MAT1252 Partial Differential Equations Final Exam Study Guide June 2025

Examination Overview: MAT1252 - Partial Differential Equations

The Final Examination for the Even Semester of the 2024/2025 academic year for the course MAT1252 - Partial Differential Equations (Persamaan Diferensial Parsial) was held on Wednesday, June 18, 2025. The examination took place from 10:30 to 12:30. The testing protocol was strictly defined as Closed Notes (Catatan Tertutup) and the use of calculators was prohibited (Tanpa Kalkulator). The document credits Ikhsan Rio Afriadi (@ikhsanrio16) and covers fundamental topics in partial differential equations, including wave equations, Fourier series, Laplace equations, and Sturm-Liouville problems.

Derivation of the One-Dimensional Wave Equation

The first requirement of the examination is to detail the steps for the derivation of the one-dimensional wave equation, expressed in the form $u_{tt} = \alpha u_{xx}$. This derivation typically involves analyzing the transverse vibrations of a flexible string. The process begins by considering a small segment of a string with constant linear density $\rho$ under a tension force $T$. By applying Newton's Second Law (\mathbf{F} = m·
\mathbf{a}) to the vertical components of the tension forces at the endpoints of a segment $[x, x + \Delta x]$, one identifies the net force acting on the segment. Assuming small angles of displacement, the vertical forces can be approximated using the slopes of the string, $u_x(x, t)$.

The mass of the segment is represented by $\rho \Delta x$, and its vertical acceleration is $u_{tt}$. Equating the mass-acceleration product to the difference in the vertical components of tension results in the equation $\rho \Delta x u_{tt} \approx T [u_x(x + \Delta x, t) - u_x(x, t)]$. Dividing by $\Delta x$ and taking the limit as $\Delta x \rightarrow 0$ leads to the second-order partial differential equation $u_{tt} = \frac{T}{\rho} u_{xx}$, where $\alpha = \frac{T}{\rho}$ represents the square of the wave propagation speed, often denoted as $c^2$.

Determination of Fourier Coefficients for Periodic Functions

The second problem focuses on a periodic function $f(x)$ with a period of $2\pi$. The function is defined piecewise as follows: $f(x) = -1$ for $-\pi \leq x < 0$, and $f(x) = x - 1$ for $0 \leq x < \pi$. The goal is to determine the Fourier coefficient $a_n$ for the general Fourier series representation $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))$.

To find $a_n$, one must evaluate the integral $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx$. This integral must be split into two parts according to the definition of $f(x)$: $a_n = \frac{1}{\pi} [ \int_{-\pi}^{0} (-1) \cos(nx) dx + \int_{0}^{\pi} (x - 1) \cos(nx) dx ]$.

The first part of the integral involves the basic integration of a cosine function, while the second part requires integration by parts for the term $x \cos(nx)$. The integration must be performed for $n = 1, 2, 3, \dots$ to determine the general formula for the coefficients.

Fourier Cosine Series of a Sine Function

The third problem asks for the Fourier Cosine Series of the function $f(x) = \sin(x)$ on the interval $0 < x < \pi$, assuming a period of $2\pi$. Because the cosine series is requested, the function must be treated as an even extension of itself over the interval $[-\pi, \pi]$. Even though the function itself is a sine, the cosine series representation for an even extension on the interval $[0, L]$ (where $L = \pi$) is given by: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(\frac{n \pi x}{L})$.

For this specific case, $L = \pi$, so the series becomes $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx)$. The coefficients are calculated using the formula $a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin(x) \cos(nx) dx$. This integral is typically solved using trigonometric product-to-sum identities: $\sin(x) \cos(nx) = \frac{1}{2} [ \sin((1+n)x) + \sin((1-n)x) ]$.

Solving the One-Dimensional Wave Equation with Specific Boundaries

The fourth problem presents a specific initial-boundary value problem (IBVP) for the wave equation: $u_{tt} = 4u_{xx}$ for the spatial domain $0 < x < 30$ and time $t > 0$.

