Advanced Search

Question: Can pdsolve provide solutions for an axially heated rod

Question:Can pdsolve provide solutions for an axially heated rod

C_R 3927 Maple 2026
pde numeric pdsolve differential-equations
July 16 2026
2

I am pretty new to pdsolve.

?pdsolve/numeric states

PDEsys - single or set or list of time-dependent partial differential equations in two independent variables

This could be interpreted in a way that problems with 3 independent variables can be handled (time + two others).

The rest of the helppage only describes two independent variables and does not mention time.

In case the above interpretation is wrong, I have two questions:

  • If only two independent variables can be treated, why is time-depencency mentioned at all?
  • Is it possible to integrate the PDE of an axially heated rod of finite length and radius with the help of pdsolve where on one end of the rod a heat flux q=q(r) is applied?

Any links to similar problems solved with Maple (not necessary with pdsolve) are very much appreciated.

Edit: Updated BCs

PDE := diff(T(r, z, t), t) = alpha*(diff(T(r, z, t), r, r)+(diff(T(r, z, t), r))/r+diff(T(r, z, t), z, z))

diff(T(r, z, t), t) = alpha*(diff(diff(T(r, z, t), r), r)+(diff(T(r, z, t), r))/r+diff(diff(T(r, z, t), z), z))

 (1)

Updated BC

BC1 := -k*(D[2](T))(r, 0, t) = q; BC2 := (D[2](T))(r, L, t) = 0; BC3 := (D[1](T))(R, z, t) = 0; BC4 := (D[1](T))(0, z, t) = 0; IC := T(r, z, 0) = 0

T(r, z, 0) = 0

 (2)

NULL

Corrected original BC (this is a different case of constant external temperature)

BC1 := -k*(D[2](T))(r, 0, t) = q; BC2 := T(r, L, t) = 0; BC3 := T(R, z, t) = 0; BC4 := (D[1](T))(0, z, t) = 0; IC := T(r, z, 0) = 0

T(r, z, 0) = 0

 (3)

NULL

Download 2d_BVP.mw
Please Wait...