2 A brief interruption to discuss boundary value/intial value problems

Transcription

1 The lecture of 1/9/ The one dimensional heat equation The punchline from the derivation of the heat equation notes (either the posted file, or equivalently what is in the text) is that given a rod of length L, such that the temperature u = u(x, t) at time t, at a point x away from the left endpoint is the same at all points of the cross section at x, then u satisfies the following PDE: u t (x, t) = ku x x(x, t) + q(x, t), where k > 0 is a constant, and q(x, t) is the heat energy generated by unit volume, unit time, at x at time t. For k to be a constant, we have to assume that the rod has constant density and constant specific heat. For this problem, to keep it one dimensional, we will always assume that the rod is perfectly insulated on the sides, that the only way heat can enter or escape is at the end-points. 2 A brief interruption to discuss boundary value/intial value problems This discussion will be done assuming three space variables x, y, z. The transition to two or one space variable should be fairly obvious. In applications one considers two types of PDE problems. A static one, in which one has to find a function u = u(x, y, z) in a region of 3-space. The region is usually an open set with a reasonably nice boundary. For example: Find u(x, y, z) such that u = 0 in x 2 + y 2 + z 2 < 1. Then there is the dynamic one in which there is an extra variable interpreted as time, usually denoted t. One has again given a region and one has to find u = u(x, y, z, t) such that u satisfies a PDE for all (x, y, z), t > 0. Example Find u(x, y, z, t) such that u t = k u for(x, y, z), t > 0. Here acts only on the space variables; the equation is u t = k ( 2 u x u y u z 2 Both from a physical as well as from a mathematical point of view, something is missing. One usually (meaning always) has to add some conditions. For the static problem the conditions to add are boundary conditions. One prescribes the behavior of u and/or some of its derivatives, on the boundary of. We call such a problem a BV problem or BVP. For the dynamic problem one prescribes not only boundary conditions but also initial conditions; what is the value of u and some of its time derivatives at time t = 0. The number of time derivatives one can (and usually must) prescribe is one less than the highest time derivative in the equation. For the example above, that is 0, one would only prescribe u. Such a problem is an IVB problem, or IVBP. Here is an example of such a problem: Find u = u(x, y, z, t) such that u tt = u, x 2 + y 2 + z 2 < 1, ). u(x, y, z, t) + u ν (x, y, z, t) = 0, x2 + y 2 + z 2 = 1, t > 0, u(x, y, z, 0) = f(x, y, z), u t (x, y, z, 0) = g(x, y, z), x 2 + y 2 + z 2 < 1, where f, g are given. Here u ν = u ν indicates the normal derivative. It is also possible to have simply IV problems; just initial values get prescribed (free boundaries, or no boundary).

