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

Short Answer

Expert verified
Answer: The steady-state solution of the given heat conduction equation, with the given boundary conditions, is u(x, t) = 0. This means that the steady-state temperature is uniformly zero throughout the entire domain and does not change with time.

Step by step solution

01

Analyze the steady-state solution

In a steady-state situation, the temperature \(u(x, t)\) does not change over time, so we can write \(u_t = 0\). The heat conduction equation becomes: $$\alpha^2 u_{xx} = 0$$.
02

Integrate twice with respect to x

In order to find the steady-state temperature function, we need to integrate the equation \(\alpha^2 u_{xx} = 0\) twice with respect to x. We start with the first integral: $$\alpha^2 u_x = C_1$$, where \(C_1\) is the constant of integration. Now, we integrate the equation again: $$\alpha^2 u(x) = C_1 x + C_2$$, where \(C_2\) is the constant of integration.
03

Apply the first boundary condition

Apply the first boundary condition, \(u(0, t) = 0\), to the equation obtained by integration: $$\alpha^2 u(0) = C_1(0) + C_2 \Rightarrow C_2 = 0$$. This means that our steady-state temperature function is now: $$\alpha^2 u(x) = C_1 x$$.
04

Apply the second boundary condition

Apply the second boundary condition, \(u_x(L, t) = 0\), to the first integral: $$\alpha^2 u_x = C_1 = 0$$, as the boundary condition states that the temperature gradient at x = L is zero. This means that the constant \(C_1\) is also equal to zero.
05

Write the final steady-state solution

Since both constants are zero, the final steady-state solution of the heat conduction equation, with the given boundary conditions, is: $$u(x, t) = 0$$. The steady-state temperature is uniformly zero throughout the whole domain, and it does not change with time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boundary Value Problems
A boundary value problem is a mathematical issue for which you need to find a solution that satisfies some specified criteria, known as boundary conditions, at the boundaries of the domain. In our exercise, the domain is the range of possible positions along a rod, and the boundary conditions are related to the temperature at specific locations. For instance, the boundary condition \(u(0, t) = 0\) stipulates that the temperature at one end of the rod (at position zero) must remain constant at zero. The other condition, \(u_x(L, t) = 0\), states that the rate of change of temperature at the other end of the rod (at position L) must be zero. The solutions to boundary value problems are crucial in fields such as physics and engineering since they can describe steady states of physical systems like the heat distribution in a rod.
Partial Differential Equations
Partial differential equations (PDEs) are equations that involve rates of change with respect to multiple variables. They are more complex than ordinary differential equations (ODEs) which only involve derivatives with respect to one variable. Our exercise features a PDE, specifically the heat conduction equation \(\alpha^2 u_{xx} = u_t\), which expresses how the temperature \(u\) changes in space (the second spatial derivative \(u_{xx}\)) and in time (the temporal derivative \(u_t\)). Solving PDEs typically requires you to consider initial conditions, which are values at the start of observation, and boundary conditions, which are values at the spatial extremities.
Mathematical Integration
Mathematical integration is a fundamental tool in solving differential equations. It allows you to find a function when its rate of change is known. In the context of the heat conduction equation from our original problem, we start with \(\alpha^2 u_{xx} = 0\) after recognizing that it's a steady-state scenario. Integrating this with respect to \(x\) yields a general form of the temperature distribution, which includes integration constants that must be determined using boundary conditions. This multiple-step process, which involves both integration and the application of boundary conditions, helps us find the specific solution to the problem.
Steady-State Solution
The steady-state solution refers to a condition where the system's behavior does not change over time. In the case of our heat conduction problem, finding the steady-state solution meant determining the temperature profile along the rod that does not change as time progresses. To achieve this, we set the time derivative of temperature \(u_t\) to zero, simplifying our original PDE. After integrating and applying the boundary conditions, we determined that the temperature remains uniform at zero across the entire length of the rod. This kind of analysis is vital when designing systems where a constant state is desired, such as in maintaining a consistent temperature in a heating system.

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Most popular questions from this chapter

assume that the given function is periodically extended outside the original interval. (a) Find the Fourier series for the given function. (b) Let \(e_{n}(x)=f(x)-s_{n}(x)\). Find the least upper bound or the maximum value (if it exists) of \(\left|e_{n}(x)\right|\) for \(n=10,20\), and 40 . (c) If possible, find the smallest \(n\) for which \(\left|e_{x}(x)\right| \leq 0.01\) for all \(x .\) $$ f(x)=x, \quad-1 \leq x<1 ; \quad f(x+2)=f(x) $$

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Consider a rod of length 30 for which \(\alpha^{2}=1 .\) Suppose the initial temperature distribution is given by \(u(x, 0)=x(60-x) / 30\) and that the boundary conditions are \(u(0, t)=30\) and \(u(30, t)=0\) (a) Find the temperature in the rod as a function of position and time. (b) Plot \(u\) versus \(x\) for several values of \(t\). Also plot \(u\) versus \(t\) for several values of \(x\). (c) Plot \(u\) versus \(t\) for \(x=12\). Observe that \(u\) initially decreases, then increases for a while, and finally decreases to approach its steady-state value. Explain physically why this behavior occurs at this point.

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