Intellectual property rights and R&D subsidies: are they complementary policies?

Abstract

IPR protection and R&D subsidy are simultaneously implemented in many economies. Are they complementary policies for improving the welfare of consumers? We address this question in a dynamic general equilibrium model with innovation-driven growth. Under concave utility, the answer is yes for two cases: (1) the economy does not begin from steady state and the elasticity of intertemporal substitution (EIS) is relatively large; (2) the economy begins from steady state with either a sufficiently small initial consumption and a relatively large EIS or a sufficiently big initial consumption and a relatively small EIS. Under linear utility, the answer is yes if the discounted lifetime utility is finite in equilibrium, no matter the economy begins from the steady state or not. As empirical evidence finds cross-country heterogeneity in EIS, they are not complementary for all economies. We also identify reasonable cases whereby they are substitute policies, so we show when it is not welfare-enhancing to simultaneously implement both policies.

Appendix: Proofs

Proof of Lemma 3.1:

For problem (19), the Hamiltonian can be written as

$$\begin{aligned} H\left[ t,C(t);\lambda (t) \right] =e^{-\rho t}u\left[ C(t) \right] +\lambda (t)\left[ rB(t)+w(t)L-C(t) \right] , \end{aligned}$$

where $\lambda (t)>0$ denotes the costate variable. Then we get the following first-order necessary conditions:

$$\begin{aligned} \frac{\partial H}{\partial C}= & {} e^{-\rho t}u'\left[ C(t) \right] -\lambda (t)=0, \end{aligned}$$

(24)

$$\begin{aligned} \frac{\partial H}{\partial B}= & {} r\lambda (t)=-\dot{\lambda }(t), \end{aligned}$$

(25)

in which (24) gives the consumption Euler equation. By using (24), we have $\lambda (0)=u'\left[ C(0) \right]$, which combined with (25) imply that $\lambda (t)=u'\left[ C(0) \right] e^{-rt}$. Plugging this $\lambda (t)$ in (24) gives the desired equation in Lemma 3.1. To complete the proof, let us take log on both sides of the consumption Euler equation, then we have $-\rho t +\ln u'\left[ C(t) \right] =\ln \lambda (t)$. Thus, under Assumption 3.1, differentiating both sides of this equation with respect to t and also applying (25), we obtain

$$\begin{aligned} -\frac{u''\left[ C(t) \right] }{u'\left[ C(t) \right] }\dot{C}(t)=r-\rho , \end{aligned}$$

by which the remaining parts of Lemma 3.1 easily follow. $\square$

Proof of Theorem 3.1:

By using Assumption 3.1 and (20), we have

$$\begin{aligned} \frac{\partial ^{2}W(\mu ,s)}{\partial \mu \partial s}&= \int _{0}^{\infty }e^{-\rho t}\left\{ u''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu } \frac{\partial C^{*}(t)}{\partial s} + u'\left[ C^{*}(t) \right] \frac{\partial ^{2} C^{*}(t)}{\partial \mu \partial s} \right\} dt \nonumber \\&= \int _{0}^{\infty }e^{-\rho t}\left\{ \frac{\partial ^{2} u\left[ C^{*}(t) \right] }{\partial \mu \partial s}\right\} dt. \end{aligned}$$

(26)

As we have from Lemma 3.1 that $u'\left[ C^{*}(t) \right] =u'\left[ C(0) \right] e^{(\rho -r)t}$, applying Implicit Function Theorem to this equation yields

$$\begin{aligned} u''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu } = u''\left[ C(0) \right] e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu }-tu'\left[ C^{*}(t) \right] \frac{\partial r}{\partial \mu } \end{aligned}$$

(27)

and

$$\begin{aligned} u''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial s} = u''\left[ C(0) \right] e^{(\rho -r)t}\frac{\partial C(0)}{\partial s}-tu'\left[ C^{*}(t) \right] \frac{\partial r}{\partial s}. \end{aligned}$$

(28)

Applying Implicit Function Theorem again and differentiating both sides of equation (27) with respect to s, we then have

$$\begin{aligned}&u'''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu } \frac{\partial C^{*}(t)}{\partial s} + u''\left[ C^{*}(t) \right] \frac{\partial ^{2} C^{*}(t)}{\partial \mu \partial s}\nonumber \\&\quad = u'''\left[ C(0) \right] e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu }\frac{\partial C(0)}{\partial s} -tu''\left[ C(0) \right] e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu }\frac{\partial r}{\partial s}\nonumber \\&\quad \quad +u''\left[ C(0) \right] e^{(\rho -r)t}\frac{\partial ^{2} C(0)}{\partial \mu \partial s} -tu''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial s} \frac{\partial r}{\partial \mu }-tu'\left[ C^{*}(t) \right] \frac{\partial ^{2} r}{\partial \mu \partial s}. \end{aligned}$$

(29)

By using (11)–(12), we get

$$\begin{aligned} \frac{\partial r}{\partial \mu }=-1<0, \ \frac{\partial r}{\partial s}=\frac{\pi }{b}(1-s)^{-2}>0 \ \ \text {and} \ \ \frac{\partial ^{2} r}{\partial \mu \partial s}=0. \end{aligned}$$

(30)

In what follows, we shall consider two cases. First, if $\zeta (0)\ne \bar{\zeta }$, then we get from (16)–(18) that

$$\begin{aligned} \frac{\partial C(0)}{\partial \mu }=\frac{\partial C(0)}{\partial s}= \frac{\partial ^{2} C(0)}{\partial \mu \partial s}=0. \end{aligned}$$

(31)

Applying (30)–(31) to (27) yields that

$$\begin{aligned} u''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu } = -tu'\left[ C^{*}(t) \right] \frac{\partial r}{\partial \mu }\ge 0 \end{aligned}$$

(32)

for $\forall t\ge 0$. Using Assumption 3.1, we thus have $\partial C^{*}(t)/\partial \mu \le 0$. Similarly, applying (30)–(31) to (28) yields that

$$\begin{aligned} u''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial s} = -tu'\left[ C^{*}(t) \right] \frac{\partial r}{\partial s}\le 0 \end{aligned}$$

(33)

for $\forall t\ge 0$. We thus get from using Assumption 3.1 that $\partial C^{*}(t)/\partial s \ge 0$. Based on these results, applying (30)–(31) to (29) shows that

$$\begin{aligned} u'''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu } \frac{\partial C^{*}(t)}{\partial s} + u''\left[ C^{*}(t) \right] \frac{\partial ^{2} C^{*}(t)}{\partial \mu \partial s} = -tu''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial s} \frac{\partial r}{\partial \mu }\le 0 \end{aligned}$$

(34)

for $\forall t\ge 0$. Solving for $\partial C^{*}(t)/\partial \mu$ from (32) and plugging it in (34) leads to

$$\begin{aligned} \frac{\partial ^{2} C^{*}(t)}{\partial \mu \partial s} = t\left( \frac{u'''\left[ C^{*}(t)\right] u'\left[ C^{*}(t)\right] }{\left\{ u''\left[ C^{*}(t)\right] \right\} ^{2}} -1 \right) \frac{\partial C^{*}(t)}{\partial s} \frac{\partial r}{\partial \mu }. \end{aligned}$$

(35)

In consequence, using (26), (32) and (35), we can obtain

$$\begin{aligned} \frac{\partial ^{2} u\left[ C^{*}(t) \right] }{\partial \mu \partial s} = \left( \frac{u'''\left[ C^{*}(t)\right] u'\left[ C^{*}(t)\right] }{\left\{ u''\left[ C^{*}(t)\right] \right\} ^{2}} -2 \right) \underset{\le 0}{\underbrace{tu'\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial s} \frac{\partial r}{\partial \mu } }} \end{aligned}$$

(36)

for $\forall t\ge 0$. Therefore, using (36) produces part (i) of Theorem 3.1.

Second, if $\zeta (0)=\bar{\zeta }$, then we get from (16)–(18) and (9) that

$$\begin{aligned} \frac{\partial C(0)}{\partial \mu }=\frac{-\eta _{1}g}{(g+\mu )^{2}}>0 \ \ \text {and} \ \ \frac{\partial C(0)}{\partial s}= \frac{\partial ^{2} C(0)}{\partial \mu \partial s}=0. \end{aligned}$$

(37)

Applying (30) and (37) to (27) gives rise to

$$\begin{aligned} u''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu } = \underset{<0}{\underbrace{u''\left[ C(0) \right] e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu }}} - \underset{\le 0}{\underbrace{tu'\left[ C^{*}(t) \right] \frac{\partial r}{\partial \mu } }}, \end{aligned}$$

