Question

Solve the given initial value problem, in which inputs of large amplitude and short duration have been idealized as delta functions. Graph the solution that you obtain on the indicated interval. \(y^{\prime \prime}=\delta(t-1)-\delta(t-3), \quad y(0)=0, \quad y^{\prime}(0)=0, \quad 0 \leq t \leq 6\)

Short answer

Question: Determine the function y(t) and graph its solution for the given initial value problem: \(y^{\prime\prime} = \delta(t-1) - \delta(t-3), y(0) = 0, y^{\prime}(0) = 0\), where \(0 \leq t \leq 6\). Answer: The function y(t) is given by \(y(t) = tH(t-1) - (t-2)H(t-3)\), where H(t) is the Heaviside step function. The graph of the solution consists of three segments: 1. For \(0 \leq t < 1\), \(y(t) = 0\). 2. For \(1 \leq t < 3\), \(y(t) = t - 1\). 3. For \(3 \leq t \leq 6\), \(y(t) = 5 - t\).

Step-by-step solution

Understand the initial conditions

The given initial value problem is: $y^{\prime \prime}=\delta(t-1)-\delta(t-3), \quad y(0)=0, \quad y^{\prime}(0)=0, \quad 0 \leq t \leq 6$ Here, \(y^{\prime \prime}\) is the second derivative of the function y(t). The problem also specifies two initial conditions: \(y(0) = 0\) and \(y^{\prime}(0) = 0\).

Integrate once

We will now integrate the given equation with respect to t once: \(\int y^{\prime\prime} \, dt = \int (\delta(t-1) - \delta(t-3)) \, dt\) Let \(y^{\prime}(t) = f(t)\), then we have: \(f(t) = \int (\delta(t-1) - \delta(t-3)) \, dt + C_1\)

Apply the first initial condition

Using the initial condition \(y^{\prime}(0) = 0\), we can find the value of the integration constant \(C_1\): \(0 = f(0) = \int_0^0 (\delta(t-1) - \delta(t-3)) \, dt + C_1\) Since the limits of integration are both 0, the integral is equal to 0, and we get: \(C_1 = 0\) The first derivative of y(t) becomes: \(y^{\prime}(t) = f(t) =\int (\delta(t-1) - \delta(t-3)) \, dt\)

Integrate once more

Now, we will integrate the first derivative of y(t) with respect to t to obtain y(t): \(y(t) = \int f(t) \, dt = \int \left( \int (\delta(t-1) - \delta(t-3)) \, dt \right) dt + C_2\)

Apply the second initial condition

Using the initial condition \(y(0) = 0\), we can find the value of the integration constant \(C_2\): \(0 = y(0) = \int_0^0 f(t) \, dt + C_2\) Similarly, since the limits of integration are both 0, the integral is equal to 0, and we get: \(C_2 = 0\) The function y(t) becomes: \(y(t) = \int \left( \int (\delta(t-1) - \delta(t-3)) \, dt \right) dt\)

Simplify y(t)

Now we can write y(t) by integrating each delta function separately. \(y(t) = \int_0^t\int_0^s \delta(u-1) \, du \, ds - \int_0^t\int_0^s \delta(u-3) \, du \, ds\) Using the property of delta functions \(\int_0^t \delta(u-a) \, du = H(t-a)\), where H(t-a) is the Heaviside step function: \(y(t) = \int_0^t H(s-1) \, ds - \int_0^t H(s-3) \, ds\) Now, by integrating the Heaviside step function, we have: \(y(t)= tH(t-1) - (t-2) H(t-3)\)

Graph the solution

To graph the solution, we will plot the function y(t) over the interval \(0 \leq t \leq 6\). The graph will consist of three segments: 1. For \(0 \leq t < 1\), \(y(t) = 0\). 2. For \(1 \leq t < 3\), \(y(t) = t - 1\). 3. For \(3 \leq t \leq 6\), \(y(t) = 5 - t\). The graph of the function will have a rising slope from \(t = 1\) to \(t = 3\), and a decreasing slope from \(t = 3\) to \(t = 6\).

Key concepts

Delta Function

The delta function, also known as the Dirac delta function, is a mathematical construct commonly used in physics and engineering to model a very short pulse or spike of infinite amplitude that occurs at a single point in time. It has properties of a limit of a series of spike functions: each spike function is zero everywhere except at the spike where it is very large, and the area under each spike is 1. In practice, the delta function isn't a function in the traditional sense, since it doesn't assign a unique output for each input in its domain. Instead, it is a distribution with the sifting property: for a continuous function f(t), the integral \(\int_{-\infty}^{\infty} f(t)\delta(t-a) \, dt\) equals f(a). This property can transform a complex problem into a much simpler one, especially when dealing with impact forces or other instantaneous changes in a system.

Heaviside Step Function

The Heaviside step function, denoted as H(t), is a discontinuous function that jumps in value from 0 to 1 at a certain point, usually denoted as the origin. It can be thought of as a mathematical switch that turns on at a specific time. In mathematical terms, it's defined as follows:
\( H(t) = \begin{cases} 0, & \text{if } t < 0 \ 1, & \text{if } t \geq 0 \end{cases} \)
This function plays a critical role in describing systems that undergo a sudden transition at a specific moment in time. For instance, in the context of the initial value problem presented, the Heaviside step function is used to integrate the delta function, as the delta function's integral from 0 to t gives the step function H(t-a), where t is the variable and a is the shift of the step.

Second Derivative

The second derivative of a function is the derivative of the derivative, denoted as \(y''\) or \(f''(t)\). It represents the rate of change of the rate of change of the function, shedding light on the function's concavity and points of inflection. When working with the second derivative in physics, it often corresponds to acceleration when the function itself represents a position with respect to time. In the given exercise, the second derivative is set equal to the delta functions, signifying that at the points t=1 and t=3, there is a sudden change in the 'acceleration', which translates to an instantaneous impulse applied to the system at these moments. The analysis of the second derivative is critical in predicting the behavior of dynamic systems and understanding their response to external stimuli.

Graphing Solutions

Graphing solutions of differential equations can be a powerful tool for visualizing and understanding the behavior of dynamical systems. The graph translates complex numerical information into a visual format, which can reveal important features such as trends, cyclical patterns, and equilibrium points. For solutions to second-order differential equations like the one in the step-by-step example, the graph typically shows how the position (or the variable of interest) changes over time. To effectively graph the solution, one needs to understand the function's distinct behaviors in different intervals, the impact of initial conditions, and the specific points where the behavior changes, as in the jumps caused by the delta functions in the given exercise. This visual representation aids in grasping the fundamental behavior of the system and predicting its future state based on its past behavior.