Question

Consider the initial value problem \(y^{\prime}=2 l-1, y(1)=0\). (a) Solve this initial value problem. (b) Suppose Euler's method is used to solve this problem numerically on the interval \(1 \leq t \leq 5\), using step size \(h=0.1\). Will the numerical solution values be the same as the exact solution values found in part (a)? That is, will \(y_{k}=y\left(t_{k}\right), k=1,2, \ldots, 40\) ? Explain. (c) What will be the answer to the question posed in part (b) if the Runge- Kutta method \((9)\) is used instead of Euler's method?

Short answer

In summary, for the given initial value problem y'=2l-1 with y(1)=0: a) The exact solution is y(t) = 2lt - t - 1. b) Using Euler's method, the numerical solution values are not equal to the exact solution values (i.e., \(y_k \neq y(t_k)\) for k=1,2,...,40). c) Using the Runge-Kutta method, the numerical solution values are equal to the exact solution values (i.e., \(y_k = y(t_k)\) for k=1,2,...,40) due to the constant slope function in this particular problem.

Step-by-step solution

Part (a): Solve the initial value problem

Given the differential equation y'=2l-1 and the initial condition y(1)=0, we need to find the general solution to the differential equation. This is a first-order linear differential equation that can be directly integrated. We can integrate both sides of the equation to find the solution: $$\int y' dt = \int (2l-1) dt$$ $$y(t) = 2lt - t + C$$ Now, we can use the initial condition y(1)=0 to determine the constant C: $$0 = 2(1) - (1) + C \implies C = -1$$ So, the exact solution for the given initial value problem is: $$y(t) = 2lt - t - 1$$

Part (b): Euler's Method

Now, we need to perform Euler's method on the given initial value problem on the interval between t=1 and t=5 with step size h=0.1 and check whether the numerical solution values are equal to the exact solution values found in part (a). Let's implement Euler's method with step size h=0.1: $$y_{k+1} = y_k + 0.1(2l - 1)$$ Given the exact solution found in part (a), we can plug in t=tₖ in: $$y(t) = 2lt - t - 1 \implies y(t_k) = 2l(t_k) - t_k - 1 \implies y_k = 2l(t_k) - t_k - 1$$ Comparing the Euler's method solution and the exact solution, we see that Euler's method uses a linear approximation that updates based on the slope at each point, while the exact function has a constant slope (2l-1). Therefore, Euler's method solution values will not be equal to the exact solution values. So, no, \(y_k \neq y(t_k)\) for k=1,2,...,40.

Part (c): Runge-Kutta Method

In this part, we need to find the answer to the question posed in part (b) but for the Runge-Kutta method instead of Euler's method. The Runge-Kutta method is more accurate than Euler's method as it gives more approximation points for the solution values. However, considering the differential equation y' = 2l-1, the slope is constant, and therefore independent of the initial value y(1)=0. For this specific problem, using the Runge-Kutta method will still result in the same numerical and exact solution values as the problem is dealing with a constant slope function. So, for Runge-Kutta method in this particular problem, the solution values will be the same as the exact solution values, i.e., \(y_k = y(t_k)\) for k=1,2,...,40.

Key concepts

Euler's method

Euler's method is a fundamental technique used for numerical approximation of solutions to initial value problems involving differential equations. Imagine you're on a path in the forest, and with each step, you can only move where a sign points. Euler's method works similarly, using the slope of the differential equation (direction of the sign) at a current point to estimate the value at the next point.

Let's break it down: first, we start with an initial condition, just as if we know our position when we start our walk in the forest. We then use the slope of the differential equation (given by the derivative) at that point to 'walk' a small step forward. By repeatedly applying this process, stepping from one point to the next using the slope as a guide, we build up a numerical solution.

The step size (denoted as 'h') is crucial. Smaller steps mean more accuracy but require more computations. In the problem at hand, the step size was chosen to be 0.1, which is a balance between the accuracy of the solution and the number of steps needed to reach the end of the interval.

Runge-Kutta method

The Runge-Kutta method is like taking Euler's method and giving it a GPS with several waypoints. Considered a more sophisticated and accurate alternative, the Runge-Kutta method mitigates Euler's method's drawbacks. It does not rely on just the initial slope; instead, it calculates multiple slopes at different points and then cleverly combines them to provide a better estimate for the next value.

Think of it as not only looking at the signpost where you are but also checking several signposts ahead and planning your direction based on an average of all that information. This method is particularly good for more complex differential equations where changes in slope are not just linear and can greatly improve the accuracy of the solution with relatively fewer steps compared to Euler's method.

Even though it usually provides a superior solution, in the context of the problem given with a constant slope, the Runge-Kutta method offers no improvement in accuracy over the exact solution—the numerical solution perfectly aligns with the analytical solution in such a linear case.

Numerical solution

A numerical solution, as the name suggests, involves finding values that approximate the solution of a problem where an analytical (exact) solution is difficult or impossible to obtain. This is the digital equivalent of approximation methods used in mathematics and is implemented on computers to solve complex problems.

The core of a numerical solution is built on discrete approximation and iterative refinement. Think of it like trying to measure the length of a curved line with a ruler; you approximate the curve using multiple straight line segments, and your measure becomes better as the number of segments increases.

In the context of differential equations, numerical solutions are vital when we're dealing with models that are too complicated to solve by hand or when the equations do not have nice, tidy analytical solutions. These solutions hinge on the choice of method (like Euler's method or Runge-Kutta method) and the trade-off between computational resources and the level of precision required.

First-order linear differential equation

A first-order linear differential equation is a relationship that involves a function and its first derivative. It's called 'first-order' because it only involves the first derivative (no square or cube of the derivative) and 'linear' because the function and its derivative are to the power of one—not multiplied or divided by each other.

Taking our forest path analogy again: if the forest is flat and you only have one simple direction to go with a consistent speed, you are following a first-order linear path. Solving such an equation analytically involves integration, a process akin to adding up small pieces to get the whole picture or path length. In the problem provided, the integration results in a linear function representing the exact solution. This simplicity yields a straightforward relationship between the initial condition and subsequent values of the function, exemplified by the fact that methods like Runge-Kutta do not distinguish themselves from simpler methods like Euler's when applied to such basic equations.