Am I doing thid correct. pdepe function

Hello to every one. I am trying to solve this problem using pdepe funcion. u_t = u_xx x in [0,1] t>=0, with boudary conditions

u_x(0,t)= pi*e^⁽-pi^2*t), and u(1,t)=0 initial condition u(x,0)=sin(pi*x)

having this matlab functions:

boundary conditios:

function [pl,ql,pr,qr] = bc1(xl,ul,xr,ur,t)

%BC1: MATLAB function M-file that specifies boundary conditions

%for a PDE in time and one space dimension.

pl = ul;

ql = pi*exp(-pi*pi*t);

pr = ur;

qr = 0;

intital conditions

function [c,b,s]=PDEeqn1(x,t,u,DuDx)

c=1;

b=DuDx;

s=0; %-DuDx+u;

end

with

x=0:0.1:1;

t=0:0.0005k:0.5;

m=0;

u = pdepe(m,@PDEeqn1,@initial1,@bc1,x,t);

My question is, if I am providing the rigth boudary conditios for my problem.

Thank to all of you.

Carlos

Answer 1

[Torsten]

x = 0:0.1:1;
t = 0:0.05:0.5;
m = 0;
u = pdepe(m,@PDEeqn1,@initial1,@bc1,x,t);
plot(x,[u(1,:);u(2,:);u(3,:);u(4,:);u(end,:)])

function u0 = initial1(x)
  u0 = sin(pi*x);
end

function [c,b,s]=PDEeqn1(x,t,u,DuDx)
  c = 1;
  b = DuDx;
  s = 0; %-DuDx+u;
end

function [pl,ql,pr,qr] = bc1(xl,ul,xr,ur,t)
  %BC1: MATLAB function M-file that specifies boundary conditions
  %for a PDE in time and one space dimension.
  pl = -pi*exp(-pi^2*t);
  ql = 1.0;
  pr = ur;
  qr = 0;
end

[Carlos Sosa Paz]

Thank you for your quick response, so to be clear, pl is for u_x(0,t) and which is for u_x(1,t) Thank you again.

[Torsten]

The boundary conditions are of the from

p + q*f = 0

where pl,ql stand for p,q at x=0 and pr,qr stand for p,q at x=1.

You set

f = DuDx

Thus the boundary condition form is

p + q*du/dx = 0

Everything else follows:

If you want to specify

u = value at x=0

, you have to set

pl = ul - value
ql = 0

If you want to specify

du/dx = value at x = 0

, you have to set

pl = -value
ql = 1

Same for the right boundary point x = 1.

[Carlos Sosa Paz]

Thank you for your answer. It clarifies a lot

[Carlos Sosa Paz]

I have this problem and when comparing it with a colleague we found different solutions I don't know if I'm doing the right thing. Thank you.

u_{t} = u_{xx} -u_{x}+ u, x in [0,1], t>=0,

boundary conditions

u_x(0,t)=2t, u(1,t)=t^2/2

initial condition

u(x,0)=sin(x)+cos(x)

matlab code:

function [c,b,s]=PDEeqn1(x,t,u,DuDx)

c=1;

b=DuDx;

s=-DuDx+u;

end

function [pl,ql,pr,qr] = bc1(xl,ul,xr,ur,t)

%BC1: MATLAB function M-file that specifies boundary conditions

%for a PDE in time and one space dimension.

pl = 2*t; % ul;%2*t;%-pi*exp(-pi*pi*t);

ql = 1;%1;

pr = t*t/2;

qr = 1;%ur;

function value = initial1(x)

%INITIAL1: MATLAB function M-file that specifies the initial condition

%for a PDE in time and one space dimension.

value = sin(x)+cos(x);%sin(pi*x); %4*x-(4*x*x); %sin(x)+cos(x);

Tank you in advance.

[Torsten]

xmesh = linspace(0,1,100);
tspan = 0:0.5:1;
m = 0;
sol = pdepe(m,@PDEeqn1,@initial1,@bc1,xmesh,tspan);
plot(xmesh,[sol(1,:);sol(2,:);sol(3,:)])

function [c,b,s]=PDEeqn1(x,t,u,DuDx)
  c = 1;
  b = DuDx;
  s = -DuDx + u;
end
function [pl,ql,pr,qr] = bc1(xl,ul,xr,ur,t)
%BC1: MATLAB function M-file that specifies boundary conditions
%for a PDE in time and one space dimension.
  pl = -2*t; 
  ql = 1;
  pr = ur - t^2/2;
  qr = 0;
end
function value = initial1(x)
%INITIAL1: MATLAB function M-file that specifies the initial condition
%for a PDE in time and one space dimension.
  value = sin(x) + cos(x);
end

[Carlos Sosa Paz]

Thank you, now I having the same output.