Question

Find the steady-state temperature in a bar whose ends are located at x = 0 and
x = 15, if these ends are kept at 200°C and 100°C respectively. (Use Fourier series)

Answer 1

There is the boundary value problem

$u_t(x, t) = ku_{xx}(x, t)$ , for $t > 0$ and $0 < x < L$ (1)

$u(0, t) = T_0$ and $u(L, t) = T_L$ , for $t > 0$

$u(x, 0) = f(x)$ , for $0 \leqslant x \leqslant L$

A steady-state temperature is one that does not depend on time.

Then u_t = 0, so the heat equation (1) simplifies u_xx = 0. Hence we are looking for a function u_s(x) defined for 0 < x < L such that

$\frac{∂^2u_s}{∂x^2}(x) = 0$ , for $0 < x < L$

$u_s(0, t) = T_0$ and $u_s(L,t) = T_L$ for $t > 0$

The solution to this boundary value problem is easily found,since the general solution

of the differential equation is u_s(x) = Ax + B; where A and B are arbitrary constants.

Then the boundary conditions reduce to

u_s(0) = B = T₀ and u_s(L) = AL + B = T_L

We conclude that B = T₀ and A = (T_L – T₀)/L, so the steady-state temperature is

$u_s(x) = (T_L – T_0)\frac{x}{L} + T_0$

$u_s(x) = (100 – 200)\frac{x}{15} + 200 = 200 – 6.66x$

Answer: 200 – 6.66x