Question

Find the solution of the given initial value problem and draw its graph. \(y^{\prime \prime}+4 y=\delta(t-\pi)-\delta(t-2 \pi) ; \quad y(0)=0, \quad y^{\prime}(0)=0\)

Short answer

For the given initial value problem involving a second-order linear differential equation with delta function forcing term, we found the Laplace transform of both sides of the equation, solved for the Laplace transform of the solution, Y(s), and took the inverse Laplace transform of Y(s) to obtain the solution, y(t). The obtained solution is a piecewise function: $ y(t)= \begin{cases} 0 & 0 \le t < \pi \\ \frac{1}{4}(\cos(2\pi) - \cos(2t)) & \pi \le t < 2\pi \\ 0 & 2\pi \le t \end{cases} $ The graph of the solution will have two horizontal line segments along the t-axis, connected by an oscillating function between 0 and 0.5.

Step-by-step solution

Taking Laplace Transform

Take the Laplace transform of both sides of the given equation: \(\mathcal{L}\{y''(t)+4y(t)=\delta(t-\pi)-\delta(t-2\pi)\}\) Using linearity of Laplace transform and properties of delta function, we have: \(s^2Y(s)-sy(0)-y'(0)+4Y(s)=e^{-\pi s}-e^{-2\pi s}\) Since \(y(0)=0\) and \(y'(0)=0\), the equation simplifies to: \((s^2+4)Y(s)=e^{-\pi s}-e^{-2\pi s}\).

Solving for Y(s)

Now, solve for Y(s) by dividing both sides by \((s^2+4)\): \(Y(s)=\frac{e^{-\pi s}-e^{-2\pi s}}{s^2+4}\)

Taking Inverse Laplace Transform

Take the inverse Laplace transform of Y(s) to get the solution y(t): \(y(t)=\mathcal{L}^{-1}\{Y(s)\}=\mathcal{L}^{-1}\{\frac{e^{-\pi s}-e^{-2\pi s}}{s^2+4}\}\) By using convolution theorem and inverse Laplace transform properties, we have: \(y(t)=\frac{1}{2}\sin(2t)*[u(t-\pi)-u(t-2\pi)]\) where \(u(t)\) is the unit step function. Now, by evaluating the convolution integral: \(y(t)=\frac{1}{2}\int_0^t \sin(2\tau)\big[u(t-\tau-\pi)-u(t-\tau-2\pi)\big]d\tau\) The solution y(t) can be described as: $ y(t)= \begin{cases} 0 & 0 \le t < \pi \\ \frac{1}{2}\int_{\pi}^t \sin(2\tau)d\tau & \pi \le t < 2\pi \\ \frac{1}{2}\int_{\pi}^{2\pi} \sin(2\tau)d\tau & 2\pi \le t \end{cases} $ We have: \(\frac{1}{2}\int_{\pi}^t \sin(2\tau)d\tau = \frac{1}{4}(\cos(2\pi) - \cos(2t))\) \(\frac{1}{2}\int_{\pi}^{2\pi} \sin(2\tau)d\tau = \frac{1}{4}(1-\cos(2\pi))=0\) So, the solution y(t) can be written as: $ y(t)= \begin{cases} 0 & 0 \le t < \pi \\ \frac{1}{4}(\cos(2\pi) - \cos(2t)) & \pi \le t < 2\pi \\ 0 & 2\pi \le t \end{cases} $

Plotting the graph of the solution

To plot the graph of the solution, use the obtained piecewise function: • For \(0 \le t < \pi\), y(t) = 0, which is a horizontal line segment along the t-axis. • For \(\pi \le t < 2\pi\), y(t) = \(\frac{1}{4}(\cos(2\pi) - \cos(2t))\), which is a function oscillating between 0 and 0.5. • For \(2\pi \le t\), y(t) = 0, which is another horizontal line segment along the t-axis starting from \(t=2\pi\). The graph will show two horizontal line segments along the t-axis with an oscillating function between them.

Key concepts

Inverse Laplace Transform

Understanding the Inverse Laplace Transform is crucial when dealing with differential equations, particularly in solving initial value problems. The Laplace transform converts a time-domain function into a complex frequency-domain representation, simplifying the process of solving differential equations. To revert our problem to the time domain and find the solution of the original differential equation, we use the inverse Laplace transform, denoted as \(\mathcal{L}^{-1}\{Y(s)\}\), which requires a contour integral in the complex plane or the usage of standard transformation tables.

Remember, just like languages have rules for translating words from one to another, mathematics has protocols for switching between domains. Inverse Laplace is essentially your mathematical 'Google Translate,' taking you from the complex frequency realm back to the time domain we live in.

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They express how a rate of change (denoted by derivatives) in one variable is related to other variables. Solving a differential equation means finding the function that satisfies the equation. These equations are essential in modeling real-world processes in physics, engineering, biology, and finance.

Differential equations can be intimidating like a tangled set of strings, but much like untying a knot, we go step-by-step. Initial value problems are a specific type of differential equation where we know the function's value and the rate of change at time zero. The exercise given deals with a second-order differential equation (involves the second derivative of the unknown function), which often represents systems with acceleration, such as spring-mass systems or electrical circuits.

Heaviside Unit Step Function

The Heaviside unit step function is like an on-off switch in mathematics. Denoted as \(u(t)\), it's equal to 0 for negative input and 1 for positive input. In the context of differential equations, it's incredibly useful for modeling situations where a force suddenly starts or stops, which can be common in electrical engineering or control systems.

In our example, two Heaviside functions are used to represent the delays of the Dirac delta function impulses. Using Heaviside functions, we can break the problem into intervals, translating a complex scenario into more manageable pieces. By understanding when each part of the system is 'active,' we can better understand and visualize the solution to the differential equation.

Convolution Theorem

If the Heaviside function is the on-off switch, think of the Convolution theorem as the blender, mixing two functions together to see how one shapes the other over time. According to the Convolution theorem, the inverse Laplace transform of the product of two Laplace transforms is the convolution of their respective inverse transforms in the time domain. In equation form: \(\mathcal{L}^{-1}\{F(s)G(s)\} = f(t) * g(t)\),

where * denotes convolution. It's an integral that essentially smooshes two functions together, accounting for the effect one has on the other over time. This process is perfectly illustrated in the step-by-step solution of our differential equation where the sinusoidal response is convolved with the shifting unit step functions to produce the final solution.

Dirac Delta Function

The Dirac delta function, \(\delta(t)\), named after physicist Paul Dirac, is a peculiar beast. It's not a function in the traditional sense but rather a 'distribution' used in the context of integrals. You can think of it as an incredibly sharp spike that occurs at a specific point in time; it's zero everywhere else, and its integral over the whole real line is one. It's incredibly useful in physics and engineering for modeling instantaneous impulses, such as kicks, strikes, or electronic pulses.

In the problem we're examining, the delta function represents two separate impulses applied at times \(t = \pi\) and \(t = 2\pi\). When the Laplace transform is applied to the delta function, it efficiently translates these impulses into exponential terms, which are much more manageable mathematically. So when faced with an unruly delta function, remember, it's just a pinch in the mathematical timeline - momentous but brief.