PYQs

Let \(\vec{q}=u \hat{i}+v \hat{\jmath}\)
Then,
\(u=-\frac{\partial \phi}{\partial x}\)
\(\implies u=-\left(2 x y-y^{2}+x^{2}\right)\)
\(\implies u=y^{2}-x^{2}-2 x y\)

Also,
\(v=-\frac{\partial \phi}{\partial y}\)
\(\Rightarrow v=-\left(x^{2}-2 x y-y^{2}\right)\)
\(\implies v=y^{2}-x^{2}-2 x y\)
We know that \(\phi+i \psi\) is an analytic function and salisfies Cauchy Riemann equations.

So, \(\frac{\partial \psi}{\partial y}=u\) and \(\frac{\partial \psi}{\partial x}=v\)
\(\Rightarrow \quad \frac{\partial \psi}{\partial x}=y^{2}-x^{2}+2 x y\)
Integrating wrt \(x\):
\(\psi=x y^{2}-\frac{x^{3}}{3}+x^{2} y+f(y)\)

Now, \(\frac{\partial \psi}{\partial y}=2 x y+x^{2}+f^{\prime}(y)\), and
\(\frac{\partial \psi}{\partial y}=-u\)

\(\Rightarrow x^{2}+2 x y+f^{\prime}(y)=x^{2}+2 x y-y^{2}\)
\(\Rightarrow f^{\prime}(y)=-y^{2}\)
\(\frac{df}{d y}=-y^{2}\)
\(\Rightarrow f=\frac{-y^{3}}{3}+c\)

So,
\(\psi=x y^{2}-\frac{\left(x^{3}+y^{3}\right)}{3}+x^{2} y+c\)
\(\psi=0\) at origin \(\implies c=0\)

So, \(\psi=x y^{2}+x^{2} y-\frac{\left(x^{3}+y^{3}\right)}{3}\)

Checking possible flow:
Since\(\frac{\partial u}{\partial x}+\frac{\partial y}{\partial y}=-2 x-2 y+2 y+2 x=0\)
\(\implies\) Equation of continuity is satisfied,
\(\implies\) It is a possible flow.

Here \(u=\frac{3 x z}{r^{5}}\), \(v=\frac{3 y z}{r^{5}}\), \(w=\frac{3 z^{2}-r^{2}}{r^{5}}=\frac{3 z^{2}}{r^{5}}-\frac{1}{r^{3}} \ldots\) (1) where
\(r^{2}=x^{2}+y^{2}+z^{2} \ldots\) (2)

From \((2), \quad \partial r / \partial x=x / r, \quad \partial r / \partial y=y / r, \quad \quad \partial r / \partial z=z / r\) From (1), ( 2 ) and (3), we have

\[\frac{\partial u}{\partial x}=3 z\left[\frac{1}{r^{5}}+(-5 x) r^{-6} \frac{\partial r}{\partial x}\right]=\frac{3 z}{r^{5}}-\frac{15 x^{2} z}{r^{7}}\] \[\frac{\partial v}{\partial y}=3 z\left[\frac{1}{r^{5}}+(-5 y) r^{-6} \frac{\partial r}{\partial y}\right]=\frac{3 z}{r^{5}}-\frac{15 y^{2} z}{r^{7}}\] \[\frac{\partial w}{\partial z}=\frac{6 z}{r^{5}}-15 z^{2} r^{-6} \frac{\partial r}{\partial z}+3 r^{-4} \frac{\partial r}{\partial z}=\frac{6 z}{r^{5}}-\frac{15 z^{2}}{r^{6}} \cdot \frac{z}{r}+\frac{3}{r^{4}} \cdot \frac{z}{r}=\frac{9 z}{r^{5}}-\frac{15 z^{3}}{r^{7}}\] \[\therefore \quad \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=\frac{15 z}{r^{5}}-\frac{15 z}{r^{7}}\left(x^{2}+y^{2}+z^{2}\right)=\frac{15 z}{r^{5}}-\frac{15 z}{r^{7}} \times r^{2}=0\]

