COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
E-mail:  mkt@jbpub.com  
Internet:  http://www.jbpub.com/  
This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 11  FOURIER SERIES and the LAPLACE TRANSFORM

a = A = b = B = c = d = s = S = t = T = u = U = v = V = w = W = x = X = y = Y = z = Z = 
                                        0; 
Clear[a,a1,a2,a3,A,A1,A2,b,b1,b2,B1,B2,B,c,d,EqA,EqB,f,f1,f2,f12,F,F1,F2,F12,F3,g,G,h,H,i,int,j,k,L,Ly,L1,L2,Lf,LDf,ℒf,ℒDf,n,PQ,PR,r,res1,res2,s,solset,s1,s2,s3,S,S9,t,table,τ,T,θ,u,U,U1,U2,U7,v,v1,v2,v3,v4,val,V,w,w1,w2,W,x,X,y,y0,y1,Y,Yi,z,z1,z2,Z]; 
Needs["Graphics`Colors`"];

Section 11.1	Fourier Series

Section 11.2	The Dirichlet Problem for the Unit Disk

Section 11.3	Vibrations in Mechanical Systems

Section 11.4	The Fourier Transform

Section 11.5	The Laplace Transform

Section 11.6	Laplace Transforms of Derivatives and Integrals

Section 11.7	Shifting Theorems and the Step Function

Section 11.8  Multiplication  and  Division  by  t

Sometimes the solutions to nonhomogeneous linear differential equations with constant coefficients involve the functions  t cos(b t),  t sin(b t), or  (t^n) (e^(a t))  as part of the solution.  We now show how the Laplace transforms of  t f (t)  and  ((f (t))/t)  are related to the Laplace transform of  f(t).  The transform of  t f (t)  will be obtained via differentiation and the transform of  ((f (t))/t)  will be obtained via integration.  To be precise, we state the following theorems.

Theorem 11.17   (Multiplication by t),  Page 531.  If  F (s)  is the Laplace transform of  f (t),  then 

			ℒ(t f (t))=- F' (s).

Proof of Theorem 11.17, see text Page 531.

Theorem 11.18  (Division by t),  Page 531.   Let both  f (t)  and  ((f (t))/t)  have Laplace transforms and let  F (s)  denote the Laplace transform of  f (t).   If  \(lim\+\(t -> \(0\^+\)\)\ \(f \((t)\)\)\/t\)  exists then  

			ℒ( ((f (t))/t))=(∫_s^∞) 
            F (σ) dσ.

Proof of Theorem 11.18, see text Page 531.

Example 11.21,  Page 532.  Use Theorem 11.17 and find  ℒ(t Cos[b t]).

Solution 11.21.

Use the fact that  ℒ(Cos[b t]) =(s/((s^2)  +  (b^2))).

Remove[b, s, t]; 
Clear[f, F]; 
f[t_]  =  Cos[b t]; 
F[s_]  = (( s)/((s^2)  +  (b^2))); 
Print["     f[t] = ", f[t]]; 
Print["  ℒ(f[t]) = ", 
        F[s]]; 
Print["     F[s] = ", F[s]]; 
Print[" "]; 
Print["   t f[t] = ", t f[t]]; 
Print[" "]; 
Print["ℒ(t f[t]) = - F'[s]"]; 
Print[" "]; 
Print["ℒ(", 
        t f[t], ") = ", - F'[s]]; 
Print["ℒ(", t f[t], ") = ", 
        Together[- F'[s]]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

Ft[s_]  =  
        LaplaceTransform[t f[t],  t,  s]; 
Print["t f[t] = ", t f[t]]; 
Print["ℒ(t f[t]) = ", 
        Ft[s]];

Example 11.22,  Page 532.  Use Theorem 11.18 and find  ℒ(Sin[t]/t).

Solution 11.22.

Use the fact that  ℒ(Sin[t])=(1/((s^2)  +  1)).

