Question

(a) As in Example 2, use (14) and (15) to solve the initial value problem $$ \begin{aligned} &u_{t}(x, t)=u_{x x}(x, t), \quad 0

Short answer

Answer: The value of the solution at (x, t) = (l/2, 1) is $$u\left(\frac{l}{2},1\right) = \sum_{n=1}^{\infty} \left(\frac{3}{2l(n - 1)}\sin n\pi - \frac{1}{2l(n - 3)}\sin n\pi\right) \sin \frac{n \pi}{2} e^{-\frac{n^2 \pi^2}{l^2}}$$ Question: What is the value of the limit of the solution as t approaches infinity? Answer: The value of the limit of the solution as t approaches infinity is 0: $$\lim _{t\rightarrow \infty} u(x, t) = 0$$

Step-by-step solution

Write down the given information

Firstly, let's write down the given equation and boundary conditions: $$ \begin{aligned} &u_{t}(x, t)=u_{x x}(x, t), \quad 0<x<l, \quad 0<t<\infty \\\ &u_{x}(0, t)=u_{x}(l, t)=0, \quad 0 \leq t<\infty \\\ &u(x, 0)=f(x), \quad 0 \leq x \leq l . \end{aligned} $$ And the function for the initial condition is given by $$f(x)=2 \cos ^{3}\left(\frac{\pi x}{l}\right)$$

Apply equations (14) and (15) to find the general solution

Recall that equations (14) and (15) state that for a given problem as described above, the solution can be found in the following form: $$u(x,t) = \sum_{n=1}^{\infty} B_n \sin \frac{n \pi x}{l} e^{-\frac{n^2 \pi^2 t}{l^2}}$$ Where the coefficients \(B_n\) are given by $$B_n = \frac{2}{l} \int_0^l f(x) \sin \frac{n \pi x}{l} dx, \quad n = 1,2,3,...$$

Compute the \(B_n\) coefficients

We know the function \(f(x)=2 \cos ^{3}\left(\frac{\pi x}{l}\right)\). To find the \(B_n\) coefficients, plug the function into the \(B_n\) formula: $$B_n = \frac{2}{l} \int_0^l 2 \cos ^{3}\left(\frac{\pi x}{l}\right) \sin \frac{n \pi x}{l} dx$$ Now, we need to compute the integral. Since it is difficult to find a closed-form expression for a complicated trigonometric expression like this, we can use the following trigonometric identity: $$\cos ^{3}x = \frac{3 \cos x + \cos 3x}{4}$$ Use this identity with \(x = \frac{\pi x}{l}\): $$B_n = \frac{4}{l} \int_0^l \left(\frac{3 \cos \frac{\pi x}{l} + \cos \frac{3 \pi x}{l}}{4}\right) \sin \frac{n \pi x}{l} dx$$ Split the integral into two parts: $$B_n = \frac{3}{l} \int_0^l \cos \frac{\pi x}{l} \sin \frac{n \pi x}{l} dx + \frac{1}{l} \int_0^l \cos \frac{3 \pi x}{l} \sin \frac{n \pi x}{l} dx$$ Now we can calculate the integrals using the following trigonometric substitution formula: $$\int \cos ax \sin bx dx = \frac{(\sin(ax) \sin(bx) - \sin(ax - bx))}{2(a - b)}$$ Apply this formula to both integrals: $$B_n = \frac{3}{2l(n - 1)}(\sin \frac{n \pi l}{l} \sin \frac{\pi l}{l} - \sin 0 \sin \frac{(n - 1) \pi l}{l}) + \frac{1}{2l(n - 3)}(\sin \frac{n \pi l}{l} \sin \frac{3 \pi l}{l} - \sin 0 \sin \frac{(n - 3) \pi l}{l})$$ Simplify the expression: $$B_n = \frac{3}{2l(n - 1)}\sin n\pi - \frac{1}{2l(n - 3)}\sin n\pi$$ Now we can write the solution for \(u(x,t)\) substituting the \(B_n\) expression: $$u(x,t) = \sum_{n=1}^{\infty} \left(\frac{3}{2l(n - 1)}\sin n\pi - \frac{1}{2l(n - 3)}\sin n\pi\right) \sin \frac{n \pi x}{l} e^{-\frac{n^2 \pi^2 t}{l^2}}$$

