Differential Equation?

The classic solution of inviscid flow around a cylinder is:

v_r = Ucosθ(1 - 1/r²)

v_θ = -Usinθ(1 + 1/r²)

Or

v_x = U*(1 - cos2θ / r²)

v_y = -Usin2θ / r²

U = free stream velocity

v_x = U*[1 + (y² - x²) / (x² + y^2)²]

v_y = -U*[2xy / (x² + y^2)²]

YOUR differential equation seems to say that

- v_y * dx + v_x * dy = 0

or v_y / v_x = dy/dx = (dy/dt) / (dx/dt)

which sort of makes sense.

Now how to actually solve it for y as a function of x???

I think it's better to try to solve r as a function of θ. It will be more manageable.

Note that

v_r / v_θ = 1/r * dr/dθ

= -(1 - 1/r²) / (1 + 1/r²) * cotθ

Separate variables to get

-cotθ dθ = (r² + 1) / [r * (r² - 1)] dr

This is much easier to solve. Integrate both sides to get

-ln|A*sinθ| = ln|(r² - 1) / r|

or r² - A*r*cscθ - 1 = 0

Use quadratic formula to get

r(θ) = B*cscθ ±√(1 + B² csc² θ) ...(1)

Alternatively you can write

A = (r² - 1) / r * sinθ

= y*(x² + y² - 1) / (x² + y²) ...(2)

This equation shows the radial distance of the stream line as a function of θ. Note that as θ approaches 0 or π, r goes to infinity which represents the upstream and downstream conditions.

For the special case where B = 0, r = 1. This refers to the streamline that goes around the circumference of the cylinder (for θ ≠ 0, π). But at θ = 0, π, the streamline coincides with the x axis (|x| ≥ 1). So this particular streamline sort of looks like a bead on a string.

This is also evident in the cartesian solution (eqn 2). When A = 0, either x² + y² = 1 or y = 0.

Here is a plot of r vs θ for the case where B = 0.1

http://i33.tinypic.com/qsqy3l.jpg

Here's what it looks like in the cartesian plane.

http://i38.tinypic.com/ap9xci.jpg

As θ --> 0, the stream line goes far away from the cylinder (r --> ∞). When θ = π/2, streamline passes over cylinder close to the circumference (r > 1). Symettry exists in all 4 quadrants.

Here's another plot this time for B = 0.1 (black) and B = 0.2 (blue).

http://i33.tinypic.com/2yv2vif.jpg

***

NOTE: It's important to remember that the complicated solution is not always the best method. In this case, using polar coordinates made the problem A LOT easier to solve. The intuitiveness of the solution is also more evident in polar than in cartesian.

*EDIT*

cotθ = cosθ / sinθ --- easy to integrate

(r² + 1) / [r * (r² - 1)] = 1/(r+1) + 1/(r-1) - 1/r --- easy to integrate

Maybe I'm wrong, but generally when there is a problem involving a circle, circular cordinates might be more handy and also more intuitive. If you're going to introduce new terms like "t" and "z", you might as well introduce terms that mean something, like "r" and "θ". In addition to this being a mathematical problem, it is also a physical problem.

Don't get me wrong Vasek, I have nothing against your solution. I think it's great. All I'm saying is that in my opinion

sinθ = Ar / (r² - 1)

is the most physically intuitive form of the solution.

dy/dx = (x^2)(8 + y) First, we separate the variables by multiplying by dx and dividing by (8 + y): dy/(8 + y) = (x^2)dx Integrate both sides, remembering the left will use a natural log and the right will have a constant: ln(8 + y) = (1/3)x^3 + C Raise e to the power of each side: 8 + y = Ae^((1/3)x^3) Subtract 8 and we have the general form: y = Ae^((1/3)x^3) - 8 Now, we have the point (0, 3), so plug these in to find A and the particular solution: 3 = Ae^((1/3)(0)^3) - 8 Solve for A: 11 = Ae^((1/3)(0)) 11 = Ae^(0) A = 11 So, the particular solution is: y = 11e^((1/3)x^3) - 8

This is a so-called exact differential equation: a(x,y)dx + b(x,y)dy = 0, where da/dy = db/dx (all partial). Therefore there is some u(x,y) such that a = du/dx and b = du/dy and the differential equation can then be rewritten do du = 0, or u(x,y) = const. The task is then, of course, to find u.

