The heat equation:
∂t∂u=α2∂x2∂2u
In steady state:
∂x2∂2u=0
Solving, we get
u=ax+b
The initial conditions, in steady–state, are
u=20,x=0
u=40,x=L
Then:
b=20,a=20/L
The temperature function in steady–state is
u(x)=20x/L+20
The boundary conditions in the transient state are
u(0,t)=50,u(L,t)=10,u(x,0)=20x/L+20
We break up the required funciton u (x,t) into two parts:
u(x,t)=us(x)+ut(x,t)
us(x) is a steady state solution, ut(x,t) is a transient solution
us(x)=ax+b
b=50,a=−40/L
us(x)=−40x/L+50
ut(x,t)=u(x,t)−us(x)
Putting boundary conditions, we have the boundary conditions relative to the transient solution :
ut(0,t)=u(0,t)−us(0)=50−50=0
ut(L,t)=u(L,t)−us(L)=10−10=0
ut(x,0)=u(x,0)−us(x)=20x/L+20+40x/L−50=60x/L−30
We have:
ut(x,t)=(Acosλx+Bsinλx)e−α2λ2t
Using boundary conditions, we get:
A=0,λ=πn/L
Then:
ut(x,t)=BsinLπnxe−L2α2π2n2t
The most general solution:
ut(x,t)=n=1∑∞BnsinLπnxe−L2α2π2n2t
Using boundary conditions:
ut(x,0)=n=1∑∞BnsinLπnx=60x/L−30,0<x<L
Bn=L20∫L(60x/L−30)sinLπnxdx=
=L2((60x/L−30)πn/L−cos(πnx/L)−L60⋅π2n2/L2−sin(πnx/L))∣0L=
=L2(−πn30Lcos(πn)−πn30L)=−πn60(cos(πn)+1)
Bn=−πn60(1+(−1)n)
Bn=0 when n is odd
Bn=−πn120 when n is even
ut(x,t)=n=2,4,6,..∑∞(−πn120sinLπnx)e−L2α2π2n2t
u(x,t)=−40x/L+50+n=2,4,6,..∑∞(−πn120sinLπnx)e−L2α2π2n2t
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