Question

find the steady-state solution of the heat conduction equation \(\alpha^{2} u_{x x}=u_{t}\) that satisfies the given set of boundary conditions. $$ u(0, t)=30, \quad u(40, t)=-20 $$

Short answer

Answer: The steady-state solution of the heat conduction equation with the given boundary conditions is: $$ u(x) = \frac{-20\alpha^2 - 30\alpha^2}{40}x + 30\alpha^2 $$

Step-by-step solution

Write down the steady-state heat conduction equation

Since the steady-state temperature distribution no longer changes with time, the time derivative becomes zero. Thus, the given heat conduction equation becomes: $$ \alpha^2 u_{xx} = 0 $$

Solve the equation subject to the boundary conditions

To solve the above equation, we'll integrate it twice with respect to \(x\): 1st integration: $$ \int \alpha^2 u_{xx} dx = \int 0 dx \\ \Longrightarrow \alpha^2 u_x = C_1 $$ 2nd integration: $$ \int \alpha^2 u_x dx = \int C_1 dx \\ \Longrightarrow \alpha^2 u(x) = C_1x + C_2 $$ Now, we need to impose the given boundary conditions: 1. \(u(0, t) = 30\): This implies that \(u(0) = 30\). Plugging this into the equation, we get: $$ \alpha^2(30) = C_1(0) + C_2 \\ \Longrightarrow C_2 = 30\alpha^2 $$ 2. \(u(40, t) = -20\): This implies that \(u(40) = -20\). Plugging this into the equation, we get: $$ \alpha^2(-20) = C_1(40) + C_2 \\ \Longrightarrow C_1 = \frac{-20\alpha^2 - 30\alpha^2}{40} $$ Thus, the steady-state solution of the heat conduction equation with the given boundary conditions is: $$ u(x) = \frac{-20\alpha^2 - 30\alpha^2}{40}x + 30\alpha^2 $$