Limitation on the method of strained coordinates for vibrations with weak grazing unilateral contact

Abstract

The nonlinear normal modes have recently been investigated with the method of strained coordinates for spring-mass models with some springs with piecewise linear stiffness. The N d.o.f. case and the mathematical validity of the method for large time were rigorously proved. The time validity is related to the nature of contact. For grazing contact, this method and also the multiscale expansion lose nonlinear features. For a small piecewise linear stiffness, we show that this method is less precise for a weak unilateral grazing contact. Thus, the validity of the asymptotic expansion for large time cannot be improved and the method has to be modified.

Appendices

Appendix 1

In this appendix, we compute the following exact expression for the period with the energy.

$$\begin{aligned} T&= \left\{ \begin{array}{cl} 2\, \pi &{} - 2 \arccos \left( |u_-|^{-1}\right) \\ + &{} 2 \sqrt{ \eta } \arccos \left( \frac{\eta }{h_0\,+\,\eta }\right) \end{array} \right. \\ u_{-}&= - \sqrt{E_0}, \qquad \qquad \qquad \eta = \frac{1}{1+\varepsilon }. \end{aligned}$$

$T(\varepsilon )$ To obtain this period formula, i.e., Formula (11) before, we use the well-known relation between the energy and the period [7]. Using the symmetry with respect to the horizontal axis in the phase space, we compute the half period:

$$\begin{aligned} \frac{T}{2} = \int _{u_-}^{u_0} \frac{\hbox {d}u }{\sqrt{E_0 - F(u)}}. \end{aligned}$$

(28)

Formula (28) comes from the relation $\dot{u}\,\,{=}\,\,{\pm } \sqrt{E- F(u)}$. For the half superior part: $\dot{u} > 0$ and since $E=E_0$, the relation becomes: $\frac{\hbox {d}u }{\hbox {d}t} =\sqrt{E_0 - F(u)}$, so $\mathrm{d}t = \frac{\hbox {d}u}{\sqrt{E_0 - F(u)}}$ which yields relation (28).

Since $F(u)$ has different expression for $u<1$ or $u>1$, we decompose the integral in (28) in two parts. The computation is explicit with the function $\arcsin$ and the following relations for $A>0$:

$$\begin{aligned} \arcsin (y)&= \int _0^y \frac{\hbox {d}u }{\sqrt{1 - u^2}}, \end{aligned}$$

(29)

$$\begin{aligned} \arcsin \left( \frac{y+ b}{\sqrt{A+b^2}} \right)&= \int _{}^y \frac{\hbox {d}u }{\sqrt{A - \left( u^2 + 2 b\, u \right) }}. \end{aligned}$$

(30)

Equality (30) means the left-hand side is an antiderivative of the right-hand side. We also recall the $\arccos$ function for $y \in [0,1]$:

$$\begin{aligned} \arccos (y) = \frac{\pi }{2} - \arcsin (y) = \int _y^1 \frac{\hbox {d}u}{\sqrt{1 - u^2}}. \end{aligned}$$

(31)

Now, we can compute $T/2$:

$$\begin{aligned} \frac{T}{2} \!=\! T_-+\! T_0 \!=\! \int _{u_-}^{1} \frac{\hbox {d}u }{\sqrt{E_0 \!-\! F(u)}} \!+\! \int _{1}^{u_0} \frac{\hbox {d}u }{\sqrt{E_0 \!-\! F(u)}}. \end{aligned}$$

Notice that $u_- = - \sqrt{E_0}$ so:

$$\begin{aligned} T_{-}&= \int _{u_-}^{1} \frac{\hbox {d}u }{\sqrt{E_0 - u^2}} \\&= \arcsin (1/\sqrt{E_0}) - \arcsin \left( u_-/\sqrt{E_0}\right) , \\&= - \arcsin (1/u_-) - \arcsin (-1)\\&\quad = \frac{\pi }{2} + \arcsin (1/|u_-|), \\&= \pi - \arccos (1/|u_-|). \end{aligned}$$

We now turn to $T_0$ with the notation $u=1+v$ and $\eta = 1/(1+\varepsilon )$:

$$\begin{aligned} T_0&= \int _{1}^{u_0} \frac{\hbox {d}u }{\sqrt{E_0 - \left( u^2 + \varepsilon \, v^2\right) }}, \\&= \int _{0}^{h_0} \frac{\hbox {d}v }{\sqrt{(1+h_0)^2+ \varepsilon \, h_0^2 - \left( (1+v)^2 + \varepsilon \, v^2\right) }},\\&= \int _{0}^{h_0} \frac{\hbox {d}v }{\sqrt{(1+\varepsilon ) (h_0^2 - v^2 )+ 2(h_0 - v)}}, \\&= \sqrt{\eta } \int _{0}^{h_0} \frac{\hbox {d}v }{\sqrt{ (h_0^2 - v^2 )+ 2\, \eta (h_0 - v)}}, \\&= \sqrt{\eta } \int _{0}^{h_0} \frac{\hbox {d}v }{\sqrt{(h_0^2 + 2 \eta h_0) - ( v^2 + 2 \eta \,v)}}, \\&= \sqrt{\eta } \left( \arcsin \left( \frac{h_0+\eta }{h_0+\eta }\right) - \arcsin \left( \frac{\eta }{h_0+\eta }\right) \right) , \end{aligned}$$

from (30) and

$$\begin{aligned} A+b^2&= h_0^2 + 2 \eta h_0 + \eta ^2= (h_0+ \eta )^2, \\&= \sqrt{\eta } \left( \frac{\pi }{2} - \arcsin \left( \frac{\eta }{h_0+\eta }\right) \right) , \\&= \sqrt{\eta }\, \arccos \left( \frac{\eta }{h_0+\eta }\right) . \end{aligned}$$

Adding $T_{-}$ and $T_{0}$ we obtain $T/2$ and then (11).

