Determine how many solutions there are for this equation

Question

Determine how many solutions there are for $$e^{\frac{1}{x}}=\lambda x^2$$ for each $\lambda \in \mathbb{R}$.

I find really difficult to face this kind of problems so I'll show you where I get stuck hoping to get help.

My attempt: Let $f(x)=\lambda x^2-e^{\frac{1}{x}}$. As standard routine, computing zeros of $f$ gives us solutions to the equation. By simple computations we have:

$$\lim_{x\to 0^+}f(x)=-\infty \qquad \lim_{x\to \pm \infty}f(x)=sgn(\lambda)\infty \qquad \lim_{x\to 0^-}f(x)=0^{\pm}$$ So we note that we should pay attention to what happens near $0^-$ as it clearly influences the number of zeros of that function. Let's compute the first derivative:

$$f'(x)=2\lambda x + \frac{e^{\frac{1}{x}}}{2x^2}$$

And that's the point I always get stuck: I can't compute the sign of the derivative as it is "non standard". Can you give me an help in understanding how to do this kind of problems? If you want to see a related exercise I've done (and its solution) you can go here. What's the approach I should have when facing these things? Thanks in advance.

Answer 1

Let assume $y=\frac1x$ then

$$e^{\frac{1}{x}}=\lambda x^2\implies y^2e^y=\lambda$$

then we can study the function $f(y)=y^2e^y \implies f'(y)=y(y+2)e^y$ and then find the number of solutions for $f(y)=\lambda$.

Answer 2

If $\lambda>0$ then $f'$ is positive. Hence $f$ is strictly increasing on $\mathbb{R}^{*}$ and there exists one solution. Using that the limit in $ 0^{-}$ is $0$, you know that the solution is in $\mathbb{R}^{*+}$. You can show that in fact the solution is given in term of Lambert function.

If $\lambda<0$ then using that $\displaystyle e^{1/x}>0$ it cant be equal to $\lambda x^2<0$.

If $\lambda =0$, there's no solution too because of the exponentiel being strictly positive.

Answer 3

Making $y = \frac{1}{x}$ we have

$$ e^y = \frac{\lambda}{y^2}\Rightarrow e^{\frac{y}{2}}=\pm\frac{\sqrt{\lambda}}{y}\Rightarrow 2\left(\frac{y}{2}\right)e^{\frac{y}{2}}=\pm\sqrt{\lambda} $$

and then using the Lambert function

$$ \frac{y}{2} = W\left(\frac{\pm\sqrt{\lambda}}{2}\right) $$

and finally

$$ x = \frac{1}{2W\left(\frac{\pm\sqrt{\lambda}}{2}\right)} $$

Attached $x \times \lambda$. In blue the negative branch and in red the positive branch.

hence for $0 < \lambda \le \frac{4}{e^2}$ we have two solutions and for $\frac{4}{e^2} < \lambda < \infty$ we have one solution