# 数学方法和建模代写|MATHEMATICAL METHODS AND MODELLING MATHS2033 University of Glasgow Assignment

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## Instructions:

Dynamical systems refer to the study of how complex systems change over time, and how these changes can be described mathematically. This can be applied to a wide range of fields, including physics, biology, economics, and engineering, among others.

Integral transforms involve transforming functions from one domain to another using an integral operation. This can be used to simplify calculations or solve differential equations, among other applications.

By learning about these topics, you will be able to build models that can help describe and predict real-world phenomena. This can be especially useful for honours students who are planning to pursue research in fields where mathematical modelling is an important tool.

Suppose that we have a hexagonal cavity filled with $\varepsilon=1$ and surrounded by perfectly conducting walls, as shown in fig. 1(i).
(a) List the symmetry operations and conjugacy classes, and give the character table of this symmetry group $\left(C_{6 \mathrm{~V}}\right)$.

The symmetry operations of the hexagonal cavity can be obtained by considering the transformations that leave the cavity invariant. These include:

1. Identity (E): No rotation or reflection is applied to the hexagonal cavity.
2. Rotation by $\frac{\pi}{3}$ (C$_{6}$): A $\frac{\pi}{3}$ counterclockwise rotation of the hexagonal cavity about its center leaves it invariant.
3. Rotation by $\frac{2\pi}{3}$ (C$_{6}^{2}$): A $\frac{2\pi}{3}$ counterclockwise rotation of the hexagonal cavity about its center leaves it invariant.
4. Rotation by $\pi$ (C$_{6}^{3}$): A $\pi$ rotation of the hexagonal cavity about its center leaves it invariant.
5. Rotation by $\frac{4\pi}{3}$ (C$_{6}^{4}$): A $\frac{4\pi}{3}$ counterclockwise rotation of the hexagonal cavity about its center leaves it invariant.
6. Rotation by $\frac{5\pi}{3}$ (C$_{6}^{5}$): A $\frac{5\pi}{3}$ counterclockwise rotation of the hexagonal cavity about its center leaves it invariant.
7. Reflection through a vertical plane ( $\sigma_{v}$): Reflection of the hexagonal cavity through a vertical plane containing two opposite vertices leaves it invariant.
8. Reflection through a horizontal plane ( $\sigma_{h}$): Reflection of the hexagonal cavity through a horizontal plane containing two opposite edges leaves it invariant.
9. Reflection through a diagonal plane ( $\sigma_{d}$): Reflection of the hexagonal cavity through a diagonal plane containing two opposite vertices and intersecting the opposite edges perpendicularly leaves it invariant.

Solve $u^{\prime}=-10 u$ by at least two of the second order methods. Stop at $t=1$, and check the accuracy (do the numbers support $(\Delta t)^2 ?$ ). How serious is the stability test on $-10 \Delta t$ ?

The differential equation $u’=-10u$ is a first-order ordinary differential equation (ODE), which can be solved using numerical methods. Here, we will use two second-order methods: the backward Euler method and the trapezoidal method.

1. Backward Euler method: The backward Euler method approximates $u(t)$ at time $t+\Delta t$ by the value $u_{n+1}$:

$u_{n+1}=u_n-10 \Delta t u_{n+1}$

Rearranging this equation gives:

$u_{n+1}=\frac{1}{1+10 \Delta t} u_n$

1. Trapezoidal method: The trapezoidal method approximates $u(t)$ at time $t+\Delta t$ by the value $u_{n+1}$:

$u_{n+1}=u_n+\frac{\Delta t}{2}\left(-10 u_n-10 u_{n+1}\right)$

Rearranging this equation gives:

$u_{n+1}=\frac{1+\frac{5}{2} \Delta t}{1-\frac{5}{2} \Delta t} u_n$

To check the accuracy of these methods, we can compare the numerical solution to the exact solution of the differential equation, which is $u(t)=e^{-10t}$.

Consider a semi-infinite highway $0 \leq x<\infty$ (with no entrances of exits other than at $x=0$ ). Show that the number of cars on the highway at time $t$ is: $$N_0+\int_0^t q(0, \tau) d \tau$$ where $N_0$ is the number of cars in the highway at time $t=0$. You may assume that $\rho(x, t) \rightarrow 0$ as $x \rightarrow \infty$. Justify the equation both: directly (by physical reasoning), as well as by using the equation $\rho_t+q_x=0$.

The equation $\rho_t + q_x = 0$ is a continuity equation, which states that the rate of change of the density of cars $\rho$ in a certain region is equal to the rate at which cars are entering or leaving that region, which is given by the flux $q$.

Now, let’s consider the region $[0, x]$ of the highway at time $t$. The number of cars in this region at time $t$ is given by the integral of the density over the region:

$N(x, t)=\int_0^x \rho(\xi, t) d \xi$

Taking the derivative with respect to time on both sides, we get:

$\frac{d N}{d t}=\int_0^x \frac{\partial \rho}{\partial t} d \xi=-\int_0^x \frac{\partial q}{\partial x} d \xi=-q(x, t)$

where we used the continuity equation. Therefore, the rate of change of the number of cars in the region $[0, x]$ is equal to the flux of cars leaving the region.

Now, let’s consider the entire highway $[0,\infty)$ at time $t$. The number of cars on the highway at time $t$ is given by $N(\infty, t)$. We can write this as:

$N(\infty, t)=N(0, t)+\int_0^{\infty} \frac{d N}{d t} d x=N(0, t)-\int_0^{\infty} q(x, t) d x$

where we used the fact that the rate of change of the number of cars on the highway is equal to the flux of cars leaving the highway, and we assumed that $\rho(x,t) \rightarrow 0$ as $x \rightarrow \infty$.

We can rearrange this equation to get:

$N(\infty, t)=N(0, t)+\int_0^t \int_0^{\infty} \frac{\partial q}{\partial t} d x d \tau$

Now, let’s use the continuity equation to express $\partial q / \partial t$ in terms of $\partial \rho / \partial t$. We have:

$\frac{\partial q}{\partial t}=-\frac{\partial(\rho q)}{\partial x}+q \frac{\partial \rho}{\partial x}$

Integrating this equation over the entire highway, we get:

$\int_0^{\infty} \frac{\partial q}{\partial t} d x=-\int_0^{\infty} \frac{\partial(\rho q)}{\partial x} d x+\int_0^{\infty} q \frac{\partial \rho}{\partial x} d x$

The first term on the right-hand side can be integrated by parts:

$-\int_0^{\infty} \frac{\partial(\rho q)}{\partial x} d x=[\rho q]_0^{\infty}=-\rho(\infty, t) q(\infty, t)$

where we used the fact that $\rho(x,t) \rightarrow 0$ as $x \rightarrow \infty$. Therefore, we can rewrite the above equation as:

$\int_0^{\infty} \frac{\partial q}{\partial t} d x=-\rho(\infty$