91影视

It is given that $\mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{1}{4\mathrm{蟺}}{鈭�}^2\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{蔚}}^2}}$and the equation $\mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}$ have to be verified. It has to be proved that as$\mathrm{蔚}鈫�0,\mathrm{D}(0,\mathrm{蔚})鈫�鈭�$as $\mathrm{蔚}鈫�0.\mathrm{D}(r,\mathrm{蔚})鈫�0$ .and the integral of the function $\mathrm{D}(\mathrm{r},\mathrm{蔚})$ over all spaces is equal to 1.

It is given that $\mathbf{D}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{蔚}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{蟺}}{\mathbf{鈭�}}^{\mathbf{2}}\frac{\mathbf{1}}{\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{蔚}}^{\mathbf{2}}}}$. The Laplacian operator with respect to r is simplified as

$$\begin{gathered} {{\mathbf{鈭�}}_{\mathbf{2}}}^{\mathbf{r}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}}\left(r\frac{鈭�}{鈭�r}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{鈭�}}^{\mathbf{2}}}{\mathbf{鈭�}{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{r}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}} \end{gathered}$$

Apply the expression role="math" localid="1654687221342" $\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{蔚}}^2}}$to the above simplified Laplacian operator as,

role="math" localid="1654689997951" $$\begin{gathered} {鈭�}^2\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{蔚}}^2}}\right)=\left[\frac{{鈭�}^2}{鈭�{\mathrm{r}}^2}+\frac{2}{\mathrm{r}}\frac{鈭�}{鈭�\mathrm{r}}\right]\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{蔚}}^2}}\right) \\ =\frac{鈭�}{鈭�{\mathrm{r}}^2}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{蔚}}^2}}\right)+\frac{2}{\mathrm{r}}\frac{鈭�}{鈭�\mathrm{r}}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{蔚}}^2}}\right) \\ =\left\{-{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-3/2}+\frac{3\mathrm{r}}{2}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}.\left(2\mathrm{r}\right)+\frac{2}{\mathrm{r}}\left(\begin{array}{c}\left(-\frac{1}{2}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-3/2}.\left(2\mathrm{r}\right)\right)\\ -\mathrm{r}{({\mathrm{r}}^2+{\mathrm{蔚}}^2)}^{-3/2}\end{array}\right)\right\} \\ =-{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}+\frac{2}{\mathrm{r}}\left(-\mathrm{r}{({\mathrm{r}}^2+{\mathrm{蔚}}^2)}^{-3/2}-\mathrm{r}{({\mathrm{r}}^2+{\mathrm{蔚}}^2)}^{-3/2}\right) \end{gathered}$$

Simplify further as

$$\begin{gathered} {鈭�}^2=\left(\frac{1}{\sqrt{r^2+{蔚}^2}}\right)=-{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}+\frac{2}{\mathrm{r}}\left(-2\mathrm{r}{({\mathrm{r}}^2+{\mathrm{蔚}}^2)}^{-3/2}\right) \\ =-{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}-4\mathrm{r}{({\mathrm{r}}^2+{\mathrm{蔚}}^2)}^{-3/2} \\ =-5{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2} \end{gathered}$$

Substitute $-5{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}$ for ${鈭�}^2=\left(\frac{1}{\sqrt{r^2+{蔚}^2}}\right)$into.

role="math" localid="1654690616425" $$\begin{gathered} \mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{1}{4\mathrm{蟺}}{鈭�}^2\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{蔚}}^2}} \\ \mathrm{D}(\mathrm{r},\mathrm{蔚})=-\frac{1}{4蟺}\left(-5{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}\right) \\ =\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2} \end{gathered}$$

Thus, the equation role="math" localid="1654690728467" $\mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}$, in part (a) is proved.

From the result of part (a)$\mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}$, .

Substitute 0 for into equation$\mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}$.

$$\begin{gathered} \mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\left(0\right)}^2+{\mathrm{蔚}}^2\right)}^{-5/2} \\ =\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\mathrm{蔚}}^2\right)}^{-5/2} \\ =\frac{3{\mathrm{蔚}}^2{\mathrm{蔚}}^{-5}}{4\mathrm{蟺}} \\ =\frac{3{\mathrm{蔚}}^{-3}}{4\mathrm{蟺}} \end{gathered}$$

Simply further as,

localid="1654691126290" $D(0,蔚)=\frac{3}{4蟺{蔚}^3}$

Substitute 0 for $蔚$, into equation $D(0,蔚)=\frac{3}{4蟺{蔚}^3}$

localid="1654691262144" $$\begin{gathered} D(0,蔚鈫�0)=\frac{3}{4蟺{\left(0\right)}^3} \\ =鈭� \end{gathered}$$

Thus, it is proved that as localid="1654692695242" $蔚鈫�0.D(0,蔚)鈫�鈭�$.

From the result of part (a),$\mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}$

Substitute 0 for $蔚$, into equation$\mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{3{\mathrm{蔚}}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2+{\mathrm{蔚}}^2\right)}^{-5/2}$

$$\begin{gathered} \mathrm{D}(\mathrm{r},\mathrm{蔚})=\frac{3{\left(0\right)}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2+{\left(0\right)}^2\right)}^{-5/2} \\ =\frac{3{\left(0\right)}^2}{4\mathrm{蟺}}{\left({\mathrm{r}}^2\right)}^{-5/2} \\ =0 \end{gathered}$$

Thus, it is proved that as role="math" localid="1654691695727" $蔚鈫�0.D(r,蔚)鈫�0$.

It is given that $\mathrm{D}(\mathrm{r},\mathrm{蔚})鈮�{\mathrm{味}}^3\left(\mathrm{r}\right)$as$蔚鈫�0$ . For all spaces the value of r ranges from $鈭�$to$-鈭�$ . Integrate the function$\mathrm{D}(\mathrm{r},\mathrm{蔚})鈮�{\mathrm{味}}^3\left(\mathrm{r}\right)$ over all values of r as,

$$\begin{gathered} 鈭�\mathrm{D}(\mathrm{r},\mathrm{蔚})\mathrm{dr}=\underset{-鈭�}{\overset{鈭�}{鈭�}}{味}^3\left(r\right)dr \\ =\underset{-鈭�}{\overset{鈭�}{鈭�}}\mathrm{味}(\mathrm{r}).\mathrm{味}\left(\mathrm{r}\right).\mathrm{味}\left(\mathrm{r}\right).\mathrm{dr} \end{gathered}$$

The multiplication of $\mathrm{味}(\mathrm{r})$with itself, any number of times, gives the delta function,$\mathrm{味}(\mathrm{r})$ . Thus above integral becomes,

role="math" localid="1654692146367" $$\begin{gathered} 鈭�\mathrm{D}(\mathrm{r},\mathrm{蔚})\mathrm{dr}=\underset{-鈭�}{\overset{鈭�}{鈭�}}\mathrm{味}\left(\mathrm{r}\right).\mathrm{味}\left(\mathrm{r}\right).\mathrm{味}\left(\mathrm{r}\right).\mathrm{dr} \\ =\underset{-鈭�}{\overset{鈭�}{鈭�}}\mathrm{味}(\mathrm{r})\mathrm{dr} \\ =1 \end{gathered}$$

Thus, it is proved that the integral of the function $D(r,蔚)$over all spaces is equal to 1.