2 3 INITIAL AND BOUNDARY CONDITIONS FOR THE ONE-DIMENSIONAL HEAT EQUATION 2 3 Initial and boundary conditions for the one-dimensional heat equation The region in this case is the interval (0, L), so the boundary consists of the points 0 and L. A typical ICBV problem is: Solve; that is find u = u(x, t) such that, u t = ku xx + q, 0 < x < L, t > 0, au(0, t) + bu x (0, t) = α(t) cu(l, t) + du x (L, t) = β(t), t > 0, u(x, 0) = f(x), 0 < x < L. The assumption here is that not both a, b are 0, and not both c, d are 0. What the BC mean. We analyze at the left end-point x = 0. The analysis for x = L is the same. We can have b = 0 so the condition is u(0, t) = α(t). That means that the temperature at the left endpoint is kept always equal to a prescribed value α(t). Common case u(0, t) = 0 for t > 0. If the rod is perfectly isolated at x = 0, then this is described as u x (0, t) = 0 for t > 0. (Case a = 0, α(t) 0. Prescribing a heat flux gives u x (0, t) = α(t) for t > 0. The rod could be in a medium of temperature U(t) at time t. Then what happens at the end-points (the only points where the medium affects the temperature of the rod) can be modelled by Newton s law of cooling: the heat flux is proportional to the difference in temperature of the rod and the medium. The heat flux is proportional to the temperature gradient; φ(x, t) = K 0 u(x, t), so this works out to K 0 u x (0, t) = H[u(0, t) U(t)], for t > 0. On the left endpoint one has to change signs: K 0 u x (L, t) = H[u(L, t) U(t)], for t > 0. Classical solutions. (Just solutions for us). A function that has all the derivatives appearing in the equation; these derivatives are continuous, and it satisfies the boundary and initial conditions. This requires some smoothness of the prescribed conditions. Example 4.7 of the textbook Suppose we have the following problem, and we can assume α, β, q are continuous. u t = ku xx + q, 0 < x < L, t > 0, u(0, t) = α(t) u(l, t) = β(t), t > 0, u(x, 0) = f(x), 0 < x < L. A solution (classical solution) is a function that is once continuously differentiable with respect to t, twice with respect to x in the strip {(x, t) : 0 < x < L, t > 0} and it must satisfy the equation at all points of this strip. It also has to satisfy that lim t 0+ u(x, t) = f(x) for all x, 0 < x < L, and lim x 0+ u(x, t) = α(t), lim x L u(x, t) = β(t) for all t > 0. What happens at the corner points x = 0, t = 0, x = L, t = 0 is left open. But, we have Theorem 4.9 of the textbook if we also can assume that u is continuous at all points of the strip with boundary {(x, t) : 0 x L, t 0}. We turn to the statement and proof of that theorem. I know that for a lot of you proof is a dirty word, so I will go very lightly with proofs in the future. But for a while, humor me. 4 auss divergence theorem and the higher dimensional heat equation What was done before in one space dimension can be done also in 2 or 3 space dimensions. I will ignore heat sources and assume we have a region in 3-space (the text does two space); E = E(x, y, z, t) is the thermal energy density at (x, y, z) at time t. The relation between E and u is E(x, y, z, t) = c(x, y, z)ρ(x, y, z)u(x, y, z, t),

3 4 AUSS DIVERENCE THEOREM AND THE HIHER DIMENSIONAL HEAT EQUATION 3 where c(x, y, z) is the specific heat at the point (x, y, z); we recall that the specific heat of a substance is defined as the amount of energy that has to be applied to raise the temperature of a unit mass of the substance by one unit of temperature. For example, it takes 4200 joules to raise the temperature of one kg of water by one degree Celsius, thus the specific heat of water is joules 4200 kg degrees celsius. The specific heat can vary from point to point if the substance is not homogeneous, ρ(x, y, z) is the density of the substance; mass/volume. The heat flux, which computes how energy flows, is assumed to obey Fourier s law which states that the amount of energy escaping across a surface patch S as in the picture is K 0 u ν where ν is the normal in the direction of the flow, indicated by ˆn in the picture. If we now consider a small little subregion, so small that everything is mostly constant in there. And now we interrupt for a few words on surface integrals, and related notions. This interruption would be unnecessary if calculus 3 were taught properly. It isn t, at least not usually at FAU, because Calculus 1 and 2 are not taught properly. All calculus courses spend inordinate amounts of time doing similar exercises again and again, and in the end there isn t enough time to develop the ideas as they should be developed. Students come out of calculus knowing a certain number of techniques they soon forget, but are not quite sure what is important and what isn t. The question on why calculus isn t taught properly could be a topic of a seminar. All this, of course, is my opinion; I could be wrong. Incidentally, a good place to brush up on this topic could be the article on auss divergence theorem in Wikipedia. We will work with regions in spaces of 1, 2, and 3 dimensions. By definition, a one dimensional region will just be an open interval, one of the following: (a, b), (a, ), (, b), or (, infty). Two dimensional regions will be planar sets bounded by a mostly smooth curve (a few corners are allowed) that does not self-intersect. It could be the inside of a circle or a square, or a region with a more complicated curve as a boundary, as shown in the picture.

