UNDERGRADUATE COMPUTATIONAL ENGINEERING AND SCIENCE

UCES

"Nonlinear Boundary Value Problems and Radiative Heat Transfer"

Course:   Methods and Analysis in UCES

Prerequisites:

  1. Math:   Calculus 3 and matrices
  2. Computing:   Maple
  3. Science:   Physics, heat diffusion

    Objectives:

    1. Math:   Newton's method in more than one variable
    2. Computing:   Solution of nonlinear problem
    3. Science:   Nonlinear thermal properties

    General Information: This is an extension of the one variable Newton method for finding roots of a single equation to finding solutions for n variables with n equations. This very important method is applied to steady state 1D space diffusion with nonlinear cooling due to radiation.

    Contact Person:   R. E. White, NCSU, white@math.ncsu.edu

    Revision Date:   8-4-94

    Copyright ©1994, R. E. White. All rights reserved.

    Return to Table of Contents


    3.7   Nonlinear Boundary Value Problems and Radiative Heat Transfer

    3.7.1.   Introduction.

    In the analysis of most of the heat transfer problems we assumed the temperature varied over a small range so that the thermal properties could be approximated by constants. This always resulted in a linear algebraic problem which could be solved by a variety of methods. Two possible difficulties are (i) nonlinear thermal properties, or (ii) larger problems which are a result of diffusion in three directions. In this section we consider the nonlinear problems.


    3.7.2.   Applied Area.

    The thermal properties of specific heat and thermal conductivity can be nonlinear. The exact nature of the nonlinearity will depend on the material and the range of the temperature variation. Usually, data is collected that reflects these properties, and a least squares curve fit is done for a suitable approximating function. Other nonlinear terms can evolve from the heat source or sink terms in either the boundary conditions, or the source term on the right side of the heat equation. We consider one such case.

    Consider a cooling fin or plate which is glowing hot, say at 900 degrees Kelvin. Here heat is being lost by radiation to the surrounding region. In this case the heat lost is not proportional, as in Newton's law of cooling, to the difference in the surrounding temperature, usur , and the temperature of the glowing mass, u. Observations indicate that the heat loss through a surface area, A, in a time interval, dt, is equal to

    dt A epsilon sigma (usur4 - u4 )
    where epsilon is the emissivity of the surface and sigma is the Stefan-Boltzman constant. One can couple this with the Fourier heat law to form various differential equations or boundary conditions which are nonlinear.

    Radiative Cooling in a Thin Wire. Consider a thin wire of length L and radius r. Let the ends of the wire have a fixed temperature of 900 and let the surrounding region be usur = 300. Suppose the surface of the wire is being cooled by the above nonlinear model. The lateral surface area of small cylindrical portion of the wire has area A = 2pi r dx. Therefore, the heat leaving the lateral surface in dt time is

    dt (2pi r dx) (epsilon sigma (usur4 - u4 )) .

    Assume steady state heat diffusion in one direction and apply the Fourier heat law to get

    0 ~ dt (2pi r dx) (epsilon sigma (usur4 - u4 )) + dt K(pi r2 )ux(x + dx/2) - dt K(pi r2 )ux(x - dx/2) .

    Now, divide by dt dx r pi and let dx go to zero to get

    0 = 2epsilon sigma (usur4 - u4 ) + (Krux )x.

    The continuum model for the heat transfer is

    Figure: Wire Segment
    -(Kux )x = c(usur4 - u4 ) where c = 2epsilon sigma /r and (1)
      u(0) = 900 = u(L) .

    The thermal conductivity will also be temperature dependent, but we will assume for simplicity that K is a constant which has been incorporated into c.


    3.7.3.   Model.

    Consider the nonlinear differential equation -uxx = f(u). The finite difference model is for h = L/(n+1) and ui ~ u(ih) with u0 = 900 = un+1

    -ui-1 + 2ui - ui+1 = h2f(ui ) for i = 1, ..., n. (2)

    This has n unknowns, ui , and n equations of the form

    Fi (u) = h2f(ui ) + ui-1 - 2ui + ui+1 = 0.

    Nonlinear Algebraic Problem.

    Find u = [u1 , ..., un ]T so that F(u) = [F1 (u), ..., Fn (u)]T = 0.

    Nonlinear problems can have multiple solutions. For example, consider the intersection the unit circle x2 + y2 - 1 = 0 and the hyperbola x2 - y2 - 1/2 = 0. Here n = 2 with u1 = x and u2 = y, and there are four solutions. In applications this can present problems in choosing the solution which most often exists in nature.


