Question

Consider the initial value problem $$ y^{\prime}=\cos 5 \pi t, \quad y(0)=1 $$ (a) Determine the solution \(y=\phi(t)\) and draw a graph of \(y=\phi(t)\) for \(0 \leq t \leq 1 .\) (b) Determine approximate values of \(\phi(t)\) at \(t=0.2,0.4,\) and 0.6 using the Euler method with \(h=0.2 .\) Draw a broken-line graph for the approximate solution and compare it with the graph of the exact solution. (c) Repeat the computation of part (b) for \(0 \leq t \leq 0.4,\) but take \(h=0.1 .\) (d) Show by computing the local truncation error that neither of these step sizes is sufftciently small. Determine a value of \(h\) to ensure that the local truncation error is less than 0.05 throughout the interval \(0 \leq t \leq 1 .\) That such a small value of \(h\) is required results from the fact max \(\left|\phi^{\prime \prime}(t)\right|\) is large.

Short answer

Answer: The exact solution for the given initial value problem is y(t) = φ(t) = (1 / (5π)) * sin(5πt) + 1. To ensure the local truncation error is less than 0.05, a step size of h = 0.0212 should be used.

Step-by-step solution

Determine the solution y = φ(t)

We are given the initial value problem $$ y^{\prime} = \cos(5 \pi t), \quad y(0) = 1 $$ To find the solution, we integrate both sides with respect to t: $$ \int y^{\prime} dt = \int \cos(5 \pi t) dt $$ $$ y(t) = \frac{1}{5\pi}\sin(5\pi t) + C $$ Now we use the initial condition y(0) = 1 to find C: $$ 1 = \frac{1}{5\pi}\sin(5\pi \cdot 0) + C \Rightarrow C=1 $$ So the solution of the given initial value problem is: $$ y(t) = \phi(t) = \frac{1}{5\pi}\sin(5\pi t) + 1 $$

Perform Euler's method with h = 0.2

We need to approximate the solution at t=0.2, 0.4, and 0.6 using Euler's method with the step size h = 0.2. The recursive formula for Euler's method is: $$ y_{n+1} = y_n + hf(t_n, y_n) $$ Here, f(t, y) = y'(t) = cos(5πt). So, we start with t0 = 0 and y0 = 1. For t1 = 0.2, $$ y_1 = y_0 + h \cos(5 \pi t_0) = 1 + 0.2 \cos(5 \pi \cdot 0) = 1 + 0.2 = 1.2 $$ For t2 = 0.4, $$ y_2 = y_1 + h \cos(5 \pi t_1) = 1.2 + 0.2 \cos(5 \pi \cdot 0.2) \approx 1.2 - 0.1936 \approx 1.0064 $$ For t3 = 0.6, $$ y_3 = y_2 + h \cos(5 \pi t_2) = 1.0064 + 0.2 \cos(5 \pi \cdot 0.4) \approx 1.0064 - 0.1936 \approx 0.8128 $$ Now, plot the exact solution graph and compare it with the broken-line graph obtained from Euler's method with h = 0.2.

Perform Euler's method with h = 0.1

Similar to the previous step, now we use a smaller step size, h = 0.1. We will perform the calculations for 0 <= t <= 0.4. For t1 = 0.1, $$ y_1 = y_0 + h \cos(5 \pi t_0) = 1 + 0.1 \cos(5 \pi \cdot 0) = 1 + 0.1 = 1.1 $$ For t2 = 0.2, $$ y_2 = y_1 + h \cos(5 \pi t_1) = 1.1 + 0.1 \cos(5 \pi \cdot 0.1) \approx 1.1 - 0.0968 \approx 1.0032 $$ For t3 = 0.3, $$ y_3 = y_2 + h \cos(5 \pi t_2) = 1.0032 + 0.1 \cos(5 \pi \cdot 0.2) \approx 1.0032 + 0.0968 \approx 1.1 $$ For t4 = 0.4, $$ y_4 = y_3 + h \cos(5 \pi t_3) = 1.1 + 0.1 \cos(5 \pi \cdot 0.3) \approx 1.1 + 0.0968 \approx 1.1968 $$ Again, plot the exact solution graph and compare it with the broken-line graph obtained from Euler's method with h = 0.1.

