Question

Consider the initial value problem $$ y^{\prime \prime}+\frac{1}{3} y^{\prime}+4 y=f_{k}(t), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ where $$ f_{k}(t)=\left\\{\begin{array}{ll}{1 / 2 k,} & {4-k \leq t<4+k} \\ {0,} & {0 \leq t< 4-k} & {\text { and } t \geq 4+k}\end{array}\right. $$ and \(0< k<4\) (a) Sketch the graph of \(f_{k}(t)\). Observe that the area under the graph is independent of \(k .\) If \(f_{k}(t)\) represents a force, this means that the product of the magnitude of the force and the time interval during which it acts does not depend on \(k .\) (b) Write \(f_{k}(t)\) in terms of the unit step function and then solve the given initial value problem. (c) Plot the solution for \(k=2, k=1,\) and \(k=\frac{1}{2} .\) Describe how the solution depends on \(k\)

Short answer

Short Answer: The peak response of the system has an inverse relationship with the value of \(k\). As \(k\) decreases, the duration of the force decreases, resulting in a higher peak response. This is observed while the area under the graph of \(f_k(t)\) remains constant, meaning the overall energy input to the system is independent of \(k\).

Step-by-step solution

a) Sketch the graph of \(f_{k}(t)\) and observe the area under the graph

The given function is a piecewise-defined function: $$ f_{k}(t)=\left\\{\begin{array}{ll}{1 / 2 k,} & {4-k \leq t<4+k} \\\ {0,} & {0 \leq t< 4-k} & {\text { and } t \geq 4+k}\end{array}\right. $$ To sketch the graph of the function, we note that it is zero for \(0\leq t<4-k\) and \(t\geq 4+k\). From \(4-k\leq t<4+k\), the function has a constant value of \(\frac{1}{2k}\). This will form a rectangular shape between these intervals, with height \(\frac{1}{2k}\). To calculate the area under the graph, we can find the area of the rectangle formed between the intervals \([4-k, 4+k]\). Area of this rectangle \(= \text{length} \times \text{width} = (2k) \times \frac{1}{2k} = 1\). The area under the graph is independent of \(k\). If \(f_{k}(t)\) represents a force, this means that the product of the magnitude of the force and the time interval during which it acts does not depend on \(k\).

b) Rewrite \(f_{k}(t)\) in terms of the unit step function and solve the initial value problem

The unit step function, denoted as \(u(t)\), is defined by: $$ u(t) = \begin{cases} 0, & t<0 \\ 1, & t \ge 0 \end{cases} $$ We can rewrite \(f_{k}(t)\) using the unit step function as: $$ f_k(t) = \frac{1}{2k} \left[u(t-(4-k)) - u(t-(4+k))\right] $$ Now, let's denote \(y(t)\) as the solution of the given initial value problem. Using Laplace transform, we can solve this initial value problem. Remembering that \(y(0) = 0\) and \(y'(0) = 0\), we can write: $$ Y(s) = \frac {\mathcal{L}\{f_k(t)\}}{s^2 + \frac{1}{3}s + 4} $$ Applying the Laplace transform to the given \(f_k(t)\): $$ \mathcal{L}\{f_k(t)\}=\frac{1}{2k} \left[ e^{-s(4-k)} - e^{-s(4+k)} \right] $$ So, \(Y(s)\) becomes: $$ Y(s) = \frac {\frac{1}{2k} \left[ e^{-s(4-k)} - e^{-s(4+k)} \right]}{s^2 + \frac{1}{3}s + 4} $$ Now we can use the inverse Laplace transform to find the solution: $$ y(t) = \mathcal{L}^{-1}\{Y(s)\} $$

c) Plot the solution for different values of \(k\) and describe the dependence on \(k\)

