Localization in Adiabatic Shear Flow Via Geometric Theory of Singular Perturbations

Abstract

We study localization occurring during high-speed shear deformations of metals leading to the formation of shear bands. The localization instability results from the competition between Hadamard instability (caused by softening response) and the stabilizing effects of strain rate hardening. We consider a hyperbolic–parabolic system that expresses the above mechanism and construct self-similar solutions of localizing type that arise as the outcome of the above competition. The existence of self-similar solutions is turned, via a series of transformations, into a problem of constructing a heteroclinic orbit for an induced dynamical system. The dynamical system is in four dimensions but has a fast–slow structure with respect to a small parameter capturing the strength of strain rate hardening. Geometric singular perturbation theory is applied to construct the heteroclinic orbit as a transversal intersection of two invariant manifolds in the phase space.

Appendices

The Loss of Hyperbolicity for $n = 0$

Consider the system (1) when $n=0$, that is, the viscoplastic effects are neglected. Then, (2) reads

$$\begin{aligned} \sigma = \tau (\theta , \gamma ) = \theta ^{-\alpha }\gamma ^m \end{aligned}$$

(65)

and (1) is written as a first-order system

$$\begin{aligned} \begin{pmatrix} v_t \\ \theta _t \\ \gamma _t \end{pmatrix} = \underbrace{ \begin{pmatrix} 0 &{} \tau _\theta (\theta , \gamma ) &{} \tau _\gamma (\theta , \gamma ) \\ \tau (\theta , \gamma ) &{} 0 &{} 0 \\ 1 &{} 0 &{} 0 \\ \end{pmatrix}}_{\triangleq B (\theta , \gamma )} \begin{pmatrix} v_x \\ \theta _x \\ \gamma _x \end{pmatrix}. \end{aligned}$$

(66)

We check hyperbolicity for (66). The characteristic speeds are the roots of

$$\begin{aligned} \det \big (B-\lambda \text {I}\big )&= -\lambda \big (\lambda ^2 - (\tau _\theta \tau + \tau _\gamma )\big ) = 0 \end{aligned}$$

The system is thus hyperbolic when $\tau _\theta \tau + \tau _\gamma > 0$ and elliptic in the t-direction when $\tau _\theta \tau + \tau _\gamma < 0$. Observe that along the evolution of (66) and under the conditions for loading of interest in our problem, we have that $\gamma$ is increasing; the equation

$$\begin{aligned} \theta _t = \tau (\theta , \gamma ) \gamma _t \end{aligned}$$

implies that

$$\begin{aligned} \tau _\theta \tau + \tau _\gamma = \frac{d}{d\gamma } \tau (\theta , \gamma ). \end{aligned}$$

We conclude that the system is hyperbolic before the maximum of the stress–strain curve and elliptic beyond the maximum point. For the constitutive law (65), a computation shows

$$\begin{aligned} \begin{aligned} \tau _\theta \tau + \tau _\gamma&= \theta ^{-\alpha } \gamma ^{m-1} \left( -\alpha \frac{\gamma ^{m+1}}{\theta ^{1 + \alpha }} + m \right) \\&= \frac{-\alpha + m}{1 + \alpha } + \frac{\alpha (1+m)}{\theta ^{1+\alpha }} \left( \frac{\theta _0(x)^{1+\alpha }}{1 + \alpha } - \frac{ \gamma _0 (x)^{1+m}}{1+m} \right) \end{aligned} \end{aligned}$$

In the region $\alpha > m$, the stress–strain curve may be initially increasing (depending on the data) but eventually decreases.

The system (66) admits the class of uniform shearing solutions

$$\begin{aligned} v_s (x) = x, \quad \gamma _s (t) = t + \gamma _0, \quad \theta _s (t) \; \text{ is } \text{ determined } \text{ by } \text{ solving } \; {\left\{ \begin{array}{ll} \frac{{\text {d}}\theta _s}{{\text {d}}t} = \tau (\theta _s, \gamma _s) &{} \\ \theta _s (0) = \theta _0 &{} \\ \end{array}\right. }\nonumber \\ \end{aligned}$$

(67)

where $\gamma _0$, $\theta _0>$ are the initial strain and temperature, respectively. We linearize around the uniform shearing solution by setting

$$\begin{aligned} v = x + {\hat{V}}(t,x), \quad \theta = \theta _s (t) + {\hat{\Theta }} (t,x), \quad \gamma = \gamma _s (t) + {\hat{\Gamma }} (t,x) \end{aligned}$$

and obtain the linearized system satisfied by the perturbation $({\hat{V}}, {\hat{\Theta }}, {\hat{\Gamma }} )$,

$$\begin{aligned} \begin{pmatrix} {\hat{V}}_t \\ {\hat{\Theta _t}} \\ {\hat{\Gamma _t}} \end{pmatrix} = B (\theta _s(t), \gamma _s (t) ) \begin{pmatrix} {\hat{V}}_x \\ {\hat{\Theta _x}} \\ {\hat{\Gamma _x}} \end{pmatrix} + \begin{pmatrix} 0 &{} 0 &{} 0 \\ 0 &{} \tau _\theta (\theta _s, \gamma _s ) &{} \tau _\gamma (\theta _s, \gamma _s ) \\ 0 &{} 0 &{} 0 \\ \end{pmatrix} \; \begin{pmatrix} {\hat{V}} \\ {\hat{\Theta }} \\ {\hat{\Gamma }} \end{pmatrix}. \end{aligned}$$

(68)

The above calculation shows that when $\alpha > m$, the linearized system loses hyperbolicity in finite time, past the maximum of the curve $\sigma _s (t) - t$.

The Equilibria of the System (S)

We discussed in Sect. 5 the equilibria $M_0$ and $M_1$ of (S). The remaining equilibria of (S) are listed below, and they are all functions of $(\alpha ,m,n,\lambda )$ that lie outside thesector

$$\begin{aligned} \mathscr {P} = \{(p,q,r,s) \; | \; p\ge 0, \, q\ge 0, \, r>0, \, s>0 \} \end{aligned}$$

in the parameter range (23). The reader will find underlined the components, indicating that the equilibrium lies outside the sector of interest: we recall the notations

$$\begin{aligned} a= & {} \frac{2+2\alpha -n}{D} + \frac{2(1 + \alpha )}{D}\lambda , \quad \\ b= & {} \frac{1+m}{D} + \frac{1+m+n}{D}\lambda , \quad D=1+2\alpha -m-n \end{aligned}$$

while t, $t_1$ and $t_2$ denote arbitrary real numbers.

$$\begin{aligned} \begin{array}{lllllll} (1) &{}\Big (0,&{}0,&{} \underline{0},&{}\underline{0}&{}\Big ), \\ (2) &{}\Big (0,&{}0,&{} \underline{0},&{}t&{} \Big ) &{}\text {provided } \lambda = \frac{-2\alpha +2m+n}{2(\alpha -m-n)},\\ (3) &{}\Big (0,&{}0,&{} \frac{n\alpha - a(\alpha -m-n)}{(1+\alpha )(m+n)}, &{}\underline{0}&{}\Big ),\\ (4) &{}\Big (0,&{}1,&{} \underline{0},&{}\underline{0}&{}\Big ), \\ (5) &{}\Big (0,&{}1,&{} \underline{0},&{}t&{} \Big ) &{}\text {provided } \lambda = \frac{2\alpha -2m-n}{1+m+n},\\ (6) &{}\Big (0,&{}1,&{} \frac{n\alpha - a(\alpha -m-n)}{(1+\alpha )(m+n)}+\frac{\lambda }{m+n}, &{}\underline{0}&{}\Big ),\\ (7) &{}\Big (t,&{}0,&{} \underline{0},&{}\underline{0}&{}\Big ) &{}\text {provided } \lambda =\frac{2+2\alpha -n}{2(\alpha -m-n)}, \\ (8) &{}\Big (t,&{}1,&{} \underline{0},&{}\underline{0}&{}\Big ) &{}\text {provided } \lambda =\frac{-2-2\alpha +n}{1+m+n}, \\ (9) &{}\Big (t_1,&{}0,&{} \underline{0},&{}t_2&{} \Big ) &{}\text {provided }1+2\alpha -m-n=0\text { and }\lambda =\frac{-1-m}{1+m+n}, \\ (10) &{}\Big (t_1,&{}1,&{} \underline{0},&{}t_2&{} \Big ) &{}\text {provided }1+2\alpha -m-n=0\text { and }\lambda =\frac{-1-m}{1+m+n}, \\ \end{array} \end{aligned}$$

