Solving Laplace’s Equation

The Laplace’s Equation V2 ф = V2y = 0, Eq. 4.24, must be solved for the single depen­dent variable, either ф or y, subject to the appropriate boundary conditions for the problem under consideration. This ensures that the solution describes the proper physical reality. The second-order Laplace’s Equation requires two such boundary conditions. For the aerodynamic problems, these are (1) the disturbance due to the presence of the body in the flow must die out far away from the body in the freestream; and (2) the velocity component normal to the body surface must be zero because the body is a solid. These two boundary conditions are expressed in terms of the scalar functions, as follows:

far from the body, u = — = — = V ; v = 0

dx dy

and

at the body surface, ^= 0, dn ds

where n and s are the directions normal and tangential to the body surface, respec­tively. In other words, the flow must be tangent to the body surface.

The Laplace’s Equation has a property that may be used to simplify enor­mously the solution—namely, that the equation is linear. Linearity means that in the equation, there are no products or terms taken to a power that contains the

are not. It

also means that no transcendental functions of the dependent variable appear. This linearity property means that superposition of solutions may be used. Thus, if ф1 and ф2 are each a solution to the Laplace’s Equation, then the linear combination ф = аф1 + Ьф2 (where a and b are arbitrary constants) also is a solution.[15] This may be verified by direct substitution into Laplace’s Equation. Of course, the appropriate boundary conditions must be satisfied by solution ф.

The far-reaching significance of superposition is that simple (or elementary) flow solutions may be added together to form solutions corresponding to more com­plicated flows. This means that it may not be necessary to solve Laplace’s Equation to carry out the solution strategy described previously. A much simpler strategy can be followed. Consider the following steps:

1. Specify the velocity components for simple flows by inspection (e. g., for a uni­form flow, the velocity components are u = VM and v = 0).

2. Integrate to find ф or у for each simple flow as in Examples 4.6 and 4.8. The con­stant of integration may be set equal to zero or to any convenient value because the final variables of interest are the velocity components, the derivatives of ф or у, and the constant of integration drops out.

3. Superpose (add) the simple (elementary) solutions for ф or у corresponding to these simple flows.

4. Differentiate this superposed solution to obtain the velocity components for a new, more complicated flow and then use these components to determine the velocity magnitude.

5. Use the Bernoulli Equation to solve for the magnitude of the corresponding static pressure.

The next section explains how elementary solutions to the Laplace’s Equation for simple flows are found. These elementary solutions then are superposed in Sec­tion 4.6 to construct more complex and realistic flow-field solutions.