(38)

thus the sign of $\partial C^{*}(t)/\partial \mu$ is not immediate right now. Also, applying (30) and (37) to (29) gives rise to

$$\begin{aligned} u'''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu } \frac{\partial C^{*}(t)}{\partial s} + u''\left[ C^{*}(t) \right] \frac{\partial ^{2} C^{*}(t)}{\partial \mu \partial s} =&-tu''\left[ C(0) \right] e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu }\frac{\partial r}{\partial s}\\&-\,tu''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial s} \frac{\partial r}{\partial \mu }, \end{aligned}$$

by which we obtain

$$\begin{aligned} u'\left[ C^{*}(t) \right] \frac{\partial ^{2} C^{*}(t)}{\partial \mu \partial s} =&- \frac{u'''\left[ C^{*}(t) \right] u'\left[ C^{*}(t) \right] }{u''\left[ C^{*}(t) \right] } \frac{\partial C^{*}(t)}{\partial \mu } \frac{\partial C^{*}(t)}{\partial s} \nonumber \\&-t\frac{u''\left[ C(0) \right] u'\left[ C^{*}(t) \right] }{u''\left[ C^{*}(t) \right] } e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu }\frac{\partial r}{\partial s}\nonumber \\&-tu'\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial s} \frac{\partial r}{\partial \mu }. \end{aligned}$$

(39)

It follows from (38) that

$$\begin{aligned} u''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu }\frac{\partial C^{*}(t)}{\partial s}= & {} \underset{<0}{\underbrace{u''\left[ C(0) \right] e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu }\frac{\partial C^{*}(t)}{\partial s}}}\nonumber \\&- \underset{\le 0}{\underbrace{tu'\left[ C^{*}(t) \right] \frac{\partial r}{\partial \mu } \frac{\partial C^{*}(t)}{\partial s}}}. \end{aligned}$$

(40)

Solving for $\partial C^{*}(t)/\partial s$ from (33) gives

$$\begin{aligned} \frac{\partial C^{*}(t)}{\partial s} = -\,t \frac{u'\left[ C^{*}(t) \right] }{u''\left[ C^{*}(t) \right] } \frac{\partial r}{\partial s}\ge 0. \end{aligned}$$

(41)

Solving for $\partial C^{*}(t)/\partial \mu$ from (38) gives

$$\begin{aligned} \frac{\partial C^{*}(t)}{\partial \mu } = \frac{u''\left[ C(0) \right] }{u''\left[ C^{*}(t) \right] } e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu }-t \frac{u'\left[ C^{*}(t) \right] }{u''\left[ C^{*}(t) \right] }\frac{\partial r}{\partial \mu }. \end{aligned}$$

(42)

Plugging (41) in (40) shows that

$$\begin{aligned} u''\left[ C^{*}(t) \right] \frac{\partial C^{*}(t)}{\partial \mu }\frac{\partial C^{*}(t)}{\partial s}= & {} -\,t\frac{u'\left[ C^{*}(t) \right] u''\left[ C(0) \right] }{u''\left[ C^{*}(t) \right] } e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu } \frac{\partial r}{\partial s}\nonumber \\&+\,t^{2}\frac{\left\{ u'\left[ C^{*}(t) \right] \right\} ^{2}}{u''\left[ C^{*}(t) \right] }\frac{\partial r}{\partial \mu }\frac{\partial r}{\partial s}. \end{aligned}$$

(43)

Using (41) and (42), we obtain

$$\begin{aligned} \frac{\partial C^{*}(t)}{\partial \mu }\frac{\partial C^{*}(t)}{\partial s}= & {} -\,t\frac{u'\left[ C^{*}(t) \right] u''\left[ C(0) \right] }{\left\{ u''\left[ C^{*}(t) \right] \right\} ^{2}} e^{(\rho -r)t}\frac{\partial C(0)}{\partial \mu } \frac{\partial r}{\partial s}\nonumber \\&+\, t^{2} \left\{ \frac{u'\left[ C^{*}(t) \right] }{u''\left[ C^{*}(t) \right] } \right\} ^{2}\frac{\partial r}{\partial \mu } \frac{\partial r}{\partial s}. \end{aligned}$$

(44)

