Chapter 4.1, Problem 2P

To determine

The Lagrengian description of the velocity of a fluid particle flowing along the surface.

Answer to Problem 2P

The Lagrengian description of the velocity of a fluid particle flowing along the surface is

$V=\frac{V_0\Delta V{e}^{-ax}}{\Delta V{e}^{-ta\left(V_0+\Delta V\right)}}$.

Explanation of Solution

Write the expression of surface velocity of the river.

    $V=V_0+\Delta V\left(1-e^{-ax}\right)$

Here, the constant initial velocity is $V_0$, the constant change in velocity is $\Delta V$, the constant displacement is $x$ and the constant acceleration is $a$.

Write the expression of velocity.

    $V=\frac{dx}{dt}$ (I)

Here, the time is $t$.

Substitute $V_0+\Delta V\left(1-e^{-ax}\right)$ for $V$ in Equation (I).

    $\begin{array}{l}{V}_0+\Delta V\left(1-e^{-ax}\right)=\frac{dx}{dt}\\ dt=\frac{dx}{V_0+\Delta V\left(1-e^{-ax}\right)}\end{array}$                                                                                   (II)

Integrate Equation (II) on both sides with respect to time $\left(t\right)$ and displacement $\left(x\right)$.

    $\begin{array}{l}{\displaystyle \int dt}={\displaystyle \int \frac{dx}{V_0+\Delta V\left(1-e^{-ax}\right)}}\\ t+c={\displaystyle \int \frac{dx}{V_0-\Delta V\left(e^{-ax}-1\right)}}\\ t+c=-{\displaystyle \int \frac{dx}{\Delta V\left(e^{-ax}-1\right)-V_0}}\end{array}$                                                                          (III)

Consider

    $u=ax$                                                                                                          (IV)

Differentiate Equation (IV) with respect to displacement.

    $\begin{array}{l}\frac{d}{dx}\left(u\right)=\frac{d}{dx}\left(ax\right)\\ \frac{du}{dx}=a\times 1\\ \frac{du}{a}=dx\end{array}$

Substitute $\frac{du}{a}$ for $dx$ and $u$ for $ax$ in Equation (III).

    $\begin{array}{c}t+c=-{\displaystyle \int \frac{\left(\frac{du}{a}\right)}{\Delta V\left(e^{-u}-1\right)-V_0}}\\ =-\frac{1}{a}{\displaystyle \int \frac{du}{\Delta V{e}^{-u}-\Delta V-V_0}}\end{array}$                                                                            (V)

Consider,

    $\frac{1}{\Delta V{e}^{-u}-\Delta V-V_0}=v$                                                                                         (VI)

Differentiate Equation (VI) with respect to $u$ on both sides.

    $\begin{array}{c}\frac{dv}{du}=\frac{d}{du}\left(\frac{1}{\Delta V{e}^{-u}-\Delta V-V_0}\right)\\ \frac{dv}{du}=-{\left(\frac{1}{\Delta V{e}^{-u}-\Delta V-V_0}\right)}^2\left(-\Delta V{e}^{-u}\right)\\ \frac{dv}{du}={\left(\frac{1}{\Delta V{e}^{-u}-\Delta V-V_0}\right)}^2\left(\Delta V{e}^{-u}\right)\\ du=\frac{{\left(\Delta V{e}^{-u}-\Delta V-V_0\right)}^{2}dv}{\Delta V{e}^{-u}}\end{array}$

Substitute $\frac{{\left(\Delta V{e}^{-u}-\Delta V-V_0\right)}^{2}dv}{\Delta V{e}^{-u}}$ for $du$ in Equation (V).

    $\begin{array}{c}t+c=-\frac{1}{a}{\displaystyle \int \frac{{\left(\Delta V{e}^{-u}-\Delta V-V_0\right)}^{2}dv}{\left(\Delta V{e}^{-u}-\Delta V-V_0\right)\Delta V{e}^{-u}}}\\ t+c=-\frac{1}{a}{\displaystyle \int \frac{\left(\Delta V{e}^{-u}-\Delta V-V_0\right)}{\Delta V{e}^{-u}}}dv\\ t+c=-\frac{1}{a}{\displaystyle \int \frac{dv}{\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-u}-\Delta V-V_0}+1}}\end{array}$                                                         (VII)

Substitute $v$ for $\frac{1}{\Delta V{e}^{-u}-\Delta V-V_0}$ in Equation (VII).

    $t+c=-\frac{1}{a}{\displaystyle \int \frac{1}{\left(V_0+\Delta V\right)v+1}}dv$                                                                        (VIII)

Consider.

