role of flux f in bcfun of pdepe for systems of PDEs

3 次查看(过去 30 天)
To solve utt=uxx, as the flux term f has a 0 entry in pdefun, the same f is in bcfun. In bcfun, there's q(x,t)f(x,t,u,ux), which appears to mean that when f=[0;ux(1)], q=[1;0] is identical to q=[0;0] because the 1st entry of q doesn't matter after multiplying the 0 in f. But when q=[1;0] is used, the program yields wrong results, why?
wave()
function wave
x=linspace(-pi,pi);t=linspace(0,2*pi);
sol=pdepe(0,@pdefun,@icfun,@bcfun,x,t);
u=sol(:,:,1);
surf(x,t,u)
function [c,f,s]=pdefun(x,t,u,ux)
c=[1;1];
f=[0;ux(1)];
s=[u(2);0];
end
function u0=icfun(x)
u0=[sin(x);0];
end
function [pl,ql,pr,qr]=bcfun(xl,ul,xr,ur,t)
% p(x,t,u)+q(x,t)f(x,t,u,ux)=0
pl=ul;pr=ur;ql=[1;0];qr=[1;0];
end
end
  2 个评论
Torsten
Torsten 2024-3-3
If you try to solve the standard-hyperbolic equation (wave equation) with a solver for parabolic-elliptic equations (for which f should always be different from 0), you shouldn't be surprised to get results that cannot be explained.
feynman feynman
feynman feynman 2024-3-4
ok, so q(x,t)f(x,t,u,ux) isn't really multiplication entry by entry?

请先登录,再进行评论。

采纳的回答

Torsten
Torsten 2024-3-4
移动:Torsten 2024-3-6
This multiplication is nowhere performed within "pdepe". But if you look at the pdepe code below, the cases where q = 0 and q different from 0 are treated differently (independent of f).
type pdepe
function varargout = pdepe(m,pde,ic,bc,xmesh,t,options,varargin) %PDEPE Solve initial-boundary value problems for parabolic-elliptic PDEs in 1-D. % SOL = PDEPE(M,PDEFUN,ICFUN,BCFUN,XMESH,TSPAN) solves initial-boundary % value problems for small systems of parabolic and elliptic PDEs in one % space variable x and time t to modest accuracy. There are npde unknown % solution components that satisfy a system of npde equations of the form % % c(x,t,u,Du/Dx) * Du/Dt = x^(-m) * D(x^m * f(x,t,u,Du/Dx))/Dx + s(x,t,u,Du/Dx) % % Here f(x,t,u,Du/Dx) is a flux and s(x,t,u,Du/Dx) is a source term. m must % be 0, 1, or 2, corresponding to slab, cylindrical, or spherical symmetry, % respectively. The coupling of the partial derivatives with respect to % time is restricted to multiplication by a diagonal matrix c(x,t,u,Du/Dx). % The diagonal elements of c are either identically zero or positive. % An entry that is identically zero corresponds to an elliptic equation and % otherwise to a parabolic equation. There must be at least one parabolic % equation. An entry of c corresponding to a parabolic equation is permitted % to vanish at isolated values of x provided they are included in the mesh % XMESH, and in particular, is always allowed to vanish at the ends of the % interval. The PDEs hold for t0 <= t <= tf and a <= x <= b. The interval % [a,b] must be finite. If m > 0, it is required that 0 <= a. The solution % components are to have known values at the initial time t = t0, the % initial conditions. The solution components are to satisfy boundary % conditions at x=a and x=b for all t of the form % % p(x,t,u) + q(x,t) * f(x,t,u,Du/Dx) = 0 % % q(x,t) is a diagonal matrix. The diagonal elements of q must be either % identically zero or never zero. Note that the boundary conditions are % expressed in terms of the flux rather than Du/Dx. Also, of the two % coefficients, only p can depend on u. % % The input argument M defines the symmetry of the