ANSWER ALL QUESTIONS The numerical solution u(x, t) of the…

Question Answered step-by-step ANSWER ALL QUESTIONS The numerical solution u(x, t) of the…  ANSWER ALL QUESTIONSThe numerical solution u(x, t) of the first-order hyperbolic partial differential equation ut + ux = 0 using the Lax-Wendroff formula as observed at t = 0, 1, 2, . . . , 20. The initial conditions are given by Equation 7.3.25 with h = 1, ?t/?x = 2 3 , and ?x = 0.02 The temperature u(r, t)/T0 within an infinitely long cylinder at various positions r/b and times a 2 t/b2 that we initially heated to the temperature T0 and then allowed to radiatively cool with hb = 1. The electrostatic potential is defined as the amount of work that must be done against electric forces to bring a unit charge from a reference point to a given point. It is readily shown4 that the electrostatic potential is described by Laplace’s equation if there is no charge within the domain Image transcription textDiscrete distribution functions 2.43.The probability function of a randomvariable X is shown in Tabl… Show more… Show moreImage transcription text2.50. Find the distribution function forthe random variable X of Problem 2.49.2.51. The distribution functi… Show more… Show moreImage transcription textChange of variables 2.62. Let X havedensity function f (x) = 1 Jer x20 lo Findthe density function of Y =… Show more… Show moreImage transcription textf(x) = 1/2 -1 1, then our scheme is unstable. This CFL criterion has its origin in the fact that if c?t > ?x, then we are asking signals in the numerical scheme to travel faster than their real-world counterparts and this unrealistic expectation leads to instability!  The behavior of u n m is determined by the values of ? given by Equation 7.5.24. If c?t/?x ? 1, then ? is real and u n m is bounded for all ? as n ? ?. If c?t/?x > 1, then it is possible to find a value of ? such that the right side of Equation 7.5.24 exceeds unity and the corresponding values of ? occur as complex conjugate pairs. The ? with the negative imaginary part produces a solution with exponential growth because n = tn/?t ? ? as ?t ? 0 for a fixed tn and c?t/?x. Thus, the value of u n m becomes infinitely large, even though the initial data may be arbitrarily small. Finally, we must check for convergence. A numerical scheme is convergent if the numerical solution approaches the continuous solution as ?x, ?t ? 0. The general procedure for proving convergence involves the evolution of the error term e n m, which gives the difference between the true solution u(xm, tn) and the finite difference solution u n m. For our first example, we resolve Equation 7.3.1 through Equation 7.3.3 and Equation 7.3.25 and Equation 7.3.26 numerically using MATLAB. The MATLAB code is clear coeff = 0.5; coeffsq = coeff * coeff % coeff = c?t/?x dx = 0.04; dt = coeff * dx; N = 100; x = 0:dx:1; M = 1/dx + 1; % M = number of spatial grid points % introduce the initial conditions via F and G F = zeros(M,1); G = zeros(M,1); for m = 1:M if x(m) >= 0.25 & x(m) <= 0.5 F(m) = 4 * x(m) - 1; end if x(m) >= 0.5 & x(m) <= 0.75 F(m) = 3 - 4 * x(m); end; end % at t = 0, the solution is: tplot(1) = 0; u = zeros(M,N+1); u(1:M,1) = F(1:M); % at t = ?t, the solution is given by Equation 7.5.16 tplot(2) = dt; for m = 2:M-1   Math MATH 133 Share QuestionEmailCopy link Comments (0)