Solving PDE of electrothermal coupling with f coefficient in stationary

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I am trying to solve a electrothermal system(joule heating) in stationary by using solvepde. The PDEs are shown below.
Equation(1) is for the E-field in stationary, where phi is voltage potential and sigma is the electrical conductivity of the material. Equation(2) is the relation between temperature and voltage potential in stationary, where T is temperature and k is thermal conductivity.
%%E field for equation (1)
Emodel = createpde;
geometryFromEdges(Emodel,g);
Emesh = generateMesh(Emodel,'Hmax',0.8); % generate mesh
% dirichlet boundary condition for E-domain
applyBoundaryCondition(Emodel,'dirichlet','Edge',8,'u',20);
applyBoundaryCondition(Emodel,'dirichlet','Edge',7,'u',0);
applyBoundaryCondition(Emodel,'neumann','Edge',1:6,'q',0,'g',0);
% laplace equation
specifyCoefficients(Emodel,'m',0,...
'd',0,...
'c',1,...
'a',0,...
'f',0);
Eresults = solvepde(Emodel);
ElectroThermal coupling
ETmodel = createpde;
geometryFromEdges(ETmodel,g);
generateMesh(ETmodel,'Hmax',0.8); % generate mesh
applyBoundaryCondition(ETmodel,'dirichlet','Edge',8,'u',273.15);
applyBoundaryCondition(ETmodel,'dirichlet','Edge',7,'u',273.15);
applyBoundaryCondition(ETmodel,'neumann','Edge',1:6,'q',1/kappa,'g',h*273.15/kappa);
f = @(region,state) sigma.*(state.ux.^2 + state.uy.^2); % <--------------u is for T here,
but what should I do for letting this u represent
the potential voltage according to result of Emodel above?
% region.ux means partial u/ partial x in this case.
specifyCoefficients(ETmodel,'m',0,...
'd',0,...
'c',kappa,...
'a',0,...
'f',f);
ETresults = solvepde(ETmodel);
I want to let f be the right part of equation 2 (let u represent "phi" instead of T). or is there some essential way to represent the coupled PDE to solve?
Thank you in advance, Xing An.

Akzeptierte Antwort

Alan Weiss
Alan Weiss am 27 Aug. 2018
If I understand your question, you would like f to represent the interpolated squared gradient of the previous solution to Emodel. So write a function to use the interpolated gradients of the solution Eresults, such as
function fout = myf(region,state,Eresults,sigma)
[gradx,grady] = evaluateGradient(Eresults,region.x,region.y)
fout = sigma.*(gradx.^2 + grady.^2);
Set this function as your f coefficient after getting Eresults and sigma into your workspace:
f = @(region,state)myf(region,state,Eresults,sigma)
Alan Weiss
MATLAB mathematical toolbox documentation
  4 Kommentare
Xing An
Xing An am 27 Aug. 2018
Alan, thank you very much. The code works well.
ADSW121365
ADSW121365 am 12 Jun. 2020
Alan, is it possible given identical meshes for the two problems to avoid the interpolation step in this approach by using results.XGradients etc?

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