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Home » Gate Study Material » Mathematics » Partial Differential Equations » The Overdetermined System

The Overdetermined System

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The Overdetermined System

The Overdetermined System $ A\vec u = \vec b$

The linear algebra aspects of Maxwell's wave operator $ \mathcal A$ are illustrated by the following problem from linear algebra:

 
Solve $ A\vec u = \vec b$ for $ \vec u$ , under the stipulation that

$\displaystyle A:$ $\displaystyle ~~R^4 \longrightarrow R^4$    
$\displaystyle \vec u_r:$ $\displaystyle ~~A\vec u_r=\vec 0\quad \textrm{so that }{\mathcal N}(A)=span \{\vec u_r \}$    
$\displaystyle \vec u_\ell^T:$ $\displaystyle ~~\vec u_\ell^T A=\vec 0\quad \textrm{so that } {\mathcal N}(A^T)=span \{\vec u_\ell \}$    
$\displaystyle \vec b:$ $\displaystyle ~~\vec b \in {\mathcal R}(A)\quad \textrm{so that }\vec u_\ell^T \vec b=0$ (653)

 

The fact that $ A$ is singular and $ \vec b$ belongs to the range of $ A$ makes the system over-determined but consistent. This means that there are more equations than there are unknowns.
One solves the problem in two steps.

 
Step I:
Let $ \{ \vec v_1,\vec v_2,\vec
v_3 \}$ be the set of eigenvectors having non-zero eigenvalues. Whatever $ A$ is, the task of finding three vectors that satisfy

  $\displaystyle \left. \begin{array}{c} A\vec v_1 c_1=~\vec v_1 \lambda_1c_1\\ A\...
..._3 \lambda_3c_3 \end{array} \right\} \lambda_i\not =0,~c_i~\textrm{are scalars}$ (654)

and


 

  $\displaystyle ~A\vec u_r c_4=\vec 0~.$ (655)

 

Being spanned by the three eigenvectors with non-zero eigenvalues, the range space of $ A$ ,

$\displaystyle {\mathcal R}(A)=span \{\vec v_1,\vec v_2,\vec v_3\}~,
$

is well-determined. However, the scalars $ c_i$ are at this stage as yet undetermined.

Step II:
Continuing the development, recall that quite generally

$\displaystyle A\vec u$
$\displaystyle \Leftrightarrow \vec b\in{\mathcal R}(A)$    
  $\displaystyle ~$  
$\displaystyle \Leftrightarrow \vec b =\vec v_1 b_1+\vec v_2 b_2+\vec v_3 b_3~,$ (656)

 

and that if

$\displaystyle \vec u$ $\displaystyle =\vec v_1 c_1+\vec v_2 c_2+\vec v_3 c_3+\vec u_4 c_4~,$    

then


 

$\displaystyle A\vec u$ $\displaystyle =~\vec v_1 \lambda_1c_1+\vec v_2 \lambda_2c_2+\vec v_3 \lambda_3c_3~.$ (657)

 

 

It is appropriate to alert the reader that in the ensuing section the vectors $ \vec v_i$ and the eigenvalues $ \lambda_i$ become differential operators which act on scalar fields $ c_i$ and that the three subscript labels will refer to the TE, TM, and TEM eletromagnetic66 vector potentials respectively.
Equating (6.56) and (6.57), one finds that the linear independence of $ \{ \vec v_1,\vec v_2,\vec
v_3 \}$ implies the following equations for $ c_1$ , $ c_2$ , and $ c_3$ :


 

Consequently, the solution is

$\displaystyle \vec u=\vec v_1 \frac{1}{\lambda_1}\,b_1+\vec v_2 \frac{1}{\lambda_2}\,b_2+\vec v_3 \frac{1}{\lambda_3}\,b_3+\vec u_4 c_4
$

where $ \vec u_4 c_4$ is an indeterminate multiple of the null space vector $ \vec u_4$ .

If one represents the stated problem $ A\vec u = \vec b$ ($ \vec u$ determines $ \vec b$ ) as an input-output process, as in Figure 6.3,

 
Figure 6.3: The matrix $ A$ defines an input-output process.
\begin{figure}\centering\epsfig{file=input-output_system.eps,scale=.5}\end{figure}
 
then its solution is represented by the inverse input-output process as in Figure 6.4.
 
Figure: The solution to $ A\vec u = \vec b$ defines an inverse input-output process.
\begin{figure}\centering\epsfig{file=inverse_input-output_system.eps,scale=.6}\end{figure}
 

In general, the task of finding the eigenvectors of a 4$ \times$ 4 matrix can be a nontrivial task. However, given the fact that the solution to

$\displaystyle \vec u_\ell^T A=\vec0
$

is already known, one finds that the associated constraints on the eigenvectors,

$\displaystyle \vec u_\ell^T \vec v_i=0
$

make the task quite easy, if not trivial.



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