Assignment-daixieTM为您提供斯坦福大学Stanford University MATH 131P Partial Differential Equations偏微分方程代写代考和辅导服务！
Instructions:
Partial Differential Equations (PDEs) are mathematical equations that involve partial derivatives of an unknown function of two or more independent variables. They are used to describe many physical phenomena, such as the behavior of fluids, the propagation of waves, and the distribution of heat.
PDEs can be classified into various types based on their characteristics, such as linear or nonlinear, homogeneous or nonhomogeneous, and elliptic, parabolic, or hyperbolic. The solution to a PDE typically involves finding a function that satisfies the equation and any given boundary or initial conditions.
There are several techniques for solving PDEs, including separation of variables, method of characteristics, finite difference methods, finite element methods, and numerical methods. PDEs also have applications in various fields, including physics, engineering, finance, and biology.
Understanding and solving PDEs is a fundamental skill in many areas of science and engineering, and has led to numerous advances in technology and scientific discovery.
Consider the time-dependent Schrödinger equation
$$
i \hbar \frac{\partial \psi(\mathbf{x}, t)}{\partial t}=-\frac{\hbar^2}{2 m} \nabla^2 \psi(\mathbf{x}, t)+V(\mathbf{x}, t) \psi(\mathbf{x}, t)
$$
Consider the ansatz
$$
\psi(\mathbf{x}, t)=\phi(\mathbf{x}, t) \exp (i m S(\mathbf{x}, t) / \hbar)
$$
where the functions $\phi$ and $S$ are real. Find the partial differential equations are $\phi$ and $S$.
Since
$$
\frac{\partial \psi}{\partial t}=
$$
and
$$
\frac{\partial^2 \psi}{\partial x_j^2}=
$$
we obtain the coupled system of partial differential equations
$$
\begin{aligned}
\frac{\partial}{\partial t} \phi^2+\nabla \cdot\left(\phi^2 \nabla S\right) & =0 \
\frac{\partial}{\partial t} \nabla S+(\nabla S \cdot \nabla) \nabla S & =-\frac{1}{m}\left(\nabla \frac{-\left(\hbar^2 / 2 m\right) \nabla^2 \phi}{\phi}+\nabla V\right) .
\end{aligned}
$$
This is the Madelung representation of the Schrödinger equation. The term
$$
-\frac{\left(\hbar^2 / 2 m\right) \nabla^2 \phi}{\phi}
$$
of the right-hand side of the last equation is known as the Bohm potential in the theory of hidden variables.
Continuing the notation of the preceding exercise, if $G_1(x)$ and $G_2(x)$ are both antiderivatives of $g(x)$, show that $u(x, t)$ does not depend on which one we choose to represent $\int g(x) d x$. [Suggestion: Write down an equation relating $G_1$ and $G_2$.]
Recall that we have the partial differential equation
$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+g(x)$
Let $G_1(x)$ and $G_2(x)$ be two antiderivatives of $g(x)$, so that $G_1′(x) = g(x)$ and $G_2′(x) = g(x)$. Then we have \begin{align*} G_1(x) – G_2(x) &= \int G_1′(x) dx – \int G_2′(x) dx \ &= \int g(x) dx – \int g(x) dx \ &= C, \end{align*} where $C$ is a constant of integration. Therefore, $G_1(x)$ and $G_2(x)$ differ by a constant, and we can write
$G_1(x)=G_2(x)+C$.
Now, let $u_1(x, t)$ and $u_2(x, t)$ be the solutions to the partial differential equation with initial conditions $u_1(x, 0) = G_1(x)$ and $u_2(x, 0) = G_2(x)$, respectively. Then we have \begin{align*} u_1(x, 0) – u_2(x, 0) &= G_1(x) – G_2(x) \ &= C, \end{align*} and \begin{align*} \frac{\partial u_1}{\partial t} – \frac{\partial u_2}{\partial t} &= \frac{\partial^2 u_1}{\partial x^2} – \frac{\partial^2 u_2}{\partial x^2} + g(x) – g(x) \ &= \frac{\partial^2}{\partial x^2} (u_1 – u_2). \end{align*} Therefore, we have the partial differential equation
$\frac{\partial}{\partial t}\left(u_1-u_2\right)=\frac{\partial^2}{\partial x^2}\left(u_1-u_2\right)$
with initial condition $(u_1 – u_2)(x, 0) = C$. This is the one-dimensional heat equation with zero boundary conditions, and it is well-known that the solution is given by
$u_1(x, t)-u_2(x, t)=\frac{1}{\sqrt{4 \pi t}} \int_{-\infty}^{\infty}\left(u_1-u_2\right)(y, 0) e^{-(x-y)^2 /(4 t)} d y$
Substituting the initial condition $(u_1 – u_2)(x, 0) = C$ and simplifying, we obtain
$u_1(x, t)-u_2(x, t)=C$.
Therefore, the solutions $u_1(x, t)$ and $u_2(x, t)$ differ by a constant, and we conclude that $u(x, t)$ does not depend on which antiderivative we choose to represent $\int g(x) dx$.
Solve the following equation
$$
u_t+u_x+y u_y=\sin t
$$
for $0 \leq t, \quad 0 \leq x, \quad-\infty<y<\infty \quad$ and with
$$
\begin{array}{ll}
u=x+y & \text { for } t=0, x \geq 0 \
u=t^2+y & \text { for } x=0, t \geq 0 .
\end{array} \quad \text { and }
$$
Proof. Rewrite the equation as $\left(x \leftrightarrow x_1, y \leftrightarrow x_2, t \leftrightarrow x_3\right)$ :
$$
\begin{aligned}
& u_{x_3}+u_{x_1}+x_2 u_{x_2}=\sin x_3 \quad \text { for } 0 \leq x_3, 0 \leq x_1,-\infty<x_2<\infty \
& u\left(x_1, x_2, 0\right)=x_1+x_2 \
& u\left(0, x_2, x_3\right)=x_3^2+x_2
\end{aligned}
$$