Since the equation of continuity is satisfied by the given values of \(u, v\) and \(w,\) the motion is possible. Let \(\phi\) be the required velocity potential. Then

\[\begin{array}{l} d \phi=\frac{\partial \phi}{\partial x} d x+\frac{\partial \phi}{\partial y} d y+\frac{\partial \phi}{\partial z} d z=-(u d x+v d y+w d z), \text { by definition of } \phi \\ =-\left[\frac{3 x}{r^{5}} d x+\frac{3 y z}{r^{5}} d y+\frac{3 z^{2}-r^{2}}{r^{5}} d z\right]=\frac{r^{2} d z-3 z(x d x+y d y+z d z)}{r^{5}} \\ \text { Thus, } \quad d \phi=\frac{r^{3} d z-3 r^{2} z d r}{\left(r^{3}\right)^{2}}=d\left(\frac{z}{r^{3}}\right), \text { using }(2) \end{array}\]

Integrating,

\[\phi=z / r^{3}\]

[Omitting constant of integration, for it has no significance in \(\phi\)]

In spherical polar coordinates \((r, \theta, \phi),\) we know that \(z=r \cos \theta .\) Hence the required potential is given by \(\phi=(r \cos \theta) / r^{3}=(\cos \theta) / r^{2}\). We now obtain the streamlines. The equations of streamlines are given by

\[\frac{d x}{u}=\frac{d y}{v}=\frac{d z}{w} \quad \text { i.e., } \quad \frac{d x}{3 x z / r^{5}}=\frac{d y}{3 y z / r^{5}}=\frac{d z}{\left(3 z^{2}-r^{2}\right) / r^{5}}\]

So,

\[\frac{d x}{3 x z}=\frac{d y}{3 y z}=\frac{d z}{3 z^{2}-r^{2}} \ldots (4)\]

Taking the first two members of (4) and simplifying, we get

\[d x / x=d y / y\] \[\implies d x / x-d y / y=0\]

Integrating, \(\log x-\log y=\log c_{1} \quad\) l.e. \(\quad x / y=c_{1}, c_{1}\) being a constant \(\ldots (5)\)

Now, each member in \((4)=\frac{x d x+y d y+z d z}{3 x^{2} z+3 y^{2} z+3 z^{3}-r^{2} z}=\frac{x d x+y d y+z d z}{3 z\left(x^{2}+y^{2}+z^{2}\right)-r^{2} z}\)

\[=\frac{x d x+y d y+z d z}{3 z\left(x^{2}+y^{2}+z^{2}\right)-z\left(x^{2}+y^{2}+z^{2}\right)}=\frac{x d x+y d y+z d z}{2 z\left(x^{2}+y^{2}+z^{2}\right)}, \text { by }(2) \ldots (6)\]

Taking the first member in (4) and (6), we get

\[\frac{d x}{3 x z}=\frac{x d x+y d y+z d z}{2 z\left(x^{2}+y^{2}+z^{2}\right)} \quad \text { or } \quad \frac{2}{3} \frac{d x}{x}=\frac{1}{2} \frac{2 x d x+2 y d y+2 z d z}{x^{2}+y^{2}+z^{2}}\]

Integrating, \((2 / 3) \times \log x=(1 / 2) \times \log \left(x^{2}+y^{2}+z^{2}\right)+\log c_{2}\) or \(x^{2 / 3}=c_{2}\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}, c_{2}\) being an arbitrary constant. \(\ldots (7)\).

The required streamlines are the curves of intersction of (5) and (7).

Given, strength of source = \(m\).

\(\implies\) The complex potential at any point \(z\) is given by, \(w=-m \log z\)

\(\begin{aligned} \therefore \frac{d w}{d z} &=\frac{-m}{z}=\frac{-m}{r e^{i \theta}}=\frac{-m}{r} e^{-i \theta} \\ \implies \frac{d w}{d z} &=\frac{-m}{r}(\cos \theta-i \sin \theta)=\frac{-m \cos \theta}{r}+\frac{i m \sin \theta}{r} \end{aligned}\) \(\implies \frac{d w}{d z}=\frac{-m x}{r^{2}}+i \frac{m y}{r^{2}}=-u_{m}+i v_{m}\)

where \(u_{m}\) and \(v_{m}\) are velocity components due to source at \(m\). \(\therefore u_{m}=\frac{m x}{r^{2}}, v_{m}=\frac{m y}{r^{2}}\)