problem. PDEFUN, ICFUN, % and BCFUN are function handles. % % [C,F,S] = PDEFUN(X,T,U,DUDX) evaluates the quantities defining the % differential equation. The input arguments are scalars X and T and % vectors U and DUDX that approximate the solution and its partial % derivative with respect to x, respectively. PDEFUN returns column % vectors: C (containing the diagonal of the matrix c(x,t,u,Dx/Du)), % F, and S (representing the flux and source term, respectively). % % U = ICFUN(X) evaluates the initial conditions. For a scalar X, ICFUN % must return a column vector, corresponding to the initial values of % the solution components at X. % % [PL,QL,PR,QR] = BCFUN(XL,UL,XR,UR,T) evaluates the components of the % boundary conditions at time T. XL and XR are scalars representing the % left and right boundary points. UL and UR are column vectors with the % solution at the left and right boundary, respectively. PL and QL are % column vectors corresponding to p and the diagonal of q, evaluated at % the left boundary, similarly PR and QR correspond to the right boundary. % 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. % % PDEPE returns values of the solution on a mesh provided as the input % array XMESH. The entries of XMESH must satisfy % a = XMESH(1) < XMESH(2) < ... < XMESH(NX) = b % for some NX >= 3. Discontinuities in c and/or s due to material % interfaces are permitted if the problem requires the flux f to be % continuous at the interfaces and a mesh point is placed at each % interface. The ODEs resulting from discretization in space are integrated % to obtain approximate solutions at times specified in the input array % TSPAN. The entries of TSPAN must satisfy % t0 = TSPAN(1) < TSPAN(2) < ... < TSPAN(NT) = tf % for some NT >= 3. The arrays XMESH and TSPAN do not play the same roles % in PDEPE: The time integration is done with an ODE solver that selects % both the time step and formula dynamically. The cost depends weakly on % the length of TSPAN. Second order approximations to the solution are made % on the mesh specified in XMESH. Generally it is best to use closely % spaced points where the solution changes rapidly. PDEPE does not select % the mesh in x automatically like it does in t; you must choose an % appropriate fixed mesh yourself. The discretization takes into account % the coordinate singularity at x = 0 when m > 0, so it is not necessary to % use a fine mesh near x = 0 for this reason. The cost depends strongly on % the length of XMESH. % % The solution is returned as a multidimensional array SOL. UI = SOL(:,:,i) % is an approximation to component i of the solution vector u for % i = 1:npde. The entry UI(j,k) = SOL(j,k,i) approximates UI at % (t,x) = (TSPAN(j),XMESH(k)). % % SOL = PDEPE(M,PDEFUN,ICFUN,BCFUN,XMESH,TSPAN,OPTIONS) solves as above % with default integration parameters replaced by values in OPTIONS, an % argument created with the ODESET function. Only some of the options of % the underlying ODE solver are available in PDEPE - RelTol, AbsTol, % NormControl, InitialStep, and MaxStep. See ODESET for details. % % [SOL,TSOL,SOLE,TE,IE] = PDEPE(M,PDEFUN,ICFUN,BCFUN,XMESH,TSPAN,OPTIONS) % with the 'Events' property in OPTIONS set to a function handle EVENTS, % solves as above while also finding where event functions g(t,u(x,t)) % are zero. For each function you specify whether the integration is