M427K Handout: PDE’s: Two-Point Boundary Value ProblemsSalman ButtApril 25, 2006High Level ViewThis chapter’s progression can be a bit confusing, but here is our plan. We want to discusssolutions to partial differential equations (PDE’s). Our method will be to change a given PDE intoa set of ODE’s subject to initial value and boundary value conditions, via a technique called sepa-ration of variables. We have handled initial value problems, but we have yet to discuss boundaryvalue problems – this is the topic of today’s class. Now our proposed solution to the PDE will bea sum (often infinite) of solutions to the ODE’s. These sums will often involve cos and sin terms,so we will also discuss these series, known as Fourier series – this will be discussed tomorrow inlecture and probably in discussion section on Thursday. Once we have all this background covered,we will be ready to actually solve some PDE’s.Two-Point Boundary Value ProblemsRecall that our previous second-order differential equations often came with two initial condi-tions:y00+ p(t)y0+ q(t)y = g(t), y(t0) = y0, y0(t0) = y00Often, especially via experimentation, instead of initial value conditions (how would you determiney0(t0) anyway?), you often have the value of y or y0at two different points. That is, instead of y(t0)and y0(t0), you have y(α) and y(β) for some points α, β. Such conditions are called boundaryconditions and a differential equation with two such boundary conditions is called a two-pointboundary value problem:y00+ p(x)y0+ q(x)y = g(x), y(α) = y0, y0(β) = y1We will look for a solution to the above differential equation in the interval α < x < β. Now homo-geneous boundary value problems are slightly different from homogeneous initial value problems.Specifically, homogeneous BVP’s are of the formOtherwise, we say the BVP is nonhomogeneous. Linear BVP’s behave like linear systems: slightchanges in the boundary conditions can change the solutions drastically; this is not really the casewith IVP’s. This is not the only similarity between these two types of problems, as we shall see.Recall some facts about the linear systemA~x =~bwhere A is an n × n matrix, b is a given n × 1 vector and ~x is the unknown n × 1 vector we wantto determine. If A is nonsingular, the system has a unique solution for any~b. But if A is singular,1then the system has either no solutions or infinitely many solutions, depending on the given~b. Nextconsider the homogeneous system got by taking~b = 0:We know one obvious solution to this homogeneous s ystem: ~x = 0, the trivial solution. If A isnonsingular, then this is the only solution, but if A is singular, then there are infinitely manynon-trivial solutions. We can state these results as follows:1.2.Consider the following examples:y00+ 2y = 0, y(0) = 1, y(π) = 0 y00+ y = 0, y(0) = 1, y(π) = aBVP’s behave like linear systems and we know the solutions to a nonhomogeneous linear syste mdepend on the homogeneous system, so consider the homogeneous BVP:y00+ p(x)y0+ q(x)y = 0, y(α) = 0, y0(β) = 0Analogous to linear systems, this equation has the trivial solution y = 0. But does it have othersolutions? In fact the same relationship holds between nonhomogeneous and homogeneous BVP’sand nonhomogeneous and homogeneous linear systems. That is,1.22.Let’s look at the corresponding homogeneous equations to our examples above:So far we have seen a strong relationship between linear syste ms and BVP’s. Does the notionof eigenvalues and eigenvectors also carry over to BVP’s? I wouldn’t be talking about this if itwasn’t the case, right? Recall the equation:This equation always has the solution ~x = 0 despite what λ may be . But for certain values of λ –the eigenvalues – there are also nonzero solutions – the corresponding eigenvectors.Let’s look at the analogous homogeneous BVPThis is just the general case of our above examples, where λ = 2 and λ = 1. Extending theterminology from linear systems, we say λ is an eigenvalue of our BVP if there exists nontrivialsolutions to the BVP, and the solution itself is called an eigenfunction. Under this terminology,λ = 2 is and λ = 1 is . Let’s try to find othereigenvalues and eigenfunctions of this BVP. We will examine the cases λ < 0, λ > 0, and λ = 0separately since the solution to the BVP is different in every one of these cases.Case 1 (λ > 0): We will use a simple trick so we avoid excessive use of radicals and set λ = µ2:This is just a homogeneous second-order diffeq, so let’s look at the characteristic polynomial:The roots of this polynomial are simply r = and we have the general solution:WMAWLOG that µ > 0. Let’s look at the boundary conditions:3So these conditions tells us that c1= andRemember that we want nontrivial solutions to this BVP, so we want to make sure c26= 0. I.e.we want the sin term to be zero. When is sin µπ = 0? Precisely when µ = .Recalling that λ = µ2, we find our eigenvalues areThe eigenfunctions are merely multiples of sin nx for n = 1, 2, 3, . . . . The eigenfunctions are de-termined only up to the multiplicative constant c2, just as in the case for eigenvectors. The bookoften sets this constant to be 1. So in our case, the eigenfunctions areCase 2 (λ < 0): As before we use a simple trick λ = −µ2and write our BVPConsidering the characteristic equation , we find the roots are r = sothe solution is of the formNote that we could have used the standard solution y = c1eµx+ c2e−µx, but we chose this solutionfor convenience when we apply the boundary conditions. Considering these conditions, we findc1= 0 andSince µ 6= 0 and recalling the definition of sinh, we see that sinh µπ 6= 0, so we know c2is forced tobe 0. Thus the only solution to this BVP is the trivial solution y = 0. Hence we know that thereare no negative eigenvalues of this BVP.Case 3 (λ = 0): Our BVP in this case just becomesWhat function integrated twice gives you 0? Well, just a polynomial of degree one:Examining the boundary conditions, we see that we forced to conclude c1= c2= 0. Hence λ = 0is not an eigenvalue.Putting all this together, we see that the full list of (real) eigenvalues is λn= n2for n =1, 2, 3, . . . with corresponding eigenfunctions sin nx (up to constant multiples). There are no com-plex eigenvalues, though we have not addressed this point (this is Problem 23 on p. 576 of thetext).These problems have real application beyond being an analogy between linear