(t)}

              n

          =  Σ Cᵢ L{fᵢ(t)}

             i=1

Example :

             Find the transform of function 

      f(t)  = 3 sin4t - 2 cos5t .

Solution :

     Applying the above theorem , we have 

L{f(t)| = 3L{sin 4t} - 2L {cos 5t}

             = 3 . 4/p²+16  -  2.p/p²+25 
by Laplace function (v) .

            =  12/p²+16  - 2p/p²+25 .

  Which is the required transformation .