Question

Show that the Euler method fails to determine an approximate solution of the initial-value problem $$ \begin{array}{rlr} y^{\prime} & =y^{\alpha}, & \alpha<1, \\ y(0) & =0 \end{array} $$ although the exact solution exists.

Short answer

Answer: The Euler method fails to approximate the solution of the given IVP because, for any α < 1, it constantly produces the same approximation point (0, 0) regardless of the step size h. This indicates that the Euler method is not able to generate an approximate solution that includes more points than just the zero-solution. In this case, although an exact solution exists, the Euler method is not suitable for approximating it.

Step-by-step solution

Recall the Euler method formula

The Euler method is a numerical method to approximate the solution of an initial-value problem of the form \(y'(t) = f(t, y(t))\) with \(y(t_0) = y_0\), where \(t_0\) and \(y_0\) are given initial conditions. The method generates a sequence of approximation points \((t_n, y_n)\) using the iterative formula: $$ y_{n+1} = y_n + hf(t_n, y_n), $$ where \(h\) is the step size and \(f(t, y)\) is the given function that describes the derivative of \(y\).

Write down the given initial-value problem (IVP)

We are given the IVP: $$ \begin{cases} y'(t) = y(t)^\alpha, \qquad \alpha < 1, \\ y(0) = 0. \end{cases} $$

Set up the Euler method for the given IVP

Using the given IVP, our Euler method iterative formula becomes: $$ y_{n+1} = y_n + h(y_n^\alpha). $$

Demonstrate the failure of the Euler method for the given IVP

Since the initial condition is \(y(0)=0\), the first approximation point is \((0, 0)\). Applying the Euler method formula: $$ y_1 = y_0 + h(y_0^\alpha) = 0 + h(0^\alpha) = 0. $$ We observe that for any \(n\), the Euler method produces the same approximation point \((t_n, 0)\), regardless of the step size \(h\). Thus, the Euler method fails to generate an approximate solution for the given IVP, which includes more points than just the zero-solution (for \(\alpha<1\)).

Show that an exact solution exists

The given IVP can be rewritten as: $$ \frac{dy}{dt} = y^\alpha. $$ If \(\alpha \neq 0\), the exact solution can be found by separating variables and integrating both sides: $$ \int \frac{dy}{y^\alpha} = \int dt \Longrightarrow y^{1-\alpha} = \frac{1-\alpha}{\alpha}t + C, $$ where \(C\) is the integration constant. To find \(C\), we can apply the initial condition: $$ 0^{1-\alpha} = \frac{1-\alpha}{\alpha} \cdot 0 + C \Longrightarrow C = 0. $$ Therefore, when \(\alpha \neq 0\), the exact solution of the IVP is: $$ y(t) = \left(\frac{1-\alpha}{\alpha}t\right)^\frac{1}{1-\alpha}. $$ For \(\alpha = 0\), the derivative becomes a constant, and the exact solution is given by \(y(t) = kt\), where \(k\) is a constant. In this case, the Euler method would correctly generate an approximate solution. In conclusion, the given IVP has an exact solution for \(\alpha < 1\), but the Euler method fails to approximate it when \(\alpha \neq 0\).

Key concepts

Understanding Initial-Value Problems

An initial-value problem (IVP) is a type of differential equation coupled with specified values, often at a starting point. It's the launch pad of the equation's trajectory. For example, when we're given an equation describing the derivative of a function and the function's value at a point, that's our IVP. It defines the starting conditions from which we can predict the behavior of the system moving forward. The IVP in our exercise is peculiar because it provides a zero at the origin which is a stationary point for the equation, making the IVP hold special significance when applying numerical methods for solution approximation, such as the Euler method.

Numerical Method Approximation

When exact analytical solutions to differential equations are hard to find or don't exist, we turn to numerical methods. These are algorithms that allow us to approximate the behavior of solutions over time. We use small, finite steps to predict the function's values, which can be highly effective for many equations, but as we see with our IVP where \(\alpha < 1\), it's not always a guarantee of success. The pitfalls these methods can face include stability issues, where the methods diverge from reality, and accuracy problems, where the approximated values aren't close to the exact solution—both situations demonstrating the limitations of numerical approximations.

Diving Into the Euler Method Formula

The Euler method is the most straightforward numerical technique for solving IVPs. It's like using tiny, incremental steps to hike up a mountain path, estimating your position at each step based on your current direction and speed. Mathematically, we say \(y_{n+1} = y_n + hf(t_n, y_n)\), where \(h\) is our step size and \(f(t, y)\) defines our slope or direction at each point. Even with its simplicity, the Euler method has limitations and isn't suitable for all types of equations, which is something we experience when \(y'(t) = y(t)^\alpha\) with \(\alpha < 1\) and \(y(0) = 0\). The method ends up stuck in place, never advancing from the initial point, thus failing to approximate the true solution.

Separation of Variables

Separation of variables is a strategy used to transform differential equations into a simpler form, often leading us to the exact solution. It's like disentangling two intertwined threads; we separate the variables \(y\) and \(t\) to opposite sides of the equation, then integrate to find a relationship between them. In the context of our problem, by rewriting \(dy/dt = y^\alpha\), we successfully use this technique to untangle \(y\) and \(t\), giving us a handle on the exact solution once we integrate. This sheds light on the unique scenarios where Euler's method might not work, despite an existing exact solution.

Existence of an Exact Solution

The presence of an exact solution is the reassurance that there's a closed-form expression satisfying the original differential equation and its initial conditions. It's what we aim to find or, in numerical methods, approximate. Our IVP, thrillingly, has such a solution that can be unearthed through the separation of variables technique. It's like having a treasure map where 'X' marks the spot of the exact solution. However, the Euler method's failure in this case highlights that the existence of a solution doesn't automatically ensure the method's success. Different methods have different key strengths and are suitable for different kinds of problems, and it's crucial to choose the right tool for the job at hand.