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. $$ u(0, t)=10, \quad u(50, t)=40 $$

Short answer

Answer: The steady-state solution for the heat conduction equation with the given boundary conditions is $u(x) = \frac{3}{5} x + 10$.

Step-by-step solution

Identify the steady-state condition

Since the steady-state solution is time-independent, we look for a solution where the temperature distribution doesn't change with time. This means that the temperature derivative with respect to time will be zero: $$ u_t = 0 $$

Rewrite the heat conduction equation with steady-state condition

Plugging in the steady-state condition from step 1 into the given heat conduction equation, we get: $$ \alpha^2 u_{xx} = u_t \\ \alpha^2 u_{xx} = 0 $$ Now we have a second-order ODE with respect to space (x) only.

Solve the ODE

We will now solve the ODE for u(x): $$ \alpha^2 u_{xx} = 0 \\ u_{xx} = 0 $$ Integrating once with respect to x, we get: $$ u_x = C_1 $$ Integrating again with respect to x, we get: $$ u(x) = C_1 x + C_2 $$

Apply boundary conditions

Now, we apply the given boundary conditions to solve for the constants C1 and C2: $$ u(0, t) = 10 \\ u(50, t) = 40 \\ u(0) = C_1 \cdot 0 + C_2 = 10 \\ u(50) = C_1 \cdot 50 + C_2 = 40 $$

Solve for the constants

From the first boundary condition, we find C2: $$ C_2 = 10 $$ Plugging in C2 into the second boundary condition, we find C1: $$ 40 = C_1 \cdot 50 + 10 \\ C_1 = \frac{30}{50} = \frac{3}{5} $$

Write the steady-state solution

Now that we have the constants C1 and C2, we can write the steady-state solution as: $$ u(x) = \frac{3}{5} x + 10 $$

Key concepts

Boundary Value Problems

Understanding the nature of boundary value problems (BVPs) is critical when solving physical phenomena like heat distribution in a solid object. Unlike initial value problems, which provide the state of a system at a single point in time and seek to describe its evolution, a BVP involves finding a solution to a differential equation that satisfies certain conditions at more than one point. The conditions specified at these points are known as boundary conditions. In the context of heat conduction, boundary conditions might represent fixed temperatures at the ends of a rod, representing how the rod interfaces with its environment. In the provided exercise, we have a BVP where the temperature values are given at two different positions along a rod, forming the essential constraints that our solution must adhere to.

Partial Differential Equations

Partial differential equations (PDEs) are used to formulate problems involving functions of several variables, and they are particularly adept at describing various physical phenomena, including the propagation of heat or sound, fluid flow, and the behavior of materials under stress. The heat conduction equation, for instance, is a PDE that represents how heat diffuses through a material. It involves partial derivatives with respect to space and time, indicating how the temperature changes in both dimensions. However, when we focus on the steady-state condition where the system does not change with time, our PDE simplifies to a second-order ordinary differential equation (ODE).

Steady-State Condition

The steady-state condition refers to a situation in a dynamic system where the variables cease to change as time progresses. In the realm of heat conduction, this implies that after a long enough time, the temperature within a material becomes uniform and no longer varies. When dealing with the heat conduction equation, the steady-state condition is represented by setting the time derivative of temperature to zero. This simplification transforms our original PDE into an ODE that solely deals with spatial variables, paving the way for straightforward integration and the application of boundary conditions.

Second-Order ODE

A second-order ordinary differential equation (ODE) contains derivatives up to the second degree and no higher. Solutions to these types of equations often describe phenomena like motion under constant acceleration and the shape of a bent beam under load. For our heat conduction problem, after imposing a steady-state, we encounter a second-order ODE, which we solve by integrating twice. The first integration gives us a first-order equation representing the rate of change of temperature across space, and the second gives us the general temperature distribution. Boundary conditions lead us to values for the integration constants, and hence, a particular solution to our physical problem, completing the model for heat distribution in steady-state conditions.