University of Florida/Egm6321/f12.team5.R1.6

Show that $c_3(Y^1,t){\ddot{Y}}^1$ is nonlinear with respect to $Y^1$

Given Equation

Equation (2) from p.5-4 gives the following expansion for the equation of motion of the wheel/magnet

$c_3(Y^1,t)=M[1-\overline{R}{u}_{,SS}^2(Y^1,t)]$

(1.6.1)

It is assumed that the term $u_{,SS}^2(Y^1,t)$ is linear.

Solution

Solved without any assistance from previous reports

From equation 1.6.1 we get,

$c_3(Y^1,t){\ddot{Y}}^1(t)=M[1-\overline{R}{u}_{,SS}^2(Y^1,t)]{\ddot{Y}}^1$

(1.6.2)

For an operator or a function to be linear, it has to satisfy the following condition:

$F(\alpha u+\beta v)=\alpha F(u)+\beta F(v),\mathrm{\forall }\alpha ,\beta \in ℝ$

(1.6.3)

This condition can be broken down into two separate conditions which have to be satisfied simultaneously,

1. The condition of homogeneity:

$F(\alpha u)=\alpha F(u)$

(1.6.4)

2. The condition of linearity

$F(u+v)=F(u)+F(v)$

(1.6.5)

As both of these conditions have to be satisfied simultaneously, an operator or function that does not satisfy any one of the two conditions above can be proved as nonlinear.

Initially, checking the condition of homogeneity (Equation 1.6.4)

$c_3(Y^1,t){\ddot{Y}}^1(t)=M[1-\overline{R}{u}_{,SS}^2(Y^1,t)]{\ddot{Y}}^1$

(1.6.2)

Now substituting $\alpha Y^1$ for $Y^1$.

$c_3(\alpha Y^1,t)\frac{{\partial }^2\alpha Y^1}{{\partial }^{2}t}=M[1-\overline{R}{u}_{,SS}^2(\alpha Y^1,t)]\frac{{\partial }^2\alpha Y^1}{{\partial }^{2}t}$

(1.6.6)

Since the term $u_{,SS}^2(Y^1,t)$ is linear, we can write:

$c_3(\alpha Y^1,t)\frac{{\partial }^2\alpha Y^1}{{\partial }^{2}t}=\alpha M[1-\overline{\alpha }R{u}_{,SS}^2(Y^1,t)]\frac{{\partial }^2{Y}^1}{{\partial }^{2}t}$

(1.6.7)

But if the term is to be homogenous then,

$c_3(\alpha Y^1,t)\frac{{\partial }^2\alpha Y^1}{{\partial }^{2}t}=\alpha M[1-\overline{R}{u}_{,SS}^2(Y^1,t)]\frac{{\partial }^2{Y}^1}{{\partial }^{2}t}$

(1.6.8)

It is evident from Equations (1.6.7) and (1.6.8) that,

$\alpha M[1-\alpha \overline{R}{u}_{,SS}^2(Y^1,t)]\frac{{\partial }^2{Y}^1}{{\partial }^{2}t}\ne \alpha M[1-\overline{R}{u}_{,SS}^2(Y^1,t)]\frac{{\partial }^2{Y}^1}{{\partial }^{2}t}$

(1.6.9)

Thus,

$c_3(\alpha Y^1,t)\frac{{\partial }^2\alpha Y^1}{{\partial }^{2}t}\ne \alpha c_3(Y^1,t)\frac{{\partial }^2{Y}^1}{{\partial }^{2}t}$

(1.6.10)

So the given term $c_3(Y^1,t){\ddot{Y}}^1$ is not homogenous with respect to $Y^1$. As it is one of the two conditions to be simultaneously satisfied for linearity, we can say that term $c_3(Y^1,t){\ddot{Y}}^1$ is also not linear with respect to $Y^1$.