Also, fluid is moving with velocity \(U \hat{i}\). \(\therefore\) Total velocity. \(\vec{q}=\left(U+\frac{m x}{r^{2}}\right) \hat{i}+\left(\frac{m y}{r^{2}}\right) \hat{j}=u \hat{i}+\hat{j} v\) Let \(\phi\) be the velocity potential. \(\therefore u=\frac{-\partial \phi}{\partial x}\)

\[\Rightarrow \frac{-\partial \phi}{\partial x}=U+\frac{m x}{r^{2}}\] \[\Rightarrow d \phi=\left(-U-\frac{m x}{r^{2}}\right) d x\]

Integrating both sides, we get \(f=-U x-m \log r+f(y)\) where, \(f(y)\) is arbitrary function of \(y\) Also, \(\frac{-\partial \phi}{\partial y}=v=\frac{m y}{r^{2}} \ldots (i)\)

\[\Rightarrow d \phi=\frac{-m y}{r^{2}} d y\] \[\implies \phi=-m \log r+g(x) \ldots (ii)\]

where \(g(x)\) is an arbitrary function of \(x .\) From (i) and (ii) \(\phi=-U x-m \log r\)

\[\Rightarrow \phi=-U r \cos \theta-m \log r\]

Now, equation of streamlines is given by:

\[\frac{d x}{u}=\frac{d y}{v}\] \[\Rightarrow \frac{d x}{u+\frac{m x}{r^{2}}}=\frac{d y}{\frac{m y}{r^{2}}}\]

\(\Rightarrow \frac{d y}{d x}=\frac{\frac{m y}{r^{2}}}{\left(U+\frac{m x}{r^{2}}\right)}\), which is the differential equation of streamlines.

Since the motion is irrotational, consequently, the corresponding velocity potential \(\phi\) exists such that:

\[\nabla^2 \phi =0 \ldots (1)\]

The boundary conditions for \(\phi\) are:

\[\left( \frac{- \partial \phi}{\partial r} \right)_{r=a} = U \cos \theta \ldots (2)\] \[\left( \frac{- \partial \phi}{\partial r} \right)_{r=b} = V \sin \theta \ldots (3)\]

The above equations suggest that \(\phi\) must involve terms containing \(\sin \theta\) and \(\cos \theta\). So, we take \(\phi\) as:

\(\phi=\left(A r+\frac{B}{r^{2}}\right) \cos \theta+\left(Cr+\frac{D}{r^{2}}\right) \sin \theta \ldots (3)\) \(\implies \frac{-\partial \phi}{\partial r}=\left(-A+\frac{2 B}{r^{3}}\right) \cos \theta+\left(-C+\frac{2 D}{r^{3}}\right) \sin \theta \ldots (4)\)

Using boundary condiitons for (1) and (2) in (4), we get:

\(\left(-A+\frac{2 B}{a^{3}}\right) \cos \theta+\left(-C+\frac{2 D}{a^{3}}\right) \sin \theta=U \cos \theta \ldots (5)\) \(\left(-A+\frac{2 B}{b^{3}}\right) \cos \theta+\left(-C+\frac{2 D}{b^{3}}\right) \sin \theta=V \sin \theta \ldots (6)\)

Comparing coefficients of \(\cos \theta\) and \(\sin \theta\) in (5) and (6), we get:

\(\frac{-A+2 B}{a^{3}}=U, \frac{-C+2 D}{a^{3}}=0 \ldots (7)\), and
\(\frac{-A+2 B}{b^{3}}=0, \frac{-C+2 D}{b^{3}}=\mathrm{V} \ldots (8)\)

Solving (7) and (8), we get:

\(A=\frac{U a^{3}}{b^{3}-a^{3}}\), \(B=\frac{U a^{3} b^{3}}{2\left(b^{3}-a^{3}\right)}\), \(C=\frac{-U b^{3}}{b^{3}-a^{3}}\), \(D=\frac{-U a^{3} b^{3}}{2\left(b^{3}-a^{3}\right)}\)

Putting these values in (3), we get:

