Potential function of a flow around a stagnation point

[happyparticle]

Homework Statement

Find the potential function around a stagnation point for a perfect incompressible fluid.

Far from the stagnation point at x = 0 ($y = y_0 >> 1$), the vertical velocity is $u_0 \hat{y}$

Relevant Equations

Laplace equation $\Delta \phi = 0$

To find the potential function, I'm starting with the laplace equation $\Delta \phi = 0$
In cartesian coordinates the solution is $\phi = \sum_m (A_m e^{mx} + B_m e^{-mx}) (C_m sin(my) + D_m cos (my))$

Using the first boundary condition.
$u = \Delta \phi = -u_0 \hat{y}$

The gradient in cartesian coordinates
$\nabla \phi = \frac{\partial \phi}{ \partial x} \hat{x} + \frac{\partial \phi}{ \partial y} \hat{y}$

Thus, $\frac{\partial \phi}{ \partial x} \hat{x} + \frac{\partial \phi}{ \partial y} \hat{y} = - u_0 \hat{y}$
This mean that $\frac{\partial \phi}{ \partial x} = 0 , \frac{\partial \phi}{ \partial y} = - u_0$

The solution is $\phi = -u_0 y_0$

So, far from the stagnation point $\phi = \sum_m (A_m e^{mx} + B_m e^{-mx}) (C_m sin(my) + D_m cos (my)) = - u_0 y_0$

Thus, m must be 0, and the potential function becomes
$\phi = (A+B) D$

So far I'm stuck here. It feels wrong. There is no dependencies on x and y. Also, I cannot use a second boundary condition since I only have constants.

Here is a scheme of the problem.

 

[pasmith]

Your sketch suggests thast $y = 0$ is a boundary of the domain. If it is, then you have the condition that $\frac{\partial \phi}{\partial y}$ vanishes there. It is also unclear to me whether the boundary condition as $y \to \infty$ applies only at $x = 0$ or applies everywhere. If the latter, then you could work in the inertial frame in which the fluid is stationary at infinity and has velocity $u_0 \hat{y}$ at the origin.

Your general solution is not appropriate here: the domain is not a finite rectangle at the edge of which you could impose a boundary condition to ontain a discrete set of eigenvalues. It also omits some solutions, such as $Ax + By + Cxy + D(x^2 - y^2)$.

 

[happyparticle]

Your general solution is not appropriate here: the domain is not a finite rectangle at the edge of which you could impose a boundary condition to ontain a discrete set of eigenvalues. It also omits some solutions,

Ah, I didn't see that. Perhaps it will be better to use the polar coordinates.

 

[happyparticle]

It is also unclear to me whether the boundary condition as y→∞ applies only at x=0 or applies everywhere.

It applies only at x=0.

According to wikipedia, the solution should be something like $2K(x^2 - y^2)$.

 

[happyparticle]

@pasmith
I couldn't just find the solution of the Laplace's equation using the separation of variables' method?

$$\phi(x,y) = X(x)Y(y)$$

Hence,

$$\frac{\partial^2 X(x)}{X(x)\partial x^2} = C$$

$$\frac{\partial^2 Y(y)}{Y(y)\partial y^2} = -C$$

 

[pasmith]

How do you determine $C$? And is $x^2 - y^2$ of that form?

The problem here is that near the origin the Taylor Series of $\phi$ must be of the form $$Axy + B(x^2 - y^2) + O(r^3) = r^2(A \sin 2\theta + B\cos 2\theta) + O(r^3)$$ but a finite velocity at infinity requires $\phi \sim r\cos \theta.$ There does not appear to be a good way to match these.

 

[happyparticle]

Is it possible to find the potential function solving the Laplace's equation?

I don't know other method than that.

I thought I could just solve the Laplace's equation with a separation of variables and then using the boundary conditions.

Maybe there is something I didn't see.
The left image is essentially what I'm trying to solve. However, here they already give the stream function.

 

[happyparticle]

After some time, I'm allowing myself to ask again, since I'm still looking for the solution of this problem.

Most of the explanations I found gives directly the components of the velocity without explaining how they get it or even if this is possible by using the Laplace's equation and the separation of variables method.

The following image is exactly what I'm looking for. However I still have no idea how to get the velocity components.

 

[renormalize]

However I still have no idea how to get the velocity components.

Are you asking about solving problem #1 as posted? The problem explicitly states that $x\text{-velocity}=u=Ax$ and $y\text{-velocity}=v=-Ay$.

 

[happyparticle]

@renormalize I posted the image as reference. Sorry for the confusion. Basically, I'm trying to find how to get the velocity in post #8 using the Laplace's equation. This is basically my question in post #1.

 

[renormalize]

Basically, I'm trying to find how to get the velocity in post #8 using the Laplace's equation. This is basically my question in post #1.

So you want to solve $\nabla^2 \phi\left(x,y\right)=0$ and get the velocity field $\vec{u}=\nabla \phi$? Don't insist on a multiplicative separation of variables; instead, consider the additive separation $\phi\left(x,y\right)\equiv\frac{1}{2}\left(Ax^2+By^2\right)$ and insert this into Laplace's equation. What relation do you get between $A,B\,$? And what do you get for the velocity field $\vec{u}\,$?

 

[happyparticle]

@renormalize By using $\phi\left(x,y\right)\equiv\frac{1}{2}\left(Ax^2+By^2\right)$ I can get the answer that I'm looking for. However, I don't know where this come from. For instance, $\vec{u}$ was given in post #8, which is equivalent to giving $\phi$, but I don't see how to get it.

 

[renormalize]

but I don't see how to get it.

