pdeval (MATLAB Functions)

Evaluate the numerical solution of a PDE using the output of pdepe

Syntax

[uout,duoutdx] = pdeval(m,xmesh,ui,xout)

Arguments

m
Symmetry of the problem: slab = 0, cylindrical = 1, spherical = 2. This is the first input argument used in the call to pdepe.

xmesh
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.

ui

A vector sol(j,:,i) that approximates component i of the solution at time and mesh points xmesh, where sol is the solution returned by pdepe.

xout

A vector of points from the interval [x0,xn] at which the interpolated solution is requested.

`m`	Symmetry of the problem: slab = `0`, cylindrical = `1`, spherical = `2`. This is the first input argument used in the call to `pdepe`.
`xmesh`	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.
`ui`	A vector `sol`(`j`,:,`i`) that approximates component `i` of the solution at time and mesh points `xmesh`, where `sol` is the solution returned by `pdepe`.
`xout`	A vector of points from the interval [`x0`,`xn`] at which the interpolated solution is requested.

Description

[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

rather than the flux

. Although the flux is continuous, the partial derivative may have a jump at a material interface.

See Also

pdepe

pdepe peaks