\(\phi=\left(\frac{U a^{3} r}{b^{3}-a^{3}}+\frac{u a^{3} b^{3}}{2\left(b^{3}-a^{3}\right) r^{2}}\right) \cos \theta+\left(\frac{-U b^{3}}{b^{3}-a^{3}} \cdot r-\frac{U a^{3} b^{3}}{2\left(b^{3}-a^{3}\right) r^{2}}\right) \sin \theta\)
\(\implies \phi=\frac{\left\{U a^{3}\left(1+\frac{b^{3}}{2 r^{3}}\right) r \cos \theta-V b^{2}\left(1+\frac{a^{3}}{2 r^{3}}\right) r \sin \theta\right\}}{\left(b^{3}-a^{3}\right)}\) \(\implies \phi=\frac{\left\{a^{3} U\left(1+\frac{b^{3} r^{-3}}{2}\right) x-b^{3} V\left(1+\frac{a^{3} r^{-3}}{2}\right) y\right\}}{b^{3}-a^{3}}\)

Now, let velocity at any point= \(\vec{q} = u\hat{i}+v\hat{j}\) where, \(u=\frac{-\partial \phi}{\partial x}\) and \(v=\frac{-\partial \phi}{\partial y}\)

Therefore,

\(u=\frac{-\partial \phi}{\partial x}=\frac{-a^{3} U\left(1+\frac{b^{3} r^{-3}}{2}\right)}{b^{3}-a^{3}}-\frac{a^{3} U x\left(\frac{-3}{2} b^{3} r^{-4} \frac{x}{r}\right)}{\left(b^{3}-a^{3}\right)}\) + \(\frac{b^{3} V y\left(-\frac{3}{2} a^{3} r^{-4}, \frac{x}{r}\right)}{b^{3}-a^{3}}\) \(\implies u=\frac{-a^{3} U\left(1+\frac{b^{3} r^{-3}}{2}\right)}{b^{3}-a^{3}}+\frac{\frac{3}{2} a^{3} b^{3} x^{2} U r^{-5}}{\left(b^{3}-a^{3}\right)}-\frac{\frac{3}{2} a^{3} b^{3} x y v r^{-5}}{b^{3}-a^{3}}\)

Also,

\[\implies u=\frac{-a^{3} u\left(1+b^{3} r^{-3}\right)}{\left(b^{3}-a^{3}\right)}+\frac{3 a^{3} b^{3} x^{2} U r^{-5}}{2\left(b^{3}-a^{3}\right)}-\frac{3 a^{3} b^{3} x y V r^{-5}}{2\left(-b^{3}-a^{3}\right)} \ldots (9)\] \[v=-\frac{\partial \phi}{\partial y}=\frac{-a^{3} U x\left(-\frac{3}{2} b^{2} r^{-4} \frac{y}{r}\right)}{b^{3}-a^{3}}+\frac{b^{3} V\left(1+\frac{1}{2} a^{3} r^{-3}\right)}{b^{3}-a^{3}} +\frac{b^{3} V y\left(-\frac{3}{2} a^{3} r^{-4} \frac{y}{r}\right)}{b^{3}-a^{3}}\] \[\implies v=\frac{-3 a^{3} b^{3} x y U r^{-5}}{2\left(b^{3}-a^{3}\right)}+\frac{b^{3} v\left(1+\frac{1}{2} a^{3} r^{-3}\right)}{b^{3}-a^{3}}-\frac{3 a^{3} b^{3} y^{2} V r^{-5}}{2\left(b^{3}-a^{3}\right)} \quad \ldots(10)\]

\(\therefore\) velocity is given by:

\(\vec{q}=u \hat{i}+\hat{j}\), where \(u\) and \(v\) are given by (9) and (10).

Given, \(\phi=\frac{1}{2} \log \left\{(x+a)^{2}+y^{2}\right]-\frac{1}{2} \log \left((x-a)^{2}+y^{2}\right) \ldots (1)\)

To determine stream lines.

