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

Problem R* 1.6

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

$\displaystyle c_{3}(Y^{1},t)=M[1-{\bar {R}}u_{,SS}^{2}(Y^{1},t)]$

(1.6.1)

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

Solution

Solved without any assistance from previous reports

From equation 1.6.1 we get,

$\displaystyle c_{3}(Y^{1},t){\ddot {Y}}^{1}(t)=M[1-{\bar {R}}u_{,SS}^{2}(Y^{1},t)]{\ddot {Y}}^{1}$

(1.6.2)

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

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

(1.6.3)

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

1. The condition of homogeneity:

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

(1.6.4)

2. The condition of linearity

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

(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)

$\displaystyle c_{3}(Y^{1},t){\ddot {Y}}^{1}(t)=M[1-{\bar {R}}u_{,SS}^{2}(Y^{1},t)]{\ddot {Y}}^{1}$

(1.6.2)

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

$\displaystyle c_{3}(\alpha Y^{1},t){\frac {\partial ^{2}{\alpha Y^{1}}}{\partial ^{2}{t}}}=M[1-{\bar {R}}u_{,SS}^{2}(\alpha Y^{1},t)]{\frac {\partial ^{2}{\alpha Y^{1}}}{\partial ^{2}{t}}}$

(1.6.6)

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

$\displaystyle c_{3}(\alpha Y^{1},t){\frac {\partial ^{2}{\alpha Y^{1}}}{\partial ^{2}{t}}}=\alpha M[1-{\bar {\alpha }}{R}u_{,SS}^{2}(Y^{1},t)]{\frac {\partial ^{2}{Y^{1}}}{\partial ^{2}{t}}}$

(1.6.7)

But if the term is to be homogenous then,

$\displaystyle c_{3}(\alpha Y^{1},t){\frac {\partial ^{2}{\alpha Y^{1}}}{\partial ^{2}{t}}}=\alpha M[1-{\bar {R}}u_{,SS}^{2}(Y^{1},t)]{\frac {\partial ^{2}{Y^{1}}}{\partial ^{2}{t}}}$

(1.6.8)

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

$\displaystyle \alpha M[1-\alpha {\bar {R}}u_{,SS}^{2}(Y^{1},t)]{\frac {\partial ^{2}{Y^{1}}}{\partial ^{2}{t}}}\neq \alpha M[1-{\bar {R}}u_{,SS}^{2}(Y^{1},t)]{\frac {\partial ^{2}{Y^{1}}}{\partial ^{2}{t}}}$

(1.6.9)

Thus,

$\displaystyle c_{3}(\alpha Y^{1},t){\frac {\partial ^{2}{\alpha Y^{1}}}{\partial ^{2}{t}}}\neq \alpha c_{3}(Y^{1},t){\frac {\partial ^{2}{Y^{1}}}{\partial ^{2}{t}}}$

(1.6.10)

So the given term $\displaystyle c_{3}(Y^{1},t){\ddot {Y}}^{1}$ is not homogenous with respect to $\displaystyle Y^{1}$. As it is one of the two conditions to be simultaneously satisfied for linearity, we can say that term $\displaystyle c_{3}(Y^{1},t){\ddot {Y}}^{1}$ is also not linear with respect to $\displaystyle Y^{1}$.