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

Short Answer

Expert verified
Question: Find the steady-state solution of the heat conduction equation with the given boundary conditions: \(u_x(0, t) = 0\) and \(u(L, t) = 0\). Answer: The steady-state solution of the heat conduction equation with the given boundary conditions is given by: $$u_s(x) = \sum_{n=0}^{\infty} A_n\cos\left(\frac{(2n+1)\pi x}{2L}\right)$$ where \(A_n\) are constants determined by solving an auxiliary problem and using orthogonality properties of the trigonometric functions.

Step by step solution

01

Rewrite the heat equation and separate variables

Let's rewrite the heat equation as a partial differential equation and separate the variables: $$\alpha^2 \frac{\partial^2 u(x,t)}{\partial x^2} = \frac{\partial u(x,t)}{\partial t}$$ We will look for a solution of the form \(u(x,t) = X(x)T(t)\). Substituting this into the PDE, we get $$\alpha^2 X''(x)T(t) = X(x)T'(t)$$ Now, we'll separate the variables by dividing both sides with \(X(x)T(t)\): $$\frac{\alpha^2 X''(x)}{X(x)} = \frac{T'(t)}{T(t)}$$ Since the left-hand side depends only on \(x\) and the right-hand side depends only on \(t\), both sides must be equal to a constant. Let's call this constant \(-k^2\), where \(k\) is a positive constant.
02

Solve the spatial ODE

Now we have two ordinary differential equations to solve: 1. \(\alpha^2 X''(x) + k^2X(x) = 0\) 2. \(T'(t) + k^2T(t) = 0\) The spatial ODE is a second-order linear homogeneous equation, and its general solution has the form: $$X(x) = A\cos(kx) + B\sin(kx)$$ where \(A\) and \(B\) are constants.
03

Apply the boundary conditions

We can now use the given boundary conditions to find the constants \(A\) and \(B\) in the spatial function \(X(x)\). Condition 1, \(u_x(0, t) = 0\), translates into \(X'(0) = 0\). Differentiating \(X(x)\) with respect to \(x\), we get: $$X'(x) = -Ak\sin(kx) + Bk\cos(kx)$$ Then, \(X'(0) = -Ak\sin(0) + Bk\cos(0) = Bk\). Since \(X'(0) = 0\), we have \(Bk = 0\). As \(k\) is non-zero, we must have \(B = 0\). So now we have: $$X(x) = A\cos(kx)$$ Condition 2, \(u(L, t) = 0\), gives us \(X(L) = 0\). Thus, we have: $$A\cos(kL) = 0$$ This equation has non-trivial solutions (i.e., solutions with \(A \neq 0\)) when \(\cos(kL) = 0\), which occurs for \(kL = (2n+1)\frac{\pi}{2}\), where \(n\) is an integer. Thus, we have: $$k_n = \frac{(2n+1)\pi}{2L}$$ So, the spatial function becomes: $$X_n(x) = A_n\cos\left(\frac{(2n+1)\pi x}{2L}\right)$$ where \(A_n\) is the constant associated with the \(n\)-th mode.
04

Obtain the steady-state solution

Since the steady-state solution involves no time dependence, we can drop the time-dependent function \(T(t)\) from our solution. Thus, the steady-state solution for the heat conduction equation is given by: $$u_s(x) = \sum_{n=0}^{\infty} A_n\cos\left(\frac{(2n+1)\pi x}{2L}\right)$$ This is the required steady-state solution of the heat conduction equation with the given boundary conditions. The constants \(A_n\) can be found by solving an auxiliary problem (usually involving initial conditions) and using orthogonality properties of the trigonometric functions.

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Key Concepts

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

Partial Differential Equations
Partial differential equations (PDEs) are fundamental to understanding phenomena where several factors vary simultaneously, such as changes in temperature over space and time in heat conduction problems. A PDE involves derivatives with respect to more than one independent variable. In the case of the steady-state heat conduction equation, \( \alpha^{2} u_{xx} = u_{t} \), the function \( u(x,t) \) depends on both position \( x \) and time \( t \), and its derivatives \( u_{xx} \) and \( u_{t} \) represent changes in space and time, respectively. Solving a PDE typically requires finding a function that satisfies the equation and the imposed boundary conditions, which specify the values of the function at the edges of the domain.

