User:EGM6321.f12.team7.Zhou/R1-5

=Problem 6=  I have solved this problem independently.

Given
The Equation of Motion of Wheel/Magnet has four terms, which are given by,
 * {| style="width:100%" border="0"

$$\displaystyle C_3(Y^1,t)\ddot Y^1+C_2(Y^1,t)(\dot Y^1)^2+C_1(Y^1,t)\dot Y^1+C_0(Y^1,t)=0$$ (6.1) The general expression of $$\displaystyle C_3(Y^1,t)\ddot Y^1$$ is given by,
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle c_3(Y^1,t)\ddot Y^1=[1-\bar{R}u_{,ss}(Y^1,t)]\ddot Y^1$$ (6.2) From the lecture note[}, we can Consider $$\displaystyle F(\cdot)$$ as an operator, $$\displaystyle F(\cdot)$$ is linear if and only if,
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle F(\alpha f+\beta g)=\alpha F(f)+\beta F(g)$$ (6.3) Where$$\displaystyle f$$ and $$\displaystyle g$$ are two possible arguments for $$\displaystyle F(\cdot)$$, and $$\displaystyle \forall \alpha, \beta \in R.$$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The derivative operator is a linear operator, which is given by,
 * {| style="width:100%" border="0"

$$\displaystyle F(\cdot)=\frac{d(\cdot)}{dx}$$ (6.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Find
$$\displaystyle C_3(Y^1,t)\ddot Y^1$$ is nonlinear with repect $$\displaystyle Y^1.$$

Solution
Suppose we have,
 * {| style="width:100%" border="0"

$$\displaystyle Y^1:[t_0,+\infty) \rightarrow R$$
 * style="width:95%" |
 * style="width:95%" |

$$\displaystyle \forall u, v:[t_0,+\infty) \rightarrow R$$

$$\displaystyle \forall \alpha, \beta\in R$$ (6.5) For the right-hand side, we have,
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle RHS:=\alpha C_3(u,t)\frac{\alpha d^2u}{dt^2}+\beta C_3(v,t)\frac{\beta d^2v}{dt^2}$$ (6.6) The left-hand side is given by,
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle LHS:=C_3(\alpha u+\beta v,t)\frac{d^2 (\alpha u+\beta v)}{dt^2}$$ (6.7) Since the derivative operator is linear operator, we can get,
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^2 (\alpha u+\beta v)}{dt^2} = \frac{\alpha d^2u}{dt^2} +\frac{\beta d^2v}{dt^2}$$ (6.8) So (6.7) turns into,
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle C_3(\alpha u+\beta v,t)\frac{d^2 (\alpha u+\beta v)}{dt^2}=C_3(\alpha u+\beta v,t)\frac{\alpha d^2u}{dt^2} +C_3(\alpha u+\beta v,t)\frac{\beta d^2v}{dt^2}$$ (6.9) If $$\displaystyle C_3(Y^1,t)$$ is nonlinear with respect to $$\displaystyle Y^1$$, by comparision between (6.5) and (6.8) we can find that,
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle LHS\neq RHS $$ (6.10) Even if $$\displaystyle C_3(Y^1,t)$$ is linear with respect to $$\displaystyle Y^1$$, on the left-hsnd side we have,
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle LHS:=[\alpha C_3(u,t)+\beta C_3(v,t)]\frac{\alpha d^2u}{dt^2}+[\alpha C_3(u,t)+\beta C_3(v,t)]\frac{\beta d^2v}{dt^2}$$ (6.11) Obviously,
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle LHS\neq RHS $$ (6.12)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }