偏微分方程(数学物理方程):抽象程度★★,计算量★★★★★。对很多本科生来说偏微分是一门每年挂科50%的杀手课,原因很简单,计算量太大。同时PDE方向的教授硬功夫很扎实,他会很自然的认为所有的拥有扎实的计算能力是理所当然的事情,这可苦了第一次学偏微分方程的学生们,计算量大然后不知道怎么算,很基础的偏微分方程课程中也会出现很多不等式估计,之前接触比较少的同学看到这些估计可能头都大了。

实际上大部分QS排名前200的数学系的本科教学计划里,一般不会设置专门的pde课(指研究Laplace算子的性质与先验估计一类的)。倒是会有一门相对浅显的课程叫数学物理方程,这门课程主要讲如何解1维2维各种标准几何体内部的波动方程和热方程,possion方程等方程,相对来说方法有迹可循,计算量在可控范围内,与正经的pde课程画风显然不一样,但是即使是这样,也是足够难的一门课.

另外一方面,如果是偏微分方程数值解这门课,由于没有任何知识储备,刚学pde数值解的时候是完全懵逼的…前面学谱方法和有限元还好,毕竟泛函刚学过还热乎着,这部分的误差估计没什么难懂的。

等到了有限差分,问题就来了。研究有限差分的误差与稳定性,势必要用先验估计的技术(Lux登场)。面对书里突如其来的洋洋洒洒十来页不等式估计,有知识断层的我感觉脑阔痛。虽然相比正经的pde课程和调和分析,这等篇幅的估计不值一提,可我当时对pde这套技术完全没基础啊QAQ人都傻了

刨去理论部分,pde数值解计算量也不小。虽说实际问题靠计算机,但是学习过程中,作业和考试里给你一个网格很粗的有限元/有限差分要求笔算,应该是基本操作吧。综合来看计算量和古典微分几何是四六开(这里说的计算是不用过脑子那种暴算,如果算上凑试验函数的过程,应该计算量可以暴打古典微分几何)

综上,pde数值解抽象不抽象?

完全不抽象

难学不难学?

不算很难学,但是要花足够多的时间!

如果只是想浅尝辄止了解一下这门课程的基本知识,但是又不想陷入复杂的计算,那么一个折中的办法就是选这门课跟着老师听一下然后exam找一个靠谱的老哥帮忙上下分,这样既可以满足体验高难度课程带来的快乐,又不会被高难度课程所带来的pass压力所困扰,而UprivateTA就可以在这门课程上提供你们需要的服务~

下面就是一场美国UCSB大学数学系偏微分方程的代考成功案例:

partial difference equations代考
Problem 1.

Let $u$ be a solution to
$$
u_{t}=u_{x x x x x x}, u(x)=\sin (x)+\cos (x)
$$
be a $[-\pi, \pi]$ periodic solution. Provide two possibilities to solve this problem. Do you have and energy estimate?

Proof .

Let $h$ be such that $\hat{h}(\xi)=e^{-\xi^{6}}$. We note that $u_{t}=u_{x x x x x x}$ implies that
$$
\frac{d}{d t} \hat{u}(\xi, t)=(i \xi)^{6} \hat{u}(\xi, t)=-\xi^{6} \hat{u}(\xi, t)
$$
That means
$$
\hat{u}(\xi, t)=e^{-t \xi^{6}} \hat{u}(\xi, 0)
$$
Then we first note that (with $\left.z=t^{1 / 6} y d z=t^{1 / 6} d y\right)$
$$
\begin{aligned}
e^{-t \xi^{6}} &=\frac{1}{\sqrt{2 \pi}} \int e^{-i t^{1 / 6} \xi y} h(y) d y \
&=t^{-1 / 6} \frac{1}{\sqrt{2 \pi}} \int e^{-i \xi z} h\left(t^{-1 / 6} z\right) d z
\end{aligned}
$$
This implies with Fourier inversion formula that
$$
\begin{aligned}
u(x, t) &=\int e^{i \xi x} \hat{u}(\xi, t) \frac{d x}{\sqrt{2 \pi}} \
&=\int e^{i \xi x} e^{-t \xi^{6}} \hat{u}(\xi, 0) \frac{d x}{\sqrt{2 \pi}} \
&=t^{-1 / 6} \iint e^{i \xi x} e^{-i \xi z} \hat{u}(\xi, 0) \frac{d x}{\sqrt{2 \pi}} h\left(t^{-1 / 6} z\right) d z \
&=t^{-1 / 6} \int u(x-z) h\left(t^{-1 / 6} z\right) \frac{d z}{\sqrt{2 \pi}}
\end{aligned}
$$
Now it suffices to plug in $u(x)=\sin (x)+\cos (x)$. For the second solution, we just use periodicity and deduce that
$$
u(x, t)=\sum_{k \in \mathbb{Z}} e^{-t k^{6}} \hat{u}(k, 0) e^{-i k x} .
$$
Note that $\sin (x)=\frac{e^{i x}-e^{-i x}}{2 i}, \cos (x)=\frac{e^{i x}+e^{i x}}{2}$ and hence
$$
\hat{u}(1,0)=\frac{1}{2}(1-i), \hat{u}(-1,0)=\frac{1}{2}(1+i)
$$
Therefore
$$
2 u(x, t)=(1-i) e^{-t(1)^{6}} e_{1}+(1+i) e^{-t(-1)^{6}} e_{-1}=e^{-t} 2 u(x, 0)
$$

implies $u(x, t)=e^{-t}(\cos (x)+\sin (x))$

Problem 2.

Let
$$
u_{t}=u_{x x}, u(x, 0)=\sin (\pi x) u(0, t)=u(1, t)=0
$$
Provide two possibilities to solve this problem. Do we have uniqueness?

