How to couple interface of two domains in MATLAB?

Hello,

Can you give me an idea/useful links/example code of similar type of problem to solve the following problems(variation of u1 and u2 along x asis)? Actually I am struggling with the interface coupling/interface boundary conditions.

Any kinds of suggestion will be appreciated!

Regards,

Anadi

[Bill Greene]

The definition of your problem is not clear to me. From your definition of S, both and must be defined over the entire domain, . But what defines the value of in the domain and in the domain ?

[Anadi Mondal]

Hello @Bill Greene,

Thank you for your reply. Actually u1 should be distributed between a and b(a<=x>=b for u1) ,and u2 should be distributed between b and c(b<=x>=c for u2). For another case, you can consider S as constant or u2 in the source (S)term as zero.

If you need more info to make a comment on this problem, let me know.

Regards,

Anadi

[Bill Greene]

In that case, you actually have a PDE with a single dependent variable and a diffusion coefficient, D, that is discontinuous at x=b. You can solve this easily with the pdepe function.

[Anadi Mondal]

Hello @Bill Greene,

Actually u1 and u2 have to be continuous at the interface. Please see the inetrafce boundary condition in the problem statement. I have to satisfy interface boundary condition anyhow, u1(b)=u2(b),-D1*u1'(b)=-D2*u2'. How can I make the variables(u1 and u2) continuous at interface?

Regards,

Anadi

[Anadi Mondal]

Hello @Bill Greene,

Can you send me a link/example code where two pde in two region have been solved with interface boundary conditions?

Regards,

Anadi

[Taj]

In this case you can defined a double node at the interface u1(b)=u2(b). But in this case you will the problem of ghost point values that youn can solve by using the second coupling condition for example after approximation your nodes e.g x_I is your interface node in the schme for the left hand boundary you will have u1(x_I+1) also for the right hand region you will have u2(x_I-1). These you can interpolate by using one sided approximation as well as the central differnce approximation.

Answer 1

[Bill Greene]

As I said, I believe that the expression for S in your definition of the problem is not correct and your explanation did not really clarify this.

Nevertheless, here is a solution using pdepe to the problem I think you may be trying to solve. As I said in my comment, only a single dependent variable (PDE) is needed to describe the problem-- not two as you show in your definition.

function twoRegionExample

a=0; b=1; c=3;

x2=linspace(b,c,20);

N1=10;

x = [linspace(a,b,N1) x2(2:end)];

t = linspace(0,.05,50);

A=10; B=2;

pdeFunc = @(x,t,u,DuDx) heatpde(x,t,u,DuDx,b);

icFunc = @(x) heatic(x, A,B, b);

bcFunc = @(xl,ul,xr,ur,t) heatbc(xl,ul,xr,ur,t,A,B);

m=0;

u = pdepe(m, pdeFunc,icFunc,bcFunc,x,t);

figure; plot(x, u(1,:)); grid on;

title 'initial value'; xlabel 'x';

figure; plot(x, u(end,:)); grid on;

title 'solution at final time'; xlabel 'x';

figure; plot(t, u(:,N1)); grid on;

title 'solution at x=b as a function of time'; xlabel 'time';

end

function [c,f,s] = heatpde(x,t,u,DuDx,b)

c = 1;

D1=10; D2=3;

if(x < b)

f = D1*DuDx;

else

f = D2*DuDx;

end

s = 0;

end

function u0 = heatic(x,T1,T2,b)

if(x < b)

u0 = T1;

else

u0 = T2;

end

end

% --------------------------------------------------------------

function [pl,ql,pr,qr] = heatbc(xl,ul,xr,ur,t,A,B)

pl = ul-A;

ql = 0;

pr = ur-B;

qr = 0;

end

[Francesco Santoni]

Hello,

how would you implement a different Neumann boundary condition, such as: ?