$$\begin{aligned} (11)\qquad \Big (&\underline{ -\tfrac{(\alpha -m-n)(1+m+n)}{(1+\alpha )(1+m)}}, \,\tfrac{2(\alpha -m-n)}{1+m}b, \,\tfrac{2(1+\alpha )}{1+2\alpha -m-n}, \,\tfrac{1+m+n}{1+\alpha } - \tfrac{n(1+2\alpha -m-n)}{2(1+\alpha )^2}\Big ), \\ (12)\qquad \bigg (&\Big ( \tfrac{2\alpha (1+m)}{D(1-m-n)} + \tfrac{2(\alpha -m-n)}{D}\lambda \Big )\Big (\tfrac{2\alpha (1+m)}{D(1-m-n)} - \tfrac{1+m+n}{D}\lambda \Big )\tfrac{1-m-n}{\lambda (2-n)}\tfrac{1-m-n}{\lambda (1+m)}, \\&\qquad \left( \tfrac{2\alpha (1+m)}{D(1-m-n)} + \tfrac{2(\alpha -m-n)}{D}\lambda \right) \left( \tfrac{1+m}{D} + \tfrac{1+m+n}{D}\lambda \right) \tfrac{1-m-n}{\lambda (1+m)}, \ \tfrac{2-n}{1-m-n}, \ \underline{0}\bigg ). \end{aligned}$$

The generic equilibria in $\mathscr {P}$ are $M_0$, $M_1$, (1), (3–4), (6), and (11–12); the rest are valid for specific parameter values.

The Linearized Problems Around $M_0$ and $M_1$

The coefficient matrix for the linearized system (S) around the equilibrium $M_0$ is

$$\begin{aligned}&\begin{pmatrix} 2 &{} 0 &{} 0 &{} 0 \\ br_0 &{} 1 &{} 0 &{} 0\\ \frac{r_0}{n}(\lambda r_0) &{} \frac{r_0}{n} &{} \frac{r_0}{n}\Big (\frac{\alpha -m-n}{\lambda (1+\alpha )} - \frac{n\alpha }{\lambda (1+\alpha )r_0}\Big ) &{} \frac{r_0}{n}(\frac{\alpha r_0}{\lambda })\\ s_0(\lambda r_0) &{} s_0 &{} s_0\Big (\frac{\alpha -m-n}{\lambda (1+\alpha )} + \frac{n}{\lambda (1+\alpha )r_0}\Big ) &{} s_0(-\frac{r_0}{\lambda }) \end{pmatrix}\\&\quad =\begin{pmatrix} 2 &{} 0 &{} 0 &{} 0 \\ br_0 &{} 1 &{} 0 &{} 0\\ \frac{r_0}{n}(\lambda r_0) &{} \frac{r_0}{n} &{} \frac{r_0}{n}\frac{1}{\lambda }\Big (1-s_0-\frac{n}{r_0}\Big ) &{} \frac{r_0}{n}(\frac{\alpha r_0}{\lambda })\\ s_0(\lambda r_0) &{} s_0 &{} s_0\frac{1}{\lambda }(1-s_0) &{} s_0(-\frac{r_0}{\lambda }) \end{pmatrix} \end{aligned}$$

The corresponding eigenvectors $X_{0j}$ are collected in the matrix $S_0$ as jth column vector, $j=1,2,3,4$.

$$\begin{aligned} \begin{aligned} S_0&= \begin{pmatrix} 1 &{} 0 &{} 0 &{} 0\\ br_0 &{} 1 &{} 0 &{} 0\\ y_1 &{} y_2 &{} 1 &{} y_4\\ z_1 &{} z_2 &{} z_3 &{}1 \end{pmatrix}, \quad \quad \begin{array}{l} \begin{pmatrix} y_1\\ z_1 \end{pmatrix} =-(\lambda +b)r_0\begin{pmatrix} \frac{ \frac{1+\alpha }{\lambda }r_0 + \frac{2}{s_0} }{ \Delta _1 }\\ \frac{ \frac{n}{r_0}\big (\frac{1}{\lambda } + 2\big ) }{ \Delta _1 } \end{pmatrix}, \quad \begin{pmatrix} y_2\\ z_2 \end{pmatrix} =-\begin{pmatrix} \frac{ \frac{1+\alpha }{\lambda }r_0 + \frac{\mu _{02}}{s_0} }{ \Delta _2 }\\ \frac{ \frac{n}{r_0}\big (\frac{1}{\lambda } + \mu _{02}\big ) }{ \Delta _2 } \end{pmatrix}\\ z_3=n\bigg (\frac{\frac{1-s_0}{\lambda }}{\frac{n r_0}{\lambda } + \frac{n\mu _{0}^+}{s_0}}\bigg ),\quad y_4=\frac{\frac{r_0}{\lambda }+\frac{\mu _0^-}{s_0}}{\frac{1-s_0}{\lambda }}, \end{array} \end{aligned} \end{aligned}$$

