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Pennsylvania State University
MATH 251

Second Order Linear Partial Differential Equations Part I Second linear partial differential equations; Separation of Variables; 2▯ point boundary value problems; Eigenvalues and Eigenfunctions Introduction We are about to study a simple type of partial differential equations (PDEs): the second order linear PDEs. Recall that a partial differential equation is any differential equation that contains two or more independent variables. Therefore the derivative(s) in the equation are partial derivatives. We will examine the simplest case of equations with 2 independent variables. A few examples of second order linear PDEs in 2 variables are: 2 α u =xx t (one▯dimensional heat conduction equation) 2 a u =xx tt (one▯dimensional wave equation) u + u = 0 (two▯dimensional Laplace/potential equation) xx yy In this class we will develop a method known as the method of Separation of Variables to solve the above types of equations. © 2008, 2012 Zachary S Tseng E▯1 ▯ 1 The One-Dimensional Heat Conduction Equation Consider a thin bar of length L, of uniform cross▯section and constructed of homogeneous material. Suppose that the side of thebar is perfectly insulated so no heat transfer could occur through it (heat could possibly still move into or out of the bar through the two ends of the bar). Thus, the movement of heat inside the bar could occur only in the x▯direction. Then, the amount of heat content at any place inside thebar, 0 < x < L, and at any time t > 0, is given by the temperature distribution function u(x, t). It satisfies the homogeneous one▯dimensional heat conduction equation: 2 α u xxu t 2 Where the constant coefficient α is the thermo diffusivity of the bar, given by α = k/ρs. (k = thermal conductivity, ρ = density, s = specific heat, of the material of the bar.) Further, let us assume that both ends of the bar are kept constantly at 0 degree temperature (abstractly, by connecting themboth to a heat reservoir © 2008, 2012 Zachary S Tseng E▯1 ▯ 2 of the same temperature; more practically, say theyare immersed in iced water). This assumption imposes explicit restriction on the bar’s ends, in this case: u(0, t) = 0, and u(L,t) = 0. t > 0 Those two conditions are called the boundary conditions of this problem. They literally specify the conditions present at the boundaries between the bar and the outside. Think them as the “environmental factors” of the given problem. In addition, there is an initial condition: the initial temperature distribution within the bar, u(x, 0). It is a snapshot of the temperature everywhere inside the bar at t = 0. Therefore, it is an (arbitrary) function of the spatial variable x only. That is, the initial condition is u(x, 0) = f(x). Hence, what we have is a problem given by: 2 (Heat conduction eq.) α u =xx t , 0 < x < L, t > 0, (Boundary conditions) u(0,t) = 0, and u(L,t) = 0, (Initial condition) u(x,0) = f(x). This is an example of what is known, formally, as an initial▯boundary value problem. Although it is still true that we will find a general solution first, then apply the initial condition to find the particular solution. A major difference now is that the general solution is dependent not only on the equation, but also on the boundary conditions. In other words, the given partial differential equation will have a general solution for each distinct set of boundary conditions. © 2008, 2012 Zachary S Tseng E▯1 ▯ 3 But before any of those boundary and initial conditions could be applied, we will first need to process the given partial differential equation. What can we do with it? There are other tools (by Laplace transforms, for example), but the most accessible method to us is called the method of Separation of Variables. The idea is to somehow de▯couple the independent variables, therefore rewrite the single partial differential equation into 2 ordinary differential equations of one independent variable each (which we already know how to solve). We will solve the 2 equations individually, and then combine their results to find the general solution of the given partial differential equation. For a reason that should become clear very shortly, the method of Separation of Variables is sometimes called the method of Eigenfunction Expansion. © 2008, 2012 Zachary S Tseng E▯1 ▯ 4 Separation of Variables 2 Start with the one▯dimensional heat conduction equation α uxx u t Suppose that its solution u(x,t) is such a function that it can be expressed as a product, u(x,t) = X(x)T(t), where X is a function of x alone and T is a function of t alone. Then, its partial derivatives can also be expressed simply by: u = X T uxx X″T ux= X′T utt XT″ ut= XT′ uxt u =tx′T′ Hence, the heat conduction equation α u =xx cantbe rewritten as 2 α X″T = XT′. 