MHB How do we get to this differential equation?

[evinda]

Hello! (Wave)

If we have the initial value problem

$$h''(t)=-\frac{R^2 g}{(h(t)+R)^2} \\ h(0)=0, h'(0)=V$$

we have the fundamental units: $T,L$ such that $T=[t], L=[h], L=[R], LT^{-1}=[V], LT^{-2}=[g]$ and we get the independent dimensionless quantities $\pi_1=\frac{h}{R}, \pi_2=\frac{t}{\frac{R}{V}}, \pi_3=\frac{V}{\sqrt{gR}}$

If we take $\pi_1=\frac{h}{R}$ and $\pi_2=\frac{t}{\frac{R}{V}}$ we have the scaling:

$$\overline{h}=\frac{h}{R}, \overline{t}=\frac{t}{\frac{R}{V}}$$

Then the initial value problem is written equivalently

$$\epsilon \overline{h}''(\overline{t})=-\frac{1}{(1+\overline{h}(\overline{t}))^2} \\ \overline{h}(0)=0, \overline{h}'(0)=1 \\ \text{ where } \epsilon=\frac{V^2}{Rg}$$

Could you explain me how we get to $\epsilon \overline{h}''(\overline{t})=-\frac{1}{(1+\overline{h}(\overline{t}))^2}$ ? (Thinking)

 

[I like Serena]

Hi! (Emo)

How about substituting the relationships between h and $\bar h$, t and $\bar t$, and g and $\bar g$ in the original equation? (Wondering)

 

[evinda]

Hi! (Emo)

How about substituting the relationships between h and $\bar h$, t and $\bar t$, and g and $\bar g$ in the original equation? (Wondering)

Is it as follows?

$h(t)= R \overline{h}\left( \frac{R}{V} \overline{t}\right) \\ h'(t)=\frac{\partial}{\partial{\overline{t}}} \left( R \overline{h} \left( \frac{R}{V} \overline{t} \right)\right) \frac{d \overline{t}}{dt} \\ h'(t)=\frac{V}{R} \frac{\partial}{\partial{\overline{t}}} \left( R \overline{h}\left( \frac{R}{V} \overline{t}\right) \right)\\ h''(t)=\frac{\partial}{\partial{\overline{t}}} \left( \frac{V}{R} \frac{\partial}{\partial{\overline{t}}}\left( R \overline{h}\left( \frac{R}{V} \overline{t}\right) \right)\right) \frac{V}{R} \\= \frac{V^2}{R^2} \frac{\partial^2}{\partial{\overline{t}}^2} \left( R \overline{h}\left( \frac{R}{V} \overline{t}\right) \right) = \frac{V^2}{R^2} R \overline{h}''\left( \frac{R}{V} \overline{t}\right) \frac{R^2}{V^2}= R \overline{h}'' \left( \frac{R}{V} \overline{t}\right)$Or have I done something wrong?

 

[I like Serena]

Is it as follows?

$h(t)= R \overline{h}\left( \frac{R}{V} \overline{t}\right) \\
h'(t)=\frac{\partial}{\partial{\overline{t}}} \left( R \overline{h} \left( \frac{R}{V} \overline{t} \right)\right) \frac{d \overline{t}}{dt}$

Or have I done something wrong?

I believe that should be:
$$h(t)= R \bar{h}( \bar{t}) \\
h'(t)=\frac{d}{d{\bar{t}}} \left( R \bar{h} ( \bar{t} )\right) \cdot \frac{d \bar{t}}{dt}
= R \bar{h}'( \bar{t} ) \cdot \d {} t \left(\!\frac{t}{\frac RV}\!\right)
= V \bar{h}'( \bar{t} )
$$
That's because $\bar h$ is a function of $\bar t$ instead of $t$. (Wasntme)

 

[evinda]

I believe that should be:
$$h(t)= R \bar{h}( \bar{t}) \\
h'(t)=\frac{d}{d{\bar{t}}} \left( R \bar{h} ( \bar{t} )\right) \cdot \frac{d \bar{t}}{dt}
= R \bar{h}'( \bar{t} ) \cdot \d {} t \left(\!\frac{t}{\frac RV}\!\right)
= V \bar{h}'( \bar{t} )
$$
That's because $\bar h$ is a function of $\bar t$ instead of $t$. (Wasntme)

Could you explain me further why this equality holds: $h(t)= R \bar{h}( \bar{t})$ ? (Thinking)

 

[I like Serena]

Could you explain me further why this equality holds: $h(t)= R \bar{h}( \bar{t})$ ? (Thinking)

In the problem statement you have that the quantity $\bar h$ is scaled as $\bar h = \frac h R$, making it dimensionless.
Now we make $\bar h$ a function of the corresponding $\bar t$ variable that is expressed in the same set of dimensionless units. That is, $\bar h = \bar h(\bar t)$.
And we already had $h = h(t)$.

Thus $\bar h(\bar t) = \frac {h(t)}{R} \Rightarrow h(t) = R\bar h(\bar t)$. (Wasntme)

 

[evinda]

In the problem statement you have that the quantity $\bar h$ is scaled as $\bar h = \frac h R$, making it dimensionless.
Now we make $\bar h$ a function of the corresponding $\bar t$ variable that is expressed in the same set of dimensionless units. That is, $\bar h = \bar h(\bar t)$.
And we already had $h = h(t)$.

Thus $\bar h(\bar t) = \frac {h(t)}{R} \Rightarrow h(t) = R\bar h(\bar t)$. (Wasntme)

So $\bar h$ is dimensionless, so we can write it as a function of $\bar t$ since the latter is also dimesnionless, right?

And so we use this equality : $\bar t=\frac{t}{\frac{R}{V}}$ if we want to differentiate? (Thinking)

 

[evinda]

So is it then as follows?

$$h'(t)=R \frac{d}{dt}(\bar h(\bar t))= R \frac{d}{d \bar t}(\bar h (\bar t)) \frac{d \bar t}{dt}=R \bar h'(\bar t) \frac{V}{R}=V \bar h'(t)$$

$$h''(t)=V \frac{d}{d \bar t}(\bar h'(\bar t)) \frac{d \bar t}{dt}=\frac{V^2}{R} \bar h''(\bar t)$$

Or am I wrong?

 

[I like Serena]

So $\bar h$ is dimensionless, so we can write it as a function of $\bar t$ since the latter is also dimesnionless, right?

And so we use this equality : $\bar t=\frac{t}{\frac{R}{V}}$ if we want to differentiate? (Thinking)

So is it then as follows?

$$h'(t)=R \frac{d}{dt}(\bar h(\bar t))= R \frac{d}{d \bar t}(\bar h (\bar t)) \frac{d \bar t}{dt}=R \bar h'(\bar t) \frac{V}{R}=V \bar h'(t)$$

$$h''(t)=V \frac{d}{d \bar t}(\bar h'(\bar t)) \frac{d \bar t}{dt}=\frac{V^2}{R} \bar h''(\bar t)$$

Or am I wrong?

Yep! (Nod)

 

[evinda]

Yep! (Nod)

Great! Thanks a lot! (Happy)