Is it possible to solve for y using non-linear methods?

  • Thread starter pjoseph98
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In summary, the conversation discusses solving a linear algebra problem involving matrices and a single unknown variable, y. The problem is not straightforward due to the non-symmetric and non-square matrices involved. Possible solutions include using brute force methods or the least squares method, and the Lambert W function may also be needed.
  • #1
pjoseph98
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Hi there,

I am no expert in linear algebra (and I don't think this problem is linear anyway).

I am trying to solve the following for y: [A]y = C

A is an 8x2 matrix (fully known)
C is an 8x1 matrix (fully known)

B is an 2x1 matrix (whose terms are a function of the single unknown y).

The two terms in are: b1*e^(b2 + y) and b3*e^(b4 + y) where b1, b2, b3, and b4 are fully known.

Is it possible to solve for y? Do I use my favorite method--brute force or is there something more elegant. The problem (as I understand it) is that the matrices are not symmetric, far less, square.

And I need to solve this at each integration point in a Finite Element Analysis mesh...with up to 10,000 integration points, so ideally a brute force method would not be my preference...

Paul
 
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  • #2
y has to be 1x1 (a scalar) - but that is what you had in mind.
Note: taking out the common factor...$$B=\begin{pmatrix}b_1e^{b_2}\\ b_3e^{b_4}\end{pmatrix}e^y$$
Put ##AB=Xe^y## so that ##Xye^y=C## (X and C will be fully known) and compare terms.
 
  • #3
The first thing I would suggest is pulling y out of

[A](*e^-y)y*e^y = C
where (*e^-y) does not depend on y

The next problem is we do not know if C is a multiple of [A](*e^-y)
if not we can use least squares
that is instead of solving
$$\mathbf{Ax=b} \\
\text{we instead solve}\\
\min_x \mathbf{\|b-Ax \|_2}$$ but we must make sure this solution is reasonable to use
otherwise we may need to reformulate the problem with y as a matrix

finally we need the Lambert W function

http://en.wikipedia.org/wiki/Linear_least_squares_(mathematics)
http://en.wikipedia.org/wiki/Moore–Penrose_pseudoinverse
http://en.wikipedia.org/wiki/Lambert_W_function
 

1. How do I solve for an unknown variable in an equation?

To solve for an unknown variable in an equation, you need to isolate the variable on one side of the equation by using inverse operations. This means performing the opposite operation of what is being done to the variable. For example, if the variable is being multiplied by a number, you would divide both sides of the equation by that number.

2. What is the order of operations when solving for an unknown variable?

The order of operations when solving for an unknown variable is the same as solving any other equation. You should first simplify any expressions inside parentheses, then perform any operations involving exponents, followed by multiplication and division from left to right, and finally addition and subtraction from left to right.

3. Can I solve for an unknown variable if there are multiple variables in the equation?

Yes, you can still solve for an unknown variable if there are multiple variables in the equation. The key is to use algebraic manipulation to isolate the variable you are trying to solve for. This may involve combining like terms, factoring, or using the distributive property.

4. How do I know if my solution for an unknown variable is correct?

To check if your solution for an unknown variable is correct, you can plug the value back into the original equation and see if it satisfies the equation. This means the left side of the equation should equal the right side when the value is substituted.

5. What should I do if I get a negative number as my solution for an unknown variable?

If you get a negative number as your solution for an unknown variable, it is important to check your work and make sure you followed the correct order of operations. You can also plug the solution back into the equation and see if it makes sense in the context of the problem. If it still seems incorrect, you may need to go back and review the steps you took to solve the equation.

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