Question

Express the solution of the given initial value problem in terms of a convolution integral. \(y^{\prime \prime}+y^{\prime}+\frac{5}{4} y=1-u_{\pi}(t) ; \quad y(0)=1, \quad y^{\prime}(0)=-1\)

Short answer

Question: Express the solution of the given initial value problem in terms of a convolution integral: \(y'' + y' + \frac{5}{4}y = \delta(t) - \delta(t-\pi)\), where \(y(0) = 1\) and \(y'(0)=-1\). Answer: The solution of the given initial value problem in terms of convolution integrals is given by: \(y(t) = \int_{0}^{t} f(\tau)g(t-\tau)d\tau - \int_{0}^{t} f(\tau)u_{\pi}(t-\tau)e^{-(t-\tau)}d\tau\), where \(f(t) = L^{-1}\left[\frac{s+1}{s^2+s+\frac{5}{4}}\right]\) and \(g(t) = L^{-1}\left[\frac{1}{s}\right]\).

Step-by-step solution

Find the Laplace Transform of the given equation

In this step, we'll find the Laplace Transform of the given differential equation. Recall that the Laplace Transform of the first and second derivatives are given by: \(L[y'] = sY(s) - y(0)\) and \(L[y''] = s^2Y(s) - sy(0) - y'(0)\) Taking the Laplace Transform of the given equation, we have: \(s^2Y(s) - sy(0) - y'(0) + sY(s) - y(0) + \frac{5}{4}Y(s) = \frac{1}{s} - \frac{1}{s}e^{-\pi s}\) Now we can plug in the initial conditions: \(s^2Y(s) - s(1) - (-1) + sY(s) - 1 + \frac{5}{4}Y(s) = \frac{1}{s} - \frac{1}{s}e^{-\pi s}\)

Solve for Y(s)

In this step, we will solve for Y(s) by combining like terms and isolating Y(s): \(Y(s)[s^2 + s + \frac{5}{4}] = 1 + s + \frac{1}{s} - \frac{1}{s}e^{-\pi s}\) Now we will divide through by the polynomial in Y(s): \(Y(s) = \frac{1+s+\frac{1}{s}-\frac{1}{s}e^{-\pi s}}{s^2+s+\frac{5}{4}}\)

Express the solution as a convolution integral

We need to split Y(s) into terms whose inverse Laplace Transforms we can easily compute. Let's rewrite Y(s) as: \(Y(s) =\frac{(s+1)(1-e^{-\pi s})}{s(s^2+s+\frac{5}{4})}\) Now let's take the inverse Laplace Transform of each term: \(y(t) = L^{-1}[Y(s)] = L^{-1}\left[\frac{s+1}{s(s^2+s+\frac{5}{4})}\right] - L^{-1}\left[\frac{(s+1)e^{-\pi s}}{s(s^2+s+\frac{5}{4})}\right]\) Now we can rewrite \(y(t)\) as a convolution integral. We get: \(y(t) = \int_{0}^{t} f(\tau)g(t-\tau)d\tau - \int_{0}^{t} f(\tau)u_{\pi}(t-\tau)e^{-(t-\tau)}d\tau\) Where \(f(t) = L^{-1}\left[\frac{s+1}{s^2+s+\frac{5}{4}}\right]\) and \(g(t) = L^{-1}\left[\frac{1}{s}\right]\). Now, the solution of the given initial value problem in terms of convolution integrals is expressed as: \(y(t) = \int_{0}^{t} f(\tau)g(t-\tau)d\tau - \int_{0}^{t} f(\tau)u_{\pi}(t-\tau)e^{-(t-\tau)}d\tau\)

Key concepts

Initial Value Problem

When grappling with differential equations, an Initial Value Problem (IVP) is one of the fundamental concepts that students encounter. It comprises a differential equation coupled with initial conditions. These conditions specify the value of the unknown function and often its derivatives at a certain point, typically when the independent variable is zero. The initial values are crucial as they allow us to find a unique solution to the differential equation, a task that could be as daunting as finding a needle in a haystack without these specific starting points.

An IVP involving second-order differential equations, as in our exercise, takes the form of \[ y'' + p(t)y' + q(t)y = g(t), \] where \(y(0)\) and \(y'(0)\) are known. Taking this conceptual understanding to our exercise, the initial values given were \(y(0) = 1\) and \(y'(0) = -1\). These values guide the solution process, which involves turning the original differential equation into its Laplace transform counterpart. By integrating the initial values into the Laplace equation, a unique solution to the IVP is pursued, which is then translated back to the time domain through the Inverse Laplace Transform.

Understanding the IVP not only aids in solving the immediate problem but also instills an appreciation for the system's behavior at the inception of the given scenario.

Convolution Integral

The Convolution Integral is a mathematical tool extensively used to merge two functions into a single function that expresses how the shape of one is 'smeared' by the other. It's a critical concept when working with Laplace Transforms because it allows us to find solutions to differential equations with non-constant, or even discontinuous, driving functions.

In our exercise, the solution to the IVP is expressed in terms of a convolution integral, reflecting the system's response to an input represented by a unit step function that shifts at time \(\pi\). Conceptually, the convolution integral is akin to translating one function along the timeline while maintaining its original orientation, and at each point, multiplying it by the value of another function reflected across the vertical axis and also translated by the same amount. The resulting products are then added together to form the final output for each point in time.

The mathematical form \( y(t) = \int_{0}^{t} f(\tau)g(t-\tau)d\tau \) shows how the two functions \( f(\tau) \) and \( g(t-\tau) \) integrate over the interval from 0 to \( t \) to produce the solution to the differential equation. This process is visualized in the solution steps where \( y(t) \) incorporates two convolutions: one with the function \( f(\tau) \) and another involving the step function that activates at time \(\pi\). By understanding how convolution works, students gain insights into the dynamics of various systems subject to different inputs over time.

Inverse Laplace Transform

The Inverse Laplace Transform is the reverse process of the Laplace Transform and is essential in reverting functions from the Laplace domain back to the time domain—essentially solving the IVP in a form that's meaningful for real-world scenarios.

In the context of our exercise, after finding the function \( Y(s) \) in the Laplace domain, the next challenge is to translate it back into the time function \( y(t) \) that we're ultimately seeking. The solution involves identifying parts of \( Y(s) \) that correspond to known Laplace Transforms and their inverses. These known inverses can often be found in tables or computed using algorithms designed for the purpose.

Interestingly, the inverse transformation process often results in the convolution of two or more time functions, indicative of the underlying processes being intertwined over time. In simple terms, this is like playing back a song in reverse to interpret the lyrics that were initially hidden within the music—we reconstruct the original sound (time function) from a modified version (Laplace domain).

In our specific problem, we decompose \( Y(s) \) into simpler fractions whose Inverse Laplace Transforms can be easily obtained, then use these pieces to construct the solution via the convolution integral. Students benefit from mastering the Inverse Laplace Transform as it enables them to convert complex frequency-domain representations into time-dependent functions that correspond to real-world phenomena.