Not enough input arguments.

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

Answer 1

[Tony Mohan Varghese]

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

[Johannes Pommerening]

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?