如果你也在 怎样代写偏微分方程Partial Differential Equations这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。偏微分方程Partial Differential Equations是一个方程，它规定了一个多变量函数的各种偏导数之间的关系。该函数通常被认为是一个有待解决的 “未知数”，类似于x被认为是代数方程（如x2-3x+2=0）中有待解决的一个未知数。因此，在现代数学和科学研究中，有大量使用计算机对某些偏微分方程的解进行数值近似的方法。
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我们提供的偏微分方程Partial Differential Equations及其相关学科的代写，服务范围广, 其中包括但不限于:
调和函数 harmonic function
椭圆方程 elliptic equation
抛物方程 Parabolic equation
双曲方程 Hyperbolic equation
非线性方法 nonlinear method
变分法 Calculus of Variations
几何分析 geometric analysis
偏微分方程数值解 Numerical solution of partial differential equations
数学代写|偏微分方程作业代写Partial Differential Equations代考|A Few Examples
A Few Examples. We begin by revisiting the example of the last section.
Example 2.4.1. Recall Burgers’s equation from Section 2.3:
$$
u_{t}+u u_{x}=0, u(x, 0)=g(x) .
$$
We want to find a solution in $\Omega$, the upper half-plane $x \in \mathbb{R}, t \geq 0$. Our data is given on $\Gamma$, the $t=0$ axis. This is certainly of the form (2.24) with
$$
a(x, t, u)=u, \quad b(x, t, u)=1, \quad c(x, t, u)=0,
$$
and, hence, the characteristic equations (2.26) become
$$
\dot{x}(s)=z, \quad \dot{t}(s)=1, \quad \dot{z}(s)=0 .
$$
As usual, we will take $t=s$ and parametrize the characteristics with time $t$. Thus we have the two equations
$$
\frac{d x}{d t}=z(t) \quad \text { and } \quad \frac{d z}{d t}=0,
$$
which we solve with initial conditions $x(0)=x_{0}$ and $z(0)=z_{0}=g\left(x_{0}\right)$.
The solutions are
$$
z(t)=g\left(x_{0}\right) \text { and } x(t)=g\left(x_{0}\right) t+x_{0} .
$$
Finally, given $(x, t)$ with $t>0$, we have
$$
u(x, t)=g\left(x_{0}\right) \quad \text { where } \quad \frac{x-x_{0}}{t}=g\left(x_{0}\right) .
$$
数学代写|偏微分方程作业代写Partial Differential Equations代考|More Than Two Independent Variables
Than Two Independent Variables. The method readily extends to quasilinear equations in any number of independent variables. The following is an example with three independent variables.
Example 2.4.3. We find a solution $u(x, y, t)$ to the initial value problem
$$
u_{t}+u u_{x}+t u_{y}=y, u(x, y, 0)=x
$$
on $\Omega={(x, y, t) \mid t \geq 0}$, the upper half-space.
Using $t$ to parametrize the characteristics, the characteristic equations become
$$
\dot{x}(t)=z(t), \quad \dot{y}(t)=t, \quad \dot{z}(t)=y(t),
$$
with initial conditions $x(0)=x_{0}, y(0)=y_{0}$, and $z(0)=x_{0}$. We solve first the $y$ equation followed by the $z$ equation to find
$$
y(t)=\frac{t^{2}}{2}+y_{0} \quad \text { and } \quad z(t)=\frac{t^{3}}{6}+y_{0} t+x_{0} .
$$
Solving the $x$ equation yields
$$
x(t)=\frac{t^{4}}{24}+y_{0} \frac{t^{2}}{2}+x_{0} t+x_{0}
$$
Finally, given $x, y$, and $t>0$, we have
$$
x_{0}=\frac{x-\frac{t^{4}}{24}-y_{0} \frac{t^{2}}{2}}{t+1} \quad \text { and } \quad y_{0}=y-\frac{t^{2}}{2},
$$
and hence (after a little algebra),
$$
u(x, y, t)=\frac{24 x-12 y t^{2}+5 t^{4}}{24(1+t)}+y t-\frac{t^{3}}{3}
$$
数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|Pausing to Reflect on the Inherent Logic Behind the Method of Characteristics and Local Solutions of Quasilinear Equations
Before continuing to general first-order equations, let us draw reference to the general flow of logic we followed in the method of characteristics for linear and quasilinear equations. This logic and approach are very common in PDEs, as discussed in Section 1.6.