    3.7.4.   Method.

    In order to derive Newton's method for n equations and n unknowns, it is instructive to review the one equation case. The idea behind Newton's algorithm is to approximate the function f(x) at a given point by a straight line. Then find the root of the equation associated with this straight line. One continues to repeat this until little change in the approximations of the root is observed. We make use of the approximation

    df ~ f'(x) dx. (3)

    Figure: Newton's Algorithm

    The equation for the straight line at the mth iteration is

    (y - f(xm ))/(x - xm ) = f'(xm ) .

    Define xm+1 so the y = 0 where the straight line intersects the x axis

    (0 - f(xm ))/(xm+1 - xm ) = f'(xm ) .

    Solve for xm+1 to obtain Newton's algorithm

    xm+1 = xm - f(xm )/f'(xm ) .

    The derivation of Newton's method for more than one equation and one unknown requires an analog of the approximation in (3). Consider Fi (u) as a function of n variables uj. If just the j component of u changes, then (3) will hold with x replaced by uj and f replaced by Fi. If all of the components change, then the net change in Fi can be approximated by sum of the (partial derivatives of Fi with respect to uj ) times the (change in uj ):

    For n = 2 this is depicted in the following figure where

    Figure: Change in Fi

    This can be put into matrix form to obtain the desired analog of (3)

    delta F ~ F'(u) delta u (4)
    where delta F = [delta F1 , ..., delta Fn]T, delta u = [delta u1 , ..., delta un]T and one defines the derivative matrix, or Jacobian, as

    Newton's method is obtained by letting

    u = um , delta u = um+1 - um and delta F = 0 - F(um ) .

    The vector approximation in (4) is replaced by an equality to get

    0 - F(um ) = F'(um ) (um+1 - um ) .

    This vector equation can be solved for um+1 , and we have the n variable Newton method

    um+1 = um - F'(um+1 )-1 F(um) .

    In practice the inverse of the Jacobian in not used, but we must find the solution, delta u, of

    0 - F(um ) = F'(um ) delta u.

    Consequently, Newton's method consists of solving a sequence of linear problems. One usually stops when either F is "small", or delta u is "small."

    Newton's Algorithm.

    
            choose initial u
            for m = 1,maxit
                    compute F(u) and F'(u)
                    solve F'(u) delta u =-F(u)
                    u = u + delta u
                    test for convergence
            endloop.
    
    

    Example. Let n = 2, F1 (u) = u12 + u22 - 1 and F2 (u) = u12 - u22 - 1/2.

    and it is nonsingular if both variables are nonzero.

    If the initial guess is near the solution in a quadrant, then Newton's method will converge to that solution.

    Example. Consider the nonlinear differential equation for the radiative heat transfer problem in (2) where

    Fi (u) = h2f(ui ) + ui-1 - 2ui + ui+1 = 0.

    The Jacobian matrix is easily computed and must be tridiagonal because each Fi only depends on ui , ui-1 and ui+1 .


    3.7.5.   Implementation.

    The following is a Maple code which uses Newton's method to solve the 1D diffusion problem with heat loss due to radiation. We have used the Maple procedure linsolve to solve each linear subproblem. One could use an iterative method, and for a larger problem where there is diffusion of heat in more than one direction this might be the best way. We have experimented with several emissitivities, and the graphs indicate the higher the emissitivity the more the cooling.

    Maple Code for Nonlinear Cooling with Diffusion.

       Input Data:
            > with(linalg):
    	> with(plots):
            > f:=(u)->.000005*(300^4 - u^4);
    		
            
            > fp:=(u)->.000005*(-4*u^3);
    		
            
            > n:=19:
            > h:=1/(n+1):
            > u0:=900:
            > eps:=.1:
    
       Procedure NEWTON is Defined:
            > NEWTON:=proc(n,h,eps,f,fp,u0):
            > FP:=matrix(n,n,0):
            > F:=vector(n,0):
            > u:=vector(n,u0):
            > for m from 1 to 10 do
            >     for i from 1 to n do
            >         if i=1 then
            >            F[i]:=f(u[i])*h*h - 2*u[i] + u[i+1] + u0:
            >            FP[i,i]:=fp(u[i])*h*h -2:
            >            FP[i,i+1]:=1:
            >         fi:	
            >         if i>1 and i<n then
            >            F[i]:=f(u[i])*h*h + u[i-1] - 2*u[i] + u[i+1]:
            >            FP[i,i]:=fp(u[i])*h*h - 2:
            >            FP[i,i-1]:=1:
            >            FP[i,i+1]:=1:
            >         fi:
            >         if i=n then
            >            F[i]:=f(u[i])*h*h - 2*u[i] + u[i-1] + u0:
            >            FP[i,i]:=fp(u[i])*h*h -2:
            >            FP[i,i-1]:=1:
            >         fi:		
            >     od:
            >     du:=linsolve(FP,F):
            >     u:=evalm(u - du):
            >     error:=norm(F):
            >     if error<eps then break fi:
            > od:
            > end:
    
