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Obtain a formula for the local truncation error for the Euler method in terms of \(t\) and the solution \(\phi\) $$ y^{\prime}=\sqrt{t+y}, \quad y(1)=3 $$

Short Answer

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Answer: The local truncation error for the Euler method, given the differential equation \(y^{\prime}=\sqrt{t+y}\) and initial condition \(y(1)=3\), can be represented by the formula \(\tau(t) = \frac{1}{4}(t+\phi)^{-\frac{1}{2}}(1+\phi^{\prime})(\Delta t)^2 + O((\Delta t)^3)\), where \(\phi\) is the exact solution of the differential equation.

Step by step solution

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01

Write down the differential equation and initial condition

We are given the differential equation: $$ y^{\prime}=\sqrt{t+y}, \quad y(1)=3 $$ This describes the change of \(y\) with respect to \(t\). The Euler method is a numerical technique used to approximate the solution of such differential equations.
02

Find the second derivative of y with respect to t

To find the local truncation error, we need to expand the solution using a Taylor series up to the second derivative. Therefore, we need to find \(y''(t)\). But before we do that, we need to rewrite the given equation to express the second derivative: $$ y^{\prime}=\sqrt{t+y} \Rightarrow \frac{dy}{dt} = \sqrt{t+y} $$ Now, differentiate both sides with respect to \(t\) to find \(y''(t)\): $$ \frac{d^2y}{dt^2} = \frac{d}{dt}(\sqrt{t+y}) = \frac{1}{2}(t+y)^{-\frac{1}{2}}(1+y^{\prime}) $$
03

Determine the local truncation error for the Euler method

The local truncation error \(\tau(t)\) is given by the difference between the exact solution \(y(t)\) and the Euler approximation \(y_{E}(t)\). If we expand the exact solution to the second-order Taylor series, we can find an approximation of \(\tau(t)\). From the Taylor series expansion, we have: $$ y(t+\Delta t) = y(t) + y^{\prime}(t)\Delta t + \frac{1}{2}y^{\prime\prime}(t) (\Delta t)^2 + O((\Delta t)^3) $$ The Euler method has the following form: $$ y_{E}(t+\Delta t) = y(t) + y^{\prime}(t)\Delta t $$ Thus the local truncation error is given by: $$ \tau(t) = y(t+\Delta t) - y_{E}(t+\Delta t) = \frac{1}{2}y^{\prime\prime}(t)(\Delta t)^2 + O((\Delta t)^3) $$
04

Substitute the value of the second derivative

Now, we need to substitute the expression we found for \(y^{\prime\prime}(t)\) in step 2 into the local truncation error: $$ \tau(t) = \frac{1}{2} \left( \frac{1}{2}(t+y)^{-\frac{1}{2}}(1+y^{\prime}) \right)(\Delta t)^2 + O((\Delta t)^3) $$
05

Final formula for the local truncation error

We can now rewrite the local truncation error in terms of \(t\) and the exact solution \(\phi\): $$ \tau(t) = \frac{1}{4}(t+\phi)^{-\frac{1}{2}}(1+\phi^{\prime})(\Delta t)^2 + O((\Delta t)^3) $$ This is the final formula for the local truncation error for the Euler method in terms of \(t\) and the solution \(\phi\).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Local Truncation Error
In the context of numerical methods like the Euler method, the local truncation error is important as it measures the error incurred when a differential equation's solution is approximated over a finite step. Specifically, this error arises from truncating the Taylor series expansion of the exact solution after a certain number of terms, making it a useful measure of an approximation's accuracy for small time steps.

When applying the Euler method to solve a differential equation, the local truncation error can be derived by comparing the exact solution's Taylor series expansion with the Euler approximation for the next time step. In our case, given the differential equation\[ y^{\prime}=\sqrt{t+y} \], the local truncation error \( \tau(t) \) can be expressed as follows:
  • First, expand the exact solution \( y(t+\Delta t) \) using a Taylor series up to the second derivative:
  • \[ y(t+\Delta t) = y(t) + y^{\prime}(t)\Delta t + \frac{1}{2}y^{\prime\prime}(t)(\Delta t)^2 + O((\Delta t)^3) \]
  • Next, apply the Euler method, which gives the approximation:
  • \[ y_{E}(t+\Delta t) = y(t) + y^{\prime}(t)\Delta t \]
  • Finally, calculate the local truncation error by subtracting the Euler approximation from the Taylor expansion:
  • \[ \tau(t) = \frac{1}{2}y^{\prime\prime}(t)(\Delta t)^2 + O((\Delta t)^3) \]

This error term captures the leading error term that results in the Euler method being first-order, emphasizing importance when choosing step sizes for accurate solutions.
Numerical Methods
Numerical methods are techniques used to approximate solutions to mathematical problems, particularly those that are difficult or impossible to solve analytically. These methods are invaluable in computing as they allow for practical solutions to complex systems that arise in fields such as engineering, physics, and finance.

