a) Determine the steady-state temperature for the system below, if one exists.
PDE: ut = uxx + 1, 0 < x < L
BCs:
(
ux(0, t) = 1
ux(L, t) = β 0 < t < ∞
IC: u(x, 0) = f (x) 0 ≤ x ≤ L
You may leave an arbitrary constant in your answer.
(b) For what β does the steady-state solution exist?

The problem involves determining the **steady-state temperature** for the given system governed by the Partial Differential Equation (PDE):

\[
u_t = u_{xx} + 1 \quad \text{for} \quad 0 < x < L
\]

with boundary conditions:

\[
u_x(0, t) = 1 \quad \text{and} \quad u_x(L, t) = \beta \quad \text{for} \quad 0 < t < \infty
\]

and initial condition:

\[
u(x, 0) = f(x) \quad \text{for} \quad 0 \leq x \leq L
\]

### Part (a): Steady-State Solution

To find the steady-state solution, we assume the system reaches a state where the temperature no longer changes with time, i.e., \( u_t = 0 \). At steady-state, the PDE simplifies to:

\[
0 = u_{xx} + 1
\]

This can be rewritten as:

\[
u_{xx} = -1
\]

Integrating this equation twice will give us the steady-state temperature profile \( u(x) \). First, integrate with respect to \( x \):

\[
u_x(x) = -x + C_1
\]

where \( C_1 \) is an arbitrary constant of integration. Integrate again to find \( u(x) \):

\[
u(x) = -\frac{x^2}{2} + C_1 x + C_2
\]

where \( C_2 \) is another arbitrary constant of integration.

### Apply Boundary Conditions

We now use the boundary conditions to determine the constants \( C_1 \) and \( C_2 \).

1. From the condition \( u_x(0, t) = 1 \), we know that at \( x = 0 \):

\[
u_x(0) = -0 + C_1 = 1 \quad \Rightarrow \quad C_1 = 1
\]

2. From the condition \( u_x(L, t) = \beta \), we have:

\[
u_x(L) = -L + C_1 = \beta
\]

Substitute \( C_1 = 1 \):

\[
- L + 1 = \beta \quad \Rightarrow \quad \beta = 1 - L
\]

Thus, the steady-state solution exists for \( \beta = 1 - L \).

### Steady-State Temperature Profile

Substituting \( C_1 = 1 \) into the equation for \( u(x) \):

\[
u(x) = -\frac{x^2}{2} + x + C_2
\]

\( C_2 \) remains undetermined because it depends on the initial condition or additional information about the temperature distribution.

Thus, the **steady-state temperature profile** is:

\[
u(x) = -\frac{x^2}{2} + x + C_2
\]

where \( C_2 \) is an arbitrary constant, and the steady-state condition requires \( \beta = 1 - L \).

### Part (b): Value of \( \beta \) for Steady-State Solution

From the boundary condition analysis, we determined that the steady-state solution exists when:

\[
\beta = 1 - L
\]

This is the value of \( \beta \) that ensures a steady-state solution exists for the given system.

### Summary:
- **Steady-state temperature**: \( u(x) = -\frac{x^2}{2} + x + C_2 \), where \( C_2 \) is arbitrary.
- **Value of \( \beta \) for steady-state**: \( \beta = 1 - L \).

Would you like more details on any step or specific clarification?

### 5 related questions:
1. How would the steady-state solution change if the source term in the PDE was different?
2. Can the initial condition \( u(x, 0) = f(x) \) affect the steady-state solution?
3. What is the physical interpretation of the constants \( C_1 \) and \( C_2 \) in this system?
4. How does the length \( L \) of the domain impact the steady-state solution and boundary condition \( \beta \)?
5. How would the solution differ if there was a time-dependent term in the PDE?

### Tip:
To check if a steady-state solution exists, always start by assuming \( u_t = 0 \), which simplifies the time-dependent PDE into an easier ordinary differential equation (ODE).

a) Determine the steady-state temperature for the system below, if one exists. PDE: ut = uxx + 1, 0 < x < L BCs: ( ux(0, t) = 1 ux(L, t) = β 0 < t < ∞ IC: u(x, 0) = f (x) 0 ≤ x ≤ L You may leave an arbitrary constant in your answer. b) For what β does the steady-state solution exist?

Explore the steady-state temperature solution for a partial differential equation (PDE) with a source term, given boundary conditions. The solution involves integrating the heat equation and applying boundary conditions to determine constants. Includes the formula for the value of β where the steady-state solution exists.

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