to % terminate at a zero and whether the direction of the zero crossing % matters. These are the three column vectors returned by EVENTS: % [VALUE,ISTERMINAL,DIRECTION] = EVENTS(M,T,XMESH,UMESH). % XMESH contains the spatial mesh and UMESH is the solution at the mesh % points. Use PDEVAL to evaluate the solution between mesh points. % For the I-th event function: VALUE(I) is the value of the function, % ISTERMINAL(I) = 1 if the integration is to terminate at a zero of this % event function and 0 otherwise. DIRECTION(I) = 0 if all zeros are to be % computed (the default), +1 if only zeros where the event function is % increasing, and -1 if only zeros where the event function is decreasing. % Output TSOL is a column vector of times specified in TSPAN, prior to % first terminal event. SOL(j,:,:) is the solution at T(j). TE is a vector % of times at which events occur. SOLE(j,:,:) is the solution at TE(j) and % indices in vector IE specify which event occurred. % % If UI = SOL(j,:,i) approximates component i of the solution at time TSPAN(j) % and mesh points XMESH, PDEVAL evaluates the approximation and its partial % derivative Dui/Dx at the array of points XOUT and returns them in UOUT % and DUOUTDX: [UOUT,DUOUTDX] = PDEVAL(M,XMESH,UI,XOUT) % NOTE that the partial derivative Dui/Dx is evaluated here rather than the % flux. The flux is continuous, but at a material interface the partial % derivative may have a jump. % % Example % x = linspace(0,1,20); % t = [0 0.5 1 1.5 2]; % sol = pdepe(0,@pdex1pde,@pdex1ic,@pdex1bc,x,t); % solve the problem defined by the function pdex1pde with the initial and % boundary conditions provided by the functions pdex1ic and pdex1bc, % respectively. The solution is obtained on a spatial mesh of 20 equally % spaced points in [0 1] and it is returned at times t = [0 0.5 1 1.5 2]. % Often a good way to study a solution is plot it as a surface and use % Rotate 3D. The first unknown, u1, is extracted from sol and plotted with % u1 = sol(:,:,1); % surf(x,t,u1); % PDEX1 shows how this problem can be coded using subfunctions. For more % examples see PDEX2, PDEX3, PDEX4, PDEX5. The examples can be read % separately, but read in order they form a mini-tutorial on using PDEPE. % % See also PDEVAL, ODE15S, ODESET, ODEGET, FUNCTION_HANDLE. % The spatial discretization used is that of R.D. Skeel and M. Berzins, A % method for the spatial discretization of parabolic equations in one space % variable, SIAM J. Sci. Stat. Comput., 11 (1990) 1-32. The time % integration is done with ODE15S. PDEPE exploits the capabilities this % code has for solving the differential-algebraic equations that arise when % there are elliptic equations and for handling Jacobians with specified % sparsity pattern. % Elliptic equations give rise to algebraic equations after discretization. % Initial values taken from the given initial conditions may not be % "consistent" with the discretization, so PDEPE may have to adjust these % values slightly before beginning the time integration. For this reason % the values output for these components at t0 may have a discretization % error comparable to that at any other time. No adjustment is necessary % for solution components corresponding to parabolic equations. If the mesh % is sufficiently fine, the program will be able to find consistent initial % values close to the given ones, so