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. $$ u(0, t)=T, \quad u_{x}(L, t)+u(L, t)=0 $$

Short answer

Define the steady-state solution for the heat conduction equation and find the temperature function u(x), given the boundary conditions u(0,t) = T and u_x(L,t) + u(L,t) = 0.

Step-by-step solution

Define the steady-state solution

Since we are looking for the steady-state solution, we know that the temperature distribution no longer changes with time. In mathematical terms, this means setting the time derivative \(u_t\) to zero: $$ \alpha^2 u_{xx} = 0 $$

Integrate the equation twice

Now we need to solve the above equation for temperature function \(u(x,t)\). Since there is no time dependence, we can write it simply as \(u(x)\). To do this, we will integrate the equation twice with respect to x: $$ \int\alpha^2 u_{xx}dx = \int 0 dx $$ This yields: $$ \alpha^2 u_x = C_1 $$ Integrating once more with respect to x: $$ \int \alpha^2 u_x dx = \int C_1 dx $$ Which results in: $$ \alpha^2 u(x) = C_1 x + C_2 $$

Apply the first boundary condition

Now we apply the boundary condition \(u(0,t) = T\), which in terms of our steady-state solution would be \(u(0) = T\). Plugging in the value of x = 0, we get: $$ \alpha^2 u(0) = C_1(0) + C_2 $$ Substituting \(u(0) = T\): $$ \alpha^2 T = C_2 $$

Take the first derivative and apply the second boundary condition

Next, we take the first derivative of the temperature function with respect to x: $$ u_x(x) = \frac{C_1}{\alpha^2} $$ Now, we apply the second boundary condition \(u_x(L,t) + u(L,t) = 0\), which in terms of our steady-state solution would be \(u_x(L) + u(L) = 0\). Plugging in \(x=L\), we get: $$ u_x(L) + u(L) = \frac{C_1}{\alpha^2} + u(L) = 0 $$ Solving for \(C_1\): $$ C_1 = -\alpha^2 u(L) $$

Substitute \(C_1\) and \(C_2\) back into the original equation and simplify

Finally, we substitute the values of \(C_1\) and \(C_2\) back into the original equation \(\alpha^2 u(x) = C_1 x + C_2\) to obtain the steady-state solution: $$ \alpha^2 u(x) = -\alpha^2 u(L) x + \alpha^2 T $$ Divide both sides by \(\alpha^2\): $$ u(x) = -u(L) x + T $$ This is the steady-state solution for the given heat conduction equation and boundary conditions.

Key concepts

Boundary Conditions

In the realm of heat conduction problems, boundary conditions are essential. They define constraints at the edges or surfaces of the material through which heat flows. In simpler terms, think of these as rules specifying the temperature or heat flow at the boundaries of your system. In the given exercise, we have two boundary conditions:

• **First Boundary Condition:** \(u(0, t) = T\) - This condition tells us that the temperature at position \(x=0\) is constant and equals \(T\). It's like saying one end of a metal rod is kept constantly warm at temperature \(T\).
• **Second Boundary Condition:** \(u_x(L, t) + u(L, t) = 0\) - This is a mixed boundary condition. It indicates a relationship between the temperature at the point \(x=L\) and the rate of heat change at that point (denoted \(u_x\)). It could represent scenarios where the end of a rod loses heat to the surroundings at certain rates.

Together, these conditions help us determine specific values for constants when solving the differential heat equation, ensuring our solution applies to a practical, real-world situation.

Heat Conduction Equation

The heat conduction equation describes how heat diffuses through a medium over time. In this scenario, the key equation is given by: \[\alpha^{2} u_{xx} = u_{t}\]This equation represents heat conduction in a one-dimensional rod. * \(\alpha\) is the thermal diffusivity of the material, which measures how quickly heat spreads.* \(u_{xx}\) is the second spatial derivative of temperature \(u\), showing how temperature changes at various points along the rod.* \(u_{t}\) is the time derivative, describing how the temperature at a specific point changes over time.
For a steady-state scenario, we set \(u_{t} = 0\) because temperature doesn't change with time. This simplifies our equation to: \[\alpha^{2} u_{xx} = 0\]By solving this simpler equation under given boundary conditions, we can find the temperature distribution in the steady state.

Steady-State Solution

The steady-state solution provides the long-term behavior of the system when it's no longer changing over time. In this context, it means finding a temperature distribution \(u(x)\) that remains constant. To achieve this, we follow these steps:

• Since the system is at steady-state, we set the time derivative \(u_t\) to zero, leading us to solve \(\alpha^{2} u_{xx} = 0\).
• We integrate twice with respect to \(x\) to find \(u(x)\), resulting in \(\alpha^2 u(x) = C_1 x + C_2\).
• Using the boundary conditions, we substitute \(u(0) = T\) to find \(C_2\) and adjust for \(u_x(L) + u(L) = 0\) to find \(C_1\).

Once we substitute these constants back, the resulting function \(u(x) = -u(L) x + T\) describes the temperature along the rod in its final, unchanging state. This solution shows how the initial conditions combine to provide a predictable temperature profile across the material.