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Explain why the local discretization error of the finite difference method is proportional to the square of the step size. Also explain why the global discretization error is proportional to the step size itself.

Short Answer

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Question: Explain the relationship between the local and global discretization errors in the finite difference method and the step size. Answer: In the finite difference method, the local discretization error, which occurs at each step of the process, is proportional to the square of the step size. The global discretization error, which represents the accumulated error over all the steps, is proportional to the step size itself. As the step size decreases, both the local and global discretization errors decrease.

Step by step solution

01

Understand Local and Global Discretization Errors

To begin, let's understand what local and global discretization errors are in the context of the finite difference method. The finite difference method is an approximate numerical technique used for solving differential equations. However, since it is an approximation, there will be errors associated with this method. Local discretization error refers to the error that occurs at each step of the finite difference method, while global discretization error represents the accumulated error over all the steps.
02

Local Discretization Error and Step Size

To derive the relationship between the local discretization error and the square of the step size, we will consider the Taylor series expansion. The Taylor series is an infinite series of terms that represent the function as a sum of its derivatives in powers of the step size. For a given function f(x), we can represent its Taylor series expansion as: \[f(x + h) = f(x) + hf'(x) + \frac{h^2}{2!}f''(x) + \frac{h^3}{3!}f'''(x) + \cdots\] where \(h = \Delta{x}\) is the step size, and \(f'(x), f''(x),\) etc. represent the first, second, and higher-order derivatives of the function. For the finite difference method, we mainly consider the first few terms for the approximation of a derivative. Now, let's take the forward difference approximation: \[f'(x)\approx\frac{f(x+h)-f(x)}{h}\] Rearranging the Taylor series expansion to obtain the error term, we have: \[\frac{f(x + h) - f(x)}{h} = f'(x)+\frac{h}{2!}f''(x) + \cdots = f'(x) + e_L\] where \(e_L\) is the local discretization error. It can be seen that the local discretization error is proportional to the square of the step size (\(h^2\)). The local discretization error decreases as the square of the step size decreases.
03

Global Discretization Error and Step Size

Now, let's focus on the global discretization error. When applying the finite difference method over an interval, we divide the interval into equal partitions with step size h. The total error will be the sum of the local discretization errors at each step. Let's denote the global discretization error as \(e_G\). Since there are N steps in total and the local discretization error at each step is proportional to \(h^2\), we have: \[e_G \propto N \times h^2 = \frac{(b - a)}{h} \times h^2 = h \times (b - a)\] Here, (b - a) represents the width of the interval, \((x_0,x_N)\). It can be seen that the global discretization error is proportional to the step size h itself. As the step size decreases, the global discretization error also decreases. In summary, the local discretization error of the finite difference method is proportional to the square of the step size, while the global discretization error is proportional to the step size itself.

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Most popular questions from this chapter

Consider a stainless steel spoon $(k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\varepsilon=0.6$ ) that is partially immersed in boiling water at \(100^{\circ} \mathrm{C}\) in a kitchen at \(32^{\circ} \mathrm{C}\). The handle of the spoon has a cross section of about $0.2 \mathrm{~cm} \times 1 \mathrm{~cm}\( and extends \)18 \mathrm{~cm}$ in the air from the free surface of the water. The spoon loses heat by convection to the ambient air with an average heat transfer coefficient of $h=13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Assuming steady one- dimensional heat transfer along the spoon and taking the nodal spacing to be \(3 \mathrm{~cm},(a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the temperature of the tip of the spoon by solving those equations, and (c) determine the rate of heat transfer from the exposed surfaces of the spoon.

What is a practical way of checking if the round-off error has been significant in calculations?

The roof of a house consists of a \(15-\mathrm{cm}\)-thick concrete slab \(\left(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=0.69 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) that is \(18 \mathrm{~m}\) wide and \(32 \mathrm{~m}\) long. One evening at $6 \mathrm{p} . \mathrm{m}$., the slab is observed to be at a uniform temperature of \(18^{\circ} \mathrm{C}\). The average ambient air and the night sky temperatures for the entire night are predicted to be \(6^{\circ} \mathrm{C}\) and \(260 \mathrm{~K}\), respectively. The convection heat transfer coefficients at the inner and outer surfaces of the roof can be taken to be $h_{i}=5 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\( and \)h_{o}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The house and the interior surfaces of the walls and the floor are maintained at a constant temperature of \(20^{\circ} \mathrm{C}\) during the night, and the emissivity of both surfaces of the concrete roof is \(0.9\). Considering both radiation and convection heat transfers and using the explicit finite difference method with a time step of \(\Delta t=5 \mathrm{~min}\) and a mesh size of $\Delta x=3 \mathrm{~cm}$, determine the temperatures of the inner and outer surfaces of the roof at 6 a.m. Also, determine the average rate of heat transfer through the roof during that night.

Hot combustion gases of a furnace are flowing through a concrete chimney \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of rectangular cross section. The flow section of the chimney is $20 \mathrm{~cm} \times 40 \mathrm{~cm}\(, and the thickness of the wall is \)10 \mathrm{~cm}$. The average temperature of the hot gases in the chimney is \(T_{i}=280^{\circ} \mathrm{C}\), and the average convection heat transfer coefficient inside the chimney is \(h_{l}=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The chimney is losing heat from its outer surface to the ambient air at $T_{0}=15^{\circ} \mathrm{C}\( by convection with a heat transfer coefficient of \)h_{o}=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and to the sky by radiation. The emissivity of the outer surface of the wall is \(\varepsilon=0.9\), and the effective sky temperature is estimated to be \(250 \mathrm{~K}\). Using the finite difference method with \(\Delta x=\Delta y=10 \mathrm{~cm}\) and taking full advantage of symmetry, \((a)\) obtain the finite difference formulation of this problem for steady two-dimensional heat transfer, (b) determine the temperatures at the nodal points of a cross section, and \((c)\) evaluate the rate of heat loss for a \(1-m\)-long section of the chimney.

Consider transient heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0,1 , \(2,3,4\), and 5 with a uniform nodal spacing of $\Delta x$. The wall is initially at a specified temperature. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary nodes for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 5) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {surr }}\)

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