       NEWTON is Called and Output of Data:
            > f:=(u)->.000005*(300^4 - u^4):
            > fp:=(u)->.000005*(-4*u^3):
            > NEWTON(n,h,eps,f,fp,u0):
            > m;
            8
            > L:=[[0,u0],seq([i*h,u[i]],i=1..n),[1,u0]]:
    
            > f:=(u)->.0000001*(300^4 - u^4):
            > fp:=(u)->.0000001*(-4*u^3):
            > NEWTON(n,h,eps,f,fp,u0):
            > m;
            6
            > LL:=[[0,u0],seq([i*h,u[i]],i=1..n),[1,u0]]:
    
            > f:=(u)->.00000005*(300^4 - u^4):
            > fp:=(u)->.00000005*(-4*u^3):
            > NEWTON(n,h,eps,f,fp,u0):
            > m;
            5
            > LLL:=[[0,u0],seq([i*h,u[i]],i=1..n),[1,u0]]:
    
            > plot({L,LL,LLL},x=0..1);
    
    
    Figure: Temperatures for Different Emissitivities

    The next calculations were done using the Matlab version of the above code, and the very rapid convergence of Newton's method is illustrated.

    Matlab Code for Nonlinear Cooling with Diffusion.

    
            uo = 900.;
            n = 19;
            h = 1./(n+1);
            FP = zeros(n);
            F  = zeros(n,1);
            u  = ones(n,1)*uo;
            for m =1:20
                for i = 1:n
                    if i==1
                       F(i) = f(u(i))*h*h + u(i+1) - 2*u(i) + uo;
                       FP(i,i) = fp(u(i))*h*h - 2;
                       FP(i,i+1) = 1;
                    elseif i<n
                       F(i) = f(u(i))*h*h + u(i+1) - 2*u(i) + u(i-1);
                       FP(i,i) = fp(u(i))*h*h - 2;
                       FP(i,i-1) = 1;
                       FP(i,i+1) = 1;
                    else
                       F(i) = f(u(i))*h*h - 2*u(i) + u(i-1) + uo;
                       FP(i,i) = fp(u(i))*h*h - 2;
                       FP(i,i-1) = 1;
                    end
                end
                du = FP\F;
                u  = u - du;
                error = norm(F);
                if error<.0001
                   break;
                end
                m
                error
            end
    
       Output Data:
    
    	m =     1
    	error =  706.1416
    
    	m =     2
    	error =  197.4837
    
    	m =     3
    	error =   49.2847
    
    	m =     4
    	error =   8.2123
    
    	m =     5
    	error =   .3967
    
    	m =     6
    	error =    .0011
    
    	m =     7
    	error =    7.3703e-09
    
    


    3.7.6.   Assessment.

    Nonlinear problems are very common, but they are often linearized. This is done because it is easier to solve one linear problem than a nonlinear problem. In the case of Newton's method we must solve a sequence on linear problems. If one tries to use analytic methods for finding closed form solutions of nonlinear boundary value problems, it may be impossible!

    Newton's method has, under some assumption on F(u), the two very important properties of local convergence and quadratic convergence. Local convergence means that if the initial guess is suitably close to the solution, then the Newton iterates will converge to the solution. Often one can use physical insight to the problem to make a good initial guess. Also, such initial guesses can significantly reduce the number of Newton iterations needed to achieve convergence. Quadratic convergence means the error at the next step will be proportional to the square of error of the present step. Thus, after the error is less than 1, the convergence is very rapid. These two properties have contributed to the wide use and many variations of the Newton's method for solving nonlinear algebraic systems. In order for the above properties to hold the components functions should be sufficiently differentiable, and the Jacobian matrix should be inevitable and well conditioned.


    3.7.7.   Homework.

    1. Apply Newton's method to the example with n = 2. Experiment with different initial guesses in each quadrant. Observe local and quadratic convergence.

    2. Apply Newton's method to the radiative heat transfer problem. Experiment with different n, eps, L and emissitivities. Observe local and quadratic convergence as well as the number of Newton iterations required for convergence.

    3. In problem 2 determine how much heat is lost through the wire per unit time.

    4. Consider the linearized verison of -uxx = c(usur4 - u4 ) = f(u) where f(u) is replaced by its first order Taylor polynomial f(usur ) + f'(usur ) (u - usur ). Compare the nonlinear and linearized solutions.

    5. Consider the 1D diffusion problem where K is not a constant. Find the nonlinear algebraic system and solve it using Newton's method.

    6. Consider a 2D cooling plate which satisfies

      -(Kux )x - (Kuy )y = c(usur4 - u4 ) .

      Use Newton's method coupled with a linear solver that uses SOR to solve this nonlinear problem.