Question

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 \(10.4 .\)

Short answer

Question: Given a uniform bar of length L with an initial temperature distribution f(x), find the fundamental solutions of the partial differential equation and boundary conditions, and then find a formal series expansion for the temperature u(x, t) that also satisfies the initial conditions. Answer: The temperature function u(x, t) for a uniform bar with the given initial and boundary conditions can be expressed using the Fourier series expansion: $$ u(x, t) = \sum_{n=1}^{\infty} \left(\frac{2}{L} \int_0^L f(x) \sin\left(\frac{(2n-1) \pi x}{2L}\right) dx\right)e^{-(2 n-1)^{2} \pi^{2} \alpha^{2} t / 4 L^{2}} \sin [(2 n-1) \pi x / 2 L] $$

Step-by-step solution

Derive the fundamental solutions

For a uniform bar with the given boundary conditions, the heat equation will be given as: $$ u_{t} = \alpha^2 u_{xx} $$ The boundary conditions are: $$ u(0, t) = 0, \quad u_{x}(L, t) = 0 $$ To solve this boundary-value problem, we will assume a separation of variables solution of the form: $$ u(x, t) = X(x)T(t) $$ Substituting into the heat equation, we get: $$ X(x)T'(t) = \alpha^2 X''(x)T(t) $$ Dividing both sides by \(\alpha^2 X(x)T(t)\), we have: $$ \frac{T'(t)}{\alpha^2 T(t)} = \frac{X''(x)}{X(x)} = -\lambda $$ Here, \(\lambda\) is the separation constant. Now, solving the simplified boundary-value problem for X(x), we have: $$ X''(x) + \lambda X(x) = 0 $$ with boundary conditions, $$ X(0) = 0, \quad X'(L) = 0 $$ The eigenfunctions and eigenvalues for this problem are given by: $$ X_n(x) = \sin\left(\frac{(2n-1) \pi x}{2L}\right), \ \lambda_n = \left(\frac{(2n-1)\pi}{2L}\right)^2, \ n=1,2,3, \ldots $$ Now, we can plug the eigenfunctions back into our separation of variables solution form, getting the fundamental solutions: $$ u_n(x, t)=e^{-(2 n-1)^{2} \pi^{2} \alpha^{2} t / 4 L^{2}} \sin [(2 n-1) \pi x / 2 L], \quad n=1,2,3, \ldots $$

Find a formal series expansion for the temperature u(x, t)

Now, we have to find the temperature function u(x, t) as a series using the Fourier series involving only odd sine functions, given the initial condition \(u(x, 0)= f(x)\). The series expansion is given by: $$ u(x, t) = \sum_{n=1}^{\infty} c_n u_n(x, t) $$ Plugging the initial condition, we have: $$ f(x) = \sum_{n=1}^{\infty} c_n \sin\left(\frac{(2n-1) \pi x}{2L}\right) $$ To find the coefficients \(c_n\), we can multiply both sides by \(\sin\left(\frac{(2m-1) \pi x}{2L}\right)\) and integrate from \(0\) to \(L\): $$ \int_0^L f(x) \sin\left(\frac{(2m-1) \pi x}{2L}\right) dx = \sum_{n=1}^{\infty} c_n\int_0^L \sin\left(\frac{(2n-1) \pi x}{2L}\right) \sin\left(\frac{(2m-1) \pi x}{2L}\right) dx $$ Since only the term with \(n = m\) will survive the integration on the right-hand side, we get: $$ c_m = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{(2m-1) \pi x}{2L}\right) dx, \quad m = 1,2,3, \ldots $$ Finally, we can plug these coefficients back into our series expansion form to obtain the temperature function u(x, t): $$ u(x, t) = \sum_{n=1}^{\infty} \left(\frac{2}{L} \int_0^L f(x) \sin\left(\frac{(2n-1) \pi x}{2L}\right) dx\right)e^{-(2 n-1)^{2} \pi^{2} \alpha^{2} t / 4 L^{2}} \sin [(2 n-1) \pi x / 2 L] $$

Key concepts

Eigenfunctions and Eigenvalues

In mathematical physics, eigenfunctions and their corresponding eigenvalues play a central role in solving boundary-value problems, such as the heat equation for a uniform bar.

Think of an eigenfunction as a special type of solution to a differential equation that behaves predictably when acted on by a differential operator. In the context of our problem, the eigenfunctions are sine functions that satisfy both the differential equation and the given boundary conditions at the endpoints of the bar. The eigenvalues are particular constants that relate to the natural frequencies at which these functions oscillate, and in our problem, they help determine the rate at which heat dissipates over time.

For the given heat problem, the solutions include an infinite set of eigenfunctions given by \(\sin\left(\frac{{(2n-1) \pi x}}{{2L}}\right)\), where 'n' can be any positive integer, and eigenvalues represented by \(\lambda_n = \left(\frac{{(2n-1)\pi}}{{2L}}\right)^2\). These eigenfunctions are crucial because they form the building blocks of the complete solution, allowing us to express the temperature distribution along the bar at any given time.

Fourier Series

The Fourier series is a mathematical tool used to represent a periodic function as an infinite sum of sine and cosine terms. In the context of the heat equation, the Fourier series acts like a mathematical X-ray, revealing the fundamental sine wave components of the initial temperature distribution through their amplitudes and phases.

In the specific exercise we're examining, the Fourier series is slightly atypical because it involves only odd multiples of sines due to the boundary condition at \(x=L\). Despite that, this series is still rich enough to approximate any reasonable initial temperature distribution \(f(x)\), as long as \(f(x)\) can be considered a periodic function with the appropriate symmetries.

The Fourier coefficients \(c_n\) are determined by projecting \(f(x)\) onto the eigenfunctions \(\sin\left(\frac{{(2n-1) \pi x}}{{2L}}\right)\). Through integration, these coefficients measure how much of each 'note' or eigenfunction is present in the initial 'melody' or temperature distribution, \(f(x)\). These coefficients are then used to reconstruct the temperature distribution over time as it evolves according to the heat equation.

Separation of Variables

The method of separation of variables is a systematic way to find solutions to partial differential equations, which often arise in physics when describing phenomena like heat flow. This technique assumes that a solution can be written as the product of functions, each depending on a single coordinate. In our heat equation example, separation of variables allows us to assume that the temperature \(u(x, t)\) at any point \(x\) along the bar and time \(t\) can be expressed as a product \(X(x)T(t)\).

This assumption simplifies the complex problem into two easier ordinary differential equations, where \(X\) only varies with position \(x\) and \(T\) only varies with time \(t\). One equation relates to the spatial part of the heat equation, leading to the sine eigenfunctions mentioned, while the other equation describes temporal decay of the temperature as determined by the heat equation and eigenvalues.

Ultimately, separation of variables is like treating the symptoms separately in a diagnosis: by focusing on a single variable at a time, we can tackle complex problems more effectively. Once we have the spatial and temporal parts, we recombine them to produce the fundamental solutions, which when summed up, give us the complete temperature distribution for the bar.