Question

Consider the ODE $$ \frac{d y}{d t}=f(t, y), \quad 0 \leq t \leq b \text { , } $$ where \(b \gg 1\). (a) Apply the stretching transformation \(t=\tau b\) to obtain the equivalent \(\mathrm{ODE}\) $$ \frac{d y}{d \tau}=b f(\tau b, y), \quad 0 \leq \tau \leq 1 $$ (Strictly speaking, \(y\) in these two ODEs is not quite the same function. Rather, it stands in each case for the unknown function.) (b) Show that applying the forward Euler method \(^{65}\) to the ODE in \(t\) with step size \(h=\Delta t\) is equivalent to applying the same method to the \(\mathrm{ODE}\) in \(\tau\) with step size \(\Delta \tau\) satisfying \(\Delta t=b \Delta \tau\). In other words, the same stretching transformation can be equivalently applied to the discretized problem.

Short answer

Question: Apply a stretching transformation on the given ODE and show the equivalence of applying the forward Euler method with a step size transformation. ODE: $$\frac{dy}{dt}=f(t, y)$$ Stretching transformation: $$t \rightarrow \tau b$$ Step size transformation: $$\Delta t = b\Delta\tau$$ Solution: 1. Apply the stretching transformation on the given ODE to get: $$b\frac{dy}{d\tau} = f(\tau b, y),\quad 0\leq\tau\leq1$$ 2. Apply the forward Euler method to both the original ODE in \(t\) and the stretched ODE in \(\tau\) using the step size transformation \(\Delta t = b\Delta\tau\): Forward Euler method in \(t\): $$y_{n+1} = y_n + b\Delta\tau f(b\tau_n , y_n)$$ Forward Euler method in \(\tau\): $$y_{k+1} = y_k + b\Delta\tau f(\tau_k b, y_k)$$ 3. Show the equivalence of the discretized problems by comparing the forward Euler methods applied to both ODEs and noting that they have the same form.

Step-by-step solution

Apply the stretching transformation

To apply the stretching transformation, we replace \(t\) with \(\tau b\) and rewrite the ODE: $$\frac{d y}{d (\tau b)} = f(\tau b, y)$$ Now, we need to rewrite the left side of the equation using chain rule: $$\frac{dy}{d\tau}\frac{d\tau}{dt}=b\frac{dy}{d\tau}$$ So, our transformed ODE is: $$b\frac{dy}{d\tau} = f(\tau b, y),\quad 0\leq\tau\leq1$$

Apply the forward Euler method to the ODE in \(t\)

To apply the forward Euler method to the ODE in \(t\) with step size \(h=\Delta t\), we have the following formula: $$y_{n+1} = y_n + hf(t_n, y_n)$$ Where, \(t_n = nh\), \(y_{n+1}=y(t_{n+1})\), and \(y_n=y(t_n)\).

Apply the forward Euler method to the ODE in \(\tau\)

Now, let's apply the forward Euler method to the transformed ODE in \(\tau\) with step size \(\Delta \tau\). We'll get the following formula: $$y_{k+1}=y_k+b\Delta\tau f(\tau_k b, y_k)$$ Where, \(\tau_k=k\Delta\tau\), \(y_{k+1}=y(\tau_{k+1}b)\), and \(y_k=y(\tau_k b)\).

Show the equivalence of the discretized problems

We need to relate the step sizes to show the equivalence of the problems. It is given that \(\Delta t = b\Delta\tau\). Also, as \(t_n = nh = b\tau_k\), we have \(k = n\). Now, let's rewrite the forward Euler method for the ODE in \(t\) as follows: $$y_{n+1}=y_n+b\Delta\tau f(b\tau_n, y_n)$$ Comparing this with the forward Euler method for the ODE in \(\tau\), we can see that both of them have the same form: $$y_{n+1} = y_n + b\Delta \tau f(b\tau_n , y_n)$$ Hence, the same stretching transformation can be equivalently applied to the discretized problem.

Key concepts

Ordinary Differential Equations (ODEs)

Ordinary Differential Equations (ODEs) are equations that involve functions of one independent variable and their derivatives. They appear frequently in various scientific and engineering disciplines to model dynamic systems. For example, they can describe how the population of species grows over time or how a substance changes concentration in a chemical reaction.

ODEs are often expressed in the form \( \frac{d y}{d t} = f(t, y) \), where \( y \) is the dependent variable, \( t \) is the independent variable (often representing time), and \( f(t, y) \) is a given function.
The goal is to find a function \( y(t) \) that satisfies the equation for given initial conditions. Solving ODEs can be complex, and analytical solutions don't always exist. That's where **numerical methods** come in handy, allowing us to approximate these solutions for practical use.

Forward Euler Method

The Forward Euler Method is a simple yet powerful tool for numerically solving Ordinary Differential Equations (ODEs). It belongs to a broader category known as **explicit methods**, meaning that the next value is calculated directly from the previous values.

To apply the method, you start at an initial condition \( y_0 \) and repeatedly apply the update formula:

• \( y_{n+1} = y_n + h f(t_n, y_n) \)

Here, \( h \) is the step size, and \( t_n = nh \). This method progresses one step ahead, estimating the solution at discrete points over the interval. The smaller the step size, the closer the approximation is to the true solution.

While it's easy to implement, its accuracy is not very high compared to more advanced methods. Therefore, it's particularly useful for problems where simplicity and implementation speed are more critical than precision.

Stretching Transformation

The concept of a stretching transformation is used in ODEs to redefine the time-scale of a problem, making it easier to analyze or simulate. Mathematically, it involves changing variables by redefining time from \( t \) to \( \tau \), often through a scaling factor that "stretches" the interval. This transformation can simplify complex problems by contracting or expanding the range over which you need to solve the equation.

For instance, if you have an ODE in the form \( \frac{d y}{d t} = f(t, y) \), and you perform a stretching transformation \( t = \tau b \), the equation transforms to \( \frac{d y}{d \tau} = b f(\tau b, y) \). This new formulation may offer better numerical stability or efficiency, especially for large intervals where \( b \gg 1 \).
Through transformations like these, disparate time scales can be handled more effectively, and often, they make the problem suitable for numerical methods.