Applications

This section illustrates applications of the Laplace transformation in solving some problems from various applications.

Example: If a drag is administrated into the bloodstream in dosages d(t) and is removed at a rate at a rate proportional to the concentration, the concentration c(t) at time t is modeled by the following initial value problem
\[ \frac{{\text d}c}{{\text d}t} = d(t) - k\,c(t), \qquad c(0) =0 , \]
where k > 0 is a positive constant. Suppose that over a 24-hour period, a drag is introduced into the bloodstream at a rate of 24/t0 for exactly t0 hours and then stopped so that
\[ d(t) = \begin{cases} 24/t_0 , & \ 0 \le t \le t_0 , \\ 0, & \ t > t_0 . \end{cases} \]
To model the above piecewise continuous function, Mathematica offers two functions: either
\[ {\bf UnitStep}[x] = \begin{cases} 1, & \ x\ge 0 , \\ 0, & \ x < 0 ; \end{cases} \]
and
\[ {\bf HeavisideTheta}[t] = \begin{cases} 1, & \ t> 0 , \\ 0, & \ t < 0 . \end{cases} \]
The difference between these two functions is what value is assigned at x = 0: UnitStep set its value to 1, while HeavisideTheta leaves it unassigned. The definition of the Heaviside function requires its value at x = 0 to be 1/2. However, for our application of Laplace transform in this example, it does not matter.

Now we solve the corresponding initial value problem with Mathematica. First, we define the input function d(t:

d[t_, t0_] = 24/t0 * UnitStep[t0 - t]
Since the given initial value problem is linear, we can find its explicit solution; here we have two options: either use the Laplace transform or apply the standard command DSolve. So we demonstrate each approach.

Application of Laplace's transform yields an algebraic equation for cL, the Laplace transform of unknown drug concentration:

\[ \lambda\,c^L (\lambda ) = \frac{24}{\lambda\,t_0} \left( 1 - e^{-\lambda\,t_0} \right) - k\,c^L (\lambda ) =0 . \]
Mathematica confirms:
LaplaceTransform[d[t, t0], t, s]
(24 (1 - E^(-s t0)) UnitStep[t0])/(s t0)
Solving for cL, we obtain
\[ c^L (\lambda ) = \frac{1}{\lambda +k} \,\frac{24}{\lambda\,t_0} \left( 1 - e^{-\lambda\,t_0} \right) . \]
The final step is to apply the inverse Laplace transform to the above function. It is convenient to introduce an auxiliary function as the inverse Laplace transform
\[ g(t) = {\cal L}^{-1} \left[ \frac{1}{\lambda (\lambda +k)} \right] = \left( \frac{1}{k} - \frac{1}{k}\, e^{-kt} \right) H(t) \]
because
InverseLaplaceTransform[1/s/(s + k), s, t]
1/k - E^(-k t)/k
As usual, H(t) denotes the Heaviside function
\[ H(t) = \begin{cases} 1, & \ t > 0 , \\ 1/2 , & \ t=0, \\ 0, & \ t < 0 . \end{cases} \]
With this in hand, we find the solution of the given initial value problem explicitly:
\[ c(t) = \frac{24}{t_0} \left[ g(t) - g(t- t_0 ) \right] . \]
To plot solutions c(t) for different values of parameters k and t0,we use DSolve and and build a table of functions c(t), with 4 hours difference in drug administration, first with k = 0.02 and then k = 0.06.
sol[k_, t0_] := DSolve[{c'[t] == d[t, t0] - k*c[t], c[0] == 0}, c[t], t][[1, 1, 2]]
plotdrug02 = Map[sol[0.02, #] &, {4, 8, 12, 16, 20}]
Plot[plotdrug02, {t, 0, 35}]
plotdrug06 = Map[sol[0.06, #] &, {4, 8, 12, 16, 20}]
Plot[plotdrug06, {t, 0, 35}]

Clear[x,y,t,z];
soln = DSolve[{x''[t] + 25 x[t] == 0, x[0] == 1, x'[0] == 0}, x[t], t]
Plot[x[t] /. soln, {t, -1, 2.5}]
s[t_] = x[t]/.soln[[1]]
Plot[s[t],{t,0,3},AxesLabel->{"t","Displacement"}]

soln = DSolve[{x''[t] + 17 x'[t] + 16 x[t] == 0, x[0] == 1, x'[0] == 5}, x[t], t];
s1[t_] = x[t] /. soln[[1]]
Out[32]= 1/5 E^(-16 t) (-2 + 7 E^(15 t))
Plot[s1[t], {t, 0, 3}, AxesLabel -> {"t", "Displacement"}]

>>>> displace1.jpg

When does the maximum excursion occur?

FindRoot[s1'[t] == 0, {t, 0}]
Out[9]= {t -> 0.101322}

What is the maximum excursion?
s1[t /. %]
Out[10]= 1.18603

soln = DSolve[{x''[t] + 2 x'[t] + 37 x[t] == 0, x[0] == -1, x'[0] == 5}, x[t], t]
Out[2]= -(1/3) E^-t (3 Cos[6 t] - 2 Sin[6 t])}}
s3[t_] = x[t] /. soln[[1]]


Clear[x,t];
soln = DSolve[{x''[t] + 2 x'[t] + 37 x[t] == 0, x[0] == -1,
x'[0] == 5}, x[t], t]
s3[t_] = x[t] /. soln[[1]]
Plot[s3[t], {t, 0, 3}, PlotRange -> {-1, 1}]

Out[19]= {{x[t] -> -(1/3) E^-t (3 Cos[6 t] - 2 Sin[6 t])}}
Out[20]= -(1/3) E^-t (3 Cos[6 t] - 2 Sin[6 t])
Out[21]=

>>>>> solution3.jpg
amplitude =
s3[t] /. c3_ Exp[c4_] (c1_ Cos[a_] + c2_ Sin[a_]) -> c3*Sqrt[c1^2 + c2^2]
Out[19]= -(Sqrt[13]/3)
Plot[{amplitude*Exp[-t], -amplitude*Exp[-t], s3[t]}, {t, 0, 3}, PlotStyle -> {Blue, Blue, Black}]

>>>>> amplitude.jpg

Forced Oscillations

 

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