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Assume, for the given differential equation, that \(y(0)=1\). (a) Use the differential equation itself to determine the values \(y^{\prime}(0), y^{\prime \prime}(0), y^{\prime \prime \prime}(0), y^{(4)}(0)\) and form the Taylor polynomial $$ P_{4}(t)=y(0)+y^{\prime}(0) t+\frac{y^{\prime \prime}(0)}{2 !} t^{2}+\frac{y^{\prime \prime \prime}(0)}{3 !} t^{3}+\frac{y^{(4)}(0)}{4 !} t^{4} $$ (b) Verify that the given function is the solution of the initial value problem consisting of the differential equation and initial condition \(y(0)=1\). (c) Evaluate both the exact solution \(y(t)\) and \(P_{4}(t)\) at \(t=0.1\). What is the error \(E(0.1)=y(0.1)-P_{4}(0.1)\) ? [Note that \(E(0.1)\) is the local truncation error incurred in using a Taylor series method of order 4 to step from \(t_{0}=0\) to \(t_{1}=0.1\) using step size \(h=0.1 .]\) \(y^{\prime}=y^{1 / 2} ; \quad y(t)=\left(1+\frac{t}{2}\right)^{2}\)

Short Answer

Expert verified
Question: Compute the local truncation error at t=0.1 for the initial value problem y'(t) = (1 + t/2)^(1/2), y(0) = 1, using the 4th-degree Taylor polynomial approximation. Answer: The local truncation error at t=0.1 for the given initial value problem is E(0.1) = -0.0025.

Step by step solution

01

Compute y'(0)

Differentiate the given function y(t) = (1 + t/2)^2 in respect to t, to find y'(t): y'(t) = 2*(1 + t/2)*1/2 = 1 + t/2. Now evaluate y'(0): y'(0) = 1.
02

Compute y''(0)

Differentiate y'(t) in respect to t, to find y''(t): y''(t) = 1/2. Now evaluate y''(0): y''(0) = 1/2.
03

Compute y'''(0)

Differentiate y''(t) in respect to t, to find y'''(t): y'''(t) = 0. Now evaluate y'''(0): y'''(0) = 0.
04

Compute y^{(4)}(0)

Differentiate y'''(t) in respect to t, to find y^{(4)}(t): y^{(4)}(t) = 0. Now evaluate y^{(4)}(0): y^{(4)}(0) = 0.
05

Form the Taylor polynomial P_4(t)

Use the derivatives found above to form the Taylor Polynomial P_4(t) = y(0) + y'(0) * t + (y''(0)/2!) * t^2 + (y'''(0)/3!) * t^3 + (y^{(4)}(0)/4!) * t^4 = 1 + t + (1/2)*t^2.
06

Verify the given function as a solution

Plug the given function into the original differential equation: y'(t) = (1 + t/2)^{1/2}. The given function is a solution to the initial value problem as it satisfies the original differential equation and the initial condition y(0) = 1.
07

Evaluate y(t) and P_4(t) at t = 0.1

Calculate the values of the exact solution y(t) and the Taylor polynomial P_4(t) at t = 0.1: y(0.1) = (1 + 0.1/2)^2 = 1.1025, P_4(0.1) = 1 + 0.1 + (1/2)*0.1^2 = 1.105.
08

Calculate the local truncation error E(0.1)

Calculate the local truncation error as the difference between the exact solution and the Taylor polynomial at t = 0.1: E(0.1) = y(0.1) - P_4(0.1) = 1.1025 - 1.105 = -0.0025.

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Key Concepts

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

Differential Equations
Differential equations involve functions and their derivatives, and they describe how a particular function behaves as its input changes. In simpler terms, they are equations that relate a function with its rate of change.
In the context of an initial value problem, a differential equation will have specific conditions, such as the value of the function at a particular point. For example, in our exercise, the given differential equation is related to how quickly the solution curve rises or falls. This particular equation, with the form \(y' = y^{1/2}\), implies that the rate at which \(y\) changes depends on the square root of \(y\).
Solving a differential equation means finding a function that fits this relationship. The real beauty of differential equations is that they allow us to model and understand real-world processes that involve change, such as motion, heat flow, and population dynamics.
Initial Value Problem
An initial value problem (IVP) is a specific type of differential equation problem where the solution is determined by the starting conditions. It's like setting the starting point for a journey, where the path is determined by the differential equation.
In our exercise, the condition \(y(0) = 1\) serves as the initial value. This means we know the value of \(y\) at \(t = 0\).
  • Knowing this initial condition is essential because it helps us determine the unique solution curve for the differential equation.
  • Without the initial value, the solution would not be unique, as there could be many functions that satisfy the differential equation itself but not the initial condition.
Verifying the function within the initial value problem ensures that the condition \(y(0)=1\) is met at \(t=0\), validating the solution's accuracy.
Local Truncation Error
Local truncation error is an important concept when using numerical methods like the Taylor series to approximate solutions to differential equations.
It measures the error made in one step of the approximation, in this case, moving from \(t_0\) to \(t_1\).
The Taylor series approximates a function locally around a point by summing a finite number of terms from its respective polynomial.
  • In our exercise, the Taylor series polynomial \(P_4(t)\) and the exact solution \(y(t)\) were evaluated at \(t = 0.1\).
  • The difference between the exact solution \(y(0.1)\) and the polynomial \(P_4(0.1)\) gave us the local truncation error.
This error, calculated as \(E(0.1) = 1.1025 - 1.105 = -0.0025\), tells us how accurate our approximation is for that step. Thus, being aware of and minimizing this error is crucial in numerical methods since it affects the reliability of the results.