Applying (39), (43) and (44) to (26), we thus have

$$\begin{aligned} \frac{\partial ^{2} u\left[ C^{*}(t) \right] }{\partial \mu \partial s} =&\left\{ \frac{u''\left[ C(0) \right] e^{(\rho -r)t}\eta _{1}g}{(g+\mu )^{2}}-tu'\left[ C^{*}(t) \right] \right\} \nonumber \\&\quad \times \left( \frac{u'''\left[ C^{*}(t)\right] u'\left[ C^{*}(t)\right] }{\left\{ u''\left[ C^{*}(t)\right] \right\} ^{2}} -2 \right) \underset{\le 0}{\underbrace{t\frac{\partial r}{\partial s} \frac{\partial r}{\partial \mu } }}\underset{<0}{\underbrace{\frac{u'\left[ C^{*}(t) \right] }{u''\left[ C^{*}(t) \right] }}} \end{aligned}$$

(45)

for $\forall t\ge 0$. Noting from Lemma 3.1 that

$$\begin{aligned} \frac{u''\left[ C(0) \right] \eta _{1}g}{(g+\mu )^{2}}-tu'\left[ C^{*}(t) \right] =\left\{ \underset{>0}{\underbrace{\frac{u''\left[ C(0) \right] }{u'\left[ C(0) \right] }\frac{\eta _{1}g}{(g+\mu )^{2}} }} -t \right\} \underset{>0}{\underbrace{u'\left[ C(0) \right] e^{(\rho -r)t} }}, \end{aligned}$$

thus this combined with (45) produces the desired results in part (ii). $\square$

Proof of Corollary 3.2:

As shown in the proof of Theorem 3.1, we just need to consider two cases. First, if $\zeta (0)\ne \bar{\zeta }$, then the proof is exactly as that of Theorem 3.1 because the corresponding proof does not rely on whether or not g is completely independent of the IPR policy variable $\mu$. Second, if $\zeta (0)=\bar{\zeta }$, then we have

$$\begin{aligned} \frac{\partial \bar{\zeta }}{\partial \mu }=\left( \frac{1}{g+\mu } \right) ^{2}\left( \mu \frac{\partial g}{\partial \mu }-g \right) <0 \end{aligned}$$

whenever $\partial g/ \partial \mu <\frac{g}{\mu }$. Therefore, similar to (37), we arrive at

$$\begin{aligned} \frac{\partial C(0)}{\partial \mu }= \underset{<0}{\underbrace{\eta _{1}}} \cdot \underset{<0}{\underbrace{\frac{\partial \bar{\zeta }}{\partial \mu }}}>0. \end{aligned}$$

Since it is easy to verify that the rest of the proof is unchanged under the current assumptions, we claim that Theorem 3.1 holds as well. $\square$

Proof of Lemma 3.2:

By applying Lemma 3.1, we see that $C^{*}(t)=C(0)e^{(r-\rho )t}$ under the log utility with $\text {EIS}=1$ assumed in part (i) and $C^{*}(t)=C(0)e^{\sigma ^{-1}(r-\rho )t}$ under the more general CRRA utility with $\text {EIS}\ne 1$ assumed in part (ii). So, given log preference and (20), we obtain

$$\begin{aligned} W(\mu ,s)=\int _{0}^{\infty }\ln C^{*}(t)dt=\ln C(0)\int _{0}^{\infty }e^{-\rho t}dt +(r-\rho )\int _{0}^{\infty }e^{-\rho t}tdt. \end{aligned}$$

Thus, making use of L’Hospital Rule and the Formula of Integration by Parts gives (22), as desired in (i). Similarly, noting that

$$\begin{aligned} W(\mu ,s)= \frac{C(0)^{1-(1/\sigma )}}{1-(1/\sigma )}\int _{0}^{\infty } \exp \left\{ \left[ (\sigma -1)(r-\rho )-\rho \right] t \right\} dt-\frac{\sigma }{(\sigma -1)\rho } \end{aligned}$$

under the more general CRRA preference, thus (23) can be accordingly derived. $\square$

Proof of Proposition 3.1:

By using (11)–(12), we get

$$\begin{aligned} \frac{\partial r}{\partial \mu }=-1<0, \ \frac{\partial r}{\partial s}=\frac{\pi }{b}(1-s)^{-2}>0 \ \ \text {and} \ \ \frac{\partial ^{2} r}{\partial \mu \partial s}=0. \end{aligned}$$

(46)

If $\zeta (0)\ne \bar{\zeta }$, then we get from (16)–(18) that

$$\begin{aligned} \frac{\partial C(0)}{\partial \mu }=\frac{\partial C(0)}{\partial s}= \frac{\partial ^{2} C(0)}{\partial \mu \partial s}=0. \end{aligned}$$

(47)

