Why do integrated stream functions have arbitrary functions of the other variable?

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influx
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Slide of fluid flow equations: stream function u_r = (1/r)∂ψ/∂θ and u_θ = -∂ψ/∂r (4a,b); velocity potential u_r = ∂φ/∂r and u_θ = (1/r)∂φ/∂θ (5a,b); integration steps giving ψ = mθ + f1(r) and ψ = c1 + g1(θ), and compatibility result ψ = mθ (6a).


How did they go from ∂Ψ/dθ = m to Ψ = mθ + f1(r) ?

I mean I understand how they got mθ but can someone explain the f1(r) term?

Similarly, how did they go from -∂Ψ/dr = 0 to Ψ = c1 + g1(θ) ?

I understand how they got c1 but why the additional term?

I can't provide further working out as I think this is supposed to be pretty simple, I'm just missing something..

Thanks
 

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Moved this thread out of the HW sections. Note that if you post in the HW sections, you must include the homework template.
influx said:
Slide of stream function and velocity potential relations in polar coordinates: u_r = (1/r)∂ψ/∂θ and u_θ = -∂ψ/∂r (4a,b); u_r = ∂φ/∂r and u_θ = (1/r)∂φ/∂θ (5a,b); integrating gives ψ = mθ + f₁(r) and ψ = c₁ + g₁(θ), leading to ψ = mθ (6a).


How did they go from ∂Ψ/dθ = m to Ψ = mθ + f1(r) ?

I mean I understand how they got mθ but can someone explain the f1(r) term?
Ψ is assumed to be a function of both θ and r, so when you integrate ∂Ψ/dθ with respect to θ, you have to include a "constant" of integration that is a function of r alone. As a check, take the partial derivative with respect to θ of the right side, and you should get what they have on the left side.
influx said:
Similarly, how did they go from -∂Ψ/dr = 0 to Ψ = c1 + g1(θ) ?

I understand how they got c1 but why the additional term?
See above, with the difference that now the integration is with respect to r.
influx said:
I can't provide further working out as I think this is supposed to be pretty simple, I'm just missing something..

Thanks
 
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Mark44 said:
Moved this thread out of the HW sections. Note that if you post in the HW sections, you must include the homework template.
Ψ is assumed to be a function of both θ and r, so when you integrate ∂Ψ/dθ with respect to θ, you have to include a "constant" of integration that is a function of r alone. As a check, take the partial derivative with respect to θ of the right side, and you should get what they have on the left side.
See above, with the difference that now the integration is with respect to r.
That makes perfect sense. One question though, in this case we know that Ψ is a function of both θ and r but what if we were not aware of this? So say we were told to integrate ∂Ψ/dθ = m without being informed that Ψ is a function of both θ and r?

Also for:

Slide titled Example 1 showing a uniform horizontal flow diagram with arrows and labelled stream function values, followed by the solution: u = U∞, v = 0, and the resulting integrated form ψ = U∞ y + f(x).


How do we know that Ψ is a function of both x and y?

Thanks
 
influx said:
That makes perfect sense. One question though, in this case we know that Ψ is a function of both θ and r but what if we were not aware of this? So say we were told to integrate ∂Ψ/dθ = m without being informed that Ψ is a function of both θ and r?
Since you are asked to find the partial derivative ∂Ψ/∂θ, you can be certain that Ψ is a function of θ and some other variable, most likely r, as the problem is evidently working with polar coordinates. The context of the problem will usually provide information about what variables any functions are in terms of.
influx said:
Also for:

Worked example finding the stream function for uniform flow: horizontal arrows labelled U∞ in an x-y plane, with four stream function expressions listed beside them, and the solution showing u = U∞, v = 0 and the resulting partial differential equations.


How do we know that Ψ is a function of both x and y?
The fact that the flows are shown in an x-y coordinate system is a strong clue.
 
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Mark44 said:
Since you are asked to find the partial derivative ∂Ψ/∂θ, you can be certain that Ψ is a function of θ and some other variable, most likely r, as the problem is evidently working with polar coordinates. The context of the problem will usually provide information about what variables any functions are in terms of.
The fact that the flows are shown in an x-y coordinate system is a strong clue.
Thank you very much. Makes perfect sense! :)