Section 8.0 Introduction to Boundary Value Problems. How do initial value problems (IVPs) and boundary value problems (BVPs) differ?

Transcription

1 Section 8.0 Introduction to Boundary Value Problems Key terms/ideas How do initial value problems (IVPs) and boundary value problems (BVPs) differ? What are boundary conditions? In what type of problems do BVPs arise? Examples where BVPs arise.

2 With IVPs we had a differential equation and we specified the value of the solution and an appropriate number of derivatives at the same point (collectively called initial conditions). For instance for a second order differential equation yꞌꞌ(t) = f(t, y(t), yꞌ(t)) the initial conditions are, y(t 0 ) = α, yꞌ(t 0 ) = β. With BVPs we will have a differential equation and we will specify the function and/or derivatives at different points, which we ll call boundary values. For second order differential equations, which will be looking at pretty much exclusively here, any of the following can be used for boundary conditions. A BVP has conditions specified at the extremes ("boundaries") of the independent variable in the equation whereas an initial value problem has all of the conditions specified at the same value of the independent variable (and that value is at the lower boundary of the domain, thus the term "initial" value). Shows a region where a differential equation is valid and the associated boundary values

3 For example, if the independent variable is time over the domain [0,1], a boundary value problem would specify values for y(t) at both t = 0 and t = 1, whereas an initial value problem would specify a value of y(t) and yꞌ(t) at time t = 0. Finding the temperature at all points of an iron bar with one end kept at absolute zero and the other end at the freezing point of water would be a boundary value problem. If the problem is dependent on both space and time (making it a partial differential equation), one could specify the value of the problem at a given point for all the time data or at a given time for all space data. To be useful in applications, a boundary value problem should be well posed. This means that given the input to the problem there exists a unique solution, which depends continuously on the input. Much theoretical work in the field of partial differential equations is devoted to proving that boundary value problems arising from scientific and engineering applications are in fact well-posed.

4 We will confine our attention to second order BVPs with ordinary DEs. A general second-order boundary value problem (BVP) asks for a solution of on the interval a t b. This illustrated in the figure. Schematic picture of a boundary value problem. In an initial value problem, the initial value ya and initial slope sa = y (a) are specified as part of the problem. In a boundary value problem, boundary values ya and yb are specified instead; sa is unknown. As we saw previously, a differential equation under typical smoothness conditions has infinitely many solutions, and extra data is needed to pin down a particular solution. In a second-order equation two extra constraints are needed. They are given as boundary conditions for the solution y(t) at a and b.

5 To aid your intuition, consider a projectile, which satisfies the second order DE y'' = -g as it moves, where y is the projectile height and g is the acceleration of gravity. Specifying the initial position and velocity uniquely determines the projectile s motion, as an initial value problem (IVP). On the other hand, a time interval [a, b] and positions y(a) and y(b) could be specified. In that case we have a boundary value problem (BVP) which in this case also has a unique solution. Problem: Find the height trajectory of a projectile that begins at the top of a 120-foot building and reaches the ground 3 seconds later. Using Newton s second Law, F = ma, where the force of gravity is F = -mg and g = 32 ft/sec 2 we obtain the DE y'' = -g where y(t) is the height at time t. The trajectory is the solution of the IVP y'' = -g y(0) = 120 y'(0) =v 0 or the BVP y'' = -g y(0) = 120 y(3) =0 Since we don t know the initial velocity we first solve the BVP. Integrating twice gives y(t) = gt + v t + y = -16t + v t + y Using the boundary conditions gives 120 = y(0) = y 0 and 0 = y(3) = v Solving for v 0 we obtain v 0 = 8 ft/sec. Thus the trajectory is y(t) = -16t 2 + 8t +120.

6 The existence and uniqueness theory of boundary value problems is more complicated than the corresponding theory for initial value problems. Seemingly reasonable BVPs may have no solutions or infinitely many solutions, a situation that was rare for IVPs. The situation is analogous to the arc of a human cannonball acting under earth s gravity. The only information needed to determine the cannonball s path is the original position and velocity. A unique solution to the initial value problem always exists. On the other hand, if the net is set beyond the range of the cannon, no solution can exist. Moreover, for any boundary condition smaller than the cannon s range, there are two solutions, a short trip and a longer trip, violating uniqueness. Ref: T. Sauer, Numerical Analysis