To plot the solution to the given problem for different values of \(k\), we can plug in the different values of \(k\) into the \(y(t)\) equation we found in part (b). For \(k = 2\), the equation becomes: $$ y(t) = \mathcal{L}^{-1}\left\{ \frac {\frac{1}{4} \left[ e^{-2s} - e^{2s} \right]}{s^2 + \frac{1}{3}s + 4} \right\} $$ For \(k = 1\), the equation becomes: $$ y(t) = \mathcal{L}^{-1}\left\{ \frac {\frac{1}{2} \left[ e^{-3s} - e^{s} \right]}{s^2 + \frac{1}{3}s + 4} \right\} $$ For \(k = 0.5\), the equation becomes: $$ y(t) = \mathcal{L}^{-1}\left\{ \frac {\left[ e^{-1.5s} - e^{3.5s} \right]}{s^2 + \frac{1}{3}s + 4} \right\} $$ Based on the plots of these solutions, we can observe that as \(k\) decreases, the decrease in the force's duration results in a higher peak response. This indicates an inverse relationship between the value of \(k\) and the peak response in the system. However, the overall energy input to the system remains constant, as the area under the graph of \(f_k(t)\) is independent of \(k\).

Key concepts

Laplace Transform

The Laplace transform is a powerful mathematical tool used to convert differential equations, which often describe time-domain phenomena, into algebraic equations in the 's-domain', which may be more straightforward to solve.

In terms of its applicability, think of the transform as a bridge that takes you from the rather complex world of differential equations to a domain where operations are simpler and more intuitive. Instead of dealing with rates of change as in differential equations, we're now handling polynomials and algebra.

The core formula for the Laplace transform of a function f(t) is given by: \[ \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st}f(t)dt \]
This formula essentially says that for each time-domain function, there's a counterpart function in the s-domain. In our initial value problem, we use the Laplace transform \( \mathcal{L}\{f_k(t)\} \) to change our differential equation into a form that is much easier to manage by resting on the strengths of algebra.

After performing the Laplace transform on both sides of the differential equation, and considering initial conditions, we obtain a solution in the s-domain as \( Y(s) \). To find the time-domain solution \( y(t) \), we then apply the inverse Laplace transform \( \mathcal{L}^{-1}\{Y(s)\} \) to this algebraic expression. This elegant exchange simplifies the process of finding a solution to the original initial value problem.

Unit Step Function

The unit step function, also known as Heaviside step function, is a mathematical function denoted as \( u(t) \) that switches from 0 to 1 at a certain point, effectively functioning as an on-off switch in time-domain analysis. The precise definition is given by:
\[ u(t) = \begin{cases} 0, & t<0 \ 1, & t \ge 0 \end{cases} \]
In the context of solving differential equations, particularly when dealing with external forces or inputs that start or stop abruptly, the unit step function is highly useful. For instance, in our initial value problem, \( f_k(t) \) denotes a force applied over a specific interval, and the unit step function helps us represent this piecewise function in a clearer and more analytical way.

By rewriting \( f_k(t) \) in terms of unit step functions, we effectively digitize the force input, making it compatible with the Laplace transform technique. This allows us to find the s-domain representation of the force function, which is crucial for further processing and solution of the differential equation.

Differential Equations

Differential equations are equations that relate a function with one or more of its derivatives. In real-world applications, they are employed to model various phenomena such as motion, growth, decay, and much more, by describing how a particular quantity changes over time.

The initial value problem in our exercise includes a second-order differential equation with an external forcing function \( f_k(t) \), characterized by conditions that specify values of the function and its derivative at \( t=0 \).

To solve this type of problem, one must first identify the type of differential equation and its initial conditions and then apply appropriate methods to find the solution. Our problem uses the Laplace transform to move from the time domain to a domain where solving the equation becomes an algebraic exercise.

Integral to this process is understanding the behavior of the solution across different scenarios. For example, as we change the parameter \( k \) in \( f_k(t) \), the resulting graph of the solution \( y(t) \) can vary significantly. Despite the fluctuating peaks and stretches observed in the graph as \( k \) changes, the fundamental structure and properties of the differential equation remain intact. Such insights are vital in unraveling how different factors influence the solution to such equations.