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Return to Part VI of the course APMA0330
This section provides a stream of examples demonstrating applications of Laplace transformation for solving initial value problems that appear in applications.
Example 1:
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
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/t_{0} for exactly t_{0} hours and then stopped so that
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 c^{L}, the Laplace transform of unknown drug concentration:
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
To plot solutions c(t) for different values of parameters k and t_{0},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.
The Laplace transformation can be successfully applied to solve an integral equation of a convolution type (also known as the Volterra integral equation):
where the function f(t) and the kernel function K(Θ) are known, while y(t) is unknown. Since the integral in \eqref{EqApp.1} is the convolution of the kernel function K(Θ) and the unknown function y(t), application of the Laplace transform yields
Example 2:
The problem of finding the curve down which a bead placed anywhere will fall to the bottom in the same amount of time is called the tautochrone problem. It was first solved by Christiaan Huygens in 1659, who proved that such curve must be a cycloid.
The name of such curve stems from
the Greek tauto, meaning ``same,'' and chronos (χρονος), meaning
``time.'' Following Niels Abel, we formulate the problem in
integral form.
Let (x,y) be the position of the bead at some intermediate time
before it reaches the origin. Let s denote the arc length along
the curve measured from the origin to (x,y). According to the
principle of conservation of energy, the gain in kinetic energy of
the bead must equal its loss of potential energy in falling from
some fixed point (x_{0},y_{0}) to (x,y). Therefore, we get
\[
\frac{1}{2}\,mv^2 = mg (y_0 -y).
\]
Since v = ds/dt is the velocity, we have
\[
\frac{{\text d}s}{{\text d}t} = -\sqrt{2g} \,\sqrt{y_0 - y} , \qquad\mbox{or}\qquad
\frac{{\text d}s}{\sqrt{y_0 - y}} = - \sqrt{2g} \,{\text d}t ,
\]
where the minus sign indicates that s is decreasing with increasing
t. Assuming that the arc length s is some yet unknown function
of y (namely, s = f(y)), we derive the differential equation
\[
\frac{f' (y) \,{\text d}y}{\sqrt{y_0 - y}} = - \sqrt{2g} \,{\text d}t .
\]
After integration over the (constant) duration T_{0} of descent, we obtain
\[
\int_0^{y_0} \frac{f' (y) \,{\text d}y}{\sqrt{y_0 - y}} = -\sqrt{2g}
\,\int_{T_0}^0 \,{\text d}t \qquad\mbox{or} \qquad T_0 = \frac{1}{\sqrt{2g}}
\int_0^{y_0} \frac{f' (y) \,{\text d}y}{\sqrt{y_0 - y}} .
\]
We can rewrite the latter equation as a convolution of the functions
y^{-1/2} and f'(y):
\[
T_0 = \frac{1}{\sqrt{2g}} \, y^{-1/2} \ast f' .
\]
With the aid of the Laplace transform, we easily solve the convolution
equation to obtain
\[
{\cal L} [T_0 ] = \frac{1}{\sqrt{2g}} \,{\cal L} \left[ y^{-1/2}
\right] \cdot {\cal L} [f' ] \qquad\Longrightarrow\qquad
{\cal L} [f' ] = \sqrt{\frac{2g}{\pi}} \,T_0 \,\lambda^{-1/2} ,
\]
because \( {\cal L} [y^{-1/2} ] = \sqrt{\pi / \lambda} \) and \( {\cal L}
[T_0 ] = T_0 /\lambda \) (see formula in Table). Note
that the Laplace transformation is taken with respect to the
variable y. Thus,
\[
f' (y) = \frac{{\text d}s}{{\text d}y} = \sqrt{1 + \left( \frac{{\text d}x}{{\text d}y} \right)^2}
= \frac{\sqrt{2g}}{\pi} \,T_0 \,y^{-1/2} .
\]
Setting \( k=(2gT_0^2 )/\pi^2 , \) we obtain the equation:
\[ \frac{{\text d}x}{{\text d}y} = \sqrt{\frac{k-y}{y}} .
\]
After the substitution
\[
y = k\,\sin^2 \frac{\theta}{2} \qquad \mbox{and} \qquad {\text d}x =
k\,\cos^2 \frac{\theta}{2}\,{\text d}\theta ,
\]
it becomes
\[
x= k\int \cos^2 \frac{\theta}{2}\,{\text d}\theta = \frac{k}{2} \int (1+
\cos\theta )\,{\text d}\theta = \frac{k}{2}\, (\theta + \sin\theta ) +C .
\]
We set the constant of integration C to zero because the curve must
pass through the origin, and y=0 when θ=0. Finally,
\[
x= \frac{k}{2}\, (\theta + \sin\theta ) , \qquad y =\frac{k}{2}\,
(1-\cos\theta ) .
\]
This curve is known as the cycloid.
