## 数学代写|微积分代写Calculus代写|MATH0220

2023年1月2日

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## 数学代写|微积分代写Calculus代写|Dirichlet Boundary Condition

For the Dirichlet boundary condition, the potential $\phi$ is specified on surfaces $S$ that are bounding surfaces of the volume $V$ for the volume integral in $\mathrm{Fr}$. . (5.5). For this case, the Green’s function should be zero on the bounding surfaces, so that only the first surface integral in Eq. (5.5), where $\phi$ is known, enters. Then the Green’s function solution to Poisson’s equation with the potential specified on all bounding surfaces can be written as
$$\phi(\mathbf{r})=\int_V G\left(\mathbf{r}, \mathbf{r}^{\prime}\right) \rho\left(\mathbf{r}^{\prime}\right) d^3 r^{\prime}-\frac{1}{4 \pi} \oint_S \mathbf{d} \mathbf{A}^{\prime} \cdot\left[\phi\left(\mathbf{r}^{\prime}\right) \nabla^{\prime} G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\right] .$$
Using the Green’s function, the solution to Poisson’s equation can be written down in two steps. The first step is immediate and straightforward. The solution is simply written down as given in terms of the Green’s function by Eq. (5.6). The second, more difficult, step is to find the Green’s function by using its two defining properties:
I. $\nabla^{\prime 2} G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)=-4 \pi \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right)$,
II. $G\left(\mathbf{r}, \mathbf{r}{\mathbf{S}}^{\prime}\right)=0$, where the notation $\mathbf{r}{\mathbf{S}}^{\prime}$ means that $\mathbf{r}^{\prime}$ is on one of the bounding surfaces.
It is helpful to put the requirements for the Dirichlet Green’s function into words:
For the solution of Poisson’s equation for an arbitrary charge distribution with the potential specified on all boundaries, the Green’s function $G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)$ is the potential at $\mathbf{r}^{\prime}$ due to a unit point charge at $\mathbf{r}$ with all surfaces acting as grounded conductors.
Note that, even if the original surfaces are not conductors (they can’t be conductors if the specified potential on them is not constant), the Green’s function is the point charge solution found as if all the surfaces were grounded conductors.
We have discussed above the solution of Poisson’s equation for the electric field. The Green’s function method would apply in the same way for the gravitational field, or for a heat flow problem where the temperature acts as the potential and heat sources as the charge distribution. For the heat flow case, the temperature on the surface of the volume would be the Dirichlet boundary condition.

## 数学代写|微积分代写Calculus代写|Surface Green’s Function

For solving Laplace’s equation, only the surface integral in Eq. (5.6) enters, and the solution can be written as
$$\phi(\mathbf{r})=-\frac{1}{4 \pi} \oint_S \mathbf{d A}^{\prime} \cdot\left[\phi\left(\mathbf{r}^{\prime}\right) \nabla^{\prime} G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\right] .$$
We introduce a surface Green’s function, defined by
$$g\left(\mathbf{r}, \mathbf{r}{\mathbf{S}}^{\prime}\right)=-\left.\frac{1}{4 \pi} \hat{\mathbf{n}}^{\prime} \cdot \nabla^{\prime} G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\right|{\mathbf{r}^{\prime}=\mathbf{r}{\mathbf{s}}^{\prime}},$$ where the unit vector $\hat{\mathbf{n}}^{\prime}$ is directed out of the surface. Then, the Green’s function solution to Laplace’s equation can be written as $$\phi(\mathbf{r})=\oint_S d A^{\prime} g\left(\mathbf{r}, \mathbf{r}{\mathbf{S}}^{\prime}\right) \phi\left(\mathbf{r}^{\prime}\right) .$$
The surface Green’s function is given by Eq. (5.10), but it can also be found directly by considering what is required in Eq. (5.11). If the surface Green’s function satisfies the two properties
I. $\quad \nabla^2 g\left(\mathbf{r}, \mathbf{r}{\mathbf{S}}^{\prime}\right)=0$ II. $\quad g\left(\mathbf{r}{\mathbf{S}}, \mathbf{r}{\mathbf{S}}^{\prime}\right)=\delta\left(\mathbf{r}{\mathbf{S}}-\mathbf{r}_{\mathbf{S}}^{\prime}\right)$,
the potential given by Eq. (5.11) will automatically satisfy Laplace’s equation, and the Dirichlet boundary condition on the surface.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Dirichlet Boundary Condition

$$\phi(\mathbf{r})=\int_V G\left(\mathbf{r}, \mathbf{r}^{\prime}\right) \rho\left(\mathbf{r}^{\prime}\right) d^3 r^{\prime}-\frac{1}{4 \pi} \oint_S \mathbf{d} \mathbf{A}^{\prime} \cdot\left[\phi\left(\mathbf{r}^{\prime}\right)\right.$$

I. $\nabla^{\prime 2} G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)=-4 \pi \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right)$,

## 数学代写|微积分代写Calculus代写|Surface Green’s Function

$$\phi(\mathbf{r})=-\frac{1}{4 \pi} \oint_S \mathbf{d A}^{\prime} \cdot\left[\phi\left(\mathbf{r}^{\prime}\right) \nabla^{\prime} G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\right]$$

$$g\left(\mathbf{r}, \mathbf{r} \mathbf{S}^{\prime}\right)=-\frac{1}{4 \pi} \hat{\mathbf{n}}^{\prime} \cdot \nabla^{\prime} G\left(\mathbf{r}, \mathbf{r}^{\prime}\right) \mid \mathbf{r}^{\prime}=\mathbf{r} \mathbf{S}^{\prime},$$

$$\phi(\mathbf{r})=\oint_S d A^{\prime} g\left(\mathbf{r}, \mathbf{r} \mathbf{S}^{\prime}\right) \phi\left(\mathbf{r}^{\prime}\right) .$$

I。 $\quad \nabla^2 g\left(\mathbf{r}, \mathbf{r} \mathbf{S}^{\prime}\right)=0$ 二。
$$g\left(\mathbf{r S}, \mathbf{r S} \mathbf{S}^{\prime}\right)=\delta\left(\mathbf{r S}-\mathbf{r}_{\mathbf{S}}^{\prime}\right),$$

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## MATLAB代写

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