Question

Estimate the local truncation error for the Euler method in terms of the solution \(y=\phi(t) .\) Obtain a bound for \(e_{n+1}\) in terms of \(t\) and \(\phi(t)\) that is valid on the interval \(0 \leq t \leq 1 .\) By using a formula for the solution obtain a more accurate error bound for \(e_{n+1} .\) For \(h=0.1\) compute a bound for \(e_{1}\) and compare it with the actual error at \(t=0.1 .\) Also compute a bound for the error \(e_{4}\) in the fourth step. $$ y^{\prime}=2 y-1, \quad y(0)=1 $$

Short answer

The Euler method provides approximations for the solution of first-order differential equations. Given a differential equation \(y' = 2y - 1\) with an initial condition \(y(0) = 1\), we first found the exact solution as \(y = -\frac{1}{2} + \frac{3}{2}e^{-2t}\). We then computed the local truncation error as \(e^{-2t}\) and used this to find error bounds for successive steps \(e_1, e_2, e_3\), and \(e_4\). We found that the actual error at \(t=0.1\) was smaller than the error bound, as expected. Lastly, we computed the error bound for \(e_4\) as \(0.4\).

Step-by-step solution

Finding the exact solution of the differential equation

First, let's find the exact solution for the given initial value problem: $$ y^{\prime}=2 y-1, \quad y(0)=1 $$ It's a first-order linear differential equation. To solve this, we need to find an integrating factor, which is given by \(IF=\exp{\left(\int P(t)~dt\right)} = \exp{\left(\int 2~dt\right)} = e^{2t}\). Now we multiply both sides with the integrating factor and then integrate both sides with respect to \(t\): $$ e^{2t}y^{\prime}-2e^{2t}y=-e^{2t} $$ Integrate LHS by parts and RHS directly: $$ \int \left(e^{2t}\left(dy - 2y~dt\right)\right) = -\int e^{2t} ~dt $$ Integrating and simplifying, we get: $$ e^{2t}y = -\frac{1}{2}e^{2t}+ C $$ Now, divide by \(e^{2t}\) and put the initial condition \(y(0) = 1\): $$ y = -\frac{1}{2}+ C\cdot e^{-2t} $$ Plugging the initial condition, we get \(C = \frac{3}{2}\), and thus, the exact solution is: $$ \phi(t) = y = -\frac{1}{2}+ \frac{3}{2}e^{-2t} $$

Finding the local truncation error of Euler's method

The local truncation error for Euler's method is given by: $$ \tau(t) = \frac{y_{n+1} - y_n}{h} - f(t, y) = \left. \frac{dy}{dt} - f(t, y) \right|_{t = t_n} $$ where \(f(t, y) = 2y - 1\). Now let's differentiate the exact solution we found in step 1: $$ \frac{d\phi(t)}{dt} = -3 e^{-2t} $$ Now, calculate the difference \(dy/dt - f(t, \phi(t))\): $$ \tau(t) = -3 e^{-2t} - (2\left(-\frac{1}{2}+\frac{3}{}-{2t}\right) - 1) = e^{-2t} $$ So, the local truncation error is dependent on t as \(e^{-2t}\).

Obtaining a bound for \(e_{n+1}\)

The error in the Euler method is given by: $$ e_{n+1} = e_n + h\tau(t) $$ Now, we have the maximum value of \(|\tau(t)|\) on the interval \(0\leq t \leq 1\) as \(|e^{-2t}| \leq e^0 = 1\). So, we can write the bound for the error as: $$ |e_{n+1}| \leq |e_n| + hM $$ where \(h\) is the step size and \(M\) is the maximum value of \(|\tau(t)|\) on the interval.