Remove[s, t, Y]; 
Clear[f, F]; 
f[t_]  =  Sin[t]; 
F[s_]  =  (1/((s^2)  +  1)); 
Y[s_]  =  
        Integrate[F[r], {r, s, ∞}, PrincipalValue -> True, 
          Assumptions -> s > 0]; 
Print["    f[t] = ", f[t]]; 
Print["ℒ(f[t]) = ", 
        F[s]]; 
Print["    F[s] = ", F[s]]; 
Print[""]; 
\(\(Print[\*""   \!\(f[t]\/t\) = "", 
        f[t]\/t];\)\ \)
Print[""]; 
\(\(Print[\*"" ℒ(f[t]/t) = (∫_s^∞)F[r]\[DifferentialD]r""];\)\ \)
\(\(Print["ℒ(", 
        f[t]\/t, \*"") = \!\(∫\_s\%∞\)"", 
        F[r], "\[DifferentialD]r"];\)\ \)
Print[""]; 
Print["ℒ(", f[t]/t, ") = ", 
        Y[s]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

Y[s_]  =  
        LaplaceTransform[f[t]/t,  t,  s]; 
\(\(Print[\*""   \!\(f[t]\/t\) = "", f[t]\/t];\)\ \)
\(\(Print[\*""ℒ(\!\(f[t]\/t\)) = "", 
        Y[s]];\)\ \)

Example 11.23,  Page 532.  Solve the initial value problem  

		t y''[t]  -  t y'[t]  -  y[t]  =  0
with
y[0] = 0,  and  y'[0] = 0

Solution 11.23.

Enter the initial conditions and the functions f[t] and F[s].  Then Laplace transform the D.E. and solve for Y[s]. Observe that a D.E. involving Y'[s] will arise in the computations.

Remove[s, t, y, Y]; 
Clear[eqn, f, F, solset, y0, y1]; 
de  =  
        t y''[t]  -  t y'[t]  -  y[t]  ==  
          f[t]; 
eqn = 
        ReplaceAll[
          de, {f[t] -> F[s], 
            y[t] -> 
              Y[s], 
t y'[
                  t] -> - s Y'[s]  -  
                Y[s], 
t y''[
                  t] -> - (s^2) Y'[s]  -  2 s Y[s]  +  
                y0}]; 
Print[de]; 
Print[eqn]; 
Print[""]; 
y0  =  0; 
y1  =  0; 
f[t_]  = 0; 
F[s_]  =  0; 
Print[" y[0] = ", y0]; 
Print[" f[t] = ", f[t]]; 
Print[" F[s] = ", F[s]]; 
Print[""]; 
eqn   =   (- (s^2) Y'[s]  -  2 s Y[s]  +  
                y0)  - (- s Y'[s]  -  Y[s])  -  
            Y[s]   ==  F[s]; 
de  =  
        t y''[t]  -  t y'[t]  -  y[t]  ==  
          f[t]; 
eqn = 
        ReplaceAll[
          de, {f[t] -> F[s], 
            y[t] -> 
              Y[s], 
t y'[
                  t] -> - s Y'[s]  -  
                Y[s], 
t y''[
                  t] -> - (s^2) Y'[s]  -  2 s Y[s]  +  
                y0}]; 
solset  =  DSolve[eqn,  Y[s], s]; 
Print[eqn]; 
Print[solset]; 
Y[s_]  =  
        (Subscript[solset, [[1, 1, 2]]]); 
Y[s_]  =  ReplaceAll[Y[s] , C[1] -> c];  
y[t_]  =  
        InverseLaplaceTransform[Y[s],  s,  
          t]; 
Print[""]; 
Print["Y[s] = ", Y[s]]; 
Print[" "]; 
Print["y[t] = ", y[t]];

Aside. We can substitute y[t] into the D.E. and verify it is a solution.

Ly  =  
        t y''[t]  -  t y'[t]  -  y[t]; 
Print["y[t] = ", y[t]]; 
Print["y'[t] = ", y'[t]]; 
Print["y''[t] = ", y''[t]]; 
Print[" "]; 
Print["t y''[t] - t y'[t] - y[t] = ", Ly]; 
Print["t y''[t] - t y'[t] - y[t] = ", 
        Expand[Ly]];

Aside. For this problem, the solution involves a family of functions. We can use Mathematica/s to plot several solutions.

funs = Table[y[t], {c, -6, 6, 1}];  
    Plot[Evaluate[funs], {t, 0, 1}, PlotRange -> {{0, 1}, {-4, 4}}, 
      PlotStyle -> {Red,  Orange, Green, Blue, Magenta, Brown}, 
      AxesLabel -> {"t", "y"}, 
      Ticks -> {Range[0, 1, 0.5], Range[-4, 4, 1]}]; 
Print["y[t] = ", funs];

Section 11.8  Exercises for Multiplication  and  Division  by  t
See textbook page 533.

Section 11.9	Inverting the Laplace Transform

Section 11.10	Convolution

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