MATLAB: Solving two-temperature model using pdepe

harrisonMATLAB

Hi,

Am not so good in programming but am trying to do a simuation of laser ablation using two-temperature model. However, am not able to define boundary conditions correctly which is giving me a false output due to very limited temperature change. Can someone please help me code the boudary conditions for the following equation;

The code am having for the boudary condition is;

function [pl,ql,pr,qr] = pdebc(xl,ul,xr,ur,t) % Boundary Conditions

pl = [0;ul(1)];

ql = [300;300];

pr = [ur(1)-300;ur(2)];

qr = [300;300];

end

and the one for my initial condition is;

function u0 = pdeic(x) % Initial Conditions

u0 = [300; 300];

end

Best Answer

You can refer following MATLAB answer to get an idea about how to define boundary conditions for the equations.

https://www.mathworks.com/matlabcentral/answers/17349-pdepe-and-bcfun?s_tid=answers_rc1-1_p1_Topic

Also the following documentation link contains multiple example that may help you with coding boundary conditions and getting more insight into pdepe.

https://www.mathworks.com/help/matlab/ref/pdepe.html

Related Solutions

MATLAB: Solving second order PDE

The reason that pdepe imposes a boundary condition of the flux equal zero at the

center is that this is required for the problem to be mathematically well-posed.

Imposing a prescribed temperature at the center would require that the flux go to

infinity there.

An easy way to understand this is to solve the problem with the left end a small distance

from the center and with a fine mesh. I have attached a short script below that shows this.

function matlabAnswers_6_27_2020
r0=1e-6;
x = linspace(r0,1,1000);
tf=1;
t = linspace(0,tf,40);
pdeFunc = @(x,t,u,DuDx) heatpde(x,t,u,DuDx);
icFunc = @(x) heatic(x);
bcFunc = @(xl,ul,xr,ur,t) heatbcDirichlet(xl,ul,xr,ur,t);
m=1;
sol = pdepe(m, pdeFunc,icFunc,bcFunc,x,t);
figure; plot(t, sol(:,end)); grid on; title 'Temperature at outer surface'
figure; plot(t, sol(:,1)); grid on; title 'Temperature at center'
figure; plot(x, sol(end,:)); grid; title 'Temperature at final time'
end
function [c,f,s] = heatpde(x,t,u,DuDx)
c = 1;
f = DuDx;
s = 0;
end
function u0 = heatic(x)
u0 = 0;
end
function [pl,ql,pr,qr] = heatbcDirichlet(xl,ul,xr,ur,t)
pl = ul-1;
ql = 0;
pr = 0;
qr = 1;
end

MATLAB: Problem with boundary conditions for PDEPE

From the doc for pdepe,

"When m > 0 and a = 0, boundedness of the solution near x = 0 requires that the flux f vanish at a = 0. pdepe imposes this boundary condition automatically and it ignores values returned in pl and ql."

So if you are using a value of m greater than 0 (i.e. cylindrical or spherical problem symmetry), then the solver always enforces zero flux on the left boundary no matter what you specify.

Best Answer

Related Solutions

MATLAB: Solving second order PDE

MATLAB: Problem with boundary conditions for PDEPE

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