Computing a bound for \(e_1\) and comparing with the actual error at \(t=0.1\)

For \(h=0.1\), we can compute a bound for \(e_1\) using the above error bound relationship: $$ |e_1| \leq |e_0| + 0.1 \times 1 = 0 $$ Here, \(e_0\)=0, since our initial value is exact. Now, let's compute the actual error at \(t=0.1\). Using the Euler method, the approximation at \(t=0.1\) is given by: $$ y_1 = y_0 + hf(t_0,y_0) = 1 + 0.1(2-1) = 1.1 $$ And, the exact value of the solution \(\phi(0.1)\) is: $$ \phi(0.1) = -\frac{1}{2}+\frac{3}{2}e^{-2(0.1)} \approx 1.104245 $$ The actual error at \(t=0.1\) is: $$ |y_1-\phi(0.1)| = |1.1 - 1.104245| \approx 0.004245 $$ Comparing the error bound with the actual error, we find that the actual error is smaller than the error bound as expected.

Computing the bound for the error \(e_{4}\) in the fourth step

We can again use the error bound relationship derived in Step 3 to compute the error bound for \(e_4\). We can recursively compute bounds for \(e_1, e_2, e_3\), and finally, \(e_4\): $$ |e_1|\leq 0.1 \times 1 = 0.1 $$ $$ |e_2|\leq 0.1 + 0.1 \times 1 = 0.2 $$ $$ |e_3|\leq 0.2 + 0.1 \times 1 = 0.3 $$ $$ |e_4|\leq 0.3 + 0.1 \times 1 = 0.4 $$ So, the bound for the error in the fourth step is \(0.4\).

Key concepts

Local Truncation Error

Local truncation error is a key concept when dealing with numerical methods like Euler's Method. It represents the error introduced in a single step of advancing the solution. When using Euler’s Method, this error occurs because we are approximating a curve with a tangent line over an interval. Essentially, it's how much our linear approximation diverges from the true solution in one step.
For Euler’s Method, local truncation error \(\tau(t)\) is calculated as the difference between the differential equation’s rate of change and the increment posed by the Euler step:

• Formula for local truncation error: \(\tau(t) = \left(\frac{y_{n+1} - y_n}{h} - f(t, y)\right)\).
• This can be simplified to: \(\tau(t) = \frac{dy}{dt} - f(t, y)\).

In the given problem, using the derivative of the exact solution and the original differential equation, we estimate \(\tau(t) = e^{-2t}\). As such, the local truncation error depends on the function of time and decreases as time increases, given the negative exponent.

Differential Equations

Differential equations form the backbone of many modeling processes and solving them is a crucial skill. They relate an unknown function to its derivatives, and their solutions provide insight into a variety of physical and theoretical scenarios.
In the problem we face, the differential equation provided is:
\[y' = 2y - 1\quad \text{with initial condition } \ y(0)=1\]
This is a first-order linear differential equation. To solve such an equation, we often apply methods like finding an integrating factor, as illustrated:

• Determine the integrating factor (\(e^{2t}\) here), which simplifies solving the equation through multiplication.
• Use integration to solve for the function \(y\).

These steps lead to the general solution form \(\phi(t) = -\frac{1}{2}+ \frac{3}{2}e^{-2t}\), which fits the given conditions. Differential equations like this exemplify how precise techniques can yield explicit solutions for unknown functions and predict behavior over time.

Error Bound Analysis

Error bound analysis is an essential part of numerical methods, ensuring that our approximations don't diverge too far from the true values. It gives us a quantifiable measure of the worst-case deviation between the numerical solution and the exact solution.
For Euler’s Method, the error bound can be established by considering both initial errors and accumulated errors across each step:

• The error at any point \(e_{n+1}\) is constructed from the previous error \(e_n\) and the product of step size \(h\) with the local truncation error bound \(M\).
• Expressed as: \(|e_{n+1}| \leq |e_n| + hM\).

In the example provided with step size \(h=0.1\), this formula is used to calculate bounds at specific points. At each progressive step, \(|e_n|\) grows, but at a predictable rate allowing us to confidently estimate errors.
This method validates that although the actual error may be smaller, the bound sets an ultra-conservative expectation, ensuring numerical methods remain reliable regardless of step changes or complexities in the function's behavior.