if it has difficulty initializing, % try refining the mesh. % Lawrence F. Shampine and Jacek Kierzenka % Copyright 1984-2022 The MathWorks, Inc. % $Revision: 1.8.4.9.12.1 $ $Date: 2013/09/27 03:10:20 $ if nargin < 7 options = []; end switch m case {0, 1, 2} otherwise error(message('MATLAB:pdepe:InvalidM')) end nt = length(t); if nt < 3 error(message('MATLAB:pdepe:TSPANnotEnoughPts')) end if any(diff(t) <= 0) error(message('MATLAB:pdepe:TSPANnotIncreasing')) end xmesh = xmesh(:); if m > 0 && xmesh(1) < 0 error(message('MATLAB:pdepe:NegXMESHwithPosM')) end nx = length(xmesh); if nx < 3 error(message('MATLAB:pdepe:XMESHnotEnoughPts')) end if any(diff(xmesh) <= 0) error(message('MATLAB:pdepe:XMESHnotIncreasing')) end % Initialize the nx-1 points xi where functions will be evaluated % and as many coefficients in the difference formulas as possible. % For problems with coordinate singularity, use 'singular' formula % on all subintervals. singular = (xmesh(1) == 0 && m > 0); xL = xmesh(1:end-1); xR = xmesh(2:end); xM = xL + 0.5*(xR-xL); switch m case 0 xi = xM; zeta = xi; case 1 if singular xi = (2/3)*(xL.^2 + xL.*xR + xR.^2) ./ (xL+xR); else xi = (xR-xL) ./ log(xR./xL); end zeta = (xi .* xM).^(1/2); case 2 if singular xi = (2/3)*(xL.^2 + xL.*xR + xR.^2) ./ (xL+xR); else xi = xL .* xR .* log(xR./xL) ./ (xR-xL); end zeta = (xL .* xR .* xM).^(1/3); end if singular xim = (zeta .^ (m+1))./xi; else xim = xi .^ m; end zxmp1 = (zeta.^(m+1) - xL.^(m+1)) / (m+1); xzmp1 = (xR.^(m+1) - zeta.^(m+1)) / (m+1); % Form the initial values with a column of unknowns at each % mesh point and then reshape into a column vector for dae. temp = feval(ic,xmesh(1),varargin{:}); if size(temp,1) ~= numel(temp) error(message('MATLAB:pdepe:InvalidOutputICFUN')) end npde = length(temp); y0 = zeros(npde,nx); y0(:,1) = temp; for j = 2:nx y0(:,j) = feval(ic,xmesh(j),varargin{:}); end % Classify the equations so that a constant, diagonal mass matrix % can be formed. The entries of c are to be identically zero or % vanish only at the entries of xmesh, so need be checked only at one % point not in the mesh. [U,Ux] = pdentrp(singular,m,xmesh(1),y0(:,1),xmesh(2),y0(:,2),xi(1)); [c,f,s] = feval(pde,xi(1),t(1),U,Ux,varargin{:}); if any([size(c,1),size(f,1),size(s,1)]~=npde) || ... any([numel(c),numel(f),numel(s)]~=npde) error(message('MATLAB:pdepe:UnexpectedOutputPDEFUN',sprintf('%d',npde))) end [pL,qL,pR,qR] = feval(bc,xmesh(1),y0(:,1),xmesh(nx),y0(:,nx),t(1),varargin{:}); if any([size(pL,1),size(qL,1),size(pR,1),size(qR,1)]~=npde) || ... any([numel(pL),numel(qL),numel(pR),numel(qR)]~=npde) error(message('MATLAB:pdepe:UnexpectedOutputBCFUN',sprintf('%d',npde))) end D = ones(npde,nx); D( c == 0, 2:nx-1) = 0; if ~singular D( qL == 0, 1) = 0; end D( qR == 0, nx) = 0; M = spdiags( D(:), 0, npde*nx, npde*nx); % Construct block-diagonal pattern of Jacobian matrix S = kron( spdiags(ones(nx,3),-1:1,nx,nx), ones(npde,npde)); % Extract relevant options and augment with new ones reltol = odeget(options,'RelTol',[]); abstol = odeget(options,'AbsTol',[]); normcontrol = odeget(options,'NormControl',[]); initialstep = odeget(options,'InitialStep',[]); maxstep = odeget(options,'MaxStep',[]); events = odeget(options,'Events',[]); % events(m,t,xmesh,umesh) hasEvents = ~isempty(events); if hasEvents eventfcn = @(t,y) events(m,t,xmesh,y,varargin{:}); else eventfcn = []; end opts = odeset('RelTol',reltol,'AbsTol',abstol,'NormControl',normcontrol,... 