User:Egm6321.f10.team6.Kim.MK/hw1

Problem Statement
The equation of motion of a wheel on flexible guide rail type problem that is consistent with what occurs when the wheel of a high speed train interacts with its guiderail is given as :

$$ {C}_{3} \left(Y^1,t \right) \overset{..}{Y^1} + {C}_{2} \left(Y^1,t \right) (\overset{.}{Y^1})^2 + {C}_{1} \left(Y^1,t \right)\overset{.}{Y^1} + {C}_{0} (Y^1, t) = 0 $$(1)

where the coefficients may be expressed as:

$$\begin{align} & {C}_{0}(Y^1, t) : = - F^1[1- \bar{R} {u^2_{,SS}}(Y^1, t)] - F^2{u^2_{,S}}(Y^1, t) - \frac{T}{R} \\ & {}+ M [[1- \bar{R}{u^2_{,SS}}(Y^1, t)[u_{,tt}^{1}(Y^1,t) -\bar{R}{u_{,Stt}^{2}}(Y^1, t)] \\ & {}+ {u_{,s}^{2}(Y^1, t) u_{,tt}^{2}(Y^1,t)}] \end{align}$$ (2)

$$ C_{1}(Y^1, t) := 2M[[1 -\bar{R}u_{,SS}^{2}(Y^1,t)][u_{,St}^1(Y^1,t)-\bar{R}u_{,SSt}^{2}(Y^1,t)]+u_{,S}^2(Y^1,t)u^2_{,St}(Y^1,t)] $$(3)

$$C_{2}(Y^1, t) := M\left[[1-\bar{R}u_{,SS}^{2}(Y^1,t)][u_{,SS}^1(Y^1,t)-\bar{R}u_{,SSS}^{2}(Y^1,t)]+u_{,S}^2(Y^1,t)u^2_{,SS}(Y^1,t)]\right]$$ (4)

$$ C_{3}(Y^1,t):= M[1-\bar{R}u_{,SS}^{2}(Y^1,t)]^2 + \frac{I_w}{R^2} $$ (5)

Required
Prove the following is non-linear:

$$ {C}_{3} \left(Y^1,t \right) \overset{..}{Y^1} $$ (6)

Solution
In general a function may be considered linear if the following conditions are met :


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$$f\left( \alpha x \right)=\alpha f\left( x \right)$$ (7)
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$$f\left( {{x}_{1}}\pm {{x}_{2}} \right)=f\left( {{x}_{1}} \right)\pm f\left( {{x}_{2}} \right)$$ (8)
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$$ {F\left({\alpha}u + {\beta}v\right) = {\alpha} F(u) + {\beta} F(v)} $$  (9)
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$$ F\left( \alpha {Y^1} \right) = \alpha {F ( {Y^1})} $$ (10)
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Consider for a moment the differential operator. The differential operator is a linear operator, and as such the above conditions will be met if the the differential operator is applied to an arbitrary function. For example:

First let us examine the first condition of linearity given in eq. 7. According to eq. 7 the result if we multiply the differential operator by some constant C should be exactly the same as if we multiply function by some constant C and then take the derivative.

$$\begin{align} & C\frac{d{f(y)}}{dy}=\frac{d{f(C\cdot y)}}{dy} \\ & C\frac{d\left( 5y \right)}{dy}=C\left( 5 \right)=5C \\ & \frac{d\left( 5Cy \right)}{dy}=5C \\ \end{align}$$

Clearly the placement of the scalar multiple C has no effect on the result, and therefore the differential operator is linear. If we were to apply the second condition of linearity it would also hold.

For this particular problem we will apply the first condition of linearity to prove that the function given in eq. 6 is non-linear. By applying the first condition of linearity using an arbitrary constant α we obtain:

$$ \left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}} {({{\alpha}{Y}^{1}},t)}} \right]}^{2}}+\frac \right)\ddot{Y}^{1} {\left( {{\alpha}{Y}^{1}},t \right)}\overset{?}{\mathop{=}}\,{\alpha}\left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}} {\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\frac \right) \ddot{Y}^{1}{\left( {{Y}^{1}},t \right)} $$ (12)

Since the differential operator is linear we can take the scalar multiply from within the derivative terms to obtain:

$$ \left( M{{\left[ 1-{\alpha}\bar{R}u{{_{,ss}^{2}} {({{Y}^{1}},t)}} \right]}^{2}}+\frac \right){\alpha}\ddot{Y}^{1} {\left( {{Y}^{1}},t \right)}\overset{?}{\mathop{=}}\,{\alpha}\left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}} {\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\frac \right) \ddot{Y}^{1}{\left( {{Y}^{1}},t \right)} $$ (13)

Doing some simple algebraic manipulation we obtain the following:


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$$ \left( {\alpha}M{{\left[ 1-{\alpha}\bar{R}u{{_{,ss}^{2}} {({{Y}^{1}},t)}} \right]}^{2}}+{\alpha}\frac \right)\ddot{Y}^{1} {\left( {{Y}^{1}},t \right)}{\neq}\left({\alpha} M{{\left[ 1-\bar{R}u{{_{,ss}^{2}} {\left( {{Y}^{1}},t \right)}} \right]}^{2}}+{\alpha}\frac \right) \ddot{Y}^{1}{\left( {{Y}^{1}},t \right)} $$ (14)
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Clearly the left and right sides of eq. 12 don't match (there is second power scalar on the right, but only a first power on the left), therefore eq. 6 is non-linear.