Question

The motion of a very viscous fluid in the two-dimensional (wedge) region \(-\alpha<\) \(\phi<\alpha\) can be described in ( \(\rho, \phi\) ) coordinates by the (biharmonic) equation $$ \nabla^{2} \nabla^{2} \psi \equiv \nabla^{4} \psi=0 $$ together with the boundary conditions \(\partial \psi / \partial \phi=0\) at \(\phi=\pm \alpha\), which represents the fact that there is no radial fluid velocity close to either of the bounding walls because of the viscosity, and \(\partial \psi / \partial \rho=\pm \rho\) at \(\phi=\pm \alpha\), which imposes the condition that azimuthal flow increases linearly with \(r\) along any radial line. Assuming a solution in separated-variable form, show that the full expression for \(\psi\) is $$ \psi(\rho, \phi)=\frac{\rho^{2}}{2} \frac{\sin 2 \phi-2 \phi \cos 2 \alpha}{\sin 2 \alpha-2 \alpha \cos 2 \alpha} $$.

Short answer

The full expression for \(\psi(\rho, \phi)\) is \( \frac{\rho^2}{2} \frac{\sin 2 \phi - 2 \phi \cos 2 \alpha}{\sin 2 \alpha - 2 \alpha \cos 2 \alpha} \).

Step-by-step solution

Biharmonic Equation and Variables

Start with the given biharmonic equation \(abla^4 \psi = 0\) in polar coordinates \((\rho, \phi)\). We'll use the method of separation of variables, assuming that \(\psi(\rho, \phi) = R(\rho) \Phi(\phi)\).

Applying Separation of Variables

Express \(\psi\) in the form of \(\psi(\rho, \phi) = R(\rho) \Phi(\phi)\). Substitute this form into the biharmonic equation to get separate equations for \(R(\rho)\) and \(\Phi(\phi)\).

Equation for \(R(\rho)\) and \(\Phi(\phi)\)

Upon separation, we get two ordinary differential equations: one for \(R(\rho)\) and one for \(\Phi(\phi)\). Solve these differential equations under the given boundary conditions.

Boundary Conditions for \(\Phi(\phi)\)

Apply the boundary conditions \(\partial \psi / \partial \phi = 0\) at \(\phi = \pm \alpha\). This means that \(\Phi(\phi)\) must satisfy \(\Phi'(\pm \alpha) = 0\).

Boundary Conditions for \(R(\rho)\)

Apply the boundary conditions \(\partial \psi / \partial \rho = \pm \rho\) at \(\phi = \pm \alpha\), which means \(R'(\rho)\) must behave linearly.

General Solution and Constants

Combine the solutions of \(R(\rho)\) and \(\Phi(\phi)\) to form the general solution for \(\psi\). Use the boundary conditions to find the constants in the solution.

Formulate \(\psi(\rho, \phi)\)

Integrate and simplify all terms to formulate the final expression for \(\psi(\rho, \phi)\). Given boundary conditions will guide toward the expression \( \psi(\rho, \phi) = \frac{\rho^2}{2} \frac{\sin 2 \phi - 2 \phi \cos 2 \alpha}{\sin 2 \alpha - 2 \alpha \cos 2 \alpha} \).

Key concepts

viscous fluid

When we talk about a 'viscous fluid,' we are referring to a fluid that exhibits high resistance to flow. Think of honey as opposed to water; honey flows much slower because it has a higher viscosity. In the given problem, we are dealing with a 'very viscous fluid,' which means that the fluid's internal friction is so high that certain simplifying assumptions can be made. These assumptions are critical for writing mathematical equations that describe the flow of such fluids. For instance, the no-slip condition states that the fluid velocity is zero at any boundary in contact with a solid. This condition leads to the boundary conditions provided in our problem.

separation of variables

The method of separation of variables is a powerful technique for solving partial differential equations. This technique assumes that the solution can be written as the product of functions, each depending on a single coordinate. For the biharmonic equation, we assume a solution of the form \(\psi\(\rho, \phi\) = R\(\rho\)\Phi\(\phi\)\). By substituting this form into our biharmonic equation \(abla^{4} \psi = 0\), we obtain two separate equations, one for \(R\(\rho\)\) and another for \(\Phi\(\phi\)\). This separation allows us to work on simpler ordinary differential equations, making the problem more manageable. The individual solutions to these ODEs can then be combined to form the general solution of our original PDE.

boundary conditions

Boundary conditions are essential for solving differential equations because they specify the behavior of the solution at the boundaries of the domain. In this problem, we have two sets of boundary conditions: one set for \(\rho\) and one for \(\phi\). Specifically, \(abla \psi / \partial \phi = 0 \) at \(\phi = \pm \alpha\), implies that there is no radial velocity at the walls because of the fluid's viscosity. Additionally, the condition \(abla \psi / \partial \rho = \pm \rho \) at \(\phi = \pm \alpha\), suggests that the azimuthal flow increases linearly with radius. These boundary conditions are used to find specific solutions to our differential equations by eliminating constant factors that don't satisfy them.

polar coordinates

Polar coordinates \((\rho, \phi)\) are particularly useful for dealing with problems involving symmetrical systems, like circular or wedge-shaped regions. In this problem, the wedge region is defined by \(-\alpha < \phi < \alpha\). Polar coordinates simplify the biharmonic equation by taking advantage of the symmetry, making it easier to apply separation of variables. The coordinate \(\rho\) represents the radial distance from the origin, while \(\phi\) is the angular position. By transforming the biharmonic equation and the boundary conditions into polar coordinates, we can more easily find solutions that fit the specific symmetry of the problem at hand.