Chapter 12  Fourier Series and the Laplace Transform

12.9  Inverting the Laplace Transform

    This section is a continuation of our development of the Laplace Transforms in Section 12.5, Section 12.6, Section 12.7 and Section 12.8.  

    So far, most of the applications involving the Laplace transform involves a transform (or part of a transform) that is expressed by

(12.36)              

where and are polynomials that have no common factors.  The inverse of is found by using its partial fraction representation and referring to Table 11.2.  We now show how the theory of complex variables can be used systematically to find the partial fraction representation.  Theorem 11.19 is an extension of Example 8.7 to n linear factors.  We leave the proof to you.

Theorem 12.19 (Nonrepeated Linear Factors).  Let be a polynomial of degree at most .  If has degree n, and has distinct complex roots  ,  then equation (12.36) becomes   

(12.37)            .  

Proof.

Theorem 12.20 (A Repeated Linear Factors).  If are polynomials of degree , respectively and   and  ,  then equation (12.36) becomes   

(12.38)            ,  

where    is the sum of all partial fractions that do not involve factors of the form  .  Furthermore, the coefficients    can be computed with the formula  .  

Proof.

Example 12.24.  Let  .  Find  .

Solution.

From Equations (12.37) and (12.37) we write

            

We calculate the coefficient by  

               

We find the coefficients , , and by using Theorem 12.20.  

In this case and   , and we get

            
            

            
            

            

Now substitute these values into the formula  , and get the partial fraction representation for :  

            ,

The solution is found with the computation

            

Explore Solution 12.24.

Theorem 12.21 (Irreducible Quadratic Factors).  Let be polynomials with real coefficients such that the degree of is at most 1 larger than the degree of .   If    does not have a factor of the form  ,  then  

            ,  
where  

(12.21)            .  

Proof.

Example 12.25.  Let  .  Find  .

Solution.

    This solution is a little tedious and illustrates how Theorem 12.21 is applied to obtain the partial fraction expansion of when two irreducible quadratic factors are involved.

Here we have and ,  and the roots of occur at and .  The derivative of     is .  Computing the residues at  at and yields

            ,  
        and
            .

The above residues are used to determine the values in Equation (12.21), and we get

              
        and
            

We find that and , which correspond to and ,  respectively.  Applying the formula in Theorem 12.21 we obtain

              
            
which becomes        

            
            
and is simplified as

            .
            
Finally   is found with the calculation

            

Explore Solution 12.25.

Example 12.26.  Let  .  Find  .

Solution.

    This solution is a little tedious and illustrates how Theorems 12.20 and 12.21 are applied to obtain the partial fraction expansion of .

The partial fraction expression for has the form

            .  

First, we calculate the coefficient .  In the denominator of the linear factor is nonrepeated, so we have

            

Second, we calculate the coefficients , and by using Theorem 12.20.  In this case and   .   In the denominator of the term is a repeated factor, so we have  

            

    
              

Third, we calculate the coefficients and   by using Theorem 12.21.  Here we have ,  ,  and  .  In the denominator of the factor is an irreducible quadratic, with roots , so that

               

Hence we have .  Or if you prefer taking limits then we can use the calculation

            

Either way, we obtain and .  

Now substitute these values into the formula  , and get the partial fraction representation for :  

            

Now we use Table 12.2 to get  

        

Explore Solution 12.26.

Example 12.27.  Use Laplace transforms to solve the system

               

                A graph of the solution.

Solution.

We let and denote the Laplace transforms of and , respectively.  Taking the transforms of the two differential equations gives  

            

which can be written as  

              

We use Cramer's rule to solve for and :

            
        
            

We obtain the desired solution by computing the inverse transforms:

            
        
            

Explore Solution 12.27.

    According to Equation (12.29), the inverse Laplace transform is given by the integral formula  

            ,  

where is any suitably chosen large positive constant.  This improper integral is a contour integral taken along the vertical line in the complex plane.  We use the residue theory in Section 8.1 to evaluate it.  We leave the cases in which the integrand has either infinitely many poles or branch points for you to research in advanced texts.  We state the following more elementary theorem.

Theorem 12.22 (Inverse Laplace Transform).  Let , where are polynomials of degree m and n, respectively, and  .  The inverse Laplace transform of    is given by  

        ,  

where the sum is taken over all the residues of the complex function  .  

Proof.

    Using the residue calculus, the English mathematician Oliver Heaviside discovered a method for inverting the Laplace transform.

Theorem 12.23 (Heaviside Expansion Theorem).  Let and be polynomials of degree m and n, respectively, where . If has n distinct simple zeros at the points  ,  then    is the Laplace transform of the function    given by

(12.44)        .  

Proof.

Example 12.28.  Let  .  Find  .

Solution.

Here we have    and    so that has simple zeros located at the points  , , and .  When we use  ,  calculation reveals that

        


        


        

Applying Equation (12.44) in Theorem 12.23 gives as

          

Explore Solution 12.28.

Exercises for Section 12.9.  The Laplace Transform: Inverting the Laplace Transform