Solve vectorial equations

Question

I'm trying to figure out a passage for reducing a vectorial equations ! for doing this somebody told me to use a program of symbolic calculation (matlab,maple, mathematical .. or python as well .. ) I never using symbolic calculation .. but i'm confiding in your knowledge for figure out how to solve my problem ! here the equation :

$$ \mathbf{v}_l + \mathbf{A} \times \mathbf{v}_l = \mathbf{B} $$

where

$$ \mathbf{A} = \frac{D}{D_0}\mathbf{s'}, $$ $$ \mathbf{B} = \frac{\rho_s k}{D_0}(\mathbf{V_s+v_s}) -\frac{D_t}{D_0}\mathbf{v_n}+\frac{D}{D_0}\mathbf{s'}\times \mathbf{v_n}+\frac{\chi}{D_0}\mathbf{s'} $$ where $\chi $ is a generic constant. The solution of the first equations is: $$\mathbf{v}_l = \frac{1}{(1+A^2)}[\mathbf{B}-\mathbf{A}\times\mathbf{B} + (\mathbf{A} \cdot \mathbf{B} ) \mathbf{A} ] $$

how this solution comes from ??? I found it in the paper paper (eq 14) I'm wait for your precious Help !! thanks in advance

I add the pdf on the web

Answer 1

Hint:

I suppose that your vectors are in $3$D space. In tis case , if $\mathbf A=(A_x,A_y,A_z)^T$, we can represent the cross product as a matrix operation as: $$ \mathbf A \times \mathbf v_l=\mathbf A_{[\times]}\mathbf v_l=\begin{pmatrix} 0&-A_z&A_y\\ A_z&0&-A_x\\ -A_y&A_x&0 \end{pmatrix} \mathbf v_l $$ So your equation becomes: $$ \left(\mathbf I+ \mathbf A_{[\times]}\right)\mathbf v_l=\mathbf B $$

and the solution is: $$ \mathbf v_l=\left(\mathbf I+ \mathbf A_{[\times]}\right)^{-1} \mathbf B $$ If you calculate the inverse matrix $$ \left(\mathbf I+ \mathbf A_{[\times]}\right)^{-1}= \begin{pmatrix} 1&-A_z&A_y\\ A_z&1&-A_x\\ -A_y&A_x&1 \end{pmatrix}^{-1} $$ I think that you can find the result.

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Using WolframAlpha I found: $$ \begin{pmatrix} 1&-A_z&A_y\\ A_z&1&-A_x\\ -A_y&A_x&1 \end{pmatrix}^{-1}= \frac{1}{A_x^2+A_y^2+A_z^2+1} \begin{pmatrix} 1+A_x^2&A_xA_y+A_z&A_xA_z-A_y\\ A_xA_y-A_z&1+A_y^2&A_x+A_yA_z\\ A_y+A_xA_z&A_yA_z-A_x&1+A_z^2\\ \end{pmatrix}= $$

$$=\frac{1}{1+A^2}\left[ \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}+ \begin{pmatrix} 0&A_z&-A_y\\ -A_z&0&A_x\\ A_y&-A_x&0\\ \end{pmatrix}+ \begin{pmatrix} A_x^2&A_xA_y&A_xA_z\\ A_xA_y&A_y^2&A_yA_z\\ A_xA_z&A_yA_z&A_z^2\\ \end{pmatrix} \right] $$

And multiplying by $\mathbf B=(B_x,B_y,B_z)^T$ you can find the result in a form that is the same as in OP. The only difference is the factor $\frac{1}{1+A^2}$. Maybe a sign typo in the question or in the solution?

Answer 2

I'll try this two different ways.

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First, let's continue the solution by Emilio Novati. There is some matrix multiplication involved here, so let's do it step by step.

First we multiply by $\mathbf B$ on the right in order to get the desired vector $\mathbf v_l$:

$$\mathbf v_l=\frac 1{1+A^2}\left[ \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}+ \begin{pmatrix}0&A_z&-A_y\\-A_z&0&A_x\\A_y&-A_x&0\end{pmatrix}+ \begin{pmatrix}A_x^2&A_xA_y&A_xA_z\\A_xA_y&A_y^2&A_yA_z\\A_xA_z&A_yA_z&A_z^2\end{pmatrix}\right]\mathbf B.$$

Matrix multiplication distributes over matrix addition, so we have $$\mathbf v_l = \frac 1{1+A^2}\left[ \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\mathbf B + \begin{pmatrix}0&A_z&-A_y\\-A_z&0&A_x\\A_y&-A_x&0\end{pmatrix} \mathbf B + \begin{pmatrix}A_x^2&A_xA_y&A_xA_z\\A_xA_y&A_y^2&A_yA_z\\A_xA_z&A_yA_z&A_z^2\end{pmatrix}\mathbf B \right].$$

Now let's take the three terms inside the brackets one at a time, denoting the three components of $\mathbf B$ like this: $$ \mathbf B = \begin{pmatrix}B_x\\B_y\\B_z\end{pmatrix}.$$

So now we have some matrix multiplications. Matrix multiplication is defined on Wolfram MathWorld among many other places; Purplemath has a nice animation of the process. In the particular formula above, every matrix multiplication is a $3\times3$ matrix by a $3\times1$ matrix, and the rule is $$ \begin{pmatrix}m_{11}&m_{12}&m_{13}\\m_{21}&m_{22}&m_{23}\\m_{31}&m_{32}&m_{33}\end{pmatrix} \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} = \begin{pmatrix}m_{11}v_1+m_{12}v_2+m_{13}v_3\\ m_{21}v_1+m_{22}v_2+m_{23}v_3\\ m_{31}v_1+m_{32}v_2+m_{33}v_3\end{pmatrix} $$

First we have multiplication by an identity matrix: $$\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} \begin{pmatrix}B_x\\B_y\\B_z\end{pmatrix} = \begin{pmatrix}B_x\\B_y\\B_z\end{pmatrix} = \mathbf B.$$

For the next term, let's write $$\begin{pmatrix}0&A_z&-A_y\\-A_z&0&A_x\\A_y&-A_x&0\end{pmatrix} = -\begin{pmatrix}0&-A_z&A_y\\A_z&0&-A_x\\-A_y&A_x&0\end{pmatrix}.$$

So now we just have to confirm that $$\begin{pmatrix}0&-A_z&A_y\\A_z&0&-A_x\\-A_y&A_x&0\end{pmatrix} \begin{pmatrix}B_x\\B_y\\B_z\end{pmatrix} = \begin{pmatrix}A_yB_z-A_zB_y\\A_zB_x-A_xB_z\\A_xB_y-A_yB_x\end{pmatrix} = \mathbf A \times \mathbf B,$$ confirming that multiplication by the matrix $\mathbf A_{[\times]}$ does indeed perform a cross product as stated by Emilio Novarti.

Finally we look at the third term: \begin{align} \begin{pmatrix}A_x^2&A_xA_y&A_xA_z\\A_xA_y&A_y^2&A_yA_z\\A_xA_z&A_yA_z&A_z^2\end{pmatrix} \begin{pmatrix}B_x\\B_y\\B_z\end{pmatrix} &= \begin{pmatrix}A_x^2B_x+A_xA_yB_y+A_xA_zB_z\\A_xA_yB_x+A_y^2B_y+A_yA_zB_z\\A_xA_zB_x+A_yA_zB_y+A_z^2B_z\end{pmatrix} \\ &= (A_xB_x+A_yB_y+A_zB_z) \begin{pmatrix}A_x\\A_y\\A_z\end{pmatrix} \\ &= (\mathbf A \cdot \mathbf B) \mathbf A. \end{align}

