Message Boards Message Boards

1D HeatTransfer problem reformulation

Hi, I would like to solve the 1D Heat Transfer problem for $ T(x,t) $ on a rod of length $ L $ that has a sinus temperature at one end and is isolated at the other end:

(IBVP 1): $ T_t - a^2 T_xx = 0 $ with
BC: $ T(x=0,t) = T_1 sin(wt), T_x(x=L,t) = 0, $ and IC: $T(x,t=0) = T_0 $

This is easily solved using the NDSolveValue command (see below).

As I am finally interested in an analytic solution of (IBVP 1), in a frist step I reformulated the initial boundary value problem using
$T(x,t) = U(x,t) + f(t),$ with $ f(t) = T_1 sin(wt) $

arriving at the inhomogeneous (IBVP 2) with homogeneous BCs for $ U(x,t):$

(IBVP 2): $ U_t - a^2 U_xx = Q $ with $ Q=- df(t)/dt,$
BC: $U(x=0,t) = 0, U_x(x=L,t) = 0, $ IC: $ U(x,t=0) = T_0$

To confirm the identity of (IBVP 1) with (IBVP 2), I implemented both with the NDSolve command (see below). Surprisingly, the solutions are clearly different. Is there an error in my arguments? Is there a problem with my implementation of the numerical solution in NDSolve?

I noticed that an additional factor of 1000 to the souce term $ Q $ makes both solutions more similar. This brings me to the question of why the source term (as the time derivative of $f(t)$) is so very small .......?

Analytically, (IBVP 2) could be further analysed by Fourier sin transformation. If there is a known analytical solution of IBVP 1, I would be very happy for a hint to a reference. I would be very grateful for any suggestions. Thank you!

Here is my implementation in Mathematica:

POSTED BY: Uwe Schlink
3 Replies
Sort By:

Many thanks to both of you for the elegant solutions! It took me a while to understand and follow the calculations. In the end, the resulting series approximates the numerical solution (NDSolve) very well. Depending on the parameters (L, a), only the first 2 to 3 terms of the series are sufficient. Thank you very much for your help!
POSTED BY: Uwe Schlink

If your equation is: $$\left\{T^{(0,1)}(x,t)-a^2 T^{(2,0)}(x,t)=0,T(0,t)=\text{T1} \sin (t w),T^{(1,0)}(L,t)=0,T(x,0)=\text{T0}\right\}$$ Mathematica can solve analytically by infinty series:

     $Version
     (*"13.2.0 for Microsoft Windows (64-bit) (November 18, 2022)"*)

       DSolve[{D[T[x, t], t] - a^2*D[T[x, t], {x, 2}] == 0, 
   = T1 Sin[w t], Derivative[1, 0][T][L, t] == 0, 
       T[x, 0] == T0}, T[x, t], {x, t}]

    (*{{T[x, t] -> 
       T1 Sin[t w] + 
        Inactive[Sum][(
         4 E^(-((a^2 \[Pi]^2 t (1 - 2 K[1])^2)/(
           4 L^2))) (-T0 + (
            4 L^2 T1 w (-a^2 \[Pi]^2 (1 - 2 K[1])^2 + 
               E^((a^2 \[Pi]^2 t (1 - 2 K[1])^2)/(
                4 L^2)) (a^2 \[Pi]^2 Cos[t w] (1 - 2 K[1])^2 + 
                  4 L^2 w Sin[t w])))/(
            16 L^4 w^2 + 
             a^4 \[Pi]^4 (1 - 2 K[1])^4)) Sin[(\[Pi] x (-1 + 2 K[1]))/(
           2 L)])/(\[Pi] - 2 \[Pi] K[1]), {K[1], 1, \[Infinity]}]}}*)

$T(x,t)=\text{T1} \sin (t w)+\underset{K[1]=1}{\overset{\infty }{\sum }}\frac{4 e^{-\frac{a^2 \pi ^2 t (1-2 K[1])^2}{4 L^2}} \left(-\text{T0}+\frac{4 L^2 \text{T1} w \left(-a^2 \pi ^2 (1-2 K[1])^2+e^{\frac{a^2 \pi ^2 t (1-2 K[1])^2}{4 L^2}} \left(a^2 \pi ^2 \cos (t w) (1-2 K[1])^2+4 L^2 w \sin (t w)\right)\right)}{16 L^4 w^2+a^4 \pi ^4 (1-2 K[1])^4}\right) \sin \left(\frac{\pi x (-1+2 K[1])}{2 L}\right)}{\pi -2 \pi K[1]}$
POSTED BY: Mariusz Iwaniuk
POSTED BY: Roland Franzius
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract