Question

Potential funtion $\varphi$ is givne as $\varphi =x^2-y^2$. What will be the stream funciton $\left(\Psi \right)$ with the condition $\Psi =0$ at $x=y=0$?

• A. $2xy$
• B. $x^2+y^2$
• C. $x^2-y^2$
• D. $2x^2{y}^2$

Answer 1

The correct option is A $2xy$

Method I :
We know that potential function $\varphi$ and stream funciton $\Psi$ together constitute an analytic function known as complex potential ${}^{\prime }{\omega }^{\prime }$ i,e.
$\omega =\varphi +i\Psi$ is analytic where $\varphi =x^2-y^2$
$\because$ Real part is given so by case I of MILNE THOMSON method.

Step 1: $\frac{\partial \varphi }{\partial x}=2x\approx {\varphi }_1(x,y)$
Step 2: ${\varphi }_1(z,0)=2z$
Step 3: $\frac{\partial \varphi }{\partial y}=-2y\approx {\varphi }_2(x,y)$
Step 4: ${\varphi }_2(z,0)=0$
Step 5: $\omega =f\left(z\right)=\int \left[{\varphi }_1\right(z,0)-i{\varphi }_2(z,0\left)\right]dz+c$
$\omega =\int \left(2z\right)dz+c=z^2+c$
$\varphi +i\Psi =(x^2-y^2)+2ixy+c$
($\because$ At $x=y=0,\Psi =0,c=0$,)
Hence stream function
$\Psi =2xy+c=2xy$
$\because \omega =\varphi +i\Psi$ is an analytic fucntion
Method II: By C - R equation,
${\varphi }_x={\Psi }_y$ & ${\varphi }_y=-{\Psi }_x$
$\because \Psi \to f(x,y)$
So by stotal derivative concept
$d\Psi =\left(\frac{\partial \Psi }{\partial y}\right)dx+\left(\frac{\partial \Psi }{\partial y}\right)dy$
$=(-\frac{\partial \varphi }{\partial y})dx+(-\frac{\partial \varphi }{\partial y})dy$ (By C - Req is S)
$=\left(2y\right)dx+\left(2x\right)dy$ {$\because \varphi =x^2-y^2$}
$d\Psi =2(xdy+ydx)=2d\left(xy\right)$
Integrating both sides
$\Psi =2xy+c$
But at $x=y=0,\Psi =0$ so $c=0$
Hence stream function $\Psi =2xy$.
Method III:
$\Phi =x^2-y^2$
${\varphi }_x=2x={\Psi }_y$ (By C.R. eqations)
$\Psi =2xy+f\left(x\right)$ ..... (i)
Again , By C.R. equations
${\varphi }_y=-{\Psi }_x$
$\Rightarrow -2y=-[2y+f^{\prime }(x\left)\right]$
$f^{\prime }\left(x\right)=0$
$f\left(x\right)=C$
By using (i), $\Psi =2xy+C$
At $x=0,y=0,\Psi =0$
$\Rightarrow C=0$
$\therefore \Psi =2xy$
Note: It will be better to use MILNE-THOMSAN method, while solving Quesitons $3,7,8,9,10$ & $11$.