I Can Nonlinear Equations be Linearized Using Free Parameters?

Euler2718
Messages
90
Reaction score
3
I am given the equations of Lorenz with respect to deterministic non-periodic flow:

\frac{dX}{dt} = Pr(Y-X), X(0)=X_{0}
\frac{dY}{dt} = -XZ + rX - Y, Y(0) = Y_{0}
\frac{dZ}{dt} = XY-bZ, Z(0) = Z_{0}

where Pr is the Prandtl number, r = Ra/Rac is the ratio of the Rayleigh number to its critical value, and b is a parameter that characterize the wave-number. I am told for the question that none of the derivation or mathematics behind it matter.

So the "question" (not really a question but merely a statement for me to figure out) is: A nonlinear quantity YZ may be linearized by replacing with Yn or nZ, where one of the original variables becomes a free parameter n. The equations above (Lorenz') can now be converted into a vector equation of the form

\dot x = A(n)x

Where x = [X,Y,Z]^{T} and \dot x means \frac{dx}{dt}

I don't think I'm interpreting the question correctly. It says I can replace any variable with a free parameter n? However I do that for instance with letting Y be a free parameter but equations two and three of Lorenz will not be linear as two variables will still be present. I'm under assumption that 'linearization' means having one variable with respect to the other, so I guess I'm at a conceptual loss here and would like to be put in the right direction.
 
Mathematics news on Phys.org
In simple words, a mathematical expression is linear if it does not contain products of any form between the unknown variables/functions/vectors/matrices (so say if x and y are unknowns the expression 5x+2y is linear but the expression 5xy+y or 5x^2+y are not linear).

In your case your equations 2 and 3 are not linear because they contain the terms XZ and XY respectively. I believe you are asked to replace these terms with nZ and nY, leaving X as it is in every other term of the equations.
 
Last edited:
  • Like
Likes Euler2718
Delta² said:
I believe you are asked to replace these terms with nZ and nY, leaving X as it is in every other term of the equations.

That would seem to be convenient, however I must point out that it says Yn or nZ . Would it be appropriate to consider this an "inclusive or" statement? I was under the assumption given the context that one or the other but not both, but what you suggest makes more sense. I just need to be certain before continuing.
 
It is an exclusive or statement. But what I do does not violate the exclusive or. I am linearizing two different terms(the term XY to nY and the term -XZ to -nZ), not the same term in two different ways.
 
  • Like
Likes Euler2718
Delta² said:
It is an exclusive or statement. But what I do does not violate the exclusive or. I am linearizing two different terms(the term XY to nY and the term -XZ to -nZ), not the same term in two different ways.

I think I see now. Many thanks.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Back
Top