Experiments With the Laplace Transform

Part 1. Introduction

1. Here is a sample transform and inverse transform pair under the restriction that the variable a be positive.

LaplaceTransform[Exp[-a*t], t, s, 
      Assumptions -> {a > 0}]

InverseLaplaceTransform[%, s, t, 
      Assumptions -> {a > 0}]

2. Verify the differentiation rule.

3. Verify the convolution property.

f[t_] :=  t;
g[t_] :=  Exp[-2*t];
h[t_] :=  Integrate[f[z]*g[t - z], {z, 0, t}];

LaplaceTransform[f[t], t, s]*LaplaceTransform[g[t], t, s]

LaplaceTransform[h[t], t, s]

Simplify[%]

Part 2. An Ordinary Differential Equation Problem

1. The Laplace transform of f.  Note that in Mathematica the Heaviside funciton is called UnitStep.

f[t_] :=  1 - UnitStep[t - 5];
Plot[f[t], {t, 0, 6}, PlotRange -> {{0, 6}, {0, 1.5}}, 
        PlotStyle -> RGBColor[1, 0, 0]];

LaplaceTransform[f[t], t, s]

Copy the output of the previous transform calculation to define F[s] below.

F[s_] :=  ?;

2. Write the explanation of the displayed formula in the input cell below:

Answer:

InverseLaplaceTransform[
        Apart[(F[s] - 2*s - 3)/(s^2 + 2*s + 5)], s, t];
TrigReduce[ExpToTrig[%]]

Copy the above output, and paste it to define the function y[t].

y[t_] :=  ?;

3. Graph y[t].

Plot[y[t], {t, 0, 10}, PlotRange -> {{0, 10}, {-2.5, 1.5}}, 
        PlotStyle -> RGBColor[1, 0, 0]];

4. Write your explanation here:

Part 3. A Partial Differential Equation Problem

1. Give your explanation here:

2. Give your explanation here:

3. Give your explanation here:

4. Give your explanation here:

5.

InverseLaplaceTransform[-1/s*Exp[-5*s], s, t]

Use the output above to define h1.

h1[t_] :=  ?;

InverseLaplaceTransform[Exp[-Sqrt[s]*x], s, t, 
      Assumptions -> {x > 0}]

h2[t_] :=  ?;

InverseLaplaceTransform[1/s, s, t]

6. Definition of u:

Evaluate the convolution integral.  Note that Mathematica will not automatically carry out the integration over [0, t].  Use the Fundamental Theorem of Calculus to obtain the result, and substitute into u[x,t].  Notice that you are evaluating an improper integral.

Clear[x]

Integrate[h1[z]*h2[t - z], z]

u[x_, 
          t_] :=  (-1 + 
                Erf[(x/(2 (Sqrt[-5 + t])))]) UnitStep[-5 + 
                t]  +  1;

Plot u[x,t] as a function of t for a range of values of x.  Change the limits on the plot as necessary.

x  =  0.2;
Plot[u[x, t], {t, 0, 20}, 
        PlotStyle -> RGBColor[1, 0, 0]];

7. Here is a part of the graph of u.

Clear[x];
Plot3D[u[x, t], {x, 0.1, 10}, {t, 1, 20}, 
        ViewPoint -> {1.3, 2.4, 2}];

Answer to the question in step 7: