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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, $t\geq 0$

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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
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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
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3 Answers

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

$$t=\frac{\ln(-A_+/A_-)}{2\sqrt{(\Gamma/2)^2-\omega_0^2}}$$

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

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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).

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@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
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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.

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