User:Egm6321.f11.team1.colsonfe/HW7

= R*7.4 Laplacian In Elliptic Coordinates =

Given
See wikipdia reference of ellipitic coordinates.

Laplacian operator:


 * $$\Delta u=\frac{1}{h_1h_2h_3}\sum^3_{i=3}\frac{\partial}{\partial\xi_i}\left[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}\right]$$

== Find ==

Verify the Laplacian in elliptic coordinates:


 * $$\nabla^2u=\frac{1}{a^2(\sinh^2\mu+\sin^2\nu)}\left(\frac{\partial^2u}{\partial\mu^2}+\frac{\partial^2u}{\partial\nu^2}\right)$$

Solution
Using the reference listed above, we find that

$$\displaystyle \xi_1=\mu$$

$$\displaystyle\xi_2=\nu$$

$$\displaystyle h_1=h_2=a\sqrt{\sinh^2\mu+\sin^2\nu}$$

Therefore, using the definition of Laplacian operator in the above section,

$$\nabla^2u=\Delta u=\frac{1}{h_1h_2}\left[\frac{\partial}{\partial\xi_1}\left(\frac{h_1h_2}{(h_1)^2}\frac{\partial u}{\partial\xi_1}\right)+\frac{\partial}{\partial\xi_2}\left(\frac{h_1h_2}{(h_2)^2}\frac{\partial u}{\partial\xi_2}\right)\right]$$

$$\nabla^2u=\frac{1}{a^2(\sinh^2\mu+\sin^2\nu)}\left[\frac{\partial}{\partial\mu}\left((1)\frac{\partial u}{\partial\mu}\right)+\frac{\partial}{\partial\nu}\left((1)\frac{\partial u}{\partial\nu}\right)\right]$$

Therefore we arrive at our solution.

= R7.5 Laplacian In Parabolic Coordinates =

Given
See wikipdia reference of parabolic coordinates.

Laplacian operator:


 * $$\Delta u=\frac{1}{h_1h_2h_3}\sum^3_{i=3}\frac{\partial}{\partial\xi_i}\left[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}\right]$$

== Find ==

Verify the Laplacian in parabolic coordinates:


 * $$\nabla^2u=\frac{1}{\mu^2+\nu^2}\left(\frac{\partial^2u}{\partial\mu^2}+\frac{\partial^2u}{\partial\nu^2}\right)$$

Solution
Using the reference listed above, we find that

$$\displaystyle \xi_1=\sigma$$

$$\displaystyle\xi_2=\tau$$

$$\displaystyle h_1=h_2=\sqrt{\sigma^2+\tau^2}$$

I will use the following definitions to match the syntax given in the problem.

$$\displaystyle \mu:=\sigma$$

$$\displaystyle \nu:=\tau$$

Therefore, using the definition of Laplacian operator in the above section,

$$\nabla^2u=\Delta u=\frac{1}{h_1h_2}\left[\frac{\partial}{\partial\xi_1}\left(\frac{h_1h_2}{(h_1)^2}\frac{\partial u}{\partial\xi_1}\right)+\frac{\partial}{\partial\xi_2}\left(\frac{h_1h_2}{(h_2)^2}\frac{\partial u}{\partial\xi_2}\right)\right]$$

$$\nabla^2u=\frac{1}{\mu^2+\nu^2}\left[\frac{\partial}{\partial\mu}\left((1)\frac{\partial u}{\partial\mu}\right)+\frac{\partial}{\partial\nu}\left((1)\frac{\partial u}{\partial\nu}\right)\right]$$

Therefore we arrive at our solution.

= R7.9 Express Vector As Linear Combination of Basis Vectors =

Given
Non-orthonormal basis:


 * $$\displaystyle{\mathbf b_1,\mathbf b_2,\mathbf b_3}$$

Orthonormal basis:


 * $$\displaystyle{\mathbf e_1,\mathbf e_2,\mathbf e_3}$$

Expressing the non-orthonormal in the orthonormal basis:


 * $$\displaystyle\mathbf b_i=A_{ij}\mathbf e_j$$


 * where


 * $$\mathbf A=[A_{ij}]=\left[\begin{matrix}5&2&3\\4&5&6\\7&8&5\end{matrix}\right]$$

Vector $$\displaystyle\mathbf v$$ in $$\displaystyle\mathbb{R}^3$$:


 * $$\displaystyle-2\mathbf e_1+4\mathbf e_2-5\mathbf e_3$$

Definition of Gram Matrix:


 * $$\boldsymbol\Gamma=[\Gamma_{ij}]=[(\mathbf b_i\cdot\mathbf b_j)]\in\mathbb R^{n\times n}$$

Finding components of $$\displaystyle\mathbf v$$


 * $$\displaystyle{v_j}=\mathbf\Gamma^{-1}\{\mathbf b_i\mathbf v\}$$

== Find ==


 * $$\displaystyle{\mathbf v_i}\in\mathbb{R}^{3\times 1}$$ such that $$\displaystyle\mathbf=v_i\mathbf b_i$$

Solution
To begin, find each entry of the Gram Matrix:

$$\Gamma_{11}=\mathbf b_1\cdot\mathbf b_1=25+4+9=38$$

$$\Gamma_{12}=\mathbf b_1\cdot\mathbf b_2=20+10+18=48$$

$$\Gamma_{13}=\mathbf b_1\cdot\mathbf b_3=35+16+25=76$$

$$\Gamma_{21}=\mathbf b_2\cdot\mathbf b_1=48$$

$$\Gamma_{22}=\mathbf b_2\cdot\mathbf b_2=16+25+36=77$$

$$\Gamma_{23}=\mathbf b_2\cdot\mathbf b_3=28+40+30=98$$

$$\Gamma_{31}=\mathbf b_3\cdot\mathbf b_1=76$$

$$\Gamma_{32}=\mathbf b_3\cdot\mathbf b_2=98$$

$$\Gamma_{33}=\mathbf b_2\cdot\mathbf b_3=49+64+25=128$$

Thus

$$\boldsymbol\Gamma=\left[\begin{matrix}38&48&76\\48&77&98\\76&98&128\end{matrix}\right]$$

And

$$\boldsymbol\Gamma^{-1}=\left[\begin{matrix}-0.017&-0.086&0.076\\-0.086&0.060&0.005\\0.076&0.005&0.041\end{matrix}\right]$$

Also we can see that

$$\displaystyle\mathbf b_1\cdot\mathbf v=-10+8-15=-17$$

$$\displaystyle\mathbf b_2\cdot\mathbf v=-8+20-30=-18$$

$$\displaystyle\mathbf b_3\cdot\mathbf v=-14+32-25=-7$$

So then

$$\displaystyle\mathbf b_i\cdot\mathbf v=\left[\begin{matrix}-17\\-18\\-7\end{matrix}\right]$$

Therefore by matrix multiplication

$$\displaystyle\Gamma^{-1}\{\mathbf b_i\cdot\mathbf v\}=\left[\begin{matrix}1.305\\0.347\\-1.095\end{matrix}\right]$$

This implies our solution is

= References =