kill(all)$

maxima: heat:heat conduction (1D),variable : k(T),version 1.0,(plane wall)
peter.vlasschaert@gmail.com,20/07/2021

variable thermalbconductivity =k(T)
=========================
depends on the conducting medium (Temperature dependent)

model :plane wall
=====

→ ode :

pa:'diff(k*'diff(T,x),x)=0;

pb:k=α*(1+β*T);

when β=0 → k=α

depends(T,x);

pc:subst([pb],pa);

≥ final ode 

pc1:ev(pc,diff);

→ bc : (b)oundary (c)onditions (L =length of the wall)

≥ x=0 (begin : wall) ,T=T1

bc1:T(0) = T1;

≥ x=L (end : wall),T=T2

bc2:T(L)=T2;

→ heat flow 'through the wall'

pd:q=-k*A*diff(T,x);

pd1:subst([pb],pd);

solve : model ode plane wall .
=======================
use :pc1

m:ode2(pc1,T,x);

m1:solve(m,T)[2];

reason : to select pos solution because :suppose ' T>0'
rem : ? why negative ,no cooling in the system.

listofvars(m);

use :{bc1,bc2}

→ bc1

bc1a:subst([x=0,T=T1],m1);

b1ca:β*bc1a;

bc1a:ev(%+1);

bc1a:expand((lhs(%))^2=(rhs(%))^2);

bc1a:lhs(%)-rhs(%)=0;

bc1a: quadratic equation  into T1

→ bc2

bc2:subst([x=L,T=T2],m1);

bc2:bc2*β;

bc2:bc2+1;

bc2:expand((lhs(%))^2=(rhs(%))^2);

bc2:lhs(%)-rhs(%)=0;

bc2: quadratic equation  into T2

see :complexity to solve system :suppose and some algebra (bc1a,bc2)

→ find{bc1a,bc2}

p1:solve([bc1a,bc2],[%k1,%k2]);

fnal solution : ode+bc
================

p2:subst(flatten(p1),m1);

when β=0

p2b:'limit(p2,β,0)=rhs(ratsimp(limit(p2,β,0)));

heat tranfer to wall:q

first step :differentation p2

p3:'diff(T,x)=diff(rhs(p2),x);

second step: heat flow 'through the wall'

p4:ev(pd1);

p4a:subst([p3],p4);

suppose : A=1, 'surface area plane wall'

A:1;

p4b:ev(p4a);

p4b1:facsum(p4b);

rem :need help find solution bc1a & bc2 ,'see above'

when β=0, 'solution good one,see below :simple result'

p4b2::'limit(p4b1,β,0)=rhs(ratsimp(limit(p4b1,β,0)));