Solve parabolic PDE problem
u1=parabolic(u0,tlist,b,p,e,t,c,a,f,d) u1=parabolic(u0,tlist,b,p,e,t,c,a,f,d,rtol) u1=parabolic(u0,tlist,b,p,e,t,c,a,f,d,rtol,atol) u1=parabolic(u0,tlist,K,F,B,ud,M) u1=parabolic(u0,tlist,K,F,B,ud,M,rtol) u1=parabolic(u0,tlist,K,F,B,ud,M,rtol,atol) u1=parabolic(___,'Stats','off')
for (x,y) ∊ Ω, or the system PDE problem
on a mesh described by p, e, and t, with boundary conditions given by b, and with initial value u0.
For the scalar case, each row in the solution matrix u1 is the solution at the coordinates given by the corresponding column in p. Each column in u1 is the solution at the time given by the corresponding item in tlist. For a system of dimension N with np node points, the first np rows of u1 describe the first component of u, the following np rows of u1 describe the second component of u, and so on. Thus, the components of u are placed in the vector u as N blocks of node point rows.
b describes the boundary conditions of the PDE problem. b can be a Boundary Condition matrix, the name of a Boundary file, or a function handle to a Boundary file. The boundary conditions can depend on t, the time. The formats of the Boundary Condition matrix and Boundary file are described in the entries on assemb and pdebound, respectively.
The geometry of the PDE problem is given by the mesh data p, e, and t. For details on the mesh data representation, see initmesh.
The coefficients c, a, d, and f of the PDE problem can be given in a variety of ways. The coefficients can depend on t, the time. They can also depend on u, the solution, and on the components of the gradient of u, namely ux and uy. For a complete listing of all options, see Scalar PDE Coefficients and Coefficients for Systems of PDEs.
atol and rtol are absolute and relative tolerances that are passed to the ODE solver.
with initial value for u being u0.
Add the Stats name-value pair at the end of any syntax to control the display of internal ODE solver statistics. Valid values for Stats are 'off' and 'on' (default).
Solve the heat equation
on a square geometry –1 ≤ x,y ≤ 1 (squareg). Choose u(0) = 1 on the disk x2 +y2 < 0.42, and u(0) = 0 otherwise. Use Dirichlet boundary conditions u = 0 (squareb1). Compute the solution at times linspace(0,0.1,20).
[p,e,t]=initmesh('squareg'); [p,e,t]=refinemesh('squareg',p,e,t); u0=zeros(size(p,2),1); ix=find(sqrt(p(1,:).^2+p(2,:).^2)<0.4); u0(ix)=ones(size(ix)); tlist=linspace(0,0.1,20); u1=parabolic(u0,tlist,'squareb1',p,e,t,1,0,0,1);
Note In expressions for boundary conditions and PDE coefficients, the symbol t is used to denote time. The variable t is often used to store the triangle matrix of the mesh. You can use any variable to store the triangle matrix, but in the Partial Differential Equation Toolbox™ expressions, t always denotes time.