    $\left(V_0+\Delta V\right)v+1=w$                                                                                           (IX)

Differentiate Equation (IX) on both sides with respect to $v$ on both sides.

    $\begin{array}{l}\frac{d}{dv}\left\{\left(V_0+\Delta V\right)v+1\right\}=\frac{d}{dv}\left(w\right)\\ \frac{dw}{dv}=\frac{d}{dv}\left(V_0+\Delta V\right)v+\frac{d}{dv}\left(1\right)\\ \frac{dw}{dv}=\left(V_0+\Delta V\right)\\ \frac{dw}{\left(V_0+\Delta V\right)}=dv\end{array}$

Substitute $\frac{dw}{\left(V_0+\Delta V\right)}$ for $dv$ and $w$ for $\left(V_0+\Delta V\right)v+1$ in Equation (VIII).

    $\begin{array}{l}t+c=-\frac{1}{a}{\displaystyle \int \frac{dw}{w\left(V_0+\Delta V\right)}}\\ t+c=-\frac{1}{a\left(V_0+\Delta V\right)}{\displaystyle \int \frac{dw}{w}}\\ t+c=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left(w\right)\end{array}$                                                                          (X)

Substitute $\left(V_0+\Delta V\right)v+1$ for $w$ in Equation (X).

    $t+c=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left\{\left(V_0+\Delta V\right)v+1\right\}$                                                (XI)

Substitute $\frac{1}{\Delta V{e}^{-u}-\Delta V-V_0}$ for $v$ in Equation (XI).

    $t+c=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left\{\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-u}-\Delta V-V_0}+1\right\}$                                      (XII)

Substitute $ax$ for $u$ in Equation (XII).

    $t+c=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left\{\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-ax}-\Delta V-V_0}+1\right\}$                                         (XIII)

Substitute $0$ for $t$ and $0$ for $x$ for initial condition in Equation (XII).

    $\begin{array}{l}0+c=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left\{\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-a\times 0}-\Delta V-V_0}+1\right\}\\ c=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left\{\frac{\left(V_0+\Delta V\right)}{-V_0}+1\right\}\\ c=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left\{-1-\frac{\Delta V}{V_0}+1\right\}\\ c=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left(-\frac{\Delta V}{V_0}\right)\end{array}$

Substitute $-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left(-\frac{\Delta V}{V_0}\right)$ for $c$ in Equation (XIII).

    $\begin{array}{l}t-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left(-\frac{\Delta V}{V_0}\right)=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left\{\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-ax}-\Delta V-V_0}+1\right\}\\ t=-\frac{1}{a\left(V_0+\Delta V\right)}\ln \left\{\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-ax}-\Delta V-V_0}+1\right\}+\frac{1}{a\left(V_0+\Delta V\right)}\ln \left(-\frac{\Delta V}{V_0}\right)\\ -ta\left(V_0+\Delta V\right)=\ln \left\{\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-ax}-\Delta V-V_0}+1\right\}-\ln \left(-\frac{\Delta V}{V_0}\right)\\ -ta\left(V_0+\Delta V\right)=\ln \left[\frac{\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-ax}-\Delta V-V_0}+1}{-\frac{\Delta V}{V_0}}\right]\end{array}$

    $\begin{array}{l}{e}^{-ta\left(V_0+\Delta V\right)}=-\frac{\left(\frac{\left(V_0+\Delta V\right)}{\Delta V{e}^{-ax}-\Delta V-V_0}+1\right)}{\left(\frac{\Delta V}{V_0}\right)}\\ e^{-ta\left(V_0+\Delta V\right)}=\left[\frac{\Delta V{e}^{-ax}}{V_0+\Delta V\left(1-e^{-ax}\right)}\right]\left(\frac{V_0}{\Delta V}\right)\end{array}$                                                  (XIV)

Substitute $V$ for $V_0+\Delta V\left(1-e^{-ax}\right)$ in Equation (XIV).

    $\begin{array}{l}{e}^{-ta\left(V_0+\Delta V\right)}=\left[\frac{\Delta V{e}^{-ax}}{V}\right]\left(\frac{V_0}{\Delta V}\right)\\ V=\frac{V_0\Delta V{e}^{-ax}}{\Delta V{e}^{-ta\left(V_0+\Delta V\right)}}\end{array}$

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is $V=\frac{V_0\Delta V{e}^{-ax}}{\Delta V{e}^{-ta\left(V_0+\Delta V\right)}}$.