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Most popular questions from this chapter

Let \(P(t)\) denote the population of a certain colony, measured in millions of members. Assume that \(P(t)\) is the solution of the initial value problem $$ P^{\prime}=0.1\left(1-\frac{P}{3}\right) P+M(t), \quad P(0)=P_{0}, $$ where time \(t\) is measured in years. Let \(M(t)=e^{-t}\). Therefore, the colony experiences a migration influx that is initially strong but soon tapers off. Let \(P_{0}=\frac{1}{2} ;\) that is, the colony had 500,000 members at time \(t=0\). Our objective is to estimate the colony size after two years. Obtain a numerical solution of this problem, using Heun's method with a step size \(h=0.05 .\) What is your estimate of colony size at the end of two years?

$$ y^{\prime \prime}+y+y^{3}=0, \quad y(0)=0, \quad y^{\prime}(0)=1 ; \quad 0 \leq t \leq 10 . $$ A nonlinear differential equation having this structure arose in modeling the motion of a nonlinear spring. We are interested in assessing the impact of the nonlinear \(y^{3}\) term on the motion. Plot the numerical solution for \(y(t)\). If the nonlinear term were not present, the initial value problem would have solution \(y(t)=\sin t\). On the same graph, plot the function \(\sin t\). Does the nonlinearity increase or decrease the period of the motion? How do the amplitudes of the motion differ?

We ask you to use the fourth order Runge-Kutta method (14) to solve the problems in Exercises \(20-23\) of Section \(7.3\). \(y^{\prime}=\frac{1}{2 y}, \quad y(0)=1 . \quad\) The exact solution is \(y(t)=\sqrt{1+t}\)

For the given initial value problem, an exact solution in terms of familiar functions is not available for comparison. If necessary, rewrite the problem as an initial value problem for a first order system. Implement one step of the fourth order Runge-Kutta method (14), using a step size \(h=0.1\), to obtain a numerical approximation of the exact solution at \(t=0.1\). \(\mathbf{y}^{\prime}=\left[\begin{array}{rr}-1 & t \\ 2 & 0\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}-1 \\\ 1\end{array}\right]\)

In most applications of numerical methods, as in Exercises 1619 , an exact solution is unavailable to use as a benchmark. Therefore, it is natural to ask, "How accurate is our numerical solution?" For example, how accurate are the solutions obtained in Exercises 16-19 using the step size \(h=0.05\) ? This exercise provides some insight. Suppose we apply Heun's method or the modified Euler's method to the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) and we use a step size \(h\). It can be shown, for most initial value problems and for \(h\) sufficiently small, that the error at a fixed point \(t=t^{*}\) is proportional to \(h^{2}\). That is, let \(n\) be a positive integer, let \(h=\left(t^{*}-t_{0}\right) / n\), and let \(y_{n}\) denote the method's approximation to \(y\left(t^{*}\right)\) using step size \(h\). Then $$ \lim _{h \rightarrow 0 \atop t^{*} \text { fixed }} \frac{y\left(t^{*}\right)-y_{n}}{h^{2}}=C, \quad C \neq 0 $$ As a consequence of this limit, reducing a sufficiently small step size by \(\frac{1}{2}\) will reduce the error by approximately \(\frac{1}{4}\). In particular, let \(\hat{y}_{2 n}\) denote the method's approximation to \(y\left(t^{*}\right)\) using step size \(h / 2\). Then, for most initial value problems, we expect that \(y\left(t^{*}\right)-\hat{y}_{2 n} \approx\left[y\left(t^{*}\right)-y_{n}\right] / 4 .\) Rework Example 1 , using Heun's method and step sizes of \(h=0.05, h=0.025\), and \(h=0.0125 .\) (a) Compare the three numerical solutions at \(t=0.05,0.10,0.15, \ldots, 0.95 .\) Are the errors reduced by about \(\frac{1}{4}\) when the step size is reduced by \(\frac{1}{2}\) ? (Since the solution becomes unbounded as \(t\) approaches 1 from the left, the expected error reduction may not materialize near \(t=1\).) (b) Suppose the exact solution is not available. How can the Heun's method solutions obtained using different step sizes be used to estimate the error? [Hint: Assuming that $$ y\left(t^{*}\right)-\hat{y}_{2 n} \approx \frac{\left[y\left(t^{*}\right)-y_{n}\right]}{4} $$ derive an expression for \(y\left(t^{*}\right)-\hat{y}_{2 n}\) that involves only \(\hat{y}_{2 n}\) and \(\left.y_{n} .\right]\) (c) Test the error monitor derived in part (b) on the initial value problem in Example \(1 .\)

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