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While solving a system of differential equations of water flow problems, to find the time for the concentration to fall below $0.00005$ kg/L, I came across the following algebraic equation $$\frac{0.025}{10\sqrt{5}}=e^{\frac{-3+\sqrt{5}}{100}t}-e^{\frac{-3-\sqrt{5}}{100}t}.$$ which I have to solve for the value of $t$. But I am facing difficulty in solving it.

I have tried solving this with Wolfram Alpha as well, but to no avail.

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    $\begingroup$ As an approximation: if the time $t$ that you're trying to find is large, then $e^{(-3-\sqrt{5})/100*t} \simeq 0$, and so you only have to solve a regular exponential equation. $\endgroup$
    – fcz
    Jul 20, 2021 at 12:13
  • $\begingroup$ WA gives two roots, $~0.025018769815917688569$ and $~889.6319702743134113$. $\endgroup$
    – g.kov
    Jul 25, 2021 at 6:37
  • $\begingroup$ For the smaller root, you could write $bc=e^{-at}\sinh(bt)\approx e^{-at}bt$ so that $-ac=(-at)e^{(-at)}$ can be solved via the Lambert-W function. Or less exactly, use $bc\approx\frac{bt}{1+at}$ which leads to a linear equation. $\endgroup$ Jul 25, 2021 at 7:57

2 Answers 2

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As @razivo showed, you cannot expect a closed form for the equation $$\frac{1}{400\sqrt 5} e^{\frac{3+\sqrt5}{100}t}=e^{\frac{\sqrt 5}{50}t}-1$$ Suppose that $t$ is large and neglect, for the time being, the $1$ in the rhs. This gives $$e^{\frac{3-\sqrt{5}}{100} t}\sim 400\sqrt 5\implies t_0 \sim \frac{100 \log \left(400 \sqrt{5}\right)}{3-\sqrt{5}}=889.632$$

Now, expand as a series around $t=t_0$ the function $$f(t)=\frac{1}{400\sqrt 5} e^{\frac{3+\sqrt5}{100}t}-e^{\frac{\sqrt 5}{50}t}+1$$ to get $$f(t)=f(t_0)+ f'(t_0)(t-t_0)+ O\left((t-t_0)^2 \right)\implies t_1=t_0-\frac {f(t_0)}{f'(t_0)}$$ which is the same as the first iteration of Newton method. But, $f'(t_0) \sim 1.45 \times 10^{15}$ so $t1\simeq t_0$.

To give you and idea, the exact solution is $889.63197027431341130$ while $ t_0=889.63197027431341199$

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  • $\begingroup$ Solution to OP is close to zero, about $0.025$. $\endgroup$
    – g.kov
    Jul 25, 2021 at 5:06
  • $\begingroup$ @g.kov. There are two roots; the other one is $0.014304$ $\endgroup$ Jul 25, 2021 at 5:19
  • $\begingroup$ I mean, the equation you've solved is not the same as that in the question. $\endgroup$
    – g.kov
    Jul 25, 2021 at 6:09
  • $\begingroup$ @g.kov. You are correct (sign error $-1$ instead of $+1$. Thanks & cheers. $\endgroup$ Jul 25, 2021 at 6:37
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not a full answer $$\frac{1}{400\sqrt{5}}=e^{\frac{-3+\sqrt{5}}{100}t}-e^{\frac{-3-\sqrt{5}}{100}t}=e^{-\frac{3}{100}t}e^{\frac{\sqrt5}{100}t}-e^{-\frac{3}{100}t}e^{-\frac{\sqrt5}{100}t}$$ Let $u=e^{-\frac{3}{100}t},v=e^{\frac{\sqrt5}{100}t}$ $$\frac{1}{400\sqrt 5}=uv-\frac{u}{v}\implies \frac{v}{400\sqrt5}=u(v^2-1)\implies \frac{v}{400\sqrt5(v^2-1)}=u$$ $$e^{-\frac{3}{100}t}=\frac{e^{\frac{\sqrt 5}{100}t}}{400\sqrt 5(e^{\frac{\sqrt 5}{50}t}-1)}$$ $$\frac{1}{400\sqrt 5} e^{\frac{3+\sqrt5}{100}t}=e^{\frac{\sqrt 5}{50}t}-1$$ this is the simplest form I was able to get it to.

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  • $\begingroup$ You put "plus" instead of "minus", it should be $\frac{1}{400\sqrt 5}=uv\mathbf{-}\frac{u}{v}$. $\endgroup$
    – g.kov
    Jul 25, 2021 at 4:56

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