The Euler method is one of the simplest and most intuitive numerical methods used to solve ordinary differential equations (ODEs). It's often introduced as an entry point to numerical analysis because of its straightforward implementation:
  • Start at an initial condition, say \( y(1) = 3 \).
  • Use the slope defined by the differential equation, \( y^{\prime} = \sqrt{t+y} \), to estimate the next value.
  • Repeat this process, stepping forward in small increments \( \Delta t \) to build an approximate solution over an interval.
Each step in the Euler method builds upon the last, progressing iteratively to approximate the entire solution. The method works well for simple problems but is susceptible to error, which can accumulate significantly with larger step sizes, emphasizing the need for careful step size selection to maintain accuracy.
Differential Equations
Differential equations are equations that involve functions and their derivatives. They describe the rate of change of a quantity and are fundamental in modeling real-world phenomena where change is a key element, such as population dynamics, heat distribution, and motion.

In our example, the differential equation given is \( y^{\prime} = \sqrt{t+y} \). This notation describes how the function \( y(t) \) changes with respect to time \( t \). The function \( \sqrt{t+y} \) dictates the rate at which \( y \) changes.
  • Initial conditions, like \( y(1) = 3 \), set the specific path the solution follows.
  • Understanding differential equations is crucial as they can be both simple, like linear equations, or complex, involving non-linear terms as shown in our case.
  • The solutions to differential equations can provide significant insights into the behavior of dynamic systems.
Solving differential equations analytically involves finding a function that satisfies the equation for all time instances, but often, especially with non-linear or complex systems, this can be challenging, necessitating the use of numerical methods like the Euler method to find approximate solutions.

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Most popular questions from this chapter

The modified Euler formula for the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) is given by $$ y_{n+1}=y_{n}+h f\left[t_{n}+\frac{1}{2} h, y_{n}+\frac{1}{2} h f\left(t_{n}, y_{n}\right)\right] $$ Following the procedure outlined in Problem \(14,\) show that the local truncation error in the modified Euler formula is proportional to \(h^{3}\).

(a) Find a formula for the solution of the initial value problem, and note that it is independent of \(\lambda\). (b) Use the Runge-Kuta method with \(h=0.01\) to compute approximate values of the solution for \(0 \leq t \leq 1\) for various values of \(\lambda\) such as \(\lambda=1,10,20,\) and 50 , and 50 , inters of the (c) Explain the differences, if any, between the exact solution and the numerical approximations. \(y^{\prime}-\lambda y=2 t-\lambda t^{2}, \quad y(0)=0\)

In each of Problems 1 through 6 find approximate values of the solution of the given initial value problem at \(t=0.1,0.2,0.3,\) and \(0.4 .\) Compare the results with those obtained by using other methods and with the exact solution (if available). (a) Use the Runge-Kutta method with \(h=0.1\) (b) Use the Runge-Kutta method with \(h=0.05\) $$ y^{\prime}=\left(t^{2}-y^{2}\right) \sin y, \quad y(0)=-1 $$

Consider the initial value problem $$ y^{\prime}=t^{2}+e^{y}, \quad y(0)=0 $$ Using the Runge-Kutta method with step size \(h,\) we obtain the results in Table \(8.5 .\) These results suggest that the solution has a vertical asymptote between \(t=0.9\) and \(t=1.0 .\) (a) Show that for \(0 \leq t \leq 1\) the solution \(y=\phi(t)\) of the problem (i) satisfies $$ \phi_{2}(t) \leq \phi(t) \leq \phi_{1}(t) $$ where \(y=\phi_{1}(t)\) is the solution of $$ y^{\prime}=1+e^{y}, \quad y(0)=0 $$ and \(y=\phi_{2}(t)\) is the solution of $$ y^{\prime}=e^{y}, \quad y(0)=0 $$ (b) Determine \(\phi_{1}(t)\) and \(\phi_{2}(t) .\) Then show that \(\phi(t) \rightarrow \infty\) for some \(t\) between \(t=\ln 2 \cong\) 0.69315 and \(t=1 .\) (c) Solve the differential equations \(y^{\prime}=e^{y}\) and \(y^{\prime}=1+e^{y},\) respectively, with the initial condition \(y(0.9)=3.4298 .\) Use the results to show that \(\phi(t) \rightarrow \infty\) when \(t \cong 0.932 .\)

Consider the initial value problem $$ x^{\prime \prime}+t^{2} x^{\prime}+3 x=t, \quad x(0)=1, \quad x^{\prime}(0)=2 $$ Convert this problem to a system of two first order equations and determine approximate values of the solution at \(t=0.5\) and \(t=1.0\) using the Runge-Kutta method with \(h=0.1\)

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