User:Egm6321.f12.team4.hong/Report1

Statement
Show $$ c_3(Y^1,t)\ddot{Y^1} $$ is nonlinear, whether $$ u^2(Y^1,t) $$ is linear or nonlinear with respect to $$ Y^1 $$.

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

Solution
Definition of Linearity (Class note: sec4-6) $$ F(\alpha x + \beta y)=\alpha F(x)+\beta F(y) \, \forall \alpha, \beta \in \mathbb R $$

Let, $$ \alpha, \beta \in \mathbb R $$ and $$ Y^1, Y^2 $$ functions of t.

To check linearity, let's substitute $$ Y^1(t) $$ with $$ \alpha Y^1(t) + \beta Y^2(t) $$ $$ c_3(\alpha Y^1(t) + \beta Y^2(t),t)( \ddot{\alpha Y^1(t) + \beta Y^2(t)} ) = M[1-\bar{R}u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t)] (\ddot{\alpha Y^1(t) + \beta Y^2(t)}) $$ $$ = \alpha M[1-\bar{R}u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t)] \ddot{Y^1(t)} +  \beta M[1-\bar{R}u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t)] \ddot{Y^2(t)} $$

Consequently, replacing $$ Y^1(t) $$ with $$ \alpha Y^1(t)$$ and replacing $$ Y^1(t) $$ with $$ \beta Y^2(t)$$ we have following equations: $$ c_3(\alpha Y^1(t),t)( \ddot{\alpha Y^1(t)} ) = M[1-\bar{R}u^2_{,SS}(\alpha Y^1(t),t)] (\ddot{\alpha Y^1(t)}) $$ $$ c_3(\beta Y^2(t),t)( \ddot{\beta Y^2(t)} ) = M[1-\bar{R}u^2_{,SS}(\beta Y^2(t),t)] (\ddot{\beta Y^2(t)}) $$

Looking at the above three equations, in order for the main function to be linear, $$ u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t) = u^2_{,SS}(\alpha Y^1(t),t) $$       (1) and, $$ u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t) = u^2_{,SS}(\beta Y^2(t),t) $$       (2)

Regardless of the linearity of $$ u^2(Y^1,t) $$ for the linearity of the equation $$ c_3(Y^1,t)\ddot{Y^1} $$, above equations must be satisfied. For equation (1) to be satisfied, $$ \beta $$ must be zero. Then the equation (2) cannot be satisfied unless $$ \alpha $$ is also a zero. Equations does not satisfy for any real values of $$ \alpha, \beta $$.

Therefore the equation $$ c_3(Y^1,t)\ddot{Y^1} $$ is nonlinear, regardless of the linearity of $$ u^2(Y^1,t) $$.

Author and References

 * Solved and Typed by -- Seong Hyeon Hong