Proof .

Solution: Uniqueness follows from the maximum principle. The first possibility is ‘Ansatz’ (meaning try and error)
$$
u(x, t)=T(t) X(x)
$$
Then the pde implies
$$
T^{\prime} X=T X^{\prime \prime}
$$
in other words $T^{\prime}=e^{-\lambda t} T(0)$ and
$$
X^{\prime \prime}=-\lambda X
$$
Our initial condition $X(x)=\sin (\pi x)$ satisfies $X^{\prime \prime}(x)=-\pi^{2} X(x)$ and hence $\lambda=\pi^{+2}$ and we are done (because we were lucky).

The systematic solution is given by extending $\phi(x)=\sin (\pi x)$ to an odd solution to the write of 0, meaning $\phi_{e x t}=\sin (\pi x)$ and then extend to a 2-periodic function. Let $f_{k}$ be an orthonormal basis for all 2 periodic solutions, indeed $f_{k}(x)=e^{i \pi k}$ will do. Then
$$
u_{t}=u_{x x}
$$
means
$$
\frac{d}{d t} \hat{u}(k, t)=e^{-t \pi^{2} k^{2}} \hat{u}(k, 0)
$$
In our case $\sin (\pi x)=\frac{f_{1}+f_{-1}}{2 i}$ and hence
$$
u(x, t)=\frac{1}{2 i} e^{-t \pi^{2}} f_{1}-\frac{1}{2 i} e^{-t \pi^{2}} f_{-1}=e^{-t \pi^{2}} \sin (\pi x)
$$

Problem 3.

For diffusion equation (on the real line say) singularities are immediately lost. However for the wave equation they are
kept. Can you explain this on the basis of the solution formulae we have?

Proof .

The solution formula for the heat equation and the $N$ –
part of the solution for the wave equation are both given by integrals and
hence ‘smoothen out’ singularities by integrating them. However, the term $\frac{\phi(x+t)+\phi(x-t)}{2}$ keeps the singularity, and the only cancellation occurs from points which are $t$ opposite with respect to $x$. However, this cancelation
can only occur for specific $x$ at a given time, and hence in a neighbourhood we still detect the singularity (think of $(x-1)^{-1}-(x-3)^{-1}$ then at $t=1=x$ we will see some cancelation).

Problem 4.

Let $L(u)=u_{x x}+u_{y y}$ on $[-\pi, \pi] \times[-\pi, \pi] .$ Find a solution operator of the form
$$
S(t)(\phi)(x, y)=\sum_{k, j} \text { somethinge }{k}(x) e{j}(y)
$$
for $u_{t}=u_{x x}+u_{y y}, u(x, y, 0)=\phi(x, y)$.

Proof .

Let $u_{k j}(x, y)=e_{k}(x) e_{j}(y) .$ Then
$$
L(u)=-\left(k^{2}+j^{2}\right) u_{k j}
$$
is an eigenvalue and hence
$$
u_{t}=L(u)
$$
is satisfied for $u(x, y, t)=e^{-t\left(k^{2}+j^{2}\right)} u_{k j} .$ This means
$$
S(t)(\phi)=\sum_{k j}\left(e_{k j}, \phi\right) e^{-t\left(k^{2}+j^{2}\right)} e_{k j}
$$
is the desired solution operator.

Problem 5.

Wants to solve
$$
u_{t t}-u_{x x}=-r u_{t}
$$
Assume that we have a periodic solution on $[-\pi, \pi]$ can you find an equation for the Fourier coefficients?

Proof .

Solution: Assume $u(x, t)=\sum_{k} \hat{u}(k, t) e_{k} .$ Then we need
$$
\frac{d^{2}}{d t^{2}} \hat{u}(k, t)+r \frac{d \hat{u}}{d t}=k^{2} \hat{u}(k, t)
$$
or
$$
f^{\prime \prime}(t)+r f^{\prime}(t)=k^{2} f(t)
$$
This ODE is solved trying $f(t)=e^{\lambda t}$ first. Then we get
$$
\lambda^{2}+r \lambda+\frac{r^{2}}{4}=k^{2}+\frac{r^{2}}{4}
$$
This means
$$
\lambda_{p m}=\frac{r}{2} \pm \sqrt{k^{2}+\frac{r^{2}}{4}}
$$
We have exactly the two solutions we are looking for and hence
$$
u(x, t)=\sum_{k \in \mathbb{Z}} a_{k} e^{\lambda_{+} t} e_{k}+\sum_{k \in \mathbb{Z}} b_{k} e^{\lambda_{-} t} e_{k}
$$
is the general form of the solution. With the help of $a_{k}$ and $b_{k}$ we can now adjust the initial conditions. Indeed,
$$
u(x, 0)=\sum_{k}\left(a_{k}+b_{k}\right) e_{k}
$$
and
$$
u_{t}(x, 0)=\sum_{k}\left(\lambda_{+}(k) a_{k}+\lambda_{-}(k) b_{k}\right) e_{k}
$$
uniquely determines $a_{k}$ and $b_{k}$. Meaning
$$
\hat{u}(k, 0)=a_{k}+b_{k}, \hat{u}{t}(k, 0)=\left(\lambda{+}(k) a_{k}+\lambda_{-}(k) b_{k}\right)
$$
has a unique solution.

推荐几本很好的PDE入门教材给大家

Partial Differential Equations, An Introduction, Walter Strauss

Partial Differential Equations, Lawrence C. Evans

Applied Partial Differential Equations: with Fourier Series and Boundary Value Problems

Functional Analysis, Sobolev Spaces and Partial Differential Equations

Sobolev Spaces,Robert Adams John Fournier

偏微分方程代考partial difference equations代考认准UpriviateTA

real analysis代写analysis 2, analysis 3请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

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