Question

Convergence of Euler's Method. It can be shown that, under suitable conditions on \(f\) the numerical approximation generated by the Euler method for the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) converges to the exact solution as the step size \(h\) decreases. This is illustrated by the following example. Consider the initial value problem $$ y^{\prime}=1-t+y, \quad y\left(t_{0}\right)=y_{0} $$ (a) Show that the exact solution is \(y=\phi(t)=\left(y_{0}-t_{0}\right) e^{t-t_{0}}+t\) (b) Using the Euler formula, show that $$ y_{k}=(1+h) y_{k-1}+h-h t_{k-1}, \quad k=1,2, \ldots $$ (c) Noting that \(y_{1}=(1+h)\left(y_{0}-t_{0}\right)+t_{1},\) show by induction that $$ y_{n}=(1+h)^{x}\left(y_{0}-t_{0}\right)+t_{n} $$ for each positive integer \(n .\) (d) Consider a fixed point \(t>t_{0}\) and for a given \(n\) choose \(h=\left(t-t_{0}\right) / n .\) Then \(t_{n}=t\) for every \(n .\) Note also that \(h \rightarrow 0\) as \(n \rightarrow \infty .\) By substituting for \(h\) in \(\mathrm{Eq}\). (i) and letting \(n \rightarrow \infty,\) show that \(y_{n} \rightarrow \phi(t)\) as \(n \rightarrow \infty\). Hint: \(\lim _{n \rightarrow \infty}(1+a / n)^{n}=e^{a}\).

Short answer

Question: Show that the numerical solution obtained from Euler's method converges to the exact solution for the given initial value problem \(y' = 1 - t + y\), \(y(t_0) = y_0\). Solution: We first found the exact solution to the initial value problem to be \(y=(y_0-t_0)e^{-t_0}e^t+t\). Then, we applied Euler's method to obtain the numerical solution. We proved by induction that \(y_n=(1+h)^n(y_0-t_0)+t_n\). Finally, we showed that as \(n\rightarrow \infty\), the numerical solution \(y_n\) converges to the exact solution \(\phi(t)\): \(\lim_{n \rightarrow \infty}y_n=\phi(t)\).

Step-by-step solution

Part (a): Determine the Exact Solution

To find the exact solution of the given initial value problem \(y^{\prime}=1-t+y,\) \(y\left(t_{0}\right)=y_{0},\) we can solve the differential equation directly. The given differential equation is a first-order linear ordinary differential equation. The integrating factor is \(e^{\int dt} = e^t\). Multiply both sides of the equation by the integrating factor, then integrate: $$e^t(y'-y)=e^t(1-t)$$ $$\frac{d(e^ty)}{dt}=e^t(1-t)$$ Integrate both sides: $$e^ty = \int e^t(1-t) dt = e^t(t-1)+C.$$ To find the constant \(C\), we use the initial condition \(y(t_0)=y_0\): $$y_0=e^{t_0}(t_0-1)+C.$$ Solve for C and substitute it back into the equation for \(y\): $$C=y_0-e^{t_0}(t_0-1)$$ $$y=(y_0-t_0)e^{-t_0}e^t+t.$$

Part (b): Show the Euler Formula

For a given step size h and an initial value problem \(y^{\prime}=f(t, y)\), Euler's method formula is given by: $$y_{k+1}=y_k+hf(t_k, y_k).$$ Now, substitute the given function, \(f(t, y) = 1-t+y\), into the Euler formula: $$y_{k+1}=y_k+h(1-t_k+y_k).$$ For k=1, we have \(y_1=y_0+h(1-t_0+y_0).\) $$y_1=(1+h)y_0+h-h t_0.$$

Part (c): Proof by Induction

We want to prove the statement \(y_{n}=(1+h)^{n}(y_0-t_0)+t_n\) for each positive integer n by induction. Base case (n=1): $$y_1=(1+h)(y_0-t_0)+t_1$$ which has already been shown in part (b). Inductive step: Assume the statement is true for n=k: $$y_k=(1+h)^k(y_0-t_0)+t_k.$$ Now, prove the statement for n=k+1: $$y_{k+1}=y_k+h(1-t_k+y_k).$$ Substitute the assumed statement of \(y_k\): $$y_{k+1}=(1+h)^k(y_0-t_0)+t_k+h(1-t_k+(1+h)^k(y_0-t_0)+t_k).$$ Simplify the expression: $$y_{k+1}=(1+h)(1+h)^k(y_0-t_0)+h+h t_k.$$ $$y_{k+1}=(1+h)^{k+1}(y_0-t_0)+t_{k+1}.$$ Thus, the statement holds for n=k+1, and by induction, it is true for all positive integers n.