Evaluate the solution at \((x, t)=(l / 2,1)\) (part b)

Plug \(x = \frac{l}{2}\) and \(t = 1\) into the solution and simplify: $$u\left(\frac{l}{2},1\right) = \sum_{n=1}^{\infty} \left(\frac{3}{2l(n - 1)}\sin n\pi - \frac{1}{2l(n - 3)}\sin n\pi\right) \sin \frac{n \pi}{2} e^{-\frac{n^2 \pi^2}{l^2}}$$

Show that \(\lim _{t\rightarrow \infty} u(x, t)\) exists and find its value (part c)

To find the limit as \(t \to \infty\), we should consider the factor \(e^{-\frac{n^2 \pi^2 t}{l^2}}\) in the solution. As \(t \to \infty\), this term will go to 0, so the solution will converge to the following expression: $$\lim_{t \to \infty} u(x,t) = \sum_{n=1}^{\infty} \left(\frac{3}{2l(n - 1)}\sin n\pi - \frac{1}{2l(n - 3)}\sin n\pi\right) \sin \frac{n \pi x}{l} (0)$$ Since the exponential term goes to 0, the entire limit expression evaluates to 0 for all values of \(x\). This means that the limit exists: $$\lim _{t\rightarrow \infty} u(x, t) = 0$$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are fundamental to modeling how systems change over time and space, making them essential tools across the sciences. An initial value problem for a differential equation specifies the state of the system at the outset (at an initial time) and asks for the state of the system at a later time. This involves solving for a function that satisfies both the differential equation and the initial conditions. In our exercise, we're focusing on a partial differential equation (PDE) of the heat equation type, which is a common form of second-order linear PDEs for modeling diffusion processes, such as the temperature distribution over time.

Understanding the nuances of the given PDE and its boundary conditions is key to determining the behavior of the physical system in question. Through the solution process, such as separation of variables and the application of boundary and initial conditions, we arrive at a series solution that describes the temperature distribution for any given time and location within the specified domain.

Boundary Value Problems

Boundary value problems (BVPs) occur when we are dealing with differential equations involving boundary conditions, as opposed to initial conditions. These problems require the solution not just to satisfy the differential equation, but also to meet certain conditions at the 'boundaries' of the domain of interest. In our exercise, the domain extends from \(0 < x < l\) with the boundary conditions involving the derivative of the function \(u_x\), set to zero at both ends \(x = 0\) and \(x = l\).

These conditions significantly influence the form of the solution. The zero-gradient condition implies symmetry and suggests that there's no heat flux through the ends. When solving BVPs, particularly for PDEs, Fourier series solutions are often used, as they provide a way to satisfy the boundary conditions precisely by representing the solution as an infinite sum of sinusoidal waves each of which individually meets the endpoint criteria.

Fourier Series Solution

Fourier series are a powerful mathematical tool for solving BVPs by expressing functions as a sum of sines and cosines with appropriate coefficients. In our heat equation scenario, we used the Fourier sine series representation because it naturally fits the boundary conditions on the spatial derivative of \(u\).

The general solution has the form \(u(x,t) = \sum_{n=1}^{\infty} B_n \sin \frac{n \pi x}{l} e^{-\frac{n^2 \pi^2 t}{l^2}}\), with the \(B_n\) coefficients determined specifically by the initial temperature distribution \(f(x)\). Calculating these coefficients requires integrating the product of \(f(x)\) and the sine basis function over the spatial domain—the intricacies of which may require using trigonometric identities for simplification, as illustrated in the exercise.

Trigonometric Identities

Trigonometric identities are relationships involving the angles and sides of triangles that result in equivalent expressions. They are invaluable for simplifying and solving problems involving trigonometric functions. In the context of our exercise, these identities help transform the cubic cosine term into a sum involving simpler cosine terms, allowing for straightforward integration.

The identity used is \(\cos^3 x = \frac{3 \cos x + \cos 3x}{4}\), which simplifies the challenge of integrating the third power of a cosine function. Utilizing such identities enables us to compute the Fourier coefficients \(B_n\) in terms of known functions whose integrals are easier to evaluate, leading to the complete solution of the initial boundary value problem.