Question

Find T(x,t) in a laterally insulated 2 m-long rod if k=10-4 m2/s and T(x,0)=100(2x-x2), T(0,t)=0=T(2,t).

Answer 1

$T(0,t)=0=T(2,t)$ ----- (i)

This is the boundary condition for laterally insulated rod having ends maintained at certain temperatures.

Governing equation for the following condition

$∂T(x,t)/∂t=k ∂^2T(x,t)/∂x^2$ ------------- (ii)

For separation of variable $T(x,t) = f(x).g(t)$ ---------- (iii)

Substituting equation (iii) in equation (ii);

we get $g'(t)/g(t) = k f''(x)/f(x) =l$ ----- (iv)

$g'(t)/g(t)=l$

$d(g(t))/g(t)=ldt$

Integrating both sides, we get

$ln (g(t)/c_2) = lt$

$g(t)/c_2 = exp(lt)$

$g(t) = c_2 exp(lt)$ ----(v)

$k f''(x)/f(x) =l$

$∂^2f(x)/∂x^2=(l/k)f(x)$

$∂^2f(x)/∂x^2 =-m^2f(x)$

$l/k= -m^2$ -----(vi)

Solution to this differential equation is

$f(x) = c_1sin(mx)+c_2 cos(mx)$

$T(x,t) =(c_2 exp(lt))(c_1sin(mx)+c_3cos(mx))$

Using equation (i)

$T(0,t) =(c_2 exp(lt))(c_1sin(0)+c_3cos(0))= 0$

$T(0,t) =c_3c_2 exp(lt)= 0$

$c_3 =0$

Using equation (i)

$T(2,t)=(c_2 exp(lt))(c_1sin(m.2)) =0$

$T(2,t)=c_1c_2exp(lt)sin(m.2)=0$

We obtain $2.m= nπ$

$m=nπ/2$

Using equation (vi)

$l/k=-m^2=-n^2π^2/4$

$l =-n^2π^2k/4$

So the equation becomes

$T(x,t) =\sum_{n=1}^{\infty}(c_2 exp(-n^2π^2kt/4))(c_1sin(nπx/2))$

$T(x,t) =\sum_{n=1}^{\infty}(c_n(sin(nπx/2) exp(-n^2π^2kt/4))$

$c_n= c_1.c_2$

We have to find c_n now

$T(x,0) =\sum_{n=1}^{\infty}(c_n(sin(nπx/2) exp(-n^2π^2k.0/4))$

$\sum_{n=1}^{\infty} (c_n(sin(nπx/2) = 100(2x-x^2)$

It is a sine Fourier series.we know;

$c_n = 2×\int_{0}^{2}(100(2x-x^2)(sin(nπx/2))dx$

solving the integration using by parts and then putting the limits we get,

$c_n = -3200/(n^3π^3)((-1)^n-1)$

Answer:

$T(x,t) = \sum_{n=1}^{\infty} (c_n(sin(nπx/2) exp(-n^2π^2kt/4))$,

where $c_n = -3200/(n^3π^3)((-1)^n-1)$