Compute the Local Truncation Error

Local truncation error is an estimation of the error made in one iteration of Euler's method. The local truncation error is given by: $$ e_{n+1} = \frac{h^2}{2} \left|\phi^{\prime \prime}(t_n)\right| $$ We know φ(t) = (1 / (5π)) * sin(5πt) + 1. Now, let's find the second derivative: $$ \phi^{\prime}(t) = \frac{d}{dt}\left(\frac{1}{5\pi}\sin(5\pi t) + 1\right) = \cos(5\pi t) $$ $$ \phi^{\prime \prime}(t) = \frac{d^2}{dt^2}\left(\cos(5\pi t)\right) = -25\pi^2 \sin(5\pi t) $$ The maximum value of the term |sin(5πt)| is 1. Thus, to ensure the truncation error is less than 0.05, we find a value for h: $$ \frac{h^2}{2} \left|-25\pi^2 \sin(5\pi t)\right| < 0.05 $$ Simplifying it, we have: $$ h < \sqrt{\frac{10 \times 0.05}{25\pi^2}} \approx 0.0212 $$ So, using a step size h = 0.0212 would ensure a local truncation error less than 0.05 throughout the interval 0 <= t <= 1.

Key concepts

Local Truncation Error

When solving differential equations numerically, the local truncation error (LTE) is a vital concept. It refers to the error introduced when using a numerical method, such as Euler's method, to approximate solutions over a single step.
This error arises because the numerical step doesn't perfectly capture the behavior of the continuous solution. Essentially, the LTE measures how far the numerical solution moves away from the exact continuous solution after one step.

The formula for local truncation error in Euler's method is:
\[ e_{n+1} = \frac{h^2}{2} \left| \phi^{\prime \prime}(t_n) \right| \]
Here, \(h\) is the step size, and \(\phi^{\prime \prime}(t)\) is the second derivative of the function. The second derivative indicates how curved the function is, and a larger curvature means a potentially larger error.

• Smaller \(h\), or step size, generally results in a smaller LTE.
• Larger values of \(\left| \phi^{\prime \prime}(t) \right|\) suggest more curvature, thus potentially larger LTE.

Ensuring that LTE remains negligible involves choosing a sufficiently small step size, balancing between precision and computational expense.

Initial Value Problem

An Initial Value Problem (IVP) in the context of differential equations involves finding a function based on its derivative and an initial condition.
In mathematical terms, you typically have a differential equation like:
\[ y^{\prime} = f(t, y)\]
accompanied by an initial condition such as \( y(t_0) = y_0 \).
This setup seeks a solution function \( y(t) \) that not only satisfies the differential equation but also passes through the initial condition.

For example, in the given exercise, our IVP is:
\[ y^{\prime} = \cos(5\pi t), \quad y(0) = 1 \]
The goal is to find \( y(t) = \phi(t) \) such that when \( t = 0 \), \( y = 1 \). By integrating the derivative, you can solve for \( y(t) \).
IVPs are crucial in modeling various real-world phenomena where you know the initial state and the dynamics at play.

Differential Equation Solution

Solving a differential equation involves finding a function that satisfies the equation. For first-order differential equations, this usually requires integrating the equation.
In the exercise example, the differential equation is:
\[ y^{\prime} = \cos(5\pi t) \]

To find \( y(t) \), you integrate as follows:
\[ \int y^{\prime} \, dt = \int \cos(5 \pi t) \, dt \]
This gives:
\[ y(t) = \frac{1}{5\pi} \sin(5\pi t) + C \]
where \(C\) is a constant determined by initial conditions.

The step of solving involves finding \( C \) such that the solution fits the initial value. Here, since \( y(0) = 1 \), you solve:
\[ y(0) = \frac{1}{5\pi} \sin(0) + C = 1 \Rightarrow C = 1 \]
Thus, the solution is:
\[ y(t) = \frac{1}{5\pi} \sin(5\pi t) + 1 \]
The solution explains how \( y(t) \) changes over time due to the influence of \( \cos(5\pi t) \), demonstrating the unique role of differential equations in depicting dynamic systems.