Understanding PDEs is crucial as they accurately describe the spread of heat, the behavior of vibrating systems, the diffusion of chemicals, and the progression of waves, among other applications. The strategies used in solving these complex equations often involve transforming them into simpler forms, such as ordinary differential equations (ODEs) through techniques like separation of variables.
Boundary Value Problems
Boundary value problems (BVPs) are scenarios where we solve PDEs or ODEs subject to certain conditions specified at the boundaries of the domain, which, in physical terms, might represent the walls of a container, the ends of a rod, or the surface of a sphere. In the textbook exercise, we are given a heat conduction problem with specified conditions at the ends of a one-dimensional domain. These conditions, expressed as \( u_{x}(0, t) = 0 \) and \( u(L, t) = 0 \), are crucial in determining the unique solution to the PDE.

When solving BVPs, it's essential to consider the physical interpretation of boundary conditions. For instance, a zero temperature gradient \( (u_{x} = 0) \) at a point might represent an insulated boundary, while a zero temperature \( (u = 0) \) could represent a point in contact with an environment at a fixed reference temperature. Solving BVPs helps us predict how physical systems will behave under various constraints.
Fourier Series
The Fourier series is a powerful mathematical tool employed to express periodic functions as an infinite sum of sine and cosine terms. It's especially useful in solving PDEs like the heat conduction equation, because it allows for the decomposition of complex waveforms into simpler, sinusoidal components. This is evident in the steady-state solution of our exercise, where we represent the temperature distribution, \( u_s(x) \), as a sum of cosine functions multiplied by constants \( A_n \).

Each term in the Fourier series corresponds to a specific frequency or harmonic of the original function, capturing different aspects of the pattern it describes. This concept plays a critical role in many areas of physics and engineering, as it provides an approach to analyze and reconstruct signals and waveforms. Moreover, the orthogonality of the sine and cosine functions, integral to Fourier series, helps in simplifying the process of finding the coefficients in the series.
Separation of Variables
Separation of variables is a method for solving PDEs, which hinges on the assumption that a multidimensional problem can be broken into separate one-dimensional problems. We assume that the solution \( u(x,t) \) can be written as the product of a function of \( x \) and another function of \( t \)—that is, \( u(x,t) = X(x)T(t) \). By substituting this form into the PDE and rearranging the terms, we can separate the variables, as different variables then appear on different sides of the equation.

This technique simplifies the original PDE into simpler ODEs, which are often more manageable. In heat conduction problems, it enables us to find solutions that fit both the spatial and temporal development of temperature. Nevertheless, the method's application is limited to problems that admit such a separable solution, which must also satisfy boundary and initial conditions to produce an accurate and meaningful physical prediction.

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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 extended function. (b) Sketch the graph of the function to which the series converge for three periods. $$ f(x)=1-x^{2}, \quad-1 \leq x \leq 1 $$

Let \(f\) first be extended into \((L, 2 L)\) so that it is symmetric about \(x=L ;\) that is, so as to satisfy \(f(2 L-x)=f(x)\) for \(0 \leq x

Consider a uniform bar of length \(L\) having an initial temperature distribution given by \(f(x), 0 \leq x \leq L\). Assume that the temperature at the end \(x=0\) is held at \(0^{\circ} \mathrm{C},\) while the end \(x=L\) is insulated so that no heat passes through it. (a) Show that the fundamental solutions of the partial differential equation and boundary conditions are $$ u_{n}(x, t)=e^{-(2 n-1)^{2} \pi^{2} \alpha^{2} t / 4 L^{2}} \sin [(2 n-1) \pi x / 2 L], \quad n=1,2,3, \ldots $$ (b) Find a formal series expansion for the temperature \(u(x, t)\) $$ u(x, t)=\sum_{n=1}^{\infty} c_{n} u_{n}(x, t) $$ that also satisfies the initial condition \(u(x, 0)=f(x)\) Hint: Even though the fundamental solutions involve only the odd sines, it is still possible to represent \(f\) by a Fourier series involving only these functions. See Problem 39 of Section \(10.4 .\)

Consider an elastic string of length \(L .\) The end \(x=0\) is held fixed while the end \(x=L\) is free; thus the boundary conditions are \(u(0, t)=0\) and \(u_{x}(L, t)=0 .\) The string is set in motion with no initial velocity from the initial position \(u(x, 0)=f(x),\) where $$ f(x)=\left\\{\begin{array}{ll}{1,} & {L / 2-12)} \\ {0,} & {\text { otherwise. }}\end{array}\right. $$ (a) Find the displacement \(u(x, t) .\) (b) With \(L=10\) and \(a=1\) plot \(u\) versus \(x\) for \(0 \leq x \leq 10\) and for several values of \(t .\) Pay particular attention to values of \(t\) between 3 and \(7 .\) Observe how the initial disturbance is reflected at each end of the string. (c) With \(L=10\) and \(a=1\) plot \(u\) versus \(t\) for several values of \(x .\) (d) Construct an animation of the solution in time for at least one period. (e) Describe the motion of the string in a few sentences.

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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