If $\zeta (0)=\bar{\zeta }$, then we get from (16)–(18), (46) and (9) that

$$\begin{aligned} \frac{\partial C(0)}{\partial \mu }=\frac{-\eta _{1}g}{(g+\mu )^{2}}=\frac{\eta _{1}g}{(g+\mu )^{2}}\frac{\partial r}{\partial \mu }>0 \ \ \text {and} \ \ \frac{\partial C(0)}{\partial s}= \frac{\partial ^{2} C(0)}{\partial \mu \partial s}=0. \end{aligned}$$

(48)

First, as we see that C(0) and r are additively separated in (22), then a direct application of (46)–(48) yields the desired assertion in part (i). For (23), if $\zeta (0)\ne \bar{\zeta }$, then we use (46) and (47) to get

$$\begin{aligned} \frac{\partial ^{2}W(\mu ,s)}{\partial \mu \partial s}=\frac{2\sigma (\sigma -1)^{2}C(0)^{1-(1/\sigma )}}{(\sigma -1)[\sigma \rho -(\sigma -1)r]^{3}} \underset{<0}{\underbrace{\frac{\partial r}{\partial s}\frac{\partial r}{\partial \mu }}}, \end{aligned}$$

by which the assertion in part (ii-a) immediately follows. If, however, $\zeta (0)=\bar{\zeta }$, then we can use (46) and (48) to get

$$\begin{aligned} \frac{\partial ^{2}W(\mu ,s)}{\partial \mu \partial s}= \left[ \underset{<0}{\underbrace{\frac{\eta _{1}g}{(g+\mu )^{2}}}}+ \underset{>0}{\underbrace{\frac{2\sigma C(0)}{\sigma \rho -(\sigma -1)r}}} \right] \frac{(\sigma -1)C(0)^{-(1/\sigma )}}{[\sigma \rho -(\sigma -1)r]^{2}} \underset{<0}{\underbrace{\frac{\partial r}{\partial s}\frac{\partial r}{\partial \mu }}}, \end{aligned}$$

in which we can show that

$$\begin{aligned} \underset{<0}{\underbrace{\frac{\eta _{1}g}{(g+\mu )^{2}}}}+ \underset{>0}{\underbrace{\frac{2\sigma C(0)}{\sigma \rho -(\sigma -1)r}}} \ge 0 \ \ \Leftrightarrow \ \ C(0) \ge \underset{>0}{\underbrace{\frac{-\eta _{1}g[\sigma \rho -(\sigma -1)r]}{2\sigma (g+\mu )^{2}}}}, \end{aligned}$$

the desired assertion in part (ii-b) accordingly follows. $\square$

Proof of Lemma 3.3:

Under Assumption 3.2, the Hamiltonian of problem (19) can be written as

$$\begin{aligned} H\left[ t,C(t);\lambda (t) \right] =e^{-\rho t} C(t)+\lambda (t)\left[ rB(t)+w(t)L-C(t) \right] , \end{aligned}$$

where $\lambda (t)>0$ denotes the costate variable. Then we get the first-order necessary conditions:

$$\begin{aligned} \frac{\partial H}{\partial C}= & {} e^{-\rho t}-\lambda (t)=0, \end{aligned}$$

(49)

$$\begin{aligned} \frac{\partial H}{\partial B}= & {} r\lambda (t)=-\dot{\lambda }(t). \end{aligned}$$

(50)

It follows from (49) that $\dot{\lambda }(t)/\lambda (t)=-\rho$, which combined with (50) produces the desired result. $\square$

Proof of Theorem 3.2:

It follows from Lemma 3.3, (4), (11) and (12) that $\pi b^{-1}(1-s)^{-1}-\mu =\rho$, by which we can apply the Implicit Function Theorem to get $\partial \mu /\partial s>0$. It is easy to see from (9), (15)–(18) and Lemma 3.3 that

$$\begin{aligned} C^{*}(t)=(\eta _{1}\bar{\zeta }+\eta _{2})e^{gt}+\eta _{1}\left[ \zeta (0)-\bar{\zeta } \right] e^{-\mu t}, \end{aligned}$$

by which, (20) and Assumption 3.2 we thus obtain

$$\begin{aligned} W(\mu , s)=\frac{\eta _{1}\bar{\zeta }+\eta _{2}}{\rho -g}+\frac{\eta _{1}\left[ \zeta (0)-\bar{\zeta } \right] }{\rho +\mu }. \end{aligned}$$

(51)

