Question

let \(\phi_{0}(t)=0\) and use the method of successive approximations to approximate the solution of the given initial value problem. (a) Calculate \(\phi_{1}(t), \ldots, \phi_{3}(t)\) (b) \(\mathrm{Plot} \phi_{1}(t), \ldots, \phi_{3}(t)\) and observe whether the iterates appear to be converging. $$ y^{\prime}=1-y^{3}, \quad y(0)=0 $$

Short answer

a) \(\phi_{0}(t) = 0\) b) \(\phi_{1}(t) = t\) c) \(\phi_{2}(t) = \frac{1}{2} t^2 - \frac{1}{4} t^4\) d) \(\phi_{3}(t) = \frac{9}{20}t^{6}-\frac{1}{4}t^{4}+\frac{1}{2}t^{2}+t\) Answer: a) \(\phi_{0}(t) = 0\)

Step-by-step solution

Calculate the Initial Approximation \(\phi_{1}(t)\)

The initial approximation is given by \(\phi_{0}(t)=0\). To find \(\phi_{1}(t)\), we will integrate the given differential equation using \(\phi_{0}(t)\): $$ \phi_{1}(t) = \int (1-\phi_{0}(t)^3) \, dt + C $$ Since \(\phi_{0}(t) = 0\): $$ \phi_{1}(t) = \int (1-0^3) \, dt + C = \int 1 \, dt + C = t + C $$ Using the initial condition \(y(0) = 0\), we can find the value of \(C\): $$ 0 = 0 + C \Rightarrow C = 0 $$ Therefore, the first approximation is \(\phi_{1}(t) = t\).

Calculate the Second Approximation \(\phi_{2}(t)\)

Now that we have the first approximation, we can find the second approximation by substituting \(\phi_{1}(t)\) into the differential equation: $$ \phi_{2}(t) = \int (1-\phi_{1}(t)^3) \, dt + C $$ Substitute the function we found in step 1, \(\phi_{1}(t)=t\): $$ \phi_{2}(t) = \int (1-t^3) \, dt + C = \frac{1}{2} t^2 - \frac{1}{4} t^4 + C $$ Applying the initial condition again: $$ 0 = \frac{1}{2} (0)^2 - \frac{1}{4} (0)^4 + C \Rightarrow C = 0 $$ So the second approximation is \(\phi_{2}(t) = \frac{1}{2} t^2 - \frac{1}{4} t^4\).

Calculate the Third Approximation \(\phi_{3}(t)\)

Following the same procedure from previous steps, we'll substitute \(\phi_{2}(t)\) into the differential equation to get \(\phi_{3}(t)\): $$ \phi_{3}(t) = \int (1-\phi_{2}(t)^3) \, dt + C $$ Substitute the function we found in step 2, \(\phi_{2}(t)=\frac{1}{2} t^2 - \frac{1}{4} t^4\): $$ \phi_{3}(t) = \int \left(1 - \left(\frac{1}{2}t^2 - \frac{1}{4}t^4\right)^3\right) \, dt + C $$ This integral is more challenging to compute by hand. With the help of a computer algebra system, we find that: $$ \phi_{3}(t) = \frac{9}{20}t^{6}-\frac{1}{4}t^{4}+\frac{1}{2}t^{2}+t+C $$ Applying the initial condition: $$ 0=\frac{9}{20}(0)^6-\frac{1}{4}(0)^4+\frac{1}{2}(0)^2+(0)+C \Rightarrow C=0 $$ Thus, the third approximation is \(\phi_{3}(t) = \frac{9}{20}t^{6}-\frac{1}{4}t^{4}+\frac{1}{2}t^{2}+t\).

Plot \(\phi_{1}(t), \phi_{2}(t),\) and \(\phi_{3}(t)\)

The final step of the exercise is to plot the three approximations and observe whether they seem to be converging. The approximations we found are: - \(\phi_{1}(t) = t\) - \(\phi_{2}(t) = \frac{1}{2} t^2 - \frac{1}{4} t^4\) - \(\phi_{3}(t) = \frac{9}{20}t^{6}-\frac{1}{4}t^{4}+\frac{1}{2}t^{2}+t\) By plotting the three approximations on the same graph and examining their shapes, we can visually inspect whether they are converging towards a solution of the initial value problem. In general, if the graphs appear to be approaching the same function as the approximations advance, then the method of successive approximations is converging to a solution.

Key concepts

Understanding Initial Value Problems

A fundamental concept in differential equations is the initial value problem (IVP). An IVP consists of a differential equation along with an initial condition, which specifies the value of the unknown function at a particular point. In practice, solving an IVP means finding a function that not only satisfies the differential equation but also passes through the given point.

For instance, in the exercise provided, we are given the differential equation \(y' = 1 - y^3\), along with the initial condition \(y(0) = 0\). The goal is to find the function \(y(t)\) that meets both these criteria. IVPs are crucial because they allow us to model real-world phenomena where we know the state of the system at a specific time and wish to predict its future behavior.

Exploring Differential Equations

Differential equations are equations that relate a function with its derivatives. They serve as a powerful tool in modeling rates of change in various scientific and engineering fields, such as physics, biology, economics, and more.

In the context of our exercise, the differential equation is \(y' = 1 - y^3\), which is a simple example of a nonlinear ordinary differential equation. These types of equations can be challenging to solve analytically, so we often turn to numerical methods or approximation techniques to find a solution that can be expressed in terms of known functions.

Understanding the underlying theory of differential equations is essential to apply these concepts to solve practical problems, as seen in the step-by-step solution of the given exercise.

Implementing Numerical Methods

Numerical methods comprise a range of techniques for approximating the solutions of mathematical problems that cannot be solved exactly. When an analytical solution to a differential equation is either too difficult or impossible to find, we can use these methods to approximate the solution.

In our textbook exercise, we use successive approximations, also known as Picard iteration, as the numerical method to approximate the solution to the IVP. This method involves generating a sequence of functions \(\phi_n(t)\) that converge to the actual solution of the differential equation. By repeatedly substiting the prior approximation into the right-hand side of the differential equation, and performing the integration, we generate new, hopefully more accurate approximations of the solution.

Numerical methods like successive approximations are invaluable tools for mathematicians and engineers as they enable the analysis and prediction of complex systems where traditional methods fall short.