Appendix 2

We compute (26): the asymptotic expansion of the period $T(\varepsilon )$ from the exact expression of the period (11).

$$\begin{aligned} T(\varepsilon )&= 2 \, \pi - \frac{7}{6} \sqrt{2}\, \varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}). \end{aligned}$$

The details of the computations first use an expansion of the function $\arccos (y)$ near $y=1$.

$$\begin{aligned} \arccos (1-h)&= \left\{ \begin{array}{l} \sqrt{ 2\, h } + \frac{\sqrt{2}}{12} \, h^{1.5} + \frac{3 \sqrt{2}}{80} \, h^{2.5} \\ +\,\mathcal {O}(h^{3.5}). \end{array} \right. \end{aligned}$$

(32)

We now prove this formula. First, by (31), we have the following:

$$\begin{aligned} \arccos (1-h)&= \int ^1_{1-h} \frac{\hbox {d}u }{\sqrt{1 - u^2}}, \\&= \int ^1_{1-h} \frac{\hbox {d}u }{\sqrt{(1 - u)(1+u)}}, \\&\hbox {with}\,u = 1- t \, h \, \hbox {we have: } \\&= \int ^1_{0} \frac{ h \, \hbox {d}t }{\sqrt{ t \, h \, (2-t\,h)}},\\&= \sqrt{h}\int ^1_{0} \frac{\hbox {d}t }{\sqrt{t (2-t\,h)}}, \\&= \sqrt{h}\, g(h). \end{aligned}$$

We now have an integral with the parameter $h$ which is not singular. Thus, $g$ is a smooth function and we have the following expansion with fractional powers of $h$:

$$\begin{aligned}&\arccos (1-h) \\&\quad = \sqrt{h}\, \left[ g(0)+ g'(0) \, h+ \dfrac{g''(0)}{2} \, h^{2}+ \mathcal {O}(h^3) \right] , \\&\quad = g(0) \sqrt{h} + g'(0)\, h^{1.5} + \dfrac{g''(0)}{2}\, h^{2.5} + \mathcal {O}(h^{3.5}). \end{aligned}$$

We compute $g(0), \, g'(0)$ and $g''(0)$ to get formula (32).

$$\begin{aligned}&g(h) = \int ^1_{0} \frac{\hbox {d}t }{\sqrt{t (2-t\,h)}}, \\&g(0) = \int ^1_{0} \frac{\hbox {d}t }{\sqrt{ 2\, t }} = \sqrt{2},\\&g'(h) = \frac{1}{2}\int ^1_{0} \frac{ \sqrt{t }\, \hbox {d}t }{ (2-t\,h)^{3/2}}, \\&g'(0) = \frac{1}{2^{5/2} }\int ^1_{0} \sqrt{t }\, \hbox {d}t = \frac{1}{6 \, \sqrt{2}} = \frac{ \sqrt{2}}{12}, \\&g''(h) = \frac{3}{4}\int ^1_{0} \frac{ t\, \sqrt{t }\, \hbox {d}t }{ (2-t\,h)^{5/2}}, \\&g''(0) = \frac{3}{2^{9/2} }\int ^1_{0} t ^{3/2}\, \hbox {d}t = \frac{3 \, \sqrt{2 }}{80 }. \end{aligned}$$

We now turn to asymptotic expansion of the exact period $T(\varepsilon )$. For this purpose, we compute the expansions of the two last terms defining the exact period $T(\varepsilon ) = 2\, \pi + T_2 + T_3$ defined by (11):

$$\begin{aligned} T_2&= -2 \arccos (|u_-|^{-1}),\\ T_3&= 2\sqrt{\eta } \arccos \left( \frac{\eta }{\varepsilon + \eta } \right) \end{aligned}$$

To expand $T_2$ with respect to $\varepsilon$, we notice that from (28), (25):

$$\begin{aligned} |u_{-}|&= \sqrt{ (1 + \varepsilon )^2 + \varepsilon ^3 }, \\&= (1 + \varepsilon ) \sqrt{ 1 + \frac{ \varepsilon ^3 }{ (1 + \varepsilon )^2 }}, \\&= (1 + \varepsilon ) \left( 1 + \frac{ \varepsilon ^3 }{ 2 } + \mathcal {O}(\varepsilon ^4) \right) , \\&= 1+ \varepsilon + \frac{\varepsilon ^3}{2} + \mathcal {O}(\varepsilon ^4), \\ \frac{1}{ | u_-|}&= 1 - \varepsilon + \varepsilon ^2 - \frac{\varepsilon ^3}{2} + \mathcal {O}(\varepsilon ^4). \end{aligned}$$

