How come $e^x-e^{-x}=0$ does not have a solution?

Question

While solving a partial differential equation following this document, they state that

$$e^{\sigma L}-e^{-\sigma L}=0$$

does not have a solution and ask why. Here $L$ is a constant and $\sigma$ is a variable.

I'm not sure why this equation does not have a solution because when I plot $e^x-e^{-x}=0$ it crosses the $x$-axis. Any help would be appreciated.

Answer 1

On page 13 they clearly state that they consider first the case

• $\lambda <0 \Rightarrow \sigma = \sqrt{-\lambda} > 0$.

For $\sigma > 0$ the given equation does not have a solution.

Note, that the given equation is equivalent to $e^{2\sigma L} = 1 \Rightarrow \sigma = 0$, which is excluded in the given case.

Answer 2

Your document assumes $\sigma > 0$ so the only real solution $\sigma = 0$ is not admissible.

Answer 3

If: $$e^x-e^{-x}=0$$ $$e^x=e^{-x}$$ so the general solution is: $$x=-x$$ $$\therefore x=0$$ However, if there was a limitation such as $x>0$ then there would be no solution (in this range)

Answer 4

The unknown is $\sigma$. Immediately, you get that $\sigma=0$ is the only solution.

Answer 5

Perhaps they mean no non-trivial solution: if you balance the equation into a left and right side and take the logarithm, you are looking for the solution of $\sigma=-\sigma$, which for $\sigma\in\mathbb{R}$ is just zero.

Answer 6

In any other instance, that equation does have a solution, $\sigma=0$. However, note that right above in your document, they stipulate that $\sigma>0$. As $0$ is not greater than itself, there is no $\sigma$ which solves that equation for the values of interest.