Answer to Question #236080 in Differential Equations for smi

Question

u_xx=4u_t , 0<x<2 , t>0

u(0,t)=5, u(2,t)=10, t>0

u(x,0)=x, 0≤x≤2

Find the transient and steady-state temperature of the above bvp of heat equation

Answer 1

Solution

Let’s reduce boundary conditions to homogeneous u(x,t) = 5(1+x/2)+U(x,t)

For U(x,t) we obtain next problem

U_xx=4U_t , 0<x<2 , t>0

U(0,t)=0, U(2,t)=0, t>0

U(x,0)=x-5(1+x/2)=-5-3x/2, 0≤x≤2

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

Substitution into equation gives 4X(x)T’(t) = X’’(x)T(t) => 4T’(t)/T(t) = X’’(x)/ X(x)

Variables t and x are independent. So 4T’(t)/T(t) = X’’(x)/ X(x) = -k²  

NOTE: variant k² at the right side is impossible because equation X’’(x) - k² X(x) = 0 with the given boundary conditions have only trivial solution.

Therefore we’ll get two ordinary differential equations

4T’(t) + k² T(t)] = 0

 X’’(x) + k² X(x) = 0

From this equations T(t) = Aexp(-(k/2)²t), X(x) = Bsin(kx) + Ccos(kx) (A, B, C – arbitrary constants)

From boundary conditions U(0,t) = U(2,t) = 0, t > 0 => X(0) = X(2) = 0 and C = 0, X(x) = Bsin(2k) = 0 => k = ±πn/2 (n = 1,2, . . .).

Solution is

$U\left(x,t\right)=\sum_{n=1}^{\infty}A_ne^{-\left(\frac{\pi n}{4}\right)^2t}sin\left(\frac{\pi nx}{2}\right)$

From initial condition

$U\left(x,0\right)=\sum_{n=1}^{\infty}A_nsin\left(\frac{\pi nx}{2}\right)=-5-\frac{3}{2}x$

Multiplying by sin(πmx/2) and integrating with bounds 0, 2

$A_n=\int_{0}^{2}U\left(x,0\right)sin\left(\frac{\pi nx}{2}\right)dx=-\int_{0}^{2}\left(5+\frac{3}{2}x\right)sin\left(\frac{\pi nx}{2}\right)dx$

$A_n=\left[\frac{10}{\pi\ n}cos\left(\frac{\pi nx}{2}\right)-\frac{6}{\pi^2n^2}\left(sin\left(\frac{\pi nx}{2}\right)-\frac{\pi nx}{2}cos\left(\frac{\pi nx}{2}\right)\right)\right]|_0^2$

$A_n=\frac{10}{\pi\ n}\left[cos\left(\pi n\right)-1\right]+\frac{6}{\pi\ n}cos\left(\pi n\right)=\frac{10}{\pi\ n}\ \left[{(-1)}^n-1\right]+\frac{6}{\pi\ n}{(-1)}^n$

With this coefficients

$u\left(x,t\right)=5\left(1+\frac{x}{2}\right)+\sum_{n=1}^{\infty}A_ne^{-\left(\frac{nn}{4}\right)^2t}sin\left(\frac{\pi nx}{2}\right)$

It is transient temperature of the above bvp of heat equation. For t→∞ sum is equal to zero. Thus  steady-state temperature of the above bvp of heat equation is u_stat(x,t) = 5(1+x/2)