Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider an equation of the form $$x(t) = A_+ e^{-\Gamma_+ \; t}+A_- e^{-\Gamma_{-}\; t}$$

$A_{\pm},\Gamma_{\pm}$ are real constants, both $\Gamma_+$ and $\Gamma_-$ are greater than zero and $t\geq 0$.

Question is to prove that it has only one solution. I tried taking the derivative and equating to zero, but couldn't see any way to prove.

Additional Note: The above question is abstracted from a physical problem. The expression I wrote for $x(t)$ denotes displacement as a function of time, and this particular form is for an "overdamped" oscillator. The actual question asks that "Prove that an overdamped oscillator crosses the equilibriam position only once." Where $x=0$ is the equilibrium position.

$A_{\pm},\Gamma_{\pm}$ are real constants. $t$ is real, and as it is time, $t\geq 0$

share|cite|improve this question
Which variable are you solving for? – Rahul Aug 15 '11 at 3:13
@Rahul t The rest are just constants – kuch nahi Aug 15 '11 at 3:14
What do you mean by a "solution"? Do you want to solve $x(t) = 0$? And what are you assuming about $A_+$ and $A_-$? – Robert Israel Aug 15 '11 at 3:15
This is way too unclear. Are you trying to show the function $x(t)$ has a unique inverse? A vanishing derivative would make that impossible, not to mention $x(t)$ is obviously not constant anyhow. If $\Gamma_{\pm}$ are integers then $x$ is a polynomial in $e^{-t}$, making $t$ generally not unique. – anon Aug 15 '11 at 3:22
@anon I have added an edit explaining the question more. – kuch nahi Aug 15 '11 at 3:25
up vote 1 down vote accepted

Set it equal to zero and then solve

$$A_+ e^{-\Gamma_+ t} +A_- e^{-\Gamma_- t}=0$$

$$ e^{(\Gamma_+-\Gamma_-)t}=- A_+/A_- $$

The right side must be positive for there to be a real solution at all, so we'll assume that. Note also how $\Gamma_+-\Gamma_-$ can be written. Then


is the unique solution, because the exponential function has a single-valued inverse when considered strictly over real numbers.

share|cite|improve this answer

I start with:

$$0 = Ae^{-Bt}+Ce^{-Dt}$$

Then we have $Ae^{-Bt} = -Ce^{-Dt}$. Suppose WLOG that $B\geq D$. Then we have $\dfrac{Ae^{-Bt}}{e^{-Dt}} = Ae^{-(B+D)t} = -C$

Can you prove that a generic exponential function is injective? (this would prevent there from being more than one solution - but it is NOT surjective, so there is no promise that there is a solution).

share|cite|improve this answer
@Christian: Yes, that is the idea. But will you remove your comment? I intended this answer as a hint, and therefore incomplete. – mixedmath Aug 15 '11 at 9:10

You mean the question is to prove that $x(t) = 0$ has only one solution. And actually, there is at most one solution.

If $A_+$ and $A_-$ have the same sign, then so do both terms in the definition of $x(t)$, and there is no solution for $x(t) = 0$.

If $A_+$ and $A_-$ have opposite signs, then you want to solve $\lvert A_+ \rvert e^{-\Gamma_+t} = \lvert A_- \rvert e^{-\Gamma_-t}$. Take logs of both sides; you get a linear equation in $t$, and you're done.

share|cite|improve this answer
Then there is the case $A_+ = A_- = 0$. Note the word "crosses" in the question... – Robert Israel Aug 15 '11 at 4:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.