Question

let \(\phi_{0}(t)=0\) and use the method of successive approximations to solve the given initial value problem. (a) Determine \(\phi_{n}(t)\) for an arbitrary value of \(n .\) (b) Plot \(\phi_{n}(t)\) for \(n=1, \ldots, 4\). Observe whether the iterates appear to be converging. (c) Express \(\lim _{n \rightarrow \infty} \phi_{n}(t)=\phi(t)\) in terms of elementary functions; that is, solve the given initial value problem. (d) Plot \(\left|\phi(t)-\phi_{n}(t)\right|\) for \(n=1, \ldots, 4 .\) For each of \(\phi_{1}(t), \ldots . \phi_{4}(t)\) estimate the interval in which it is a reasonably good approximation to the actual solution. $$ y^{\prime}=-y / 2+t, \quad y(0)=0 $$

Short answer

In summary, the method of successive approximations was used to solve the given initial value problem. A recursive formula for the sequence of functions \(\phi_{n}(t)\) was determined, and the actual solution was found to be \(\phi(t) = 2e^{-t/2}\). By plotting the differences between the actual solution and each of the approximate solutions, one can estimate the interval in which each approximation is reasonably good.

Step-by-step solution

The method of successive approximations involves the following recursive formula: $$ \phi_{n+1}(t) = \int_{0}^{t} \left(-\frac{1}{2}\phi_{n}(\tau) + \tau\right) d\tau + \phi_0(t), $$ where the initial function \(\phi_0(t) = 0\). Our goal is to determine \(\phi_{n}(t)\) for an arbitrary value of \(n\). #Step 2: Determine the recursive formula for \(\phi_{n}(t)\)#

We will compute the first few steps of the iteration to find a pattern. From the above formula, we have: For \(n=0\), using \(\phi_0(t) = 0\): $$ \phi_{1}(t) = \int_{0}^{t} \tau d\tau = \frac{1}{2}t^2. $$ For \(n=1\), using \(\phi_1(t) = \frac{1}{2}t^2\): $$ \phi_{2}(t) = \int_{0}^{t} \left(-\frac{1}{4}\tau^2 + \tau\right) d\tau = -\frac{1}{12}t^3 + \frac{1}{2}t^2. $$ Based on the calculations for the first two iterations, we notice a pattern that the function \(\phi_{n}(t)\) is a polynomial of degree \(n+1\). Hence, the general formula for the nth approximation is: $$ \phi_{n}(t) = \sum_{k=0}^{n} a_{k}t^{k+1}, $$ where \(a_{k}\) is the coefficient of the \(t^{k+1}\) power. #Step 3: Solve for the actual solution \(\phi(t)\) in terms of elementary functions#

As we have seen from the previous steps, each successive approximation in the sequence is a polynomial of increasing degree, and the coefficients are getting smaller. Hence, it's reasonable to expect that the limit as n approaches infinity is an exponential function. Let's assume \(\phi(t) = Ce^{kt}\) and substitute it into the original ODE: $$ ke^{kt} = -\frac{1}{2}Ce^{kt} + t, $$ where \(y(0) = 0\) implies \(C = 0\). So, we have: $$ ke^{kt} = -\frac{1}{2}(0)e^{kt} + t. $$ Comparing powers of \(t\) and solving for \(k\) and \(C\), we get \(k = -\frac{1}{2}\) and \(C = 2\). Therefore, the actual solution is: $$ \phi(t) = 2e^{-t/2}. $$ #Step 4: Plot the differences \(\left|\phi(t)-\phi_{n}(t)\right|\) for \(n=1, \ldots, 4\)#

For this step, you can use a graphing tool or software like Desmos to plot the differences between the actual solution \(\phi(t)\) and each of the approximate solutions \(\phi_{n}(t)\) for \(n=1, \ldots, 4\). Based on these plots, you can estimate the interval in which each approximation is reasonably good.

Key concepts

Initial Value Problem

In mathematics, an initial value problem is a differential equation that comes with a specified value, called the initial condition. This condition tells us the value of the function at a particular point. For example, in our problem, the initial value is given by: \( y(0) = 0 \). This means the solution to the differential equation must pass through this point. Initial value problems are like solving a maze where you know your starting point and you have to find the path that follows certain rules. Understanding this starting point is crucial for finding the right solution path using methods like successive approximations.

Polynomial Approximation

Polynomial approximation involves using polynomials to estimate a function. The successive approximations method helps us generate a sequence of polynomials that converge to the actual solution. In our step-by-step solution, each approximation \( \phi_n(t) \) is derived as a polynomial function. We notice a pattern in these approximations, where the degree of the polynomial is \( n+1 \). This method is invaluable because polynomials are often easier to work with than other functions. They provide a simplified model that closely mimics the target function within a certain range, making them a powerful tool in numerical analysis.

Convergence Analysis

Convergence analysis is the study of whether a sequence approaches a particular value as it progresses towards infinity. In the successive approximations method, our goal is for \( \phi_n(t) \) to converge towards the true solution \( \phi(t) \). We observe the polynomial terms diminishing in size as \( n \) increases. This diminishing pattern suggests that the sequence is converging. Understanding convergence ensures that our approximate solutions become more accurate. Essentially, it tells us if the series we've calculated is zeroing in on the correct answer, which in mathematical terms is known as the limit.

Ordinary Differential Equations

Ordinary differential equations (ODEs) involve functions of only one variable and their derivatives. They are fundamental in describing phenomena such as physics, engineering, and many other fields. Our original equation, \( y' = -\frac{1}{2}y + t \), is an example of a first-order ODE. These equations specify a relationship between a function and its rate of change. Solving an ODE like our example involves finding a function that satisfies the equation across its domain. Methods such as successive approximations are often employed to find solutions, especially for equations that cannot be solved analytically.