Modifying values to lie within existence domain

  • Context: MHB 
  • Thread starter Thread starter Siron
  • Start date Start date
  • Tags Tags
    Domain Existence
Click For Summary
SUMMARY

This discussion focuses on modifying parameters \(s\) and \(w\) to ensure that the derived values \(a\), \(b\), and \(c\) meet specific constraints in a mathematical model. The parameters are defined as \(a= \frac{(k-3)^2 \sqrt{v}}{s}\), \(b = \frac{v}{s}(w-10s)\), and \(c = s \sqrt{v}\). The conditions \(a<0\), \(0 PREREQUISITES

  • Understanding of mathematical modeling and parameter constraints
  • Familiarity with quartic polynomials and their roots
  • Knowledge of numerical algorithms for solving equations
  • Proficiency in manipulating algebraic expressions involving multiple variables
NEXT STEPS
  • Explore numerical methods for solving nonlinear equations, such as the Newton-Raphson method
  • Study quartic polynomial root-finding techniques to understand the behavior of \(q\)
  • Learn about optimization techniques to adjust parameters in mathematical models
  • Investigate the implications of parameter sensitivity in mathematical modeling
USEFUL FOR

Mathematicians, data scientists, and engineers who are involved in mathematical modeling, optimization, and parameter adjustment in complex systems.

Siron
Messages
148
Reaction score
0
Given three parameters:
$$a= \frac{(k-3)^2 \sqrt{v}}{s}, \ \ b = \frac{v}{s}(w-10s), \ \ c = s \sqrt{v}.$$
which exact values I know (that is, I know $v,s,k$ and $w$ exactly). I need to guarantee that $a<0$ (this is always satisfied in my calculations!) and
$$0<b<1.08148a^2$$
For instance, if $b>1.08148a^2$ then I will replace $w$ by $w'$ such that $b(v,s,w') = 1.08148a^2$ and hence $b(v,s,w') - \epsilon$ satisfies the constraint. Here $\epsilon>0$ is an arbitrary small number.Consider another parameter $c = s \sqrt{v}$. I now have to guarantee that:
$$a<0, \quad 0<b<1.08148a^2, \quad q < c < 0, \qquad (*)$$
where $q$ is the greatest real root of the quartic polynomial
$$(48a^2+16b)x^4 - (40a^3+168 ab)x^3+(-45a^4+225a^2b + 72b^2)x^2+(27a^3b-162 ab^2)x+27b^3.$$
The explicit expression (which is a function of $a$ and $b$) for $q<0$ is quite horrendous. However, to guarantee that $q<0$ is real, I need that $a<0$ and $0<b<1.08148a^2$. So my question is, suppose I know the values of $a,b$ and $c$ where $b$ and $c$ do not satisfy the constraints. How can I modify $w$ and $s$ (similarly as above) **beforehand**, such that the constraints are satisfied? An ideal solution would be: replace $(s,w)$ by $(s',w')$, which gives me modified values $a',b'$ and $c'$, such that
$$0<b'<1.08148(a')^2, \qquad q(a',b')<c'<0.$$Thanks in advance!
 
Physics news on Phys.org
UPDATE:

If for instance $b>1.08148a^2$ then I replace $w$ by $w'$ such that $b(v,s,w') = 1.08148a^2$ and then I extract $\epsilon>0$ to guarantee that the new value $b'(v,s,w')<1.0848a^2$ as required. On the other hand I also need to guarantee that $q(a,b)<c<0$. Since I adjusted $b$, $q(a,b)$ also changes and hence I need to guarantee that $q(a,b(v,s,w')) = q(a,1.08148a^2-\epsilon)<c<0$. However, to that end I have to adjust $s$ since $c(s,v) = s\sqrt{v}$. Since I adjust $s$ the value for $a$ and $b$ will be different as well, because they are functions of $s$. Therefore, replacing $(s,w)$ beforehand by $(s',w')$ such that the constraints are satisfied, comes down to solving the following system of equation for $(s',w')$
$$\begin{cases} b(v,s',w') = 1.08148(a(v,s',k))^2 \\ q(a',1.08148(a')^2-\epsilon )= s' \sqrt{v} \end{cases},$$
which is equivalent with
$$\begin{cases} \frac{v}{s'}(w'-10s') = 1.0848 \frac{(k-3)^2 v}{(s')^2} \\ q(a',1.0848(a')^2-\epsilon) = s' \sqrt{v} \end{cases}$$

Now, solving the first equation of the system for $w'$ is straightforward. The value for $s'$ should be computed from second equation. However, the expression for $q(a',1.08148(a')^2-\epsilon)$ is probably quite ugly (but it is possible to compute it). Maybe I should use a numerical algorithm here? Any suggestions?

Thanks!
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K