(69)

where $\Delta _1 = \frac{1-s_0}{\lambda }\big (\frac{1+\alpha }{\lambda }r_0 + \frac{2}{s_0}\big ) -\frac{n}{r_0} \big ( \frac{1}{\lambda } + 2\big )\big (\frac{r_0}{\lambda } + \frac{2}{s_0}\big )$ and $\Delta _2 = \frac{1-s_0}{\lambda }\big (\frac{1+\alpha }{\lambda }r_0 + \frac{1}{s_0}\big ) -\frac{n}{r_0} \big ( \frac{1}{\lambda } + 1\big )\big (\frac{r_0}{\lambda } + \frac{1}{s_0}\big )$. We find that $y_1,y_2,y_4<0$; $z_1,z_2,z_3 \sim {O}(n)$, provided n is sufficiently small.

Next, the coefficient matrix for the linearized system around $M_1$ is

$$\begin{aligned}&\begin{pmatrix} -\frac{1+m+n}{\alpha -m-n} &{}\quad 0 &{}\quad 0 &{}\quad 0\\ (b-\lambda )r_1 &{}\quad -1 &{}\quad 0 &{}\quad 0\\ \frac{r_1}{n}(\lambda r_1) &{}\quad \frac{r_1}{n} &{}\quad \frac{r_1}{n}\Big (\frac{\alpha -m-n}{\lambda (1+\alpha )} - \frac{n\alpha }{\lambda (1+\alpha )r_1}\Big ) &{}\quad \frac{r_1}{n}(\frac{\alpha r_1}{\lambda })\\ s_1(\lambda r_1) &{}\quad s_1 &{}\quad s_1\Big (\frac{\alpha -m-n}{\lambda (1+\alpha )} + \frac{n}{\lambda (1+\alpha )r_1}\Big ) &{}\quad s_1(-\frac{r_1}{\lambda }) \end{pmatrix}\\&\quad =\begin{pmatrix} -\frac{1+m+n}{\alpha -m-n} &{}\quad 0 &{}\quad 0 &{}\quad 0\\ (b-\lambda )r_1 &{}\quad -1 &{}\quad 0 &{}\quad 0\\ \frac{r_1}{n}(\lambda r_1) &{}\quad \frac{r_1}{n} &{}\quad \frac{r_1}{n}\frac{1}{\lambda }\Big (1-s_1-\frac{n}{r_1}\Big ) &{}\quad \frac{r_1}{n}(\frac{\alpha r_1}{\lambda })\\ s_1(\lambda r_1) &{}\quad s_1 &{}\quad s_1\frac{1}{\lambda }(1-s_1) &{}\quad s_1(-\frac{r_1}{\lambda }) \end{pmatrix} \end{aligned}$$

In what follows, we examine all possible cases: except for the case $\mu _{11}=\mu _{12}=-1$, four linearly independent eigenvectors are attained. In the exceptional case $\mu _{11}=\mu _{12}=-1$, the repeated eigenvalue $-1$ has geometric multiplicity which is strictly less that its algebraic multiplicity.

As to the eigenvectors, notice that the eigenvalues for $M_1$ (differently from those for $M_0$) have the chance to be repeated. The analysis below shows that, unless $\mu _{11}=\mu _{12}=-1$, four linearly independent eigenvectors are attained. If the exceptional case takes place, then we will supplement precisely one generalized eigenvector for the repeated eigenvalue $-1$.

