Question

This exercise treats the simple initial value problem \(y^{\prime}=\lambda y, y(0)=y_{0}\), where we can see the behavior of the numerical solution as the step size \(h\) approaches zero. (a) Show that the solution of the initial value problem is \(y=e^{2 x} y_{0}\) - (b) Apply Euler's method to the initial value problem, using a step size \(h\). Show that \(y_{n}\) is related to the initial value by the formula \(y_{n}=(1+h \lambda)^{n} y_{0}\). (c) Consider any fixed time \(t^{*}\), where \(t^{*}>0\). Let \(h=t^{*} / n\) so that \(t^{*}=n h\). (The exact solution is \(y(t)=y_{0} e^{3^{2}}\).) Show that letting \(h \rightarrow 0\) and \(n \rightarrow \infty\) in such a way that \(f^{*}\) remains fixed leads to $$ \lim _{h \rightarrow 0 \atop t \rightarrow \infty} y_{n}=y\left(t^{*}\right) $$

Short answer

Answer: The limit of the numerical solution when the step size approaches zero is the exact solution of the initial value problem, \(y(t^{*}) = y_{0}e^{\lambda t^{*}}\).

Step-by-step solution

Solve the initial value problem analytically

Given the initial value problem \(y'(x)=\lambda y(x)\) with initial condition \(y(0)=y_{0}\). We can solve it by separation of variables: $$ \frac{dy}{y} =\lambda dx $$ Integrate both sides with respect to x: $$ \int\frac{1}{y} \, dy=\int\lambda\, dx $$ This gives us: $$ \ln|y|=\lambda x + C $$ Exponentiate both sides and apply the initial value: $$ y(x)=e^{\lambda x+C}=e^{\lambda x}e^{C}\implies y(0)=y_0=e^{C} $$ So, the exact solution is: $$ y(x)=y_0e^{\lambda x} $$

Apply Euler's method

To apply Euler's method, we use the iterative formula: $$ y_{n+1}=y_{n}+h\cdot y'_{n} $$ Since \(y'(x)=\lambda y(x)\), we get: $$ y_{n+1}=y_{n}+h\cdot \lambda y_{n} $$ We can rewrite this as: $$ y_{n+1}=(1+h\lambda)y_{n} $$ Starting with the initial condition \(y(0)=y_{0}\), we can find that the numerical solution is given by: $$ y_{n}=(1 + h\lambda)^n y_{0} $$

Show the limit when the step size approaches zero

We are given that \(t^{*} \gt 0\), and let \(h=\frac{t^{*}}{n}\), where \(t^{*}=n\cdot h\). We are asked to show that: $$ \lim_{\substack{h\rightarrow 0 \\ n \rightarrow \infty}}y_{n}=y(t^{*}) $$ Substitute the formula for \(y_{n}\) and replace \(h\) with \(\frac{t^{*}}{n}\): $$ \lim_{\substack{h\rightarrow 0 \\ n \rightarrow \infty}}\left(1 + \frac{\lambda t^{*}}{n}\right)^{n}y_{0} $$ This expression approaches the definition of the exponential function when \(n \rightarrow \infty\): $$ \lim_{n \rightarrow \infty}\left(1 + \frac{\lambda t^{*}}{n}\right)^{n} = e^{\lambda t^{*}} $$ Therefore, $$ \lim_{\substack{h\rightarrow 0 \\ n \rightarrow \infty}}y_{n}=y_{0}\cdot e^{\lambda t^{*}} $$ Which is the exact solution of the initial value problem at \(t=t^{*}\).

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation accompanied by a specific condition, known as the initial condition. In our exercise, the differential equation is given by \( y' = \text{lambda} y \), and the initial condition is \( y(0) = y_0 \). The initial condition specifies the value of the unknown function at a particular point, and it's crucial because it determines the unique solution to the differential equation.

Solving an initial value problem typically involves integrating the differential equation to find a general solution and then applying the initial condition to find the particular solution that fits the problem. As we can see in the step-by-step solution, after integrating and applying the initial value, the solution to our initial value problem is \( y(x) = y_0e^{\text{lambda} x} \), which provides the exact behavior of the function over time.

Numerical Solution

In contrast to exact solutions, numerical solutions approximate the answer to a problem using computational methods. This approach is often necessary when dealing with complex differential equations that cannot be solved analytically. Euler's method is one such numerical technique for approximating solutions to initial value problems.

This method estimates the function's value at discrete points by moving in small steps from the initial condition and using the function's derivative to update the function's value iteratively. The formula \( y_{n+1} = (1 + h\text{lambda} )y_n \) represents one iteration of Euler's method, where \( h \) is the step size. Smaller step sizes generally result in more accurate approximations, as the method can more closely follow the function's actual path.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. These powerful tools allow us to model the rates of change in various scientific and engineering fields. In our exercise, the differential equation \( y' = \text{lambda} y \) portrays an exponential growth or decay process, depending on the sign of \( \text{lambda} \), and it's a first-order linear differential equation.

Separation of Variables

The step-by-step solution begins with separating the variables, in this case, \( y \) and \( x \), and integrating both sides. This method is commonly used for solving simple first-order differential equations and provides the exact solution, illustrating the equation's long-term behavior and how solutions evolve over time.

Limit

The concept of a limit is fundamental in calculus and is used to describe the behavior of a function as the argument approaches a particular value. Limits can explain how functions behave near points of discontinuity or infinity, and are also essential in defining derivatives and integrals.

In the context of Euler's method, we examine the limit as the step size \( h \) approaches zero and the number of steps \( n \) approaches infinity while keeping \( t^* \) fixed. This approach demonstrates the convergence of the numerical solution toward the exact solution. As \( h \) gets smaller, the Euler approximation improves and theoretically should match the exact solution, illustrated by \( e^{\text{lambda} t^*} \), when \( h \) is infinitesimally small, demonstrating the fundamental connection between the discrete world of numerical approximation and the continuous world described by exact solutions.