Simultaneous Equations (How to test for redundancy)

[miniradman]

Homework Statement

I have three equations:

$F = ρw(x_0 y^2_0 - x_1 y^2_1) + \frac{1}{2} γ w (x^2_0 - x^2_1)$ ----- 1

$y_0 = y_1 \frac{x_1}{x_0}$ ----- 2

$\frac{y^2_0}{2} + gx_0 = \frac{y^2_1}{2} + gx_1$ ------ 3

Homework Equations

N/A

The Attempt at a Solution

My goal is to have $F$ expressed without either $y^2_0$ or $y^2_1$ involved in the equation. My problem is that equations 2 and 3 involve both $y^2_0$ and $y^2_1$ so when it comes to eliminate one of them in equation 1, I re-introduce the other in the equation (and vice-versa). My question is whether or not it's legal to simultaneously solve equations 2 and 3 in two different ways (have $y^2_0$ and $y^2_1$ as subjects), then substitute each equation back in equation 1? Does solving the same simultaneous equation twice to obtain two equations with different subjects make them redundant?

 

[Ray Vickson]

Homework Statement

I have three equations:

$F = ρw(x_0 y^2_0 - x_1 y^2_1) + \frac{1}{2} γ w (x^2_0 - x^2_1)$ ----- 1

$y_0 = y_1 \frac{x_1}{x_0}$ ----- 2

$\frac{y^2_0}{2} + gx_0 = \frac{y^2_1}{2} + gx_1$ ------ 3

Homework Equations

N/A

The Attempt at a Solution

My goal is to have $F$ expressed without either $y^2_0$ or $y^2_1$ involved in the equation. My problem is that equations 2 and 3 involve both $y^2_0$ and $y^2_1$ so when it comes to eliminate one of them in equation 1, I re-introduce the other in the equation (and vice-versa). My question is whether or not it's legal to simultaneously solve equations 2 and 3 in two different ways (have $y^2_0$ and $y^2_1$ as subjects), then substitute each equation back in equation 1? Does solving the same simultaneous equation twice to obtain two equations with different subjects make them redundant?

If you substitute $y_0$ from eq. (2) into eq. (3), that will give you an equation involving $x_0, x_1, y_1$ alone, which you can then solve for $y_1$, That will give you an expression for $y_1$ in terms of $x_0,x_1$ only; substituting that into eq. (2) will give you $y_0$ in terms of $x_0,x_1$. Messy, but do-able.

 

[miniradman]

If you substitute $y_0$ from eq. (2) into eq. (3), that will give you an equation involving $x_0, x_1, y_1$ alone, which you can then solve for $y_1$, That will give you an expression for $y_1$ in terms of $x_0,x_1$ only; substituting that into eq. (2) will give you $y_0$ in terms of $x_0,x_1$. Messy, but do-able.

Hi Ray, thanks for the response.

I understand I can get equation 2 in terms of $x_0 , x_1$, however my ultimate goal is to get $F$ in terms of $x_0, x_1$ . I'm worried that using the same equation twice will create redundancy. Would I be able to use the resulting $y_0$ and $y_1$ in equation 1 with no hassle?

 

[Ray Vickson]

Hi Ray, thanks for the response.

I understand I can get equation 2 in terms of $x_0 , x_1$, however my ultimate goal is to get $F$ in terms of $x_0, x_1$ . I'm worried that using the same equation twice will create redundancy. Would I be able to use the resulting $y_0$ and $y_1$ in equation 1 with no hassle?

There is no redundancy in the method I suggested (which, by the way, is 100% standard). Eq (2) gives you $y_0$ in terms of $x_0,x_1,y_1$. Putting that into eq. (3) allows you to get $y_1$ in terms of $x_0,x_1$: $y_1 = Y_1(x_0,x_1)$ for some explicit function $Y_1$. If you have numerical values for the inputs $g,w,\rho, \gamma$, I can give you arbitrary numerical values for $x_0,x_1$, and you can use your function $Y_1$ to get a unique, well-defined numerical value for $y_1$. (Actually, there are two unique values with opposite signs, depending on which square root of $y_1^2$ you choose.) You can then take your now-known numerical values of $x_0, x_1,y_1$ and use eq. (2) to get the numerical value for $y_0$. Now you can put all those value into your $F$. Except for the "sign" issue, where is the redundancy in any of that? If you have some reason for choosing, say, the positive square root for $y_1$ then all uncertainty and ambiguity disappears.