Problem 8 - Advanced Engineering Mathematics

Question

Using \(u=X T\) and \(-\lambda\) as a separation constant we obtain \\[ \begin{array}{c} X^{\prime \prime}+\lambda X=0 \\ X^{\prime}(0)=0 \\ X^{\prime}(L)=0 \end{array} \\] and \\[ \begin{array}{c} T^{\prime \prime}+\lambda a^{2} T=0 \\ T^{\prime}(0)=0 \end{array} \\] Solving the differential equations we get $$X=c_{1} \sin \frac{n \pi}{L} x+c_{2} \cos \frac{n \pi}{L} x \quad \text { and } \quad T=c_{3} \cos \frac{n \pi a}{L} t+c_{4} \sin \frac{n \pi a}{L} t$$ for \(n=1,2,3, \ldots .\) The boundary and initial conditions, together with the fact that \(\lambda=0\) is an eigenvalue with eigenfunction \(X(x)=1,\) give $$u=A_{0}+\sum_{n=1}^{\infty} A_{n} \cos \frac{n \pi a}{L} t \sin \frac{n \pi}{L} x$$ Imposing \\[ u(x, 0)=x=A_{0}+\sum_{n=1}^{\infty} A_{n} \cos \frac{n \pi}{L} x \\] gives \\[ A_{0}=\frac{1}{L} \int_{0}^{L} x d x=\frac{L}{2} \\] and \\[ A_{n}=\frac{2}{L} \int_{0}^{L} x \cos \frac{n \pi}{L} x d x=\frac{2 L}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \\] for \(n=1,2,3, \dots,\) so that \\[ u(x, t)=\frac{L}{2}+\frac{2 L}{\pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2}} \cos \frac{n \pi a}{L} t \cos \frac{n \pi}{L} x \\]

Step-by-Step Solution

Answer

The solution is \(u(x,t) = \frac{L}{2} + \frac{2L}{\pi^2} \sum_{n=1}^{\infty} \frac{(-1)^n - 1}{n^2} \cos\left(\frac{n\pi a}{L}t\right) \cos\left(\frac{n\pi}{L}x\right)\).

Step 1: Set Up the Problem

We start with two differential equations obtained from separation of variables: \(X'' + \lambda X = 0\) with boundary conditions \(X'(0)=0\) and \(X'(L)=0\), and \(T'' + \lambda a^2 T = 0\) with \(T'(0) = 0\). The solution involves finding \(X\) and \(T\) in terms of sine and cosine functions because of the nature of the boundary conditions.

Step 2: Solve for X

To solve \(X'' + \lambda X = 0\), we propose solutions of the form \(X = c_1 \sin\left(\frac{n\pi}{L}x\right) + c_2 \cos\left(\frac{n\pi}{L}x\right)\). Applying the boundary conditions gives \(c_1 = 0\), hence \(X(x) = c_2 \cos\left(\frac{n\pi}{L}x\right)\). The even index \(n\) accounts for \(n=1, 2, 3, \ldots\). The condition also specifies an eigenvalue of zero with eigenfunction \(X(x) = 1\).

Step 3: Solve for T

Solving \(T'' + \lambda a^2 T = 0\) similarly yields \(T(t) = c_3 \cos\left(\frac{n\pi a}{L}t\right) + c_4 \sin\left(\frac{n\pi a}{L}t\right)\) under the condition \(T'(0) = 0\). Hence \(c_4 = 0\), and \(T(t) = c_3 \cos\left(\frac{n\pi a}{L}t\right)\) remains.

Step 4: Construct the General Solution

Combining \(X(x)\) and \(T(t)\), we write the solution for \(u(x, t)\) as a sum of separated solutions: \(u = A_0 + \sum_{n=1}^{\infty} A_n \cos\left(\frac{n\pi a}{L}t\right) \sin\left(\frac{n\pi}{L}x\right)\). \(A_0\) results from \(\lambda = 0\), and the rest of the series is built from non-zero \(\lambda\).