\(-\frac{\partial \phi}{\partial x}=u=-\frac{\partial \psi}{\partial y},-\frac{\partial \phi}{\partial y}=v=\frac{\partial \psi}{\partial x}\) Hence \(\frac{\partial \phi}{\partial x}=\frac{\partial \psi}{\partial y}, \frac{\partial \phi}{\partial y}=-\frac{\partial \psi}{\partial x}\)

Now, \(\frac{\partial \phi}{\partial y}=\frac{x+a}{(x+a)^{2}+y^{2}}-\frac{x-a}{(x-a)^{2}+y^{2}}\)

Integrating wrt \(y\), \(\psi=\tan ^{-1} \frac{y}{x+a}-\tan ^{-1} \frac{y}{x-a}+F(x) \ldots (2)\)

where \(F(x)\) is constant of integration. To determine\(F(x)\),

\[\frac{\partial \psi}{\partial x}=-\frac{\partial \phi}{\partial y}=-\frac{y}{(x+a)^{2}+y^{2}}+\frac{y}{(x-a)^{2}+y^{2}}\]

By \((4),\) \(\frac{\partial \psi}{\partial x}=-\frac{y}{(x+a)^{2}+y^{2}}+\frac{y}{(x-a)^{2}+y^{2}}+F^{\prime}(x)\) Equating (5) to (6), \(F^{\prime}(x)=0\). Integrating this \(F(x)=\) absolute const and hence neglected. Since it has no effect on the fluid motion.

Now (2) becomes

\(\psi=\tan ^{-1} \frac{y}{x+a}-\tan ^{-1} \frac{y}{x-a}\) \(=\tan ^{-1} \frac{-2 a y}{x^{2}-a^{2}+y^{2}}\) \(\therefore\) Stream lines are given by \(\psi=\) const. i.e., \(\tan ^{-1}\left[\frac{-2 a y}{x^{2}-a^{2}+y^{2}}\right]= const. or ,\frac{y}{x^{2}-a^{2}+y^{2}}=\operatorname{const.}\) If we take const. \(=0,\) then we get \(y=0,\) i.e., \(x\) -axis. If we take conts.\(=\infty ,\) then we get circle \(x^{2}-a^{2}+y^{2}=0\).

From the fig, the \(n\) vortices are at \({A}_{0}\), \({A}_{1}\), \({A}_{2} \ldots\), \({A}_{n-1}\) such that \(\angle A_{0} O A_{1}=\angle A_{1} O A_{2}=\ldots=\angle A_{n-1} O A_{1}=\frac{2 \pi}{n}\) The coordinates of the points \(A_{r}\) are given by: \(z=z_{r}=a e^{(2 \pi / n) ir}\) where \(r=0,1,2 \cdots n-1\). These are \(n\) roots of the equation \(z^{n}-a^{n}=0\)

Now, \([\left.z^{n}-a^{n}=0 \Rightarrow z^{n}=a^{n} e^{2 \pi r i}\right]\) Hence, \(z^{n}-a^{n} =\left(z-z_{0}\right)\left(z-z_{1}\right)-\cdots\left(z-z_{n-1}\right)\) The complex potential due to \(n\) vortices at \(P\) is given by:

\[\begin{aligned}w&=\frac{i k}{2 \pi}\left[\log \left(z-z_{0}\right)+\log \left(z-z_{1}\right)+\cdots+\log \left(z-z_{n-1}\right)\right]\\ &= \frac{i k}{2 \pi} \log (z-z_0)\left(z-z_{1}\right) \cdots\left(z-z_{n-1}\right) \\ &=\frac{i k}{2 \pi} \log \left(z^{n}-a^{n}\right) \quad \ldots \quad (1)\end{aligned}\]

For the point \(A_0\), \(z=a\), so that \(r=a\) and \(\theta=0\).