2 Dividing both sides by α X T : ′′ ′ X = T X α T2 (α2 is a constant, so it could go to either side of the equation, but it is usually, and more conveniently, moved to the t side.) The equation is now “separated”, as all the x▯terms are on the left and t▯terms are on the right. Note: The above step would not have been possible if either X = 0 or T = 0. However, if either pa2t is zero, then u = XT = 0, which will trivially satisfy the given equation α u xxu . this constant zero solution is called the trivial solution of the equation. We know this is going to be the case. Therefore, we are now looking only for the nonzero solutions. © 2008, 2012 Zachary S Tseng E▯1 ▯ 5 But how do we completely pull it apart into 2 equations? The critical idea here is that, because the independent variables x and t can, and do, vary independently, in order for the above equation to hold for all values of x and t, the expressions on both sides of the equation must be equal to the same constant. Let us call the constant −λ. It is called the constant of separation. (The negative sign is optional, of course, since λ is an arbitrary number and it could be either positive or negative or even zero. But putting a negative sign here right now makes our later calculation a little easier.) Thus, X ′′ T′ = 2 =−λ . X α T Why must the two sides be the same constant? Well, think what would happen if one of the sides isn’t a constant. The equation would not be true for all x and t if that were the case, because then one side/variable could be held at a fixed value while the other side/variable changes. Next, equate first the x▯term and then the t▯term with −λ. We have X ′=−λ X → X″ = −λX → X″ + λX = 0, and, T′ 2 =−λ → T′ = −α λT → T′ + α λT = 0. α T Consequently, the single partial differential equation has now been separated into a simultaneous system of 2 ordinary differential equations. They are a second order homogeneous linear equation in terms of x, and a first order linear equation (it is also a separable equation) in terms of t. Both of them can be solved easily using what we have already learned in this class. © 2008, 2012 Zachary S Tseng E▯1 ▯ 6 Lastly, now that the partial differential equation becomes two ordinary differential equations, we need to similarly rewrite the boundary conditions. The boundary conditions can be rewritten as: u(0,t) = 0 → X(0)T(t) = 0 → X(0) = 0 or T(t) = 0 u(L,t) = 0 → X(L)T(t) = 0 → X(L) = 0 or T(t) = 0 If we choose T(t) = 0, both conditions would be satisfied. However, it would mean that the temperature distribution function, u(x, t) = X(x)T(t), would be the constant zero function (the trivial solution). That is a totally uninteresting solution that would not give us the general solution (it could not satisfy any initial condition, except when it is also constant zero). Hence, we have to let the new boundary conditions to be: X(0) = 0 and X(L) = 0. Therefore, at the end of this process, we have two ordinary differential equations, together with a set of two boundary conditions that go with the equation of the spatial variable x: X″ + λX = 0, X(0) = 0 and X(L) = 0 , 2 T′ + α λT = 0 . The general solution (that satisfies the boundary conditions) shall be solved from this system of simultaneous differential equations. Then the initial condition u(x,0) = f(x) could be applied to find the particular solution. © 2008, 2012 Zachary S Tseng E▯1 ▯ 7 Example: Separate t u + x u3 = 0 into an equation of x and an equation xx tt of t. Let u(x,t) = X(x)T(t) and rewrite the equation in terms of X and T: 3 3 t X″T + x XT″ = 0 , t X″T = − x XT″ . Divide both sides byX″ T″, we have separated the variables: 3 3 t T = − x X T ′′ X ′′ . Now insert a constant of separation: t T − x X = = −λ T ′′ X ′′ . Finally, rewrite it into 2 equations: t T = −λT″ → λT″ + t T = 0, − x X = −λX″ → λX″ − x X = 0. © 2008, 2012 Zachary S Tseng E▯1 ▯ 8 Example: Separate ux+ 2u tx− 10u tt 0, u(0, t) = 0, x (L, t) = 0. Let u(x,t) = X(x)T(t) and rewrite the equation in terms of X and T: X′T + 2X′T′ − 10XT″ = 0 , X′T + 2X′T′ = 10XT″ . Divide both sides X′ T″, and insert a constant of separation: T + 2T ′ 10X = =−λ . T ′′ X ′ Rewrite it into 2 equations: T +2T′ = −λT″ → λT″ + 2T′ + T = 0, 10X = −λX′ → λX′ + 10X = 0. The boundary conditions also must be separated: u(0,t) = 0 → X(0)T(t) = 0 → X(0) = 0 or T(t) = 0 ux(L,t) = 0 → X′(L)T(t) = 0 → X′(L) = 0 or T(t) = 0 As before, setting T(t) = 0 would result in the constant zero solution only. Therefore, we must choose the two (nontrivial) conditions in terms of x: X(0) = 0, and X′(L) = 0. © 2008, 2012 Zachary S Tseng E▯1 ▯ 9 The Two-Point Boundary Value Problems What we have done thus far is to separate the heat conduction equation, with 2 independent variables, into 2 equations of one variable each. Meanwhile we have also rewritten the boundary conditions, so that they now associate with the spatial variable x only. (1) X″ + λX 2 0, X(0) = 0 and X(L) = 0, (2) T′ + α λT = 0. The next task is to solve this system of two simultaneous ordinary differential equations, one of them with boundary conditions. We will look the 2 equations one at a time, and consolidate their solutions at the end. We will start off by solving the more interesting (and more complex) of the two, namely the second order linear equation where x is the independent variable. It is our first taste of a boundary value problem (BVP). (It is not an initial value problem; as mentioned earlier in this course, an initial value problem requires that both of its data points be taken at the same time/place, but here we have 2 data points taken at different instances of x: at x = 0 and x = L.) A little background first: unlike an initial value problem of a second order linear equation, whose solution’s existence and uniqueness (under certain well▯understood conditions) are guaranteed, there is no such guarantee for a boundary value problem. Indeed, take an arbitrarypairing of a differential equation and a set of boundary conditions, the odds are good that there is not a solution satisfying them, or that there are multiple solutions satisfying them. Notice that the boundary value problem in (1) arises from a homogeneous linear differential equation, which always has at least one solution (the constant zero solution, or the trivial solution) which would also satisfy the homogeneous boundary conditions given. Thus, (in this case, at least) the existence of a solution is not the issue. The constant zero solution, X(t) = 0, however, is not usable for us. Because, if X(t) = 0, then u(x,t) = X(x)T(t) = 0, which is not the general solution. (Why not?) Hence, what we are looking to find presently is a second, nonzero, solution to the given boundary value problem. That is, we are looking for instance(s) where the uniqueness of solution fails to hold. © 2008, 2012 Zachary S Tseng E▯1 ▯ 10 Somewhat fortunately for us, that in (1) above while the boundary conditions are fixed, the equation itself is not − note that the coefficient λ, which was just the arbitrary constant of separation, has not been determined. Thus, what we need to do here is somewhat a reverse of what you’d have expected to be doing. Namely, we will start with the fixed boundary conditions and try to find an equation (by finding an appropriate coefficient λ) that has a nonzero solution satisfying the given boundary conditions. (In other words, we are starting with the answer and then go looking for the correct question that would give that answer!) This kind of reversed boundary value problems is called an Eigenvalue problem. The specific value(s) of λ that would give a solution of the boundary value problem is called an eigenvalue of the boundary value problem. The nonzero solution that arises from each eigenvalue is called a corresponding eigenfunction of the boundary value problem. © 2008, 2012 Zachary S Tseng E▯1 ▯ 11 Eigenvalues and Eigenfunctions of a Two-Point BVP Hence, the next g
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