How can the Heat Equation be solved for a periodic heating scenario?

  • Thread starter Thread starter TobyDarkeness
  • Start date Start date
  • Tags Tags
    Heat Heat equation
TobyDarkeness
Messages
38
Reaction score
0
thanks allot they worked out fine, just another quick question if could help.
A semi-infinite bar 0<x<infinity is subject to periodic heating at x=0; the temperature at x=0 is T_0cos\omegat and is zero at x=infinity. By solving the heat equation show that

T(x,t)= T_0exp(\alphax)cos(\omegat-x\sqrt{\omega})

where alpha is a constant to be determined.

I know we have to separate the variables and solve the t dependence first, but its not really working. Any advice on how to tackle this question appropriately.
 
Physics news on Phys.org
my attempt so far

∂T/∂t= 1/2*(∂^2T/∂x^2)

T(x,t)=X(x)T(t)

∂/∂t*[X(x)T(t)]=1/2*[(∂^2)/(∂x^2)]*(X(x)T(t))

X(x)*[∂T(t)/∂t]=1/2*T(t)*[∂^2X(x)]/[∂x^2]

dividing through by 1/[X(x)T(t)]


1/[T(t)]*[∂T(t)/∂t]=1/2*[1/X(x)]*(∂^2 X(x))/∂x^2


2/T(t)*∂T(t)/∂t=1/X(x)*[(∂^2X(x))/(∂x^2)]

boundary conditions
T(x,t) =T_0exp(αx)cos(ωt − x sqrtω)

T(0,t)=T_0cos(ωt)

T(infinity,0)=0
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top