Case 1$-\frac{1+m+n}{\alpha -m-n}\ne -1$; or$-\frac{1+m+n}{\alpha -m-n}= -1$but$b=\lambda$. This case yields four linearly independent eigenvectors. The eigenvectors $X_{1j}$ are collected in the matrix $S_1$ as jth column vector, $j=1,2,3,4$, and in the case of repeated eigenvalues the corresponding eigenvectors are understood as a basis for the associated subspace:

$$\begin{aligned} S_1&= \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ x_1 &{}\quad 1 &{}\quad 0 &{}\quad 0\\ y_1 &{}\quad y_2 &{}\quad 1 &{}\quad y_4\\ z_1 &{}\quad z_2 &{}\quad z_3 &{}\quad 1 \end{pmatrix}, \quad \quad \begin{array}{l} x_1= {\left\{ \begin{array}{ll} \frac{(b-\lambda )r_1}{1+\mu _{11}} &{} \text {if }\mu _{11}\ne -1,\\ 0 &{} \text {otherwise,} \end{array}\right. }\\ z_3=n\bigg (\frac{\frac{1-s_1}{\lambda }}{\frac{n r_1}{\lambda } + \frac{n\mu _{1}^+}{s_1}}\bigg ), \quad y_4=\frac{\frac{r_1}{\lambda }+\frac{\mu _1^-}{s_1}}{\frac{1-s_1}{\lambda }},\\ \end{array} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \begin{pmatrix} y_1\\ z_1 \end{pmatrix} =&{\left\{ \begin{array}{ll} -(\lambda r_1 + x_1)\begin{pmatrix} \frac{\lambda }{1-s_1}\\ 0 \end{pmatrix} &{} \text {if }\mu _{14}=\mu _{11},\\ -(\lambda r_1 + x_1) \begin{pmatrix} \frac{ \frac{1+\alpha }{\lambda }r_1 + \frac{\mu _{11}}{s_1} }{ \Delta _3 }\\ \frac{ \frac{n}{r_1}\big (\frac{1}{\lambda } + \mu _{11}\big ) }{ \Delta _3 } \end{pmatrix}&\text {otherwise,} \end{array}\right. } \quad \\ \begin{pmatrix} y_2\\ z_2 \end{pmatrix} =&{\left\{ \begin{array}{ll} -\begin{pmatrix} \frac{\lambda }{1-s_1}\\ 0 \end{pmatrix} &{} \text {if }\mu _{14}=\mu _{12},\\ -\begin{pmatrix} \frac{ \frac{1+\alpha }{\lambda }r_1 + \frac{\mu _{12}}{s_1} }{ \Delta _4 }\\ \frac{ \frac{n}{r_1}\big (\frac{1}{\lambda } + \mu _{12}\big ) }{ \Delta _4 } \end{pmatrix}&\text {otherwise,} \end{array}\right. } \end{aligned} \end{aligned}$$

(70)

where

$$\begin{aligned} \Delta _3&= \frac{1-s_1}{\lambda }\left( \frac{1+\alpha }{\lambda }r_1 + \frac{\mu _{11}}{s_1}\right) -\frac{n}{r_1} \left( \frac{1}{\lambda } + \mu _{11}\right) \left( \frac{r_1}{\lambda } + \frac{\mu _{11}}{s_1}\right) \\&=\frac{-n}{r_1s_1}\det \left[ \begin{pmatrix} \frac{r_1}{n}\left( \frac{1-s_1}{\lambda }-\frac{n}{\lambda r_1}\right) &{} \frac{r_1}{n}\frac{\alpha r_1}{\lambda }\\ s_1\frac{1-s_1}{\lambda } &{} -s_1\frac{r_1}{\lambda } \end{pmatrix} -\mu _{11}\text {I}\right] \ne 0, \\ \Delta _4&= \frac{1-s_1}{\lambda }\left( \frac{1+\alpha }{\lambda }r_1 + \frac{\mu _{12}}{s_1}\right) -\frac{n}{r_1} \left( \frac{1}{\lambda } + \mu _{12}\right) \left( \frac{r_1}{\lambda } + \frac{\mu _{12}}{s_1}\right) \\&=\frac{-n}{r_1s_1}\det \left[ \begin{pmatrix} \frac{r_1}{n}\big (\frac{1-s_1}{\lambda }-\frac{n}{\lambda r_1}\big ) &{} \frac{r_1}{n}\frac{\alpha r_1}{\lambda }\\ s_1\frac{1-s_1}{\lambda } &{} -s_1\frac{r_1}{\lambda } \end{pmatrix} -\mu _{12}\text {I}\right] \ne 0, \end{aligned}$$

respectively, for the corresponding cases.

