I have been trying to solve this diff. equation for quite some time now but haven't been able to do it correctly. It describes the lost height of the water $h$ at a certain time $t$ of a leaking reservoir. I have obtained this equation by using Torricelli's law as well as the law of continuity. At $t=0$, the seal is removed and the reservoir starts leaking. Moreover, $A_O > A_G$ and to be less precise it can be assumed $A_O >> A_G$, however, I would like to solve it as precisely as possible. The equation is:
$$\frac{2gA_O^2}{A_O^2 – A_ G^2}h(t)+\frac{A_O^2}{A_G^2}\bigg(\frac{dh(t)}{dt}\bigg)^2 – \frac{2gA_O^2}{A_O^2 – A_G^2}H_0=0$$ with $h(0)=0$. Additionally, $h(t)$ is bounded above by $H_0$, is a strictly increasing function and thus also $\lim_{t \to \inf} h(t)=H_0$
The constants:
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$A_O$ is the surface of the top of the reservoir.
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$A_G$ is the surface area of the hole.
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$g$ is the gravitational acceleration.
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$H_0$ is the height of the water at t=0.
The simplified formula would be:
$$\lambda h(t)+\mu\left(\frac{dh(t)}{dt}\right)^2-\lambda H_0=0$$
And just to be clear $\bigg(\frac{dh(t)}{dt}\bigg)^2$ is simply the first derivative squared not the second derivative.
I really hope someone can help me solve this!
If there is anything unclear or you want more information please ask I will check this post regularly. Thanks in advance!
Best Answer
First, write $h(t)=H_0+f(t)$ to get a slightly simpler equation $$ \frac{2gA_O^2}{A_O^2 - A_ G^2}f(t)+\frac{A_O^2}{A_G^2}\bigg(\frac{df(t)}{dt}\bigg)^2=0. $$ Writing $r=A_O/A_G$, we can simplify the constants and find $$ \frac{2g}{r^2 - 1}f(t)+\bigg(\frac{df(t)}{dt}\bigg)^2=0. $$ If $f'(t)=0$, the equation tells you $f(t)=0$.
I'll assume $r>1$ so that the equation forces $f\leq0$. This holds as $A_0>A_G$. The other case will be similar but I will stick to this choice as in the updated question.
If $f'(t)\neq0$, you get $$ f'(t) = \pm\sqrt{-\frac{2g}{r^2 - 1}f(t)} $$ or $$ \frac{df}{-\sqrt{\frac{2g}{r^2 - 1}f}} = \pm dt. $$ Integrating gives $$ \sqrt{\frac{r^2 - 1}{8g}} \sqrt{-f} = t_0\pm t $$ for some $t_0$, so $$ f(t) = -\frac{8g}{r^2 - 1} (t_0\pm t)^2. $$ As the time difference is squared anyway, you can rewrite this general solution as $$ f(t) = -\frac{8g}{r^2 - 1} (t-t_0)^2. $$ This gives you two kinds of solutions, and for $r<1$ you will get something similar.
Be careful with solutions when $f=0$ (which is at $t=t_0$). There the solution fails to be unique because the function can stop at the zero level. For example, $$ f(t) = \begin{cases} -\frac{8g}{r^2 - 1} t^2,&t<0\\ 0,&0\leq t\leq 1\\ -\frac{8g}{r^2 - 1} (t-1)^2,&t>1 \end{cases} $$ is a solution.
Your definition of $H_0$ requires that $f(0)=0$, as $f=0\iff h=H_0$. With your update I think you mean $h(0)=0$ instead of $h(0)=H_0$.
Let us then see what happens with your added assumption that $h(t)\leq H_0$, $h$ is strictly increasing and $\lim_{t\to\infty}h(t)=H_0$. With my notation this gives $f(t)\leq0$, $f'(t)\geq0$, and $\lim_{t\to\infty}f(t)=0$. The updated initial condition seems to be $f(0)=-H_0$. Let's start with $$ f(t) = -\frac{8g}{r^2 - 1} (t-t_0)^2. $$ Putting in the initial condition gives $$ -H_0 = f(0) = -\frac{8g}{r^2 - 1} t_0^2, $$ so $$ f(t) = -a (t-\sqrt{H_0/a})^2, $$ where $a=\frac{8g}{r^2 - 1}$. As $f$ should be increasing, we have $$ f(t) = \begin{cases} -a (t-\sqrt{H_0/a})^2 ,& 0\leq t\leq \sqrt{H_0/a}\\ 0 ,& t>\sqrt{H_0/a}. \end{cases} $$ The function is increasing but not strictly increasing; it reaches it's final value in finite time. Without the increasing assumption the solution can decide to "bounce back down" starting at any time after $t_0=\sqrt{H_0/a}$.