source: issm/trunk-jpl/examples/LcurveAnalysis/runme.m@ 23752

step=[1];
if any(step==1)
        % Generate observations
        md = model;
        md = triangle(md,'DomainOutline.exp',100000);
        md = setmask(md,'all','');
        md = parameterize(md,'Square.par');
        md = setflowequation(md,'SSA','all');
        md.cluster = generic('np',2);
        md = solve(md,'Stressbalance');
        plotmodel(md,'axis#all','tight','data',md.materials.rheology_B,'caxis',[ 1.3 1.9]*10^8,'title','"True" B',...
                'data',md.results.StressbalanceSolution.Vel,'title','"observed velocities"')
        save model1 md
end
if any(step==2)
        % Modify rheology, now constant
        loadmodel('model1.mat');
        md.materials.rheology_B(:) = 1.8*10^8;

        %results of previous run are taken as observations
        md.inversion=m1qn3inversion();
        md.inversion.vx_obs  = md.results.StressbalanceSolution.Vx;
        md.inversion.vy_obs  = md.results.StressbalanceSolution.Vy;
        md.inversion.vel_obs = md.results.StressbalanceSolution.Vel;

        md = solve(md,'Stressbalance');
        plotmodel(md,'axis#all','tight','data',md.materials.rheology_B,'caxis',[ 1.3 1.9]*10^8,'title','B first guess',...
                'data',md.results.StressbalanceSolution.Vel,'title','modeled velocities')
        save model2 md
end
if any(step==3)
        % Perform L-curve analysis for ice rigidity inversion
        loadmodel('model2.mat');

        % Set up inversion parameters
        maxsteps = 20;
        md.inversion.iscontrol = 1;
        md.inversion.control_parameters = {'MaterialsRheologyBbar'};
        md.inversion.maxsteps = maxsteps;
        md.inversion.cost_functions = [101 502];
        md.inversion.cost_functions_coefficients = ones(md.mesh.numberofvertices,length(md.inversion.cost_functions));
        md.inversion.min_parameters = cuffey(273)*ones(md.mesh.numberofvertices,1);
        md.inversion.max_parameters = cuffey(200)*ones(md.mesh.numberofvertices,1);
   md.verbose = verbose('solution',false,'control',true);
        
        % Starting L-curve analysis:
        %
        % J = Jo + alpha*R.
        % J: total cost function to be minimized.
        % Jo: sum of the objective cost function(s) (ex.: 101, or 101+102, or 101+102+103).
        % R: regularization term (ex.: 502).
        % alpha: weight of the regularization term.
        %
        % L-curve analysis is a method to find the best value for alpha.
        % Basicaly, it loops over different values of alpha. A plot can be generated for each
        % respective value of Jo and R (R versus Jo).
        %
        min_alpha   = 1.e-20;
        max_alpha   = 1.e-11;
        nstep_alpha = 30;
        log_step    = (log10(max_alpha)-log10(min_alpha))/nstep_alpha;
        log_alphas  = [log10(min_alpha):log_step:log10(max_alpha)];
        alphas      = 10.^log_alphas;
        J           = zeros(length(alphas),length(md.inversion.cost_functions)+1);
        % Loop over the alphas
        for i=1:length(alphas),
                disp('------------------------------------------------------------');
                disp(['      alpha iteration: ' int2str(i) '/' int2str(length(alphas)) ', alpha value: ' num2str(alphas(i))]);
                disp('------------------------------------------------------------');
                md.inversion.cost_functions_coefficients(:,end) = alphas(i);
                md = solve(md,'Stressbalance');
                J(i,:) = md.results.StressbalanceSolution.J(end,:); % J comes in [Jo, alphaR, J]. In this example: [101, alpha*502, 101+alpha*502]
        end

        % Plot the L-curve (log-log)
        Jo = zeros(length(alphas),1);
        for i=1:size(J,2)-2,
                Jo = Jo + J(:,i); % sum of the cost functions (no regularization term). In this example, only 101
        end
        R  = J(:,end-1)./alphas(:); % only the regularization term
        
        % Tip:
        % A rescale in the axes may be useful to visualize the L-curve.
        %
        % Remember: J = Jo + alpha*R
        %
        % Apply a linear transformation on the original axis (Jo, R): 
   %
        % |   1       alpha | | Jo  |   | Jo + alpha*R |   |    J    |
        % |                 | |     | = |              | = |         |
        % | 1/alpha     1   | |  R  |   | Jo/alpha + R |   | J/alpha |
        %
        % Then, use:
        % Jo2 = J(:,end);
        % R2  = J(:,end)./alphas(:);
        % loglog(Jo2,R2,... 
        %
        loglog(Jo,R,'-s','Color',[.3 .8 .4],'MarkerSize',6,'MarkerFaceColor','m','MarkerEdgeColor','k','LineWidth',2)
        voffset=0.1*R;
        hoffset=0.1*Jo;
        text(Jo+hoffset,R+voffset,[repmat('\alpha = ',length(alphas),1) num2str(alphas(:),'%2.0e')],...
            'FontSize',10,'HorizontalAlignment','left','VerticalAlignment','Middle')
        xlabel('$\mathrm{log}(\mathcal{J}_0$)','Interpreter','latex')
        ylabel('$\mathrm{log}(\mathcal{R})$','Interpreter','latex')
end
if any(step==4)
        %invert for ice rigidity
        loadmodel('model2.mat');

        %Set up inversion parameters
        maxsteps = 20;
        md.inversion.iscontrol = 1;
        md.inversion.control_parameters = {'MaterialsRheologyBbar'};
        md.inversion.maxsteps = maxsteps;
        md.inversion.cost_functions = [101 502];
        md.inversion.cost_functions_coefficients      = ones(md.mesh.numberofvertices,1);
        md.inversion.cost_functions_coefficients(:,2) = 4.e-17*ones(md.mesh.numberofvertices,1); % here you can use the best value found for alpha
        md.inversion.min_parameters    = cuffey(273)*ones(md.mesh.numberofvertices,1);
        md.inversion.max_parameters    = cuffey(200)*ones(md.mesh.numberofvertices,1);

        %Go solve!
        md.verbose=verbose(0);
        md=solve(md,'Stressbalance');
        plotmodel(md,'axis#all','tight','data',md.results.StressbalanceSolution.MaterialsRheologyBbar,'caxis',[ 1.3 1.9]*10^8,'title','inferred B',...
                'data',md.results.StressbalanceSolution.Vel,'title','modeled velocities')
end