Case 2$-\frac{1+m+n}{\alpha -m-n}= -1$and$b\ne \lambda$: for this case, $\mu _{11} = \mu _{12} = -1$ has algebraic multiplicity two, but its geometric multiplicity is one, so we replace the first column of $S_1$ by the generalized eigenvector $\big (\frac{1}{(b-\lambda )r_1}, 0, y_1', z_1'\big )^T$, where

$$\begin{aligned} \begin{aligned} \begin{pmatrix} y_1'\\ z_1' \end{pmatrix} ={\left\{ \begin{array}{ll} \begin{pmatrix} -\frac{\lambda }{1-s_1}\big (\frac{\lambda }{b-\lambda } -\frac{n}{r_1}z_2\big )\\ 0 \end{pmatrix} &{} \text {if }\mu _{14}=-1,\\ -\frac{\lambda }{b-\lambda } \begin{pmatrix} \frac{ \frac{1+\alpha }{\lambda }r_1 + \frac{\mu _{11}}{s_1} }{ \Delta _3 }\\ \frac{ \frac{n}{r_1}\big (\frac{1}{\lambda } + \mu _{11}\big ) }{ \Delta _3 } \end{pmatrix} + \frac{n}{r_1} \begin{pmatrix} \frac{ y_2\left( \frac{r_1}{\lambda } + \frac{\mu _{11}}{s_1}\right) + z_2\frac{\alpha r_1}{\lambda } }{ \Delta _3 }\\ \frac{ y_2\left( \frac{1-s_1}{\lambda }\right) + z_2\left( -\frac{1-s_1}{\lambda }+\frac{n}{r_1}\left( \frac{1}{\lambda }+\mu _{11}\right) \right) }{ \Delta _3 } \end{pmatrix}&\text {otherwise.} \end{array}\right. } \end{aligned} \end{aligned}$$

(71)

$$\begin{aligned} 0=&Mat_1 \begin{pmatrix} w\\ x\\ y\\ z \end{pmatrix} -\mu \begin{pmatrix} w\\ x\\ y\\ z \end{pmatrix}\\ =&\begin{pmatrix} (\mu _{11}-\mu ) w\\ (b-\lambda )r_1 w +(\mu _{12}-\mu )x\\ \frac{r_1}{n} \left[ \lambda r_1 w + x + \big (\frac{1-s_1}{\lambda } - \frac{n}{r_1}\big (\frac{1}{\lambda }+\mu \big )\big )y + \frac{\alpha r_1}{\lambda }z\right] \\ s_1 \left[ \lambda r_1 w + x + \big (\frac{1-s_1}{\lambda }\big )y -\big (\frac{r_1}{\lambda }+\frac{\mu }{s_1}\big )z\right] \end{pmatrix},\\ A\triangleq&\begin{pmatrix} \frac{1-s_1}{\lambda } - \frac{n}{r_1}\big (\frac{1}{\lambda }+\mu \big ) &{} \frac{\alpha r_1}{\lambda }\\ \frac{1-s_1}{\lambda } &{} -\big (\frac{r_1}{\lambda }+\frac{\mu }{s_1}\big ) \end{pmatrix} \begin{pmatrix} y\\ z \end{pmatrix} = -(\lambda r_1 w +x)\begin{pmatrix} 1\\ 1\end{pmatrix}\\ A^{-1} =&\frac{1}{\Delta } \begin{pmatrix} \big (\frac{r_1}{\lambda }+\frac{\mu }{s_1}\big ) &{} \frac{\alpha r_1}{\lambda }\\ \frac{1-s_1}{\lambda } &{} -\frac{1-s_1}{\lambda } + \frac{n}{r_1}\left( \frac{1}{\lambda }+\mu \right) \end{pmatrix}, \quad \\ \Delta =&\frac{1-s_1}{\lambda }\left( \frac{1+\alpha }{\lambda }r_1 + \frac{\mu }{s_1}\right) - \frac{n}{r_1}\left( \frac{1}{\lambda } +\mu \right) \left( \frac{r_1}{\lambda }+\frac{\mu }{s_1}\right) \end{aligned}$$