Not enough input arguments.

Question

Johannes Pommerening 2018-7-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/408400-not-enough-input-arguments

评论： Johannes Pommerening 2018-7-11

在 MATLAB Online 中打开

I want to solve an equation with second derivatives of t and r and got These Problems.

"Not enough input arguments.

Error in PDE_lin_z>pde1DParams (line 70) c = DwDt*B/A;

Error in pdepe (line 246) [c,f,s] = feval(pde,xi(1),t(1),U,Ux,varargin{:});

Error in PDE_lin_z (line 20) sol = pdepe(m,@pde1DParams,@pde1DInitialCondition,@pde1DBoundaryCondition,r,t);"

% <a href="https://de.mathworks.com/help/matlab/ref/pdepe.html">PDEPE 1D Example on Mathworks</a>
m = 1; 
r = linspace(0,1,50);
t = linspace(0,5,100);
sol = pdepe(m,@pde1DParams,@pde1DInitialCondition,@pde1DBoundaryCondition,r,t);
w = sol(:,:,1);
surf(r,t,w)
colorbar
title(['Numerical solution computed with ' num2str(length(r)) ' mesh points.']) 
xlabel('Radius r') 
ylabel('Time t')
function [c,f,s] = pde1DParams(r,t,w,DwDr,DwDt)
E = 70;
rho = 2.6989;
ny = 0.34;
A = (0.5-ny);
B = rho*(1-2*ny)*(1+ny)/E;
    c = DwDt*B/A;
    f = DwDr+1;
    s = 0;
end
function [w0r,w0t] = pde1DInitialCondition(r,t)
      w0r = sin(pi*r);
      w0t = 0; % ?
end
function [pl, ql, pr, qr] = pde1DBoundaryCondition(rl, wl, rr, wr, t)
      pl = wl;
      ql = 0;
      pr = pi * exp(-t); % z.B
      qr = 1;
end

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Tony Mohan Varghese 2018-7-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/408400-not-enough-input-arguments#answer_328368

在 MATLAB Online 中打开

Please refer to the example for Partial Differential Equations. The pdefun and icFun in the PDEPE function signatures are in the form of:

[c,f,s] = pdefun(x,t,u,dudx)
u = icfun(x)

Reference: Partial differential Equations

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Johannes Pommerening 2018-7-11

Thanks. I know this. But I need the second derivative of time. That's the reason why I implemented the extra DwDt. So is this not possible? Other Ideas?

请先登录，再进行评论。