Chapter 12  Fourier Series and the Laplace Transform

12.10  Convolution for 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, Section 12.8 and Section 12.9.

    If we let    denote the transforms of  ,  respectively, then the inverse of the product    is given by the function  .  It is called the convolution of    and can be regarded as s generalized product of   .  Convolution will assist us in solving integral equations.

Theorem 12.24 (Convolution Theorem).  Let and denote the Laplace transforms of and , respectively.  Then the product    is the Laplace transform of the convolution of    and ,  and is denoted by  ,  and has the integral representation  

                

Proof.

Example 12.29.  Show that  .  

Solution.

    If we let  ,  and  ,  then  ,  and  ,  respectively.  Now, applying the convolution theorem, we get

          

Explore Solution 12.29.

Example 12.30.  Use the convolution theorem to solve the integral equation  .  

                                              A graph of the solution.

Solution.

    Letting    and using    in the convolution theorem, we obtain  

            
        
            .   

Solving for  ,  we have  ,  and then  , and then  , and then we get

              .

Finally, the solution    is obtained using facts from Table 12.2:    and  ,  and the computation  

            

Explore Solution 12.30.

    Engineers and physicists sometimes consider forces that produce large effects but that are applied over a very short time interval.  The force acting at the time an earthquake starts is an example.  This phenomenon leads to the idea of a unit impulse function  .  Let's consider the small positive constan a.  The function    is defined by

            .     

    The unit impulse function is obtained by letting the interval width go to zero, or

            .  

    Figure 11.29 shows the graph    for  .  Although is called the  Dirac  Delta function,  it is not an ordinary function.  To be precise it is a distribution, and the theory of distributions permits manipulations of as though it were a function.  Here, we will treat as a function and investigate its properties.

                                  Figure 12.29.  Graphs of    for  .

Example 12.31.  Show that  .  

Solution.

    By definition, the Laplace transform of is

              

Letting    in the above equation and using L'Hôpital's rule, we obtain  

              

Explore Solution 12.31.

    We now turn to the unit impulse function.  First, we consider the function    obtained by integrating  :  

              

Taking the limit as     results in the important fact that

            ,

where  is the unit step function that was introduced in Section 12.7.  The situation is illustrated in Figure 12.30.

                   Figure 12.30.  The integral of is , which becomes when for .

    We demonstrate the response of a system to the unit impulse function in Example 12.32.

Example 12.32.   Solve the initial value problem

              
            with  
            .  

Solution.

    Taking transforms results in    so that

            ,  
and the solution is

            

                              Figure 12.31.  A graph of the solution  .

Remark.  The condition    is not satisfied by the "solution" .  Recall that all solutions involving the use of the Laplace transform are to be considered zero for values of  ,  hence the graph of   is given above in Figure 12.31.  Note that    has a jump discontinuity of magnitude    at the origin.  This discontinuity occurs because either or   must have a jump discontinuity at the origin whenever the Dirac delta function, , occurs as part of the input or driving function.

Explore Solution 12.32.

    The convolution method can be used to solve initial value problems.  The tedious mechanical details of problem solving can be facilitated with computer software such as  Maple,  Matlab,  or  Mathematica.

Theorem 12.25 (Initial value Problem - IVP convolution method). The unique solution to the initial value problem  

            ,  

with    and   ,  is given by  

            ,  

where    is the solution to the homogeneous equation  

            ,  

with  ,  and  

              has the Laplace transform given by  .  

Proof.

Example 12.33. Use convolution to solve the initial value problem

              
            with
            .  

                         
                                    A graph of the solution.

Solution.

    First, we obtain the portion    of the solution by solving    with and .  Taking the Laplace transform yields  ,  and we can rearrange the terms to obtain  .  Solving for gives  ,  and it follows that  

              

Second, we observe that    and  ,  and also that  .  We compute the portion    of the solution with a convolution:   

              

Now we can compute the solution using the convolution method in Theorem 12.25:  

              

Explore Solution 12.33.