- We started by assuming that a solution $u$ exists to the PDE and auxiliary condition. This was required in order to write down the $z$ equation for the method of characteristics.
- Noting that the PDE is a statement about directional derivatives of $u$ in certain directions, we wrote down the ODEs for
(i) the characteristics in the domain of $u$ which follow these directions and
(ii) the value of $u$ itself along these characteristics.
The only result that this analysis actually proves is the following statement: If $u$ solves the PDE and auxiliary solution and $(x(s), y(s))$ solve the appropriate characteristic ODEs, then $z(s)$, defined to be $u(x(s), y(s))$, must solve the characteristic ODE for $z$. However, these steps give us a way of synthesizing the solution via the following procedure. We attempt to uniquely solve the characteristic ODEs with the hope that the projected characteristics both fill up the domain in question and do not intersect. This will be the case if we can invert the solutions for the projected characteristics in order to solve uniquely for the variables used to parametrize the characteristics (e.g., $x_{0}$ and $s$ ) in terms of the independent variables (e.g., $x$ and $y$ ). Our canonical example of Burgers’s equation illustrates that this inversion step can fail. However, in cases where one can invert these equations and find an explicit formula for $u$, we can (of course) check that it does indeed solve the PDE (though we usually skip this last step).
A natural general question to ask is: Assuming smoothness of all the involved ingredients – the coefficient functions of the PDE, the data curve (where the data is specified), and the data itself – does there at least exist a local solution (a solution defined in a neighborhood of the data curve)? The answer is yes assuming the data is noncharacteristic in the sense that the transversality condition holds. For the general quasilinear equation (2.24) in two independent variables with data $g$ on the curve $\Gamma$ described in parametric form by
$$
\left(x_{0}(\tau), y_{0}(\tau)\right),
$$
the transversality condition is
$$
\operatorname{det}\left[\begin{array}{ll}
a\left(x_{0}(\tau), y_{0}(\tau), g\left(x_{0}(\tau), y_{0}(\tau)\right)\right) & x_{0}^{\prime}(\tau) \
b\left(x_{0}(\tau), y_{0}(\tau), g\left(x_{0}(\tau), y_{0}(\tau)\right)\right) & y_{0}^{\prime}(\tau)
\end{array}\right] \neq 0 .
$$
The key tool in the proof is the Inverse Function Theorem.
偏微分方程代写
数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|A FEW EXAMPLES
几个例子。我们首先回顾上一节的例子。
示例 2.4.1。回忆一下第 2.3 节中的 Burgers 方程：
$$
u_{t}+u u_{x}=0, u(x, 0)=g(x) .
$$
We want to find a solution in $\Omega$, the upper half-plane $x \in \mathbb{R}, t \geq 0$. Our data is given on $\Gamma$, the $t=0$ axis. This is certainly of the form (2.24) with
$$
a(x, t, u)=u, \quad b(x, t, u)=1, \quad c(x, t, u)=0,
$$
and, hence, the characteristic equations (2.26) become
$$
\dot{x}(s)=z, \quad \dot{t}(s)=1, \quad \dot{z}(s)=0 .
$$
As usual, we will take $t=s$ and parametrize the characteristics with time $t$. Thus we have the two equations
$$
\frac{d x}{d t}=z(t) \quad \text { and } \quad \frac{d z}{d t}=0,
$$
which we solve with initial conditions $x(0)=x_{0}$ and $z(0)=z_{0}=g\left(x_{0}\right)$.
The solutions are
$$
z(t)=g\left(x_{0}\right) \text { and } x(t)=g\left(x_{0}\right) t+x_{0} .
$$
Finally, given $(x, t)$ with $t>0$, we have
$$
u(x, t)=g\left(x_{0}\right) \quad \text { where } \quad \frac{x-x_{0}}{t}=g\left(x_{0}\right) .
$$
数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|MORE THAN TWO INDEPENDENT VARIABLES
比两个自变量。该方法很容易扩展到任意数量的自变量中的拟线性方程。以下是具有三个自变量的示例。
示例 2.4.3。我们找到解决方案在(X,是,吨)初值问题
$$
u_{t}+u u_{x}+t u_{y}=y, u(x, y, 0)=x
$$
on $\Omega={(x, y, t) \mid t \geq 0}$, the upper half-space.