The Boundary Conditions (SB) are homogeneous Dirichlet conditions: $u(0, t) = 0$ and $u(30, t) = 0$ for all $t > 0$.

The Initial Conditions (SA) are given by:

• The initial displacement $u(x, 0) = f(x)$.
• The initial velocity $u_t(x, 0) = 0$.

The function $f(x)$ is defined as a triangle-like distribution: $f(x) = \frac{x}{10}$ for $0 < x \leq 10$. $f(x) = \frac{30 - x}{20}$ for $10 < x \leq 30$.

Since the initial velocity is zero, the solution takes the form of a separation of variables solution: $u(x, t) = \sum_{n=1}^{\infty} A_n \sin(\frac{n \pi x}{30}) \cos(\frac{2 n \pi t}{30})$. The coefficients $A_n$ must be determined by the Fourier Sine Series of the initial displacement function $f(x)$ on the interval $[0, 30]$, calculated as $A_n = \frac{2}{30} \int_{0}^{30} f(x) \sin(\frac{n \pi x}{30}) dx$.

Laplace's Equation on a Semi-Infinite Rectangular Plate

The fifth problem requires solving Laplace's equation $u(x, y)$ on a rectangular plate defined by the boundaries $0 < x < 50$ and $0 < y < \infty$ (a semi-infinite strip). The boundary conditions are:

• $u(0, y) = 0$ and $u(50, y) = 0$ for $0 < y < \infty$ (lateral boundaries held at zero).
• $u(x, \infty) = 0$ (implied by the requirement for a physically bounded solution as $y$ approaches infinity).
• The bottom boundary condition is given by $u(x, 0) = f(x)$, where: $f(x) = x$ for $0 < x \leq 25$. $f(x) = 50 - x$ for $25 \leq x \leq 50$.

The general solution for Laplace's equation in this geometry is $u(x, y) = \sum_{n=1}^{\infty} C_n \sin(\frac{n \pi x}{50}) e^{-\frac{n \pi y}{50}}$. The coefficients $C_n$ are found by evaluating the Fourier Sine Series of $f(x)$ on the interval $[0, 50]$: $C_n = \frac{2}{50} \int_{0}^{50} f(x) \sin(\frac{n \pi x}{50}) dx$.

Eigenvalues and Eigenfunctions of a Sturm-Liouville Problem

The final problem investigates a Sturm-Liouville problem defined by the differential equation $y'' + \lambda y = 0$ on the interval 0 \leq x
\leq 1. The boundary conditions are mixed:

• At $x = 0$, the derivative is zero: $y'(0) = 0$.
• At $x = 1$, there is a boundary condition involving both the function and its derivative: $y'(1) + y(1) = 0$.

To solve this, one must analyze three cases for the eigenvalue $\lambda$.

1. If $\lambda < 0$ (let $\lambda = -k^2$), the solution follows the form $y(x) = c_1 \cosh(kx) + c_2 \sinh(kx)$. Setting $y'(0) = 0$ leads to $c_2 = 0$. The second condition leads to a transcendental equation that must be checked for non-trivial solutions.
2. If $\lambda = 0$, the solution is linear: $y(x) = c_1 x + c_2$. Setting $y'(0) = 0$ implies $c_1 = 0$, meaning $y(x) = c_2$. The second condition $y'(1) + y(1) = 0$ results in $0 + c_2 = 0$, yielding only the trivial solution.
3. If $\lambda > 0$ (let $\lambda = k^2$), the solution is $y(x) = c_1 \cos(kx) + c_2 \sin(kx)$. The condition $y'(0) = 0$ forces $c_2 = 0$. Applying the second condition, $y'(1) + y(1) = 0$, results in the equation $-k \sin(k) + \cos(k) = 0$, or $\tan(k) = \frac{1}{k}$. The values of $k$ that satisfy this transcendental equation define the eigenvalues $\lambda_n = k_n^2$, and the corresponding eigenfunctions are $y_n(x) = \cos(k_n x)$.