Part (d): Convergence of the Numerical Solution

To show the convergence of the numerical solution, consider a fixed point \(t>t_0\). Choose \(h=\frac{t-t_0}{n}\), which implies \(t_n = t\) for every \(n\). As \(n\rightarrow\infty\), we have \(h\rightarrow 0\). From the induction result, we can write the numerical solution \(y_n\) as: $$y_n=(1+h)^n(y_0-t_0)+t_n$$ Substitute \(h=\frac{t-t_0}{n}\) and \(t_n=t\): $$y_n=\left(1+\frac{t-t_0}{n}\right)^n(y_0-t_0)+t$$ Now, as \(n\rightarrow \infty\), we have: $$\lim_{n \rightarrow \infty}y_n=\lim_{n \rightarrow \infty}\left(1+\frac{t-t_0}{n}\right)^n(y_0-t_0)+t.$$ Using the hint \(\lim_{n \rightarrow \infty}(1+\frac{a}{n})^n = e^a\), we get: $$\lim_{n \rightarrow \infty}y_n=e^{t-t_0}(y_0-t_0)+t=\phi(t).$$ Thus, we have shown that the numerical solution \(y_n\) converges to the exact solution \(\phi(t)\) as \(n\rightarrow\infty\).

Key concepts

Convergence of Numerical Solutions

Understanding the convergence of numerical solutions is crucial when working with numerical methods for solving differential equations. Convergence refers to the idea that as we refine the process (usually by making the step size smaller), the numerical solution produced by a method like Euler's gets closer and closer to the true solution of the differential equation.

In the exercise, you saw how the convergence of Euler's Method was demonstrated. As the step size, denoted by h, becomes smaller, the numerical solution approximates the exact solution more accurately. This is a foundation of numerical analysis—ensuring that the methods we use are not just computationally efficient but also yield results that reflect the reality modeled by the differential equation.

When you're working with numerical methods, you should always consider whether your method converges, and under what conditions. The exercise showcases that Euler's method does have this desirable property, known as convergence, given that the step size h decreases appropriately. But remember, the rate of convergence can vary from one method to another, and analyzing this rate often requires a deeper layer of mathematical understanding and investigation.

• Convergence is essential for the credibility and utility of numerical solutions.
• Step size is a critical factor affecting the convergence rate in methods like Euler's.
• The smaller the step size, typically, the closer the numerical solution is to the true solution.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation along with a specified value, called the initial condition, at a specific point. Most real-life scenarios that require modeling—from physics to economics—are described by IVPs. The form is typically given by y'(t) = f(t, y), y(t_0) = y_0, where t_0 is the initial point and y_0 is the initial value at that point.

In our exercise, we dealt with such a problem where we were given the initial condition y(t_0) = y_0 which is crucial in determining the unique solution of the differential equation. Without this initial condition, we could not proceed with methods like Euler's.

These initial conditions serve as the starting inputs for numerical methods that advance step by step—such as Euler's method—from that known point. The accurate solution of an IVP heavily depends on the precision with which we can compute these numerical steps.

• Every IVP has a differential equation and an initial condition at a specific point.
• The initial condition is indispensable for determining the unique solution to the equation.
• Numerical methods begin solving IVPs using this initial condition and proceed iteratively.

Ordinary Differential Equation

An ordinary differential equation (ODE) is an equation involving derivatives of a function with respect to one independent variable. It is 'ordinary' because it involves derivatives with respect to only one variable, as opposed to partial differential equations that involve derivatives with respect to multiple variables.

In the context of Euler's method for solving ODEs, as shown in the exercise, the ODE provides the basis for the iterative computation process in which each step is computed based on the derivative at the previous step. The ODE in our exercise was a first-order linear ODE, expressed as y' = f(t, y), which described the rate of change of the function y(t).

Solving ODEs can sometimes be done analytically with exact solutions, but when that's not possible or practical, numerical methods like Euler's method provide an alternative that can approximate the solution with arbitrary precision. The balance between precision and computational cost is a common theme in the numerical solution of ODEs.

• ODEs describe dynamic processes in various fields with a single independent variable.
• Euler's method uses the rate of change from ODEs to approximate solutions iteratively.
• ODEs can have analytical solutions, but numerical methods provide flexibility for more complicated cases.