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Implementing symmetry boundary condition for the diffusion equation

  1. Jul 23, 2017 #1
    The following lines of codes implements 1D diffusion equation on 10 m long rod with fixed temperature at right boundary and right boundary temperature varying with time.
    Code (Text):
        xsize   =   10;               % Model size, m
    xnum    =   10;                 % Number of nodes
    xstp    =   xsize/(xnum-1);     % Grid step
    tnum    =   504;                 % number of timesteps
    kappa   =   833.33;            %     k/rhocp;            % Thermal diffusivity, m^2/s
    dt     =   300;                      % Timestep
    x       =   0:xstp:xsize;         %Creating vector for nodal point positions
    tlbc = sin(linspace(0.1,2.9,tnum));      % left boundary condition
    %Define initial temperature profile
    tback   =   0;               % background temperature, K

    for i=1:1:xnum
        % Background profile  
        t0imp(i)    =   tback;      % profile for implicit solving
        end
    end
    % Time cycle
    timesum=0; % Elapsed time
    for t=1:1:tnum

        % Matrix of coefficients initialization for implicit solving
        L       =       sparse(xnum,xnum);
        % Vector of right part initialization for implicit solving
        R       =       zeros(xnum,1);

        % Implicit solving of 1D temperature equation: dT/dt=kappa*d2T/dx2
        % Composing matrix of coefficients L()
        % and vector (column) of right parts R()
        % First point: T=tback
        L(1,1)  =       1;
        R(1,1)  =       tlbc(t);
        % Intermediate points
        for i=2:1:xnum-1
            % dT/dt=kappa*d2T/dx2
            % Tnew(i)/dt-kappa*(Tnew(i-1)-2*Tnew(i)+Tnew(i+1))/dx^2=Told(i)/dt
            %%% Eq 10.6
            L(i,i-1)    =   -kappa/xstp^2;
            L(i,i)      =   1/dt+2*kappa/xstp^2;
            L(i,i+1)    =   -kappa/xstp^2;
            R(i,1)      =   t0imp(i)/dt;
        end
        % Last point:T=tback
        L(xnum,xnum)    =   1;

        R(xnum,1)       =  tback;
        % Obtaining solution for implicit temperature profile
        t1imp           =   L\R;

    end
     
    My question is how can I implement a symmetry boundary condition. I mean the temperature calculated for 9th node is assigned to 10th node? Also I want to know how can I implement infinite boundary condition?
     
  2. jcsd
  3. Jul 23, 2017 #2
    Node10=Node9 ...?
    I don't understand why you would want to do this though. Can you explain how this relates to the diffusion problem stated?
    For numerical solvers, you can't do this in general. I'm also not aware of any physical problem which would require an infinite boundary condition. If you want an infinite boundary for whatever reason, the best you can do numerically is assign a very large value at the needed boundary.
     
  4. Jul 24, 2017 #3
    I want to implement a boundary condition where presence of boundary has no impact . I am using diffusion equation to model fluid flow in porous medium and want to correlate the numerical model with an analytical model where flux at right boundary is zero i.e. ## \frac{\partial u}{\partial x} = 0 ## . Consider u as fluid head instead of temperature. I think in this case, having zero flux at right boundary means that right boundary is having no impact on fluid head/heat flow in x direction.
     
  5. Jul 24, 2017 #4
    Zero flux is equivalent to one end having an impermeable barrier, and would definitely have an impact on the solution. If your goal is to model the absence of a boundary or a boundary at infinity, then your boundary condition can be taken as the diffusion equation at the end point,
    $$\frac{\partial u}{\partial t}=D\frac{\partial^{2}u}{\partial x^{2}}\bigg |_{x=10}$$
    Depending on how your algorithm works, this can be done with a one-sided finite difference at the boundary.
     
  6. Jul 24, 2017 #5
    Could you please explain this that in the given case, how can this infinity boundary condition be implemented here?
     
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