■
Example 3:
Let us consider a boundary value problem
Since we don't know the value of the derivative of the unknown function y(x) at the leftend x = 0, we denote it by \( y' (0) = c . \) Then the Laplace transform of the required function y(x) is
Example 4:
A uniformly loaded beam of length ℓ is supported at both ends. The deflection y(x) is a function of horizontal position x and obeys the differential equation
where E is Young’s modulus, I is the moment of inertia and q(x) is the load per unit length at point x. We assume in this problem that q(x) = q (a constant). The boundary conditions are (i) no deflection at x = 0 and x = ℓ; (ii) no curvature of the beam at x = 0 and x = ℓ.
In addition to being subject to a uniformly distributed load, a beam is supported so that there is no deflection and no curvature of the beam at its ends. Since the curvature of the function y(x) is
Applying a Laplace transform to the differential equation (4.1), we find that the Laplace transform of the deflection, y^{L}, satisfies the algebraic equation
The derivatives evaluated at the origin are unknown constants, but they can be determined by using the remaining two boundary conditions (4.2) at x = ℓ. Hence, we get two equations for their determination:
Example 5:
The principle of financial Ponzi scheme is to attract potential inverters promissing high rate of return, which is impossible to deliver. The only way this financial piramid can work is when new investement will be sufficient to pay the promised interest. This financial scheme was carried out from 1869 to 1872 by Adele Spitzeder in Germany and by Sarah Howe in the United States in the 1880s through the "Ladies' Deposit".
In the 1920s, Charles Ponzi (1882--1949) carried out this piramid and became well known throughout the United States because of the huge amount of money that he took in. His original scheme was based on the legitimate arbitrage of international reply coupons for postage stamps, but he soon began diverting new investors' money to make payments to earlier investors and to himself.
Recent example of this financial piramod provides
the Chara bank operated more than a year in Moscow (Russia). However, the largest Ponzi scheme operated in the United States, where its owner, Bernard Madoff, was arrested in 2008 for allegedly running a multigillion-dollar that caused a lost of $50 billion dollars in investments. Mardoff's fund offered 10% return for investors, while Chara bank paid about 20%.
We present a mathematical model following
Artzrouni, M., The mathematics of Ponzi schemes, Mathematical Social Sciences, 2009, pp. 190--201.
Assume that the fund starts at t = 0 with an initial deposite K ≥ 0 followed bya continuous cash inflow s(t). Let us denote by r_{p} a promised rate of return and by r_{n} a nominal interest rate corresponding to the market value. We consider the case when r_{p} > r_{n}, thw fund oromised moew than it can deliver. The promised rate r_{p} may be called the "Ponzi rate" and is equal to 0.10 in Madoff's case when ewalistic rate was 0.02.
To attract investors, all Ponzi scheme followers offered customers a possibility to withdraw the interest. For simplicity, we assume that the withdrawal rate r_{w} is constant. Then initial investors expect to gain (according to continuous compound interest) the amount
\( \displaystyle K\, e^{t \left( r_p - r_w \right)} \) in accumilated capital at time t.
Withdrawals by those who added to the Ponzi fund between times 0 and t are obtained by calculating withdrawals for those who invested at time u and then summing for u between 0 and t. The quantity s( u) is invested at time u with an expected growth rate
r_{p} − r_{w} for the duration t − u. The expected accumilated value at time t is then
\( \displaystyle s(u)\, e^{\left( r_p - r_w \right) (t-u)} {\text d} u . \) The density of withdrawals at time t is then
\( \displaystyle r_u s(u)\, e^{\left( r_p - r_w \right) (t-u)} \) with a sum
\( \displaystyle r_u \int_0^t s(u)\,e^{\left( r_p - r_w \right) (t-u)} {\text d} u \) of withdrawals W(t) is then the sum of withdrawals by initial investors and by those who invested between times 0 and t:
If S(t) is the ampount in the fund at time t, then S(t+dt) is obtained by adding to S(t) the nominal interest r_{n}S(t), the inflow of fresh money s(t)dt and subtracting the withdrawals W(t)dt:
where C is the initial value, which may or may not be equal to K, the initial deposit made by customers.
A simple model for inflow investment is the Malthus law:
\[
s(t) = s_0 e^{r_i t} ,
\]
wheres_{0} is the initial density of the deposits and r_{i} is the investment (malthus) rate. The integral in Eq.(5.1) is the convolution of two functions:
is the Laplace transform of s(t). As usual, we denote by S^{L} the Laplace transform of the amount of the fund, we obtain from the initial value problem (5.2) a linear equation:
Using the relations of the inverse Laplace transform
\begin{align*}
{\cal L}^{-1} \left[ \frac{1}{\lambda -a} \right] &= e^{at} H(t) ,
\\
{\cal L}^{-1} \left[ \frac{1}{\left( \lambda -a \right) \left( \lambda - b \right)} \right] &= \frac{1}{a-b} \left[ e^{at} - e^{bt} \right] H(t) ,
\\
{\cal L}^{-1} \left[ \frac{1}{\left( \lambda -a \right) \left( \lambda - b \right)\left( \lambda - c \right)} \right] &= \frac{e^{at}}{\left( a - b \right)\left( a - c \right)} - \frac{e^{bt}}{\left( a - b \right)\left( b - c \right)} + \frac{e^{ct}}{\left( a - c \right)\left( b - c \right)} ,
\end{align*}
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