If \(w^{\prime}\) is the complex potential at \(A_0\), then

\[\begin{aligned}w^{\prime} &=w-\frac{i k}{2 \pi} \log (z-a) \\ &=\frac{i k}{2 \pi}\left[\log \left(z^{n}-a^{n}\right)-\log (z-a)\right]\end{aligned}\] \[\phi^{\prime}+i \psi^{\prime}=\frac{i k}{2 \pi}\left[\log \left(r^{n} e^{i n \theta}-a^{n}\right)-\log \left(r e^{i \theta}-a\right)\right]\] \[\psi^{\prime}=\frac{k}{4 \pi}\left[\log \left(r^{2 n}+a^{2 n}-2r^{n} a^{n} \cos n \theta\right)-\log \left(r^{2}+a^{2}-2 r a \cos \theta\right)\right]\] \[\frac{\partial \psi^{\prime}}{\partial r}=\frac{k}{4 \pi}\left[\frac{2n r^{2 n-1}-2 n r^{n-1} a^{n} \cos n \theta}{r^{2 n}+a^{2 n}-2 r^{n} a^{n} \cos n \theta}-\frac{2 r-2 a \cos \theta}{r^{2}+a^{2}-2 r a \cos \theta}\right]\] \[\frac{\partial \psi^{\prime}}{\partial \theta}=\frac{k}{4 \pi}\left[\frac{2 n r^{n} a^{n} \sin n \theta}{r^{2 n}-2 r^{n} a^{n} \cos n \theta+a^{2 n}}-\frac{2 r a \cos \theta}{r^{2}+a^{2}-2 r a \cos \theta}\right]\] \[\left(\frac{\partial \psi^{\prime}}{\partial r}\right)_{r=a}=\frac{k}{4 \pi a}\left[n\left(\frac{1-\cos n \theta}{1-\cos u \theta}\right)-\left(\frac{1-\cos \theta}{1-\cos \theta}\right)\right]=\frac{k}{4 \pi a}(n-1)\] \[\begin{aligned} \left(\frac{\partial \psi^{\prime}}{\partial \theta}\right)_{r=a}&=\frac{k}{4 \pi}\left[\frac{n \sin n \theta}{1-\cos n \theta}-\frac{\sin \theta }{1- \cos \theta}\right] \\ &=\frac{k}{4 \pi}\left[\frac{n^{2} \cos n \theta}{n \sin n \theta}-\frac{\cot \theta}{\sin \theta}\right] \operatorname{as} \theta \rightarrow 0 \text{ (using L'Hospital's Rule)} \\ &=\frac{k}{4 \pi}\left[\frac{-n^{3} \sin n \theta}{n^{2} \cos n \theta}-\frac{(-\sin \theta)}{\cos \theta}\right] \text{ as } \theta \rightarrow 0 \text{ (using L'Hospital's Rule)} \\ &=\frac{k}{4 \pi}[0+0] \\ &=0 \end{aligned}\]

Finally, \(\frac{\partial \psi^{\prime}}{\partial \theta}\)=\(\frac{k}{4 \pi a}(n-1)= 0\), as \(r \rightarrow a, \theta \rightarrow 0\).

Consequently, the velocity \(q_0\) of the vertex \(A_0\) is given by:

\[q_{0}=\left[\left(\frac{\partial \psi^{\prime}}{\partial r}\right)^{2}+\frac{1}{r}\left(\frac{\partial \psi^{\prime}}{\partial \theta}\right)^{2}\right]^{1 / 2}=\frac{k(n-1)}{4 \pi a}\]

It is given that velocity potential, \(\phi=x f(r)\).

For motion of fluid to be possible, \(\nabla^{2} \phi=0\)

\[\Rightarrow \frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0\]

\(\left[\frac{x}{r} f^{\prime}(r)+\frac{x^{3}}{r^{2}} f^{\prime \prime}(r)+\frac{2 x}{r} f^{\prime}(r)-\frac{x^{3}}{r^{3}} f^{\prime}(r)\right]\)+\(\left[\frac{x}{r} f^{\prime}(r)-\frac{x y^{2}}{r^{3}} f^{\prime}(r)+\frac{x y^{2}}{r^{2}} f^{\prime \prime}(r)\right]\) +\(\left[\frac{x}{r} f^{\prime}(r)-\frac{x z^{2}}{r^{3}} f^{\prime}(r)-\frac{x^{3}}{r^{2}} f^{\prime \prime}(r)\right]=0\)

\(\Rightarrow x f^{\prime \prime}(r)+\frac{4 x f^{\prime}(r)}{r}=0\) \(\frac{f^{\prime \prime}(r)}{f^{\prime}(r)}=\frac{-4}{r}\) \(\log f^{\prime}(r)=-4 \log r+c_{1}\) \(f^{\prime}(x)=\frac{c_{1}}{r^{4}} \ldots (1)\) \(f=\frac{- c_{1}}{3 r^{3}}+C_{2} \ldots (2)\)

