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MATH421课程简介
Description
Prerequisites: MATH 224A and MATH 312. This course introduces three main types of partial differential equations (PDEs): parabolic, elliptic, and hyperbolic as well as mathematical and computational tools for solving PDEs. It balances mathematical rigor, computational techniques, and real-world applications. Topics include heat equation, method of separation of variables, Laplace’s equation, Fourier series, wave equation, finite difference/element methods, and highdimensional PDEs.
Credits
3
Prerequisites
Prerequisite(s): MATH 420 with a grade of C- or better. Partial differential equations arise in the mathematical modeling of many physical, chemical, and biological phenomena. They play a crucial role in diverse subject areas, such as fluid dynamics, electromagnetism, material science, astrophysics, financial modeling, and hydrogeology, for example. This course is an introduction to partial differential equations with emphasis on the wave, diffusion and Laplace equations. The focus will be on understanding the physical meaning and mathematical properties of solutions of partial differential equations. Methods of solutions include separation of variables using orthogonal series, transform methods, method of characteristics, and some numerical methods. 3 hours lecture.
MATH421 Partial Differential Equations HELP（EXAM HELP， ONLINE TUTOR）
Consider the quadratic polynomial: $p(x, y)=2 x^2+6 x y+2 y^2$.
(a) Will the level sets of $p$ be ellipses, hyperbolas, parabolas, or lines?
(b) Write $p$ as a quadratic form with a symmetric matrix.
(a) The level sets of $\$ p \$$ will be ellipses because $\$ p \$$ is a quadratic polynomial with positive leading coefficients in both $\$ x \$$ and $\$ y \$$.
(b) We can write $\$ \mathrm{p} \$$ as a quadratic form with a symmetric matrix as follows:
$$
p(x, y)=\left(\begin{array}{ll}
x & y
\end{array}\right)\left(\begin{array}{ll}
2 & 3 \
3 & 2
\end{array}\right)\left(\begin{array}{l}
x \
y
\end{array}\right)
$$
(c) Diagonalize the matrix.
(d) Find a linear change of coordinates that removes the $x y$ “cross term” from $p$.
(c) To diagonalize the matrix, we first find the eigenvalues by solving the characteristic equation: \begin{align*} \begin{vmatrix} 2-\lambda & 3 \ 3 & 2-\lambda \end{vmatrix} &= (2-\lambda)^2 – 3^2 \ &= \lambda^2 – 4\lambda – 5 \ &= (\lambda – 5)(\lambda + 1) \end{align*} So the eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = -1$. To find the eigenvectors, we solve the system of equations $(A – \lambda_i I) \mathbf{v}i = \mathbf{0}$ for each eigenvalue. For $\lambda_1 = 5$, we have: \begin{align*} (A – \lambda_1 I) \mathbf{v}1 &= \begin{pmatrix} -3 & 3 \ 3 & -3 \end{pmatrix} \begin{pmatrix} v{11} \ v{12} \end{pmatrix} = \mathbf{0} \ \Rightarrow v_{11} &= v_{12} \end{align*} So an eigenvector corresponding to $\lambda_1$ is $\begin{pmatrix} 1 \ 1 \end{pmatrix}$. For $\lambda_2 = -1$, we have: \begin{align*} (A – \lambda_2 I) \mathbf{v}2 &= \begin{pmatrix} 3 & 3 \ 3 & 3 \end{pmatrix} \begin{pmatrix} v{21} \ v_{22} \end{pmatrix} = \mathbf{0} \ \Rightarrow v_{21} &= -v_{22} \end{align*}
So an eigenvector corresponding to $\$ \backslash$ lambda_ $2 \$$ is $\$ \backslash$ begin ${$ pmatrix $} 1 \backslash-1$ lend{pmatrix}\$. Therefore, we can diagonalize $\$ A \$$ as follows:
$$
A=P \Lambda P^{-1}=\left(\begin{array}{cc}
1 & 1 \
1 & -1
\end{array}\right)\left(\begin{array}{cc}
5 & 0 \
0 & -1
\end{array}\right)\left(\begin{array}{cc}
1 / 2 & 1 / 2 \
1 / 2 & -1 / 2
\end{array}\right)
$$
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