Step 5: Apply Initial Conditions and Calculate A_0

The initial condition \(u(x,0) = x\) implies \(A_0 + \sum_n A_n \cos\left(\frac{n\pi}{L}x\right) = x\). To find \(A_0\), integrate: \(A_0 = \frac{1}{L}\int_0^L x\, dx = \frac{L}{2}\).

Step 6: Calculate Coefficients A_n

For \(A_n\), using the Fourier coefficients from the series representation, we have \(A_n = \frac{2}{L}\int_0^L x \cos\left(\frac{n\pi}{L}x\right) dx = \frac{2L}{n^2\pi^2}((-1)^n - 1)\). This formula results from integrating each term with respect to the cosine function and solving for the coefficients.

Step 7: Write the Final Solution

Substitute \(A_0\) and \(A_n\) into the series solution: \(u(x,t) = \frac{L}{2} + \frac{2L}{\pi^2} \sum_{n=1}^{\infty} \frac{(-1)^n - 1}{n^2} \cos\left(\frac{n\pi a}{L}t\right) \cos\left(\frac{n\pi}{L}x\right)\). This is the final expression for the solution.

Key Concepts

Separation of Variables

Separation of variables is a powerful technique used to solve partial differential equations (PDEs). It simplifies complex equations by breaking them into simpler, separate parts that are easier to solve. The general idea is to assume the solution can be written as a product of functions, each depending only on a single variable.
For example, in this exercise, the solution for the function \( u(x,t) \) is assumed to be the product of a function \( X(x) \) depending only on \( x \) and a function \( T(t) \) depending only on \( t \). Mathematically, it's expressed as \( u = X(x) \cdot T(t) \).
This method helps reduce the PDE to ordinary differential equations (ODEs), making them manageable. This simplification leads to two separate equations: one for \( X \, (X'' + \lambda X = 0) \) and another for \( T \, (T'' + \lambda a^2 T = 0) \).
This approach works well when boundary conditions help define how the function behaves at the edges of the problem's domain. Separation of variables is particularly useful because it sets a groundwork to find particular solutions, which can then be combined to form a general solution.

Boundary Value Problems

Boundary value problems involve determining a function that satisfies a differential equation and specific conditions at the boundaries. They are central to understanding physical phenomena where boundaries play a crucial role, such as vibrations or heat flow.
In this situation, we looked at the boundaries given by the conditions \( X'(0)=0 \) and \( X'(L)=0 \). Essentially, these conditions specify that the slope or derivative of the function \( X \) must be zero at \( x = 0 \) and \( x = L \).
By applying these boundary conditions, we ensure the solution \( X(x) \) correctly models the physical situation, such as maintaining equilibrium or symmetry at the ends. These conditions often dictate the form of the solutions considerably

• They necessitate certain forms for \( X \) such as cosine functions in this exercise which inherently satisfy these derivative conditions.
• The zero eigenvalue condition also plays a crucial role, producing a constant solution representing a steady solution at these boundaries.

Thus, solving boundary value problems helps find solutions that are not only mathematically correct but also physically meaningful.

Partial Differential Equations

Partial differential equations (PDEs) involve functions of multiple variables and their partial derivatives. They form the mathematical foundation for describing various physical systems and processes, such as heat, sound, fluid flow, and quantum mechanics.
In this exercise, we focused on two linked PDEs through separation of variables, where \( u(x,t) \) is decomposed into parts \( X(x) \) and \( T(t) \). The challenge was solving them under specific conditions to model a physical scenario accurately.
PDEs can be complex because they consider different rates of change simultaneously, whether it's over time or across space. Solving them often requires methods that simplify these interdependent changes.

• The solution often involves finding eigenvalues \( \lambda \) which indicate modes of system behavior, each associated with a particular sinusoidal function form.
• The physical context, from the boundary conditions to initial states, frequently guides the form these solutions take.

Understanding PDEs is crucial for both theoretical insights and practical applications, with methods like Fourier series offering structured ways to reach solutions fitting given constraints. In essence, solving PDEs provides a way to visualize and predict the system's response over multiple dimensions.