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数学代写|偏微分方程代考Partial Differential Equations代写|Elementary Solutions of Laplace’s Equation

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数学代写|偏微分方程代考Partial Differential Equations代写|Elementary Solutions of Laplace’s Equation

If we take the function $\psi$ to be given by the equation
$$\psi=\frac{q}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=\frac{q}{\sqrt{\left(x-x^{\prime}\right)^2+\left(y-y^{\prime}\right)^2+\left(z-z^{\prime}\right)^2}}$$
where $q$ is a constant and $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ are the coordinates of a fixed point, then since
$$\begin{gathered} \frac{\partial \psi}{\partial x}=-\frac{q\left(x-x^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3}, \text { etc. } \ \frac{\partial^2 \psi}{\partial x^2}=-\frac{q}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3}+\frac{3 q\left(x \cdots x^{\prime}\right)^2}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^5}, \text { etc. } \end{gathered}$$
it follows that
$$\nabla^2 \psi=0$$
showing that the function (1) is a solution of Laplace’s equation except possibly at the point $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$, where it is not defined.

From what we have said in $(c)$ of Sec. 1 it follows that the function $\psi$ given by equation (1) is a possible form for the electrostatic potential corresponding to a space which, apart from the isolated point $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$, is empty of electric charge. To find the charge at this singular point we make use of Gauss’ theorem (Problem 1 of Sec. 1). If $S$ is any sphere with center $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$, then it is easily shown that
$$\int_S \frac{\partial \psi}{\partial n} d S=-4 \pi q$$
from which it follows, by Gauss’ theorem, that equation (1) gives the solution of Laplace’s equation corresponding to an electric charge $+q$.

数学代写|偏微分方程代考Partial Differential Equations代写|Separation of Variables

We shall now apply to Laplace’s equation the method of separation of variables outlined in Sec. 9 of Chap. 3 .

In spherical polar coordinates $r, \theta, \phi$ Laplace’s equation takes the form
$$\frac{\partial^2 \psi}{\partial r^2}+\frac{2}{r} \frac{\partial \psi}{\partial r}+\frac{1}{r^2} \frac{\partial^2 \psi}{\partial \theta^2}+\frac{\cot \theta}{r^2} \frac{\partial \psi}{\partial \theta}+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2 \psi}{\partial \phi^2}=0$$
and it was shown in Example 5 of Sec. 9, Chap. 3 that this equation is separable with solutions of the form
$$\left{A_n r^n+\frac{B_n}{r^{n+1}}\right} \Theta(\cos \theta) e^{ \pm i m \phi}$$
where $A_n, B_n, m$ are constants and $\Theta(\mu)$ satisfies Legendre’s associated equation
$$\left(1-\mu^2\right) \frac{d^2 \Theta}{d \mu^2}-2 \mu \frac{d \Theta}{d \mu}+\left{n(n+1)-\frac{m^2}{1-\mu^2}\right} \Theta=0$$
If we take $m=0$, we see that equation (3) reduces to Legendre’s equation
$$\left(1-\mu^2\right) \frac{d^2 \Theta}{d \mu^2}-2 \mu \frac{d \Theta}{d \mu}+n(n+1) \Theta=0$$
In the applications we wish to consider we assume that $n$ is a positive integer. In that case it is readily shown ${ }^1$ that this equation has two independent solutions given by the formulas
$$\begin{gathered} P_n(\mu)=\frac{1}{2^n n !} \frac{d^n}{d \mu^n}\left(\mu^2-1\right)^n \ Q_n(\mu)=\frac{1}{2} P_n(\mu) \log \frac{\mu+1}{\mu-1}-\sum_{s=0}^p \frac{2 n-4 s-1}{(2 s+1)(n-s)} P_{n-2 s-1}(\mu) \end{gathered}$$
where $p=\frac{1}{2}(n-1)$ or $\frac{1}{2} n-1$ according as $n$ is odd or even. The general solution of equation (4) is thus
$$\Theta=C_n P_n(\mu)+D_n Q_n(\mu)$$

偏微分方程代写

数学代写|偏微分方程代考Partial Differential Equations代写|Elementary Solutions of Laplace’s Equation

$$\psi=\frac{q}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=\frac{q}{\sqrt{\left(x-x^{\prime}\right)^2+\left(y-y^{\prime}\right)^2+\left(z-z^{\prime}\right)^2}}$$

$$\begin{gathered} \frac{\partial \psi}{\partial x}=-\frac{q\left(x-x^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3}, \text { etc. } \ \frac{\partial^2 \psi}{\partial x^2}=-\frac{q}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3}+\frac{3 q\left(x \cdots x^{\prime}\right)^2}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^5}, \text { etc. } \end{gathered}$$

$$\nabla^2 \psi=0$$

$$\int_S \frac{\partial \psi}{\partial n} d S=-4 \pi q$$

数学代写|偏微分方程代考Partial Differential Equations代写|Separation of Variables

$$\frac{\partial^2 \psi}{\partial r^2}+\frac{2}{r} \frac{\partial \psi}{\partial r}+\frac{1}{r^2} \frac{\partial^2 \psi}{\partial \theta^2}+\frac{\cot \theta}{r^2} \frac{\partial \psi}{\partial \theta}+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2 \psi}{\partial \phi^2}=0$$

$$\left{A_n r^n+\frac{B_n}{r^{n+1}}\right} \Theta(\cos \theta) e^{ \pm i m \phi}$$

$$\left(1-\mu^2\right) \frac{d^2 \Theta}{d \mu^2}-2 \mu \frac{d \Theta}{d \mu}+\left{n(n+1)-\frac{m^2}{1-\mu^2}\right} \Theta=0$$

$$\left(1-\mu^2\right) \frac{d^2 \Theta}{d \mu^2}-2 \mu \frac{d \Theta}{d \mu}+n(n+1) \Theta=0$$

$$\begin{gathered} P_n(\mu)=\frac{1}{2^n n !} \frac{d^n}{d \mu^n}\left(\mu^2-1\right)^n \ Q_n(\mu)=\frac{1}{2} P_n(\mu) \log \frac{\mu+1}{\mu-1}-\sum_{s=0}^p \frac{2 n-4 s-1}{(2 s+1)(n-s)} P_{n-2 s-1}(\mu) \end{gathered}$$

$$\Theta=C_n P_n(\mu)+D_n Q_n(\mu)$$

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