# 数学方法|MA30059 Mathematical methods 2代写

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ToTopics will be chosen from the following: Elliptic equations in two independent variables: Harmonic functions. Mean value property. Maximum principle (several proofs). Dirichlet and Neumann problems. Representation of solutions in terms of Green’s functions.

$$|\Gamma(z)| \leq \Gamma(x), \quad \operatorname{Re}(z)=x>0$$
If $f(z)$ is a complex function of $z$, then writing $f(z)$ as
$$f(z)=u(x, y)+i v(x, y),$$
it follows that the Cauchy-Riemann equations [7] must be satisfied, i.e.,
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$
For $f(z)=\Gamma(z)$, we have, using the relation,
\begin{aligned} t^{z} &=e^{z \ln t}=e^{x \ln t+i y \ln t} \ &=e^{x \ln t}[\cos (y \ln t)+i \sin (y \ln t)] \end{aligned}

## MA30059 COURSE NOTES ：

Therefore, we conclude that
$$R(m, n) \equiv \int_{0}^{1} x^{m}(\ln x)^{n} d x=\frac{(-1)^{n} n !}{(m+1)^{n+1}}$$
Note that for $n=0$, we obtain
$$R(m, n)=\int_{0}^{1} x^{m} d x=\frac{1}{m+1}$$
and for $m=0$,
$$R(0, n)=\int_{0}^{1}(\ln x)^{n} d x=(-1)^{n} n !$$

# 数学方法2|MA30059/MA40059/MA50059 Mathematical Methods 2代写

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In some sense, the unifying theme of the unit is partial differential equations (we shall see that
although variational principles such as those mentioned in Section 1.1.3 are not differential equations, they are intimately linked to them). You have met several PDEs over the last few years, but
we shall (hopefully) study PDEs in a different manner to how you may have done so up to this
point.

Fix $\mathbf{x} \in \mathbb{R}^{m}$.
(i) A distribution $T: C_{c}^{\infty}\left(\mathbb{R}^{m}\right) \rightarrow \mathbb{R}$ is called a fundamental solution of Laplace’s equation with respect to the point $\mathbf{x} \in \mathbb{R}^{m}$ if
$$\Delta T=\delta_{\mathrm{x}},$$
that is, equality as distributions. Here $\delta_{\mathbf{x}}$ is the shifted Dirac Delta distribution defined by $\delta_{\mathbf{x}}(\phi)=\phi(\mathbf{x})$ for all $\phi \in C_{c}^{\infty}\left(\mathbb{R}^{m}\right)$.
(ii) Let $\phi \in C_{c}^{\infty}\left(\mathbb{R}^{m}\right)$. As $\phi$ is compactly supported, there exists $\rho>0$ such that $|\mathbf{x}|<\rho$ and $\phi(\mathbf{y})=0$ for all $\mathbf{y} \in \mathbb{R}^{m}$ with $|\mathbf{y}| \geq \rho$. Choose $\Omega:=B(\mathbf{0}, \rho+1)$, so that
$\phi, \frac{\partial \phi}{\partial n}=0 \quad$ on $\partial \Omega$,
and $\mathbf{x} \in \Omega$. Since evidently $\phi \in C^{2}(\bar{\Omega})$, Green’s Integral Representation reduces to
$$\int_{\mathbb{R}^{m}} N_{\mathbf{x}}(\mathbf{y}) \Delta \phi(\mathbf{y}) \mathrm{d} \mathbf{y}=\phi(\mathbf{x}) .$$
As $\phi$ was arbitrary, in terms of distributions, this reads
$$\Delta T_{N_{\mathrm{x}}}=\delta_{\mathrm{x}}$$
where $T_{N_{\mathbf{x}}}$ is the distribution corresponding to $N_{\mathbf{x}}$, as required.

## MA30059/MA40059/MA50059 COURSE NOTES ：

We need to find a $v$ satisfying $(1)-(3)$ above. Then we set $G(\mathbf{x}, \mathbf{y}):=N_{\mathbf{x}}(\mathbf{y})+v(\mathbf{x}, \mathbf{y})$ If $\mathbf{x}=0$, then (3) becomes
$$v(\mathbf{0}, \mathbf{y})=\frac{1}{4 \pi} \quad \forall y \in \partial \Omega_{0},$$
so we can just choose
$$v(\mathbf{0}, \mathbf{y})=\frac{1}{4 \pi} \quad \forall y \in \bar{\Omega}{0},$$ which satisfies Laplace’s equation in $\Omega$ (and is in $C^{2}(\bar{\Omega})$ ) since it is constant. We now consider the case $\mathbf{x} \neq 0$. In light of the key property of $\mathbf{r}(\mathbf{x})$, namely, $$|\mathbf{x}| \cdot|\mathbf{y}-\mathbf{r}(\mathbf{x})|=|\mathbf{y}-\mathbf{x}| \quad \forall \mathbf{x} \in \Omega{0} \backslash{\mathbf{0}}, \forall \mathbf{y} \in \partial \Omega_{0},$$
the requirement (3) becomes
$$v(\mathbf{x}, \mathbf{y})=\frac{1}{4 \pi} \frac{1}{|\mathbf{x}-\mathbf{y}|}=\frac{1}{4 \pi} \frac{1}{|\mathbf{x}| \cdot|\mathbf{y}-\mathbf{r}(\mathbf{x})|} \quad \forall \mathbf{y} \in \partial \Omega_{0}$$
Thus, we let $v$ equal the function on the right hand side of the above for all $\mathbf{y} \in \Omega_{0}$, which is well defined since $\mathbf{r}(\mathbf{x})-\mathbf{y} \neq \mathbf{0}$ for all $\mathbf{y} \in \bar{\Omega}{0}$ and $\mathbf{x} \in \Omega{0} \backslash{\mathbf{0}}$. Furthermore, since
$$v(\mathbf{x}, \mathbf{y})=-\frac{1}{|\mathbf{x}|} N_{\mathbf{r}(\mathbf{x})}(\mathbf{y}) \quad \forall \mathbf{y} \in \bar{\Omega}_{0},$$