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[uout,duoutdx] = pdeval(m,x,ui,xout)
Symmetry of the problem: slab = 0, cylindrical = 1, spherical = 2. This is the first input argument used in the call to pdepe. | |
A vector [x0, x1, ..., xn] specifying the points at which the elements of ui were computed. This is the same vector with which pdepe was called. | |
A vector sol(j,:,i)
that approximates component i of the solution at time
| |
A vector of points from the interval [x0,xn] at which the interpolated solution is requested. |
[uout,duoutdx] = pdeval(m,x,ui,xout) approximates
the solution
and its
partial derivative
at points from the interval
[x0,xn]. The pdeval function
returns the computed values in uout and duoutdx,
respectively.
Note
pdeval evaluates the partial derivative
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![]() | pdepe | peaks | ![]() |
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