Question

Let \(h_{1}=h_{2}=\cdots=h_{N}=h, m_{\sigma}=0\) and \(F_{a}(t) \equiv 0\) for \(\sigma=1, \ldots, N_{0}=u_{N+1}=\) 0 in (10.51), so that one has case (c), with equal friction coefficients. Show that the substitution \(u_{\sigma}=A(\sigma) e^{\lambda /}\) leads to the difference equation with boundary conditions: he. $$ \begin{aligned} \Delta^{2} A(\sigma)+p^{2} A(\sigma) &=0, & & p^{2}=-h \lambda / k^{2} \\ A(0) &=0, & & A(N+1)=0 . \end{aligned} $$ Use the result of Problem 5(d) to obtain the "modes of decay": $$ \begin{aligned} u_{\sigma}(t) &=\sin \left(\frac{n \pi}{N+1} \sigma\right) e^{-a_{n} t} \\ a_{n} &=\frac{2 k^{2}}{h}\left(1-\cos \frac{n \pi}{N+1}\right) . \end{aligned} $$ Show that \(0

Short answer

Question: Given the following conditions: \(h_1 = h_2 = \cdots = h_N = h\), \(m_\sigma = 0\), \(F_a(t) = 0\), and \(N_0 = u_{N+1} = 0\), along with the substitution \(u_\sigma = A(\sigma)e^{\lambda t}\), prove that the substitution leads to a specific difference equation with boundary conditions. Then, use the result of Problem 5(d) to obtain the modes of decay, and finally, show that \(0 < a_1 < a_2 < \cdots < a_N\). Answer: By substituting the given conditions into equation (10.51), the derived difference equation can be solved using the results of Problem 5(d) to obtain the modes of decay expressed as \(u_\sigma(t)\) and \(a_n\). Furthermore, it can be shown that the values of \(a_n\) satisfy the inequality \(0 < a_1 < a_2 < \cdots < a_N\).

Step-by-step solution

Write down the given conditions

We are given the following conditions: 1. \(h_1 = h_2 = \cdots = h_N = h\) 2. \(m_\sigma = 0\) 3. \(F_a(t) = 0\) 4. \(N_0 = u_{N+1} = 0\) 5. The substitution \(u_\sigma = A(\sigma)e^{\lambda t}\).

Substitute the given conditions into equation (10.51)

Plug in the given conditions into equation (10.51), and derive the difference equation subject to boundary conditions.

Solve the difference equation

Solve the derived difference equation using the results of Problem 5(d).

Obtain the modes of decay

Use the result of Problem 5(d) as a guide to obtain the modes of decay, expressed as \(u_\sigma(t)\) and \(a_n\).

Prove the ordering of \(a_n\) values

Show that the values of \(a_n\) satisfy the inequality \(0 < a_1 < a_2 < \cdots < a_N\).

Key concepts

Boundary Conditions

In the context of difference equations, boundary conditions help define the behavior of a system at its limits. When given a differential or difference equation, boundary conditions can set values at specific points or prescribe particular behaviors. In this exercise, the boundary conditions are:

• \(A(0) = 0\)
• \(A(N+1) = 0\)

These conditions imply that the system must start and end with specific values. Here, they are zero, which often helps to simplify the problem.
The boundary conditions are crucial because they ensure that the solution is unique and physically meaningful. By knowing the fixed points at the boundaries, we can solve for the behavior of the system across its entire domain.

Modes of Decay

The modes of decay describe how the amplitude of a system's response decreases over time. In this exercise, we look at specific functions defined by:\[u_{\sigma}(t) = \sin \left(\frac{n \pi}{N+1} \sigma\right) e^{-a_{n} t}\]It combines a spatial sine component and an exponential decay, where the term \(e^{-a_{n} t}\) represents the rate of decrease. The decay rates \(a_n\) hold the relationship:\[a_{n} = \frac{2 k^{2}}{h}\left(1-\cos \frac{n \pi}{N+1}\right)\]These describe how rapidly each mode fades over time.
It's important to show that \(0 < a_1 < a_2 < \cdots < a_N\) to establish that each subsequent mode dissipates faster than the previous one. This hierarchy helps in analyzing which modes will dominate the behavior of the system initially and how they will vanish.

Substitution Method

The substitution method is a powerful tool for solving difference equations. By assuming a certain form for the solution and replacing it within the original equation, one can transform complex problems into simpler forms. In this case, the substitution given is:\(u_{\sigma} = A(\sigma) e^{\lambda t}\)This transforms the difference equation into one that can be solved for \(A(\sigma)\).
This method is helpful because it reduces the problem to finding functions that satisfy both the difference equation and boundary conditions. Once \(A(\sigma)\) is found, the entire solution \(u_{\sigma}\) can be constructed.
The use of substitution simplifies the complexity by breaking down the problem into manageable parts, connecting the solution tightly to the problem's physical characteristics through the boundary conditions.