Using $t$ to parametrize the characteristics, the characteristic equations become
$$
\dot{x}(t)=z(t), \quad \dot{y}(t)=t, \quad \dot{z}(t)=y(t),
$$
with initial conditions $x(0)=x_{0}, y(0)=y_{0}$, and $z(0)=x_{0}$. We solve first the $y$ equation followed by the $z$ equation to find
$$
y(t)=\frac{t^{2}}{2}+y_{0} \quad \text { and } \quad z(t)=\frac{t^{3}}{6}+y_{0} t+x_{0} .
$$
Solving the $x$ equation yields
$$
x(t)=\frac{t^{4}}{24}+y_{0} \frac{t^{2}}{2}+x_{0} t+x_{0}
$$
Finally, given $x, y$, and $t>0$, we have
$$
x_{0}=\frac{x-\frac{t^{4}}{24}-y_{0} \frac{t^{2}}{2}}{t+1} \quad \text { and } \quad y_{0}=y-\frac{t^{2}}{2},
$$
and hence (after a little algebra),
$$
u(x, y, t)=\frac{24 x-12 y t^{2}+5 t^{4}}{24(1+t)}+y t-\frac{t^{3}}{3}
$$
数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|PAUSING TO REFLECT ON THE INHERENT LOGIC BEHIND THE METHOD OF CHARACTERISTICS AND LOCAL SOLUTIONS OF QUASILINEAR EQUATIONS
在继续讨论一般一阶方程之前，让我们参考我们在线性和拟线性方程的特征方法中遵循的一般逻辑流程。如第 1.6 节所述，这种逻辑和方法在 PDE 中非常常见。
- 我们首先假设一个解决方案在存在于偏微分方程和辅助条件。这是为了写下和特征方法的方程。
- 注意到 PDE 是关于方向导数的陈述在在某些方向上，我们写下了 ODE
一世领域的特征在遵循这些指示和
一世一世的价值在本身就沿着这些特点。
该分析实际证明的唯一结果是以下陈述：如果在求解 PDE 和辅助解和(X(s),是(s))求解适当的特征 ODE，然后和(s)，定义为在(X(s),是(s))，必须求解特征 ODE 为和. 但是，这些步骤为我们提供了一种通过以下过程合成解决方案的方法。我们尝试对特征 ODE 进行唯一求解，希望投影特征既能填满所讨论的域，又不会相交。如果我们可以反转投影特征的解，以便对用于参数化特征的变量进行唯一求解，就会出现这种情况和.G.,$X0$一种nd$s$就自变量而言和.G.,$X$一种nd$是$. Burgers 方程的典型例子说明了这个反演步骤可能会失败。但是，如果可以反转这些方程并找到一个明确的公式在， 我们可以这FC这在rs和检查它是否确实解决了 PDE吨H这在GH在和在s在一种ll是sķ一世p吨H一世sl一种s吨s吨和p.
一个自然的一般性问题是：假设所有相关成分的平滑度 – PDE 的系数函数，数据曲线在H和r和吨H和d一种吨一种一世ssp和C一世F一世和d，以及数据本身——是否至少存在本地解决方案一种s这l在吨一世这nd和F一世n和d一世n一种n和一世GHb这rH这这d这F吨H和d一种吨一种C在r在和? 答案是肯定的，假设数据在横向条件成立的意义上是非特征性的。对于一般拟线性方程2.24在两个带有数据的自变量中G在曲线上Γ以参数形式描述
(X0(τ),是0(τ)),
横向条件是
这[一种(X0(τ),是0(τ),G(X0(τ),是0(τ)))X0′(τ) b(X0(τ),是0(τ),G(X0(τ),是0(τ)))是0′(τ)]≠0.
证明中的关键工具是反函数定理。
数学代写|偏微分方程作业代写Partial Differential Equations代考 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。