'InitialStep',initialstep,'MaxStep',maxstep,'Events',eventfcn,... 'Mass',M,'JPattern',S); % Call DAE solver tfinal = t(end); try if hasEvents [t,y,te,ye,ie] = ode15s(@pdeodes,t,y0(:),opts); else [t,y] = ode15s(@pdeodes,t,y0(:),opts); end catch ME if strcmp(ME.identifier,'MATLAB:daeic12:IndexGTOne') error(message('MATLAB:pdepe:SpatialDiscretizationFailed')) else rethrow(ME); end end % Verify and process the solution if t(end) ~= tfinal nt = length(t); if ~hasEvents || (te(end) ~= t(end)) % did not stop on a terminal event warning(message('MATLAB:pdepe:TimeIntegrationFailed',sprintf('%e',t(end)))); end end sol = zeros(nt,nx,npde); for i = 1:npde sol(:,:,i) = y(:,i:npde:end); end varargout{1} = sol; if hasEvents % [sol,t,sole,te,ie] = pdepe(...) varargout{2} = t(:); sole = zeros(length(te),nx,npde); for i = 1:npde sole(:,:,i) = ye(:,i:npde:end); end varargout{3} = sole; varargout{4} = te(:); varargout{5} = ie(:); end %--------------------------------------------------------------------------- % Nested functions %--------------------------------------------------------------------------- function dudt = pdeodes(tnow,y) %PDEODES Assemble the difference equations and evaluate the time derivative % for the ODE system. u = reshape(y,npde,nx); up = zeros(npde,nx); [U,Ux] = pdentrp(singular,m,xmesh(1),u(:,1),xmesh(2),u(:,2),xi(1)); [cL,fL,sL] = feval(pde,xi(1),tnow,U,Ux,varargin{:}); % Evaluate the boundary conditions [pL,qL,pR,qR] = feval(bc,xmesh(1),u(:,1),xmesh(nx),u(:,nx),tnow,varargin{:}); % Left boundary if singular denom = cL; denom(denom == 0) = 1; up(:,1) = (sL + (m+1) * fL / xi(1)) ./ denom; else up(:,1) = pL; idx = (qL ~= 0); denom = (qL(idx)/xmesh(1)^m) .* (zxmp1(1)*cL(idx)); denom(denom == 0) = 1; up(idx,1) = ( pL(idx) + (qL(idx)/xmesh(1)^m) .* ... (xim(1)*fL(idx) + zxmp1(1)*sL(idx))) ./ denom; end % Interior points for ii = 2:nx-1 [U,Ux] = pdentrp(singular,m,xmesh(ii),u(:,ii),xmesh(ii+1),u(:,ii+1),xi(ii)); [cR,fR,sR] = feval(pde,xi(ii),tnow,U,Ux,varargin{:}); denom = zxmp1(ii) * cR + xzmp1(ii-1) * cL; denom(denom == 0) = 1; up(:,ii) = ((xim(ii) * fR - xim(ii-1) * fL) + ... (zxmp1(ii) * sR + xzmp1(ii-1) * sL)) ./ denom; cL = cR; fL = fR; sL = sR; end % Right boundary up(:,nx) = pR; idx = (qR ~= 0); denom = -(qR(idx)/xmesh(nx)^m) .* (xzmp1(nx-1)*cL(idx)); denom(denom == 0) = 1; up(idx,nx) = ( pR(idx) + (qR(idx)/xmesh(nx)^m) .* ... (xim(nx-1)*fL(idx) - xzmp1(nx-1)*sL(idx))) ./ denom; dudt = up(:); end % pdeodes % -------------------------------------------------------------------------- end % pdepe
  7 个评论
Torsten
Torsten 2024-3-5
移动:Torsten 2024-3-6
But your question is answered.
As far as I remember, you asked why pdepe handles the cases q = [0 0] and q = [1 0] differently. And this happens because up(1) is computed differently.
If you have a new question, you ahould accept the answer given here and open a new post.
By the way: Paragraph 2.3 of the paper explicitly explains how the equations in the boundary points are derived.
feynman feynman
feynman feynman 2024-3-6
移动:Torsten 2024-3-6
I appreciate your help a lot, but you put all answers in comments so I can't accept your answer by clicking any button.

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Geometry and Mesh 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by