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For the second way, we take the formula $$\frac1{1+A^2}[\mathbf B - \mathbf A\times\mathbf B + (\mathbf A \cdot \mathbf B) \mathbf A],$$ which is proposed as the value of $\mathbf{v}_l,$ and try plugging it into the formula $$ \mathbf{v}_l + \mathbf{A} \times \mathbf{v}_l$$ to see if we actually come up with the result $\mathbf{B}$ as desired. First let's work on the term $\mathbf A\times\mathbf v_l.$ Using various rules of vector arithmetic, we have \begin{align} \mathbf A\times\mathbf v_l &= \mathbf A\times\left[\frac1{1+A^2}[\mathbf B - \mathbf A\times\mathbf B + (\mathbf A \cdot \mathbf B) \mathbf A]\right] \\ &= \frac1{1+A^2}[\mathbf A\times\mathbf B - \mathbf A\times(\mathbf A\times\mathbf B) + \mathbf A\times(\mathbf A \cdot \mathbf B) \mathbf A] \\ &= \frac1{1+A^2}[\mathbf A\times\mathbf B - \mathbf A\times(\mathbf A\times\mathbf B)], \end{align} because $$ \mathbf A\times(\mathbf A \cdot \mathbf B) \times\mathbf A = (\mathbf A \cdot \mathbf B) [\mathbf A\times\mathbf A] = \mathbf 0. $$ Therefore the proposed substitution for $\mathbf v_l$ says that \begin{align} \mathbf v_l + \mathbf A \times \mathbf v_l &= \frac1{1+A^2}[\mathbf B - \mathbf A\times\mathbf B + (\mathbf A \cdot \mathbf B) \mathbf A] + \frac1{1+A^2}[\mathbf A\times\mathbf B - \mathbf A\times(\mathbf A\times\mathbf B)] \\ &= \frac1{1+A^2}[\mathbf B + (\mathbf A \cdot \mathbf B) \mathbf A - \mathbf A\times(\mathbf A\times\mathbf B)]. \end{align}

Next, let's see how the vectors $(\mathbf A \cdot \mathbf B) \mathbf A$ and $\mathbf A\times(\mathbf A\times\mathbf B)$ relate to the plane containing $\mathbf A$ and $\mathbf B.$ Defining $A = \|\mathbf A\|$ and $B = \|\mathbf B\|,$ if $\theta$ is the angle between $\mathbf A$ and $\mathbf B$ then $\mathbf A\times\mathbf B$ is a vector of length $AB\sin\theta$ perpendicular to the plane containing $\mathbf A$ and $\mathbf B.$ But $\mathbf A\times(\mathbf A\times\mathbf B)$ is perpendicular to $\mathbf A\times\mathbf B,$ so it's in the plane containing $\mathbf A$ and $\mathbf B,$ and it's also perpendicular to $\mathbf A.$ By a couple of applications of the right-hand rule for the cross product, we find that $\mathbf A\times(\mathbf A\times\mathbf B)$ is on the side of $\mathbf A$ opposite from $\mathbf B,$ so $-\mathbf A\times(\mathbf A\times\mathbf B)$ is on the same side as $\mathbf B$ and makes an angle $\frac\pi2 - \theta$ with $\mathbf B.$ Moreover, its length is $$A(AB\sin\theta)\sin\frac\pi2 = A^2B\sin\theta$$

Meanwhile, $\mathbf A \cdot \mathbf B = AB\cos\theta,$ so $(\mathbf A \cdot \mathbf B) \mathbf A$ is a vector of length $A^2B\cos\theta$ parallel to $\mathbf A.$ So now we have two perpendicular vectors, one of length $A^2B\cos\theta$ parallel to $\mathbf A$ and the other of length $A^2B\sin\theta$ perpendicular to $\mathbf A.$ The vector sum of these vectors is a vector of length $A^2B$ at an angle $\theta$ from $\mathbf A$; in fact it is exactly in the same direction as $\mathbf B,$ and it is the vector $A^2 \mathbf B.$ That is, $$ (\mathbf A \cdot \mathbf B) \mathbf A - \mathbf A\times(\mathbf A\times\mathbf B) = A^2 \mathbf B.$$

So now we have $$ \mathbf{v}_l + \mathbf{A} \times \mathbf{v}_l = \frac1{1+A^2}[\mathbf B + A^2 \mathbf B] = \mathbf B, $$ showing that $$\mathbf{v}_l = \frac1{1+A^2}[\mathbf B - \mathbf A\times\mathbf B + (\mathbf A \cdot \mathbf B) \mathbf A]$$ is indeed a solution of that equation.