# Fluids PDE Problem: Understanding the Elimination of c_1 in Boundary Condition

• member 428835
In summary, the conversation discusses a problem involving fluid flow in 2D passing a cylinder of radius a. The resulting PDE is solved in polar coordinates, yielding an infinite series solution. In the far wake, it is suggested that the velocity in the x direction is constant, leading to a simplified solution of psi = Uy. The question then arises as to why there is no constant term in this solution and whether the natural logarithm can be ignored due to the growth of r.

#### member 428835

Hi PF!

So my book has boiled the problem down to $$\psi(r,\theta) = c_1 \ln \frac{r}{a}+\sum_{n=1}^\infty A_n\left(r^n-\frac{a^{2n}}{r^n}\right)\sin n\theta$$ subject to ##\psi \approx Ur\sin\theta## as ##r## get big. The book then writes
$$\psi(r,\theta) = c_1 \ln \frac{r}{a}+U\left(r-\frac{a^2}{r}\right) \sin \theta$$ I understand ##A_n=0\, \forall\, n \geq 2## and ##A_1=U## but why isn't ##c_1## eliminated too? To me it seems the natural log does not allow the "boundary condition" to be satisfied.

Whats the context of the problem?

Fluid flow in 2D passing a cylinder of radius ##a##. The flow satisfies continuity, so ##\nabla \cdot \vec{v} = 0##. Define ##\psi \equiv \nabla \times \vec{v}## so that continuity is satisfied. Then ##v_x = \psi_y## and ##v_y=-\psi_x##. The resulting PDE is then ##\nabla^2 \psi = 0##. Solving this in polar coordinates yields the above solution I posted, the infinite series. In the far wake, velocity in the ##x## direction is constant ##U##, which suggests ##v_x = \psi_y = U \implies \psi = Uy## (I guess I don't understand why there is no constant too, which is to say why isn't this true: ##\psi = Uy + C_0(x)##).

Any ideas? My thought process is since ##r## grows much larger than ##\ln r## we may leave the natural logarithm? Can anyone confirm this?

## 1. What is a fluid PDE problem?

A fluid PDE (partial differential equation) problem involves studying the behavior of a fluid (such as water or air) using mathematical equations and computational methods. These equations describe how the fluid's properties, such as velocity and pressure, change over time and space.

## 2. Why is the elimination of c1 important in boundary conditions?

The elimination of c1 in boundary conditions is important because it allows for a unique solution to the fluid PDE problem. Boundary conditions are constraints that define the behavior of the fluid at the boundaries of the problem domain. By eliminating c1, we can ensure that the solution satisfies these boundary conditions and accurately represents the behavior of the fluid.

## 3. How is c1 eliminated in boundary conditions?

C1 can be eliminated in boundary conditions by using the method of characteristics. This involves transforming the PDE into a system of ordinary differential equations (ODEs) and solving for the unknown coefficients, including c1. Once c1 is eliminated, the ODEs can be solved to obtain the unique solution to the fluid PDE problem.

## 4. What is the significance of the elimination of c1 in fluid dynamics?

The elimination of c1 is significant in fluid dynamics because it allows for the accurate prediction of fluid behavior in various situations. Without eliminating c1, the solution to the PDE problem may not satisfy the boundary conditions, leading to incorrect or unrealistic results. By eliminating c1, we can ensure that the solution accurately describes the fluid's behavior and can be used for practical applications.

## 5. Are there any limitations to the elimination of c1 in boundary conditions?

Yes, there can be limitations to the elimination of c1 in boundary conditions. In some cases, it may not be possible to completely eliminate c1 without making assumptions about the fluid's behavior or the problem domain. Additionally, the elimination of c1 may only result in an approximate solution, rather than an exact one. These limitations should be considered when using the elimination of c1 in fluid PDE problems.