Lecture 38. Boundary Value Problems and Eigenfunctions1. Boundary ConditionsDifferential EquationsLECTURE 38Boundary Value Problems and Eigenfunctions1. Boundary ConditionsAt this point, we’re essentially finished discussing ordinary differential equations. We’re goingto begin discussing partial differential equations. Partial differential equations are much morecomplicated than ordinary differential equations: we’ll need to specify the way we want solutionsto behave on the boundary of the region our equation is valid on. Such data are called boundaryvalues or boundary conditions, and a combination of a differential equation and some boundaryconditions is called a boundary value problem.Boundary conditions depend on the region our equation is valid on: if we have an ordinarydifferential equation, this will be some interval; for a partial differential equation, this might alsobe an interval, or it might be on a square in the plane. To try to get a better feel for how boundaryvalues work, let’s see how they affect solutions to ordinary differential equations.Example 38.1. Let’s consider the second order differential equation y00+ y = 0. Specifyingboundary conditions for this equation involves specifying the values of the solution (or its deriva-tives) at two points. Let’s say we fix y(0) = 0 and y(2π) = 0. We know that solutions to theequation have the formy(x) = A cos(x) + B sin(x).Applying the first boundary condition tell us that 0 = y(0) = A. Applying the second conditiongives 0 = y(2π) = B sin(2π). Now, sin(2π) = 0, so this condition doesn’t tell us anything about B;it can be anything and this condition will still be satisfied. So the solutions to this boundary valueproblem are any functions of the formy(x) = B sin(x). Example 38.2. Consider y00+ y = 0 with boundary values y(0) = y(6) = 0. This seems similarto the previous problem; the solutions have the formy(x) = A cos(x) + B sin(x)and the first condition still tells us A = 0. The second condition tells us that 0 = y(6) = B sin(6).Now, sin(6) 6= 0, so we must have B = 0 and the entire solution is y(x) = 0. Boundary value problems are very natural things. We can interpret Examples 38.1 and 38.2physically. We know the equation y00+ y = 0 models an oscillator; something like a rock hangingfrom a Slinky1This rock oscillates with a frequency of12π. The condition y(0) = 0 just means thatwhen we start observing, we want the rock to be at the equilibrium spot. If we specify y(2π) = 0,this is no problem for any solution, because it will automatically happen: the motion is 2π-periodic.On the other hand, it’s impossible for the rock to return to the equilibrium point after 6 seconds.It will come back in 2π seconds, which is a bit more than 6. So the only possible way the rockcan be at equilibrium after 6 seconds is if it doesn’t leave, which is why the only solution is the 0solution.1It’s Slinky, it’s Slinky, for fun it’s a wonderful toy. It’s Slinky, it’s Slinky, it’s fun for a a girl and a boy!1Differential Equations Lecture 38: Boundary Value Problems and EigenfunctionsExamples 38.1 and 38.2 are examples of homogeneous boundary value problems. We say thata boundary value problem is homogeneous if the equation is homogeneous and the two boundaryconditions involve zero. That is, homogeneous boundary conditions might be of the following types:y(x1) = 0 y(x2) = 0y0(x1) = 0 y(x2) = 0y(x1) = 0 y0(x2) = 0y0(x1) = 0 y0(x2) = 0.On the other hand, if the equation is nonhomogeneous or any of the boundary conditions don’tinvolve zero, we say that the boundary value problem is nonhomogeneous. Let’s look at someexamples of nonhomogeneous boundary value problems.Example 38.3. Take y00+ 9y = 0 with boundary conditions y(0) = 2 and yπ6= 1. Thegeneral solution to the differential equation isy(x) = A cos(3x) + B sin(3x).The two conditions give2 = y(0) = A1 = yπ6= Bso that the solution isy(x) = 2 cos(3x) + sin(3x). Example 38.4. Take y00+ 9y = 0 with boundary conditions y(0) = 2 and y (2π) = 2. Thegeneral solution to the differential equation isy(x) = A cos(3x) + B sin(3x).The two conditions give2 = y(0) = A2 = y (2π) = A.In other words, this time the second condition didn’t give us any new information, like in Example38.1 and B doesn’t affect whether or not the solution satisfies the boundary conditions or not. Wethen have infinitely many solutions,y(x) = 2 cos(3x) + B sin(3x). Example 38.5. Take y00+ 9y = 0 with boundary conditions y(0) = 2 and y(2π) = 4. Thegeneral solution to the differential equation isy(x) = A cos(3x) + B sin(3x).The two conditions give2 = y(0) = A4 = y (2π) = A.On the one hand, we must have A = 2. On the other, we have to have A = 4. This isn’t possible,and this boundary value problem in fact has no solutions. These examples hopefully illustrate that a small change to the boundary conditions can dra-matically change the problem, unlike with initial value
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