Expressing solution to an inequality with Lambert W function

I'm new to Lambert functions, any ideas on how to solve this are welcome: $$\theta \rho^{\theta}+r \theta>v$$ where $\theta \in \mathbb{R}^{+}, -1<r,v<1, \ 0<\rho<1$. I've tried setting $\theta=\frac{r \theta}{r}$ to get

$$\theta \big( \rho^{r \theta} \big)^{\frac{1}{r}} + r \theta >v$$ Then set $\rho^{r \theta}=b,\theta=\frac{\log b}{r \log \rho}$ to get: $$b^{\frac{1}{r}} \log b + r \log b >r v \log \rho$$ Didn't get me anywhere though. I know solutions to equations of the type $a^z +z =b_1$ and $z a^z=b_2$. but this one is quite tricky for me.

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Since the left side is a continuous function of $\theta$ the solutions, if any, will be some open intervals whose endpoints can only be $0$, $\infty$ or solutions of $\theta \rho^\theta + r \theta = v$. That equation doesn't appear to be solvable using LambertW: at least, Maple 16 and Wolfram Alpha don't find closed form solutions.

For small $v$ there appears to be a power series solution:

$$\theta= \frac{v}{1+r}-{\frac {\ln \left( \rho \right) }{ \left( 1+r \right) ^{3}}}{v}^{2}-{\frac { \left( \ln \left( \rho \right) \right) ^{2} \left( r-3 \right) }{2 \left( 1+r \right) ^{ 5}}}{v}^{3}-{\frac { \left( \ln \left( \rho \right) \right) ^{3 } \left( -13\,r+16+{r}^{2} \right) }{6 \left( 1+r \right) ^{7}}}{v}^{4} -{\frac { \left( \ln \left( \rho \right) \right) ^{4} \left( - 39\,{r}^{2}+171\,r+{r}^{3}-125 \right) }{24 \left( 1+r \right) ^{9}}}{v} ^{5}+O \left( {v}^{6} \right)$$

Note that in this solution $\theta/v$ appears to be a function only of $r$ and $v \ln(\rho)$. This can be seen from the fact that under the scaling $\theta = v t$, $\rho = \exp(z/v)$, the equation simplifies to $t \exp(z t) + r t = 1$, and the series is the Taylor series of the solution of this in powers of $z$.

EDIT: The function $f(\theta) = \theta \rho^\theta + r \theta$ is concave for $0 \le \theta \le 2/\ln(1/\rho)$ and convex for $\theta \ge 2/\ln(1/\rho)$. $f'(0) = 1+r > 0$, while $f'(2/\ln(1/\rho)) = r - e^{-2}$ and $\lim_{\theta \to \infty} f(\theta)$ is $-\infty$, $0$ or $+\infty$ depending on whether $r < 0$, $r=0$ or $r > 0$. So there are the following cases:

• If $-1 < r < 0$, the graph of $f(\theta)$ looks like this:

If $v \le 0$, there is one positive solution $\theta_1$ of $f(\theta) = v$, and $f(\theta) > v$ if and only if $\theta < \theta_1$. This $\theta_1$, by the way, is not the power series solution I obtained above, which would be negative for small negative $v$.

If $0 < v < v_{max}$ (the maximum value of $f(\theta)$), there are two positive solutions $\theta_0 < \theta_1$ of $f(\theta)=v$, and $f(\theta) > v$ if $\theta_0 < \theta < \theta_1$. The power series solution I obtained above should be $\theta_0$, at least for small $v$.

If $v \ge v_{max}$, $f(\theta) > v$ has no solution.

• If $r = 0$, the graph looks like this:

Then if $v \le 0$, $f(\theta) > v$ for all $\theta > 0$, while if $0 < v < v_{max}$ we again have $f(\theta) > v$ for $\theta_0 < \theta < \theta_1$, and no solution if $v \ge v_{max}$.

• If $0 < r < e^{-2}$, the graph looks like this:

Here there is a local minimum value $v_{min}$ as well as a local maximum value $v_{max}$. There are up to three solutions $\theta_0$, $\theta_1$ and $\theta_2$ to $f(\theta) = v$.

If $v \le 0$, $f(\theta) > v$ for all $\theta > 0$.

If $0 < v \le v_{min},$f(\theta) > v$for$\theta > \theta_0$. If$v_{min} < v < v_{max}$,$f(\theta) > v$for$\theta_0 < \theta < \theta_1$or$\theta > \theta_2$. If$v > v_{max}$,$f(\theta) > v$for$\theta > \theta_2$. • If$r > e^{-2}$, the graph is increasing for all$\theta > 0$; if$v > 0$there is one solution$\theta_0$to$f(\theta) = v$, and$f(\theta) > v$for$\theta > \theta_0$. - The problem is, both$v$and$z$can be negative – Alex Oct 23 '12 at 1:06 Not if$0 < \rho < 1$. – Robert Israel Oct 23 '12 at 4:57 Can you give a bit more explanation if possible? I cn't get my head around, since this is an inequality and the answer should be something like 'LHS>RHS if$\theta > \varphi(\rho, r, v)\$. –  Alex Oct 23 '12 at 8:29
Sorry, didn't see the edit) now it is better –  Alex Oct 26 '12 at 5:26