We next analyze two cases. First, we consider the simpler case with $\zeta (0)=\bar{\zeta }$, then we get from (17), (51) and Assumption 3.2 that

$$\begin{aligned} \text {sgn}\left\{ \frac{\partial ^{2} W(\mu , s)}{\partial \mu \partial s} \right\} = - \text {sgn}\left\{ \frac{\partial ^{2} \bar{\zeta }}{\partial \mu \partial s} \right\} . \end{aligned}$$

(52)

Noting that

$$\begin{aligned} \frac{\partial ^{2} \bar{\zeta }}{\partial \mu \partial s} = \frac{2g}{(g+\mu )^{3}}\frac{\partial \mu }{\partial s}>0, \end{aligned}$$

(53)

thus the desired assertion immediately follows from applying (52). Second, if $\zeta (0)\ne \bar{\zeta }$, then we can rewrite (51) as

$$\begin{aligned} W(\mu , s)=\underset{<0}{\underbrace{\frac{\eta _{1}(\mu +g)}{(\rho -g)(\rho +\mu )}}}\bar{\zeta }+ \frac{\eta _{2}}{\rho -g}+\frac{\eta _{1} \zeta (0)}{\rho +\mu }, \end{aligned}$$

where we have used (17) and Assumption 3.2. We thus have

$$\begin{aligned} \frac{\partial W}{\partial \mu }=\frac{\eta _{1}\bar{\zeta }}{(\rho -g)(\rho +\mu )} - \frac{\eta _{1}\bar{\zeta }(g+\mu )}{(\rho -g)(\rho +\mu )^{2}} +\frac{\eta _{1}(g+\mu )}{(\rho -g)(\rho +\mu )}\frac{\partial \bar{\zeta }}{\partial \mu } - \frac{\eta _{1}\zeta (0)}{(\rho +\mu )^{2}}, \end{aligned}$$

by which we arrive at:

$$\begin{aligned} \frac{\partial ^{2} W}{\partial \mu \partial s} \ \ =&\ \ \frac{\eta _{1}}{(\rho -g)(\rho +\mu )}\frac{\partial \bar{\zeta }}{\partial s} - \frac{\eta _{1}\bar{\zeta }}{(\rho -g)(\rho +\mu )^{2}}\frac{\partial \mu }{\partial s} - \frac{\eta _{1}(g+\mu )}{(\rho -g)(\rho +\mu )^{2}}\frac{\partial \bar{\zeta }}{\partial s} \nonumber \\&- \frac{\eta _{1}\bar{\zeta }}{(\rho -g)(\rho +\mu )^{2}}\frac{\partial \mu }{\partial s} + 2 \frac{\eta _{1}\bar{\zeta }(g+\mu )}{(\rho -g)(\rho +\mu )^{3}}\frac{\partial \mu }{\partial s}\nonumber \\&+\frac{\eta _{1}}{(\rho -g)(\rho +\mu )}\frac{\partial \mu }{\partial s}\frac{\partial \bar{\zeta }}{\partial \mu } + \frac{\eta _{1}(g+\mu )}{(\rho -g)(\rho +\mu )}\frac{\partial ^{2} \bar{\zeta }}{\partial \mu \partial s}\nonumber \\&- \frac{\eta _{1}(g+\mu )}{(\rho -g)(\rho +\mu )^{2}}\frac{\partial \mu }{\partial s}\frac{\partial \bar{\zeta }}{\partial \mu }+2\frac{\eta _{1}\zeta (0)}{(\rho +\mu )^{3}}\frac{\partial \mu }{\partial s}. \end{aligned}$$

(54)

Substituting

$$\begin{aligned} \frac{\partial \bar{\zeta }}{\partial s}= -\frac{g}{(g+\mu )^{2}}\frac{\partial \mu }{\partial s} \ \ \text {and} \ \ \frac{\partial \bar{\zeta }}{\partial \mu }= -\frac{g}{(g+\mu )^{2}} \end{aligned}$$

into equation (54) and rearranging the algebra give rise to

$$\begin{aligned} \frac{\partial ^{2} W}{\partial \mu \partial s} \ = \ \underset{<0}{\underbrace{\frac{2\eta _{1}}{(\rho +\mu )^{3}}}}\cdot \underset{>0}{\underbrace{\frac{\partial \mu }{\partial s}}} \cdot \underset{>0}{\underbrace{\left[ \frac{g}{\rho -g}+\zeta (0) \right] }}, \end{aligned}$$

as desired. $\square$