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. $$y^{\prime}=t y+1, \quad y(0)=0$$

Short answer

Answer: The method of successive approximations is an iterative method used to solve first-order initial value problems by refining an initial guess using the integral equation \(\phi_{n+1}(t) = y_0 + \int_{t_0}^t f(\tau, \phi_n(\tau)) d\tau\). For the given initial value problem \(y^\prime = ty + 1\) and \(y(0) = 0\), the approximations can be obtained by iteratively integrating the expressions \(\phi_n(t)\) for \(n = 1, 2, \ldots\). The first few iterations are \(\phi_1(t) = t\), \(\phi_2(t) = \frac{t^3}{3} + t\), \(\phi_3(t) = \frac{t^5}{20} + \frac{t^3}{3} + t\), and \(\phi_4(t) = \frac{t^7}{504} + \frac{t^5}{20} + \frac{t^3}{3} + t\). Plotting these iterations can help visually analyze the convergence of the solution, although other methods may be required to rigorously prove convergence.

Step-by-step solution

Understanding the method of successive approximations

The method of successive approximations is an iterative method used to solve first-order initial value problems of the form: $$ y^\prime = f(t, y), \quad y(t_0) = y_0 $$It involves finding an approximate solution by making an initial guess and then refining the guess by repeatedly applying the integral equation: $$ \phi_{n+1}(t) = y_0 + \int_{t_0}^t f(\tau, \phi_n(\tau)) d\tau $$In our case, we have the initial value problem: $$ y^\prime = ty + 1, \quad y(0) = 0 $$So, \(f(t, y) = ty + 1\), \(t_0 = 0\), and \(y_0 = 0\).

Find the general formula for \(\phi_n(t)\)

First, we define \(\phi_0(t) = 0\). Now, we want to find an expression for \(\phi_n(t)\) for any positive integer n: Note that: $$ \phi_1(t) = \int_0^t f(\tau, \phi_0(\tau)) d\tau = \int_0^t (\tau \cdot 0 + 1) d\tau $$Now, for \(n = 1, 2, \ldots,\) let's find \(\phi_n(t)\) in terms of \(\phi_{n-1}(t):\) $$ \phi_{n+1}(t) = \int_0^t f(\tau, \phi_n(\tau)) d\tau = \int_0^t (\tau \phi_n(\tau) + 1) d\tau $$

Plot \(\phi_n(t)\) for \(n = 1, 2, 3, 4\)

To determine if the successive approximations converge, we will plot the first few iterations \(\phi_n(t)\) for \(n = 1, \ldots, 4\): To find the functions, we'll now integrate the previous expressions: $$ \phi_1(t) = \int_0^t (1) d\tau = t $$ $$ \phi_2(t) = \int_0^t (\tau \phi_1(\tau) + 1) d\tau = \int_0^t (\tau^2 + 1) d\tau = \frac{t^3}{3} + t $$ $$ \phi_3(t) = \int_0^t (\tau \phi_2(\tau) + 1) d\tau = \int_0^t (\tau (\frac{\tau^3}{3} + \tau) + 1) d\tau = \frac{t^5}{20} + \frac{t^3}{3} + t $$ $$$$ \phi_4(t) = \int_0^t (\tau \phi_3(\tau) + 1) d\tau = \int_0^t (\tau (\frac{\tau^5}{20} + \frac{\tau^3}{3} + \tau) + 1) d\tau = \frac{t^7}{504} + \frac{t^5}{20} + \frac{t^3}{3} + t $$Now you can plot these functions using any graphing tool, such as Desmos or Wolfram Alpha, to observe if the iterates are converging. Keep in mind that this is just for visual analysis and to properly prove convergence or divergence, you may need to use other methods such as fixed-point theorems.

Key concepts

Initial Value Problem

An initial value problem (IVP) is a fundamental concept in differential equations, where the solution to a differential equation is determined by the value of the unknown function at a specific point. In essence, an initial value problem requires us to solve a differential equation subject to a given initial condition, usually denoted as \(y(t_0) = y_0\).

In the iterative problem-solving process, the initial guess often begins with a simple function that satisfies the given initial condition. For our example, the initial guess is \(\phi_0(t) = 0\), which respects the condition \(y(0) = 0\). The subsequent approximations \(\phi_n(t)\) further refine this guess, aiming to make the values of \(\phi_n(t)\) better match the true solution as \(n\) increases.

Differential Equations

Differential equations are mathematical tools that describe relationships involving rates of change. Specifically, they represent how a particular quantity changes over time or another variable. The equation \(y' = ty + 1\) from the exercise is a first-order differential equation because it involves the first derivative of \(y\) with respect to \(t\). Solving differential equations requires finding a function \(y(t)\) that satisfies the equation for all values of \(t\).

The presence of the \(t\) variable as a coefficient in this equation implies that the rate of change of \(y\) depends on both \(t\) and \(y\) itself, which makes the equation nonlinear and thus more complex to solve analytically.

Iterative Methods

Iterative methods are techniques that build successive approximations to a desired solution. These methods are particularly useful when an exact solution to an equation is difficult to find. Instead of arriving at the solution in one step, an iterative method begins with an initial estimate and repeatedly applies calculations that bring the estimate closer to the actual solution. Iterative methods are crucial for handling complex equations that defy simple analytical solutions.

In our example, the method of successive approximations is used. It's an iterative approach where each approximation \(\phi_{n+1}(t)\) is calculated based on the previous one \(\phi_n(t)\) using the integral equation provided. This process builds a sequence of functions \(\phi_n(t)\), with each function intended to be a better estimate of the actual solution than the last.

Integral Equation

The integral equation plays a central role in the method of successive approximations for solving differential equations. It is derived from the original differential equation and integrates it with respect to time or another independent variable. In our case, the integral equation formula is \(\phi_{n+1}(t) = y_0 + \int_{t_0}^t f(\tau, \phi_n(\tau)) d\tau\), where \(\int\) denotes the integral sign, \(t_0\) is the initial time, \(y_0\) is the initial value, and \(f(\tau, \phi_n(\tau))\) is the right-hand side of the original differential equation evaluated using the nth approximation.

The integral equation transforms the problem from finding the derivative that satisfies the differential equation to finding a function that satisfies an equation involving integrals. This is particularly helpful as integration can often be handled more effectively than differentiation, especially in an iterative context.