Let $h$ be defined as:

$$\begin{aligned} h&= 1 - |u_-|^{-1}= \varepsilon - \varepsilon ^2 + \frac{\varepsilon ^3}{2} + \mathcal {O}(\varepsilon ^4), \\ \sqrt{h}&= \sqrt{\varepsilon } \sqrt{ 1 - \varepsilon + \frac{\varepsilon ^{2}}{2} + \mathcal {O}(\varepsilon ^{3})}, \\&= \sqrt{\varepsilon } \left( 1 -\frac{ \varepsilon }{2}+ \frac{\varepsilon ^{2}}{4} + \frac{3}{8} \varepsilon ^2 + \mathcal {O}(\varepsilon ^{3}) \right) , \\&= \sqrt{\varepsilon } - \frac{\varepsilon ^{1.5}}{2} + \frac{5}{8}\varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}),\\ h^{1.5}&= \varepsilon ^{1.5} - \frac{3}{2} \varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}), \\ h^{2.5}&= \varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}). \end{aligned}$$

We now can compute $T_2$:

$$\begin{aligned} - T_2&= 2 \arccos ( 1 - h), \\&= 2 \sqrt{ 2 \, h} + \frac{\sqrt{2}}{6} h^{1.5} + \frac{3\, \sqrt{2}}{80} h^{2.5} + \mathcal {O}\left( h^{3.5}\right) , \\&= 2 \sqrt{ 2} \sqrt{\varepsilon } - \frac{5 \sqrt{2}}{6} \varepsilon ^{1.5} + \frac{83 \sqrt{2}}{80} \varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}). \end{aligned}$$

For the third term, we have

$$\begin{aligned} \frac{\eta }{\varepsilon + \eta }&= (1 + \varepsilon + \varepsilon ^2)^{-1} \\&= 1 - \varepsilon + \varepsilon ^3 + \mathcal {O}(\varepsilon ^{4}) = 1 - h. \end{aligned}$$

Now we expand $h, \, h^{1.5}$ and $h^{2.5}$:

$$\begin{aligned}&h = \varepsilon - \varepsilon ^3 + \mathcal {O}(\varepsilon ^{4}) = \varepsilon \left( 1 - \varepsilon ^2 + \mathcal {O}(\varepsilon ^{3}) \right) , \\&\sqrt{h } = \sqrt{\varepsilon } - \frac{\varepsilon ^{2.5}}{2} + \mathcal {O}(\varepsilon ^{3.5}), \\&h^{1.5} = \varepsilon ^{1.5}+ \mathcal {O}(\varepsilon ^{3.5}), \\&h^{2.5} = \varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}), \end{aligned}$$

These yield to the expansion of $T_3$:

$$\begin{aligned} 2 \arccos \left( \frac{\eta }{\varepsilon + \eta } \right)&= 2 \sqrt{2 } \sqrt{\varepsilon } + \frac{\sqrt{2}}{6}\varepsilon ^{1.5} \\&- \frac{191\sqrt{2}}{240} \varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}), \end{aligned}$$

$$\begin{aligned} \sqrt{\eta }&= (1+\varepsilon )^{-1/2} = 1 - \frac{\varepsilon }{2} + \frac{3}{8} \varepsilon ^2 + \mathcal {O}(\varepsilon ^3), \\ T_3&= 2 \sqrt{2} \sqrt{\varepsilon } - \frac{5\sqrt{2}}{6} \varepsilon ^{1.5} - \frac{31\, \sqrt{2}}{240} \varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}). \end{aligned}$$

Finally, adding the expansions for $T_2$ and $T_3$, we obtain

$$\begin{aligned} T - 2\, \pi = - \frac{280\, \sqrt{2}}{240} \varepsilon ^{2.5} + \mathcal {O}(\varepsilon ^{3.5}), \end{aligned}$$

and then (26).

Appendix 3

We compute the exact solution to compare it numerically with some asymptotic expansions. Since the ODE (7) is piecewise linear, we have after some computations:

$$\begin{aligned} \begin{array}{cll} u_\varepsilon (t) &{}= \eta \varepsilon {+} (1 + \eta \varepsilon ^2) \cos (\sqrt{1+\varepsilon }\, t ), &{} \quad \!\! 0 < t < \tau , \\ u_\varepsilon (t) &{}= u_- \cos \left( t -\dfrac{T}{2} \right) , &{} \tau < t < T -\tau \!, \\ u_\varepsilon (t)&{}= u_\varepsilon (T- t ), &{} T -\tau < t<T\!, \end{array} \end{aligned}$$

where $u_-$ and $\eta$ are defined in Sect. 3, $T=T(\varepsilon )$ and

$$\begin{aligned} \tau = \tau (\varepsilon )= \dfrac{\arccos \left( \dfrac{1- \eta \varepsilon }{1+\eta \varepsilon ^2}\right) }{\sqrt{1+\varepsilon }}. \end{aligned}$$