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# 数学代写|偏微分方程作业代写Partial Differential Equations代考|The Method of Characteristics, Part II: Quasilinear Equations

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## 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.

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|A FEW EXAMPLES

$$u_{t}+u u_{x}=0, u(x, 0)=g(x) .$$
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

$$u_{t}+u u_{x}+t u_{y}=y, u(x, y, 0)=x$$
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

• 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.

(X0(τ),是0(τ)),