# Not enough input arguments.

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Johannes Pommerening on 2 Jul 2018
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.'])
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

Tony Mohan Varghese on 11 Jul 2018
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)
Johannes Pommerening on 11 Jul 2018
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?

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