Now, \(\vec{q}=\nabla \phi=f(r) \hat{i}+\frac{x}{r} f^{\prime}(r) \hat{r}\)

\[\implies \bar{q}=\left(\frac{- c_{1}}{3 r^{3}}+c_{2}\right) \hat{\imath}+\left(\frac{x c_{1}}{r^{5}}\right) \hat{r} \text{ [using(1) and (2)]}\] \[\begin{aligned} \text { Since } \bar{q} & \rightarrow 0 \text { as } \lambda \rightarrow \infty \Rightarrow c_{2}=0 \\ & \Rightarrow \bar{q}=\frac{c_{1}}{3 r^{3}}\left(\hat{\imath}-\frac{3 x}{r^{2}} \hat{r}\right) \end{aligned}\]

\(\text { Surfaces of constant speed are given by }q^{2}=\text { Constant }\) \(\begin{aligned}\Rightarrow \vec{q}.\vec{q}&=\text{constant} \\ &=\left(\frac{c_{1}}{3 r^{3}}\right)^{2}\left[\left(\hat{i}-\frac{3 x}{r^{2}} \hat{r}\right) \cdot\left(\hat{i}-\frac{3 x}{r^{2}} \hat{r}\right)\right] \\ &=\frac{c_{1}^{2}}{9r^{6}}\left(1+\frac{3 x^{2}}{r^{2}}\right) \\ &=\frac{c_{1}^{2}}{9 r^{8}}\left(r^{2}+3 x^{2}\right)\end{aligned}\)

\(\therefore\) Surfaces of constant speed are given by \(\frac{\left(r^{2}+3 x^{2}\right)}{r^{8}}=\operatorname{Constant}\)

Let \(\phi\) velocity potential of sphere and \(U\) be velocity of fluid along \(z-axis\). Then \(\phi\) satifies Laplace’s equation, thus \(\frac{\partial^{2} \phi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \phi}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} \phi}{\partial \theta^{2}}+\frac{\cot \theta}{r^{2}} \frac{\partial \phi}{\partial \theta}=0\)

\(\Rightarrow \phi\) is of the form \(f(r) \cos \theta\) Let \(\phi=\left(A{r}+\frac{B}{r^{2}}\right) \cos \theta\)

At boundary of sphere, i.e, \(r=a\), normal velocity of sphere=velocity of fluid at that point.

\[\begin{aligned} \Rightarrow-\frac{\partial \phi}{\partial r} &=0 \\ \Rightarrow-\left(A-\frac{2 B}{a^{3}}\right) \cos \theta &=0 \\ \Rightarrow A&=\frac{2 B}{a^{3}} \ldots (1) \\ \implies \phi &= \left[r\left(\frac{2 B}{a^{3}}\right)+\frac{B}{r^{3}}\right] \cos \theta \quad [using \space (1)] \end{aligned}\] \[- \frac{\partial \phi}{\partial r}=-\left[\frac{2 B}{a^{3}}-\frac{2 B}{r^{3}}\right] \cos \theta\] \[\begin{aligned} & \text{As } r \rightarrow \infty, \text { velocity }=U \cos \theta\left(=\frac{\partial \phi}{\partial x}\right) \\ & \Rightarrow U \cos \theta=-\left[\frac{2 B}{a^{3}}-\frac{2 B}{r^{3}}\right] \cos \theta \end{aligned}\]

\(\Rightarrow B=\frac{-a^{3} U}{2} \Rightarrow A=-U \quad[using \space (1)]\) \(\Rightarrow \phi=(-U)\left(r+\frac{a^{3}}{2 r^{2}}\right) \cos \theta\) Now, velocity components are given by:

\[q_{r}=-\frac{\partial \phi}{\partial r}=U\left(1-\frac{a^{3}}{r^{3}}\right) \cos \theta\] \[q_{\theta}=\frac{-1}{r} \frac{\partial \phi}{d \theta}=(-u)\left(1+\frac{a^{3}}{2 r^{3}}\right)\]