1. Assume an additive (not multiplicative) separation-of-variables for the velocity-potential: $\phi\left(x,y\right)=f\left(x\right)+g\left(y\right)\,$.
2. Substitute $\phi$ into Laplace's equation: $0=\nabla^{2}\phi\left(x,y\right)=f^{\prime\prime}\left(x\right)+g^{\prime\prime}\left(y\right)\,$.
3. Integrate twice to find: $\phi\left(x,y\right)=c_{1}+c_{2}\,x+c_{3}\,y+c_{4}\left(x^{2}-y^{2}\right)\,$.
4. Choose the arbitrary integration constants so as to satisfy the required boundary conditions.
5. And you're done.

 

[happyparticle]

How did you know that we had to use the additive separation?

Related to the question above. When using the polar coordinates and multiplication separation of variables, I get the general solution $\phi(r, \theta) = (Ar^k + Br^{-k}) (C cos(k \theta) + D sin(k \theta))$

Because $\phi \rightarrow \infty$ at $r \rightarrow 0$, B must be 0.

Also,
at $x = 0 , y \gg1$ which means $\phi = \pi/2$ and $r \gg 1$, $\vec{u} = - u_0 \hat{y} = -u_0 (sin \theta \hat{r} + cos \theta \hat{\theta})$

Thus, $\vec{u}(r \gg1 , \pi/2) = \nabla \phi = - u_0 \hat{y} = -u_0 (sin \theta \hat{r} + cos \theta \hat{\theta})$. Which mean that $\phi = - u_0 r sin \theta = -u_0 r$

To get this result at $r \gg 1$ and $\theta = \pi/2$ from the general solution.
$k = 1$ and $A \cdot D = - u_0$

Finally, $\vec{u} \cdot \hat{n} = \nabla \phi \cdot \hat{n} = 0$ at $\theta = 0$

Which give $A \cdot C = 0$

The solution is then:

$\phi(r,\theta) = -u_0 r sin(\theta)$ which is not the same as $\phi\left(x,y\right)\equiv\frac{1}{2}\left(Ax^2+By^2\right)$
Which is the right answer.

Yet, the boundary conditions seems to work.

 

[renormalize]

How did you know that we had to use the additive separation?

Just like multiplicative-separation, the possibility of additive separation-of-variables is something you should always check for when solving PDEs. See, e.g., section 1.1 of this reference: https://www.reed.edu/physics/courses/P342.S10/Physics342/page1/files/Lecture.1.pdf.

When using the polar coordinates and multiplication separation of variables, I get the general solution $\phi(r, \theta) = (Ar^k + Br^{-k}) (C cos(k \theta) + D sin(k \theta))$

Note that for your solution $B=0,k=1$ we have $\phi\left(r,\theta\right)=A\,r\left(C\text{cos}\theta+D\text{sin}\theta\right)=AC\,x+AD\,y$ so that $\vec{u}\equiv\nabla\phi=\frac{\partial\phi}{\partial x}\hat{i}+\frac{\partial\phi}{\partial y}\hat{j}=AC\,\hat{i}+AD\,\hat{j}\,$. Thus, $\vec{u}$ is an everywhere constant vector and can't possibly satisfy the required boundary conditions at both $y=0$ and $y\rightarrow\infty$.

 

[happyparticle]

Thank you for the link. To be honest, I had never used the additive separation of variables.

I thought it would be possible to find the potential function using the polar coordinates. I have a few problem to solve with different walls (surfaces), like a flow inside a sharp corner of angle $\beta$ and I thought I could transpose the technic using in this problem to different cases.

 

[renormalize]

I thought it would be possible to find the potential function using the polar coordinates.

It is indeed possible to use multiplicative separation-of-variables in polar coordinates to get the same answer as in cartesian coordinates. Start from your polar solution to Laplace's equation $\left(ar^{n}+br^{-n}\right)\left(c\cos n\theta+d\sin n\theta\right)$. Because you want a solution that's finite at the origin, we drop the $r^{-n}$ term and then define the velocity potential $\phi_{012}$ that is the superposition of the three solutions with $n=0,1,2\,$:$$\phi_{012}\left(r,\theta\right)=c_{1}+r\left(c_{2}\cos\theta+c_{3}\sin\theta\right)+r^{2}\left(c_{4}\cos2\theta+c_{5}\sin2\theta\right)\tag{1}$$Now change from polar-coordinates $\left(r,\theta\right)$ to cartesian-coordinates $\left(x,y\right)$ via $r=\sqrt{x^{2}+y^{2}}\,,\theta=\tan^{-1}\left(\frac{y}{x}\right)$ to find:$$\phi_{012}\left(x,y\right)=c_{1}+c_{2}\,x+c_{3}\,y+c_{4}\left(x^{2}-y^{2}\right)+2c_{5}\,xy\tag{2}$$This is exactly the additive-separation solution I gave in post #13 superposed with an additional multiplicative-solution $xy$.

 

[happyparticle]

Thank you very much! I have been struggling with this problem for long time.

One more thing. why you stopped at $n=2$? Also, in the problem with the flow inside a sharp corner. The solution seems to be n=3. However, I'm not sure to fully understand.

 

[renormalize]

One more thing. why you stopped at $n=2$? Also, in the problem with the flow inside a sharp corner. The solution seems to be n=3. However, I'm not sure to fully understand.

I went up to $n=2$ simply because that's the number required to duplicate the additive solution I gave in post #13. You are of course free to superpose any number of solutions (even infinitely many). But there's no practical reason to add more terms than are necessary to satisfy the boundary conditions of the particular problem you're trying to solve.

 

[happyparticle]

I was busy last week. Thank you for everything!