Both the functions are singular at (0,0), so let's choose another reference point for integrating it. I would go with (0,1) for example. The integration is done like this: to get u(x,y), you choose some curve beginning at this point and going only along horizontal and vertical lines (if you are able to, to simplify the integration as much as possible). Then you just "normally" integrate the given parts with fixed x or y by one variable.

Starting at the aforementioned point (0,1), we can continue along y axis to (0,y) (the x coordinate is fixed at zero here) and then finish by a horizontal line to (x,y) (now y is fixed). The value of u will be

u(0,1) + int(1+(t^2-0^2)/(0^2+t^2)^2, t=1..y) + int(2*t*y/(t^2+y^2)^2, t=0..x).

The first integral is

int(1 + t^2/t^4, t=1..y) = int(1+1/t^2, t=1..y) = [t-1/t] from 1 to y = y - 1 - 1/y + 1 = y - 1/y,

for the latter:

int(2*t*y / (t^2+y^2)^2, t=0..x)

substitute for z = t^2+y^2, dz = 2t dt

= int(y/z^2 dz, z=y^2 .. x^2+y^2) = [-y/z] from y^2 to x^2+y^2 = 1/y - y/(y^2+x^2)

Putting all together,

u(x,y) = u(0,1) + y - 1/y + 1/y - y/(x^2+y^2) = y - y/(x^2+y^2) + constant.

We can check back:

du / dx = 2xy / (x^2+y^2)^2 ... OK

du / dy = 1 + (2y^2 - x^2 - y^2) / (x^2+y^2)^2 = 1 + (y^2-x^2) / (x^2+y^2)^2 ... OK

There are many ways of writing the implicit result: for example, if you want it to end with "= 0", you can get all the family of solutions putting various c's in

y - y/(x^2+y^2) + c = 0.

Hope it helps!

N.B.: You may have noticed my (0,1)--(0,y)--(x,y) path can't reach points with negative y coordinate without crossing the singular point in origin. However, we can extend our solution to be valid on all R^2 - O since we checked that the differential equation is satisfied anywhere. Or we can choose another integration part.

Moreover, the function found above has a removable discontinuity at the origin, its limit in this point is c.

*** Thank you, Dr. D, for answering my question (others: I cut accidentally it but was addressing the complexity of integrals involved in our two approaches as opposed to intuitiveness). I'm wondering now why _both_ of us got so many thumbs down...

[2xy/(x²+ y²)²]dx + [1+ ((y²- x²)/(x²+ y²)²) ]dy= 0

Polar coordinates:

x = rcos(θ) → dx = sin(θ)dr + rcos(θ)dθ

y = rsin(θ) → dy = cos(θ)dr - rsin(θ)dθ

[r²sin(2θ)/(r^4)][sin(θ)dr + rcos(θ)dθ] + [1+ (-r²cos(2θ))/(r^4)][cos(θ)dr - rsin(θ)dθ] = 0

[r²sin(2θ)][sin(θ)dr + rcos(θ)dθ] + [r^4 + (-r²cos(2θ))][cos(θ)dr - rsin(θ)dθ] = 0

[sin(2θ)][sin(θ)dr + rcos(θ)dθ] + [r² - cos(2θ)][cos(θ)dr - rsin(θ)dθ] = 0

sin(2θ)sin(θ)dr + rcos(θ)sin(θ)dθ + r²cos(θ)dr - r³sin(θ)dθ - cos(2θ)cos(θ)dr + rcos(2θ)sin(θ)dθ = 0

[sin(2θ)sin(θ) + r²cos(θ) - cos(2θ)cos(θ)]dr

+ [rcos(θ)sin(θ) - r³sin(θ) + rcos(2θ)sin(θ)]dθ = 0

cos(θ)[sin²(θ) + r² - cos(2θ)]dr

+ rsin(θ)[cos(θ) - r² + cos(2θ)]dθ = 0

This is as far as I can get. I'll star it.