4 4 AUSS DIVERENCE THEOREM AND THE HIHER DIMENSIONAL HEAT EQUATION 4 In 3 dimensions it would be a solid bounded by a closed surface that is mostly smooth, for example a ball, a cube, a torus (doughnut), or try to imagine a three dimensional analog of the planar region above. If the symbol for the region is, we use the symbol to denote its boundary; the boundary consisting of all points P so that there are points inside the region and outside of the region infinitely close to P. For example: In one dimension, if I = (0, 5) (the interval from 0 to 5), then I consists of just the two points 0 and 5; I = {0, 5}. In two dimensions, if = {(x, y) : x 2 + y 2 < 4} (disk of radius 2 centered at the origin), then = {(x, y) : x 2 + y 2 = 4} (circle of radius 2 centered at the origin). In two dimensions, if is the red region of the figure above, then consists of all points of the black curve surrounding the figure. In three dimensions, if = {(x, y, z) : (x 1) 2 + (y 2) 2 + z 2 < 9} (ball of radius 3 centered at (1, 2, 0)), then = {(x, y, z) : (x 1) 2 + (y 2) 2 + z 2 = 9} (sphere of radius 3 centered at (1, 2, 0)). Or one could say that the boundary is what you think the boundary should be. At least for two and three dimensional regions, one can integrate functions defined on the boundary. The integral of f over the boundary is sort of a sum of all the infinite values of S, weighted by that amount of boundary length or area at which those values are assumed. I cannot go into details here, that s what Calculus 3 is supposed to do. If is a two or three dimensional region, we ll denote by fds or, if a variable is to be specified, by f(x)ds x the integral of a function f defined (and continuous) at all points of over. Calculus 3 texts usually use ds for the integral over curves, such as the boundary of a planar region (arc length) and ds for integration on surfaces, such as the boundary of solids. I ll use ds for both. One has, if f(x) 1, { length of the boundary of in the case of two dimensions, ds = area of the boudary of in the case of three dimensions. The interpretation of the boundary integral is that it is a sort of sum of all values of f weighted by the length or area of the set of points where they are assumed. In the days of Newton and friends ds x really meant something, for each x on the curve or surface, it was the length (curve) or area (area) of an infinitesimal portion of the curve or surface, respectively. Then f(x)ds x was an actual product of the value of f at x by ds x ; also an infinitesimal. But when you added up all these infinitesimals along the curve or surface, you got a finite, perceptible value; the integral of f over the boundary. At almost every point x of the boundary of a nice region there is a unique vector of length 1 perpendicular to the boundary and pointing outward. The exception are corners, and there shouldn t be too many of them. This vector is called the outward normal and I ll denote it by ν or by ν(x) if the point x is to be emphasized. If F is a vector field defined (and continuous) on the region and its boundary, then auss divergence theorem states that F(x)dx = F ds. This is a more or less sophisticated way of stating it, in which x stands for a two or three dimensional vector, and can be a single integral, a double integral, or a triple integral. The way Calculus books do it doesn t allow, as it should, for a unified treatment. If one uses x, y for two dimensions, x, y, z for three, then the breakdown is The two dimensional version: In this case F = f, g and it is a convention (an annoying one 1, I think) of Calculus texts to use da for integrals over planar regions, so the theorem looks like ( ) f g (x, y) + (x, y) da = (f(x, y)ν 1 (x, y) + g(x, y)ν 2 (x, y)) ds. x y 1 it really defeats the purpose of having the d symbol, which should indicate which are the variables of integration.

5 4 AUSS DIVERENCE THEOREM AND THE HIHER DIMENSIONAL HEAT EQUATION 5 The three dimensional version: In this case F = f, g, h and the Calculus texts (silly) convention is to use dv for integrals over spatial regions, so the theorem looks like ( f x (x, y, z) + g y ) h (x, y, z) + (x, y, z) dv = (f(x, y, z)ν 1 (x, y, z) + g(x, y, z)ν 2 (x, y, z) + h(x, y, z)ν 3 (x, y, z)) ds. z This also works in the one dimensional case, in which is an interval (a, b). In this case = {a, b}. The integral of f over the boundary is the sum of its values on the boundary; that is, it is just f(a) + f(b). The normal ν(a) at a is the vector < 1 >; an arrow of length 1 pointing to the left at a; ν(b) =< 1 >, and arrow of length 1 pointing to the right. A one-dimensional vector field on the interval is just a function f(x) defined for all values of x in [a, b] and f(x) = f (x). Also f ν at b is < f(b) > < 1 >= f(b), at a, < f(a) > < 1 >= f(a). The formula F dx = F(x) ν(x) ds reduces to b in other words, the fundamental theorem of Calculus. a f (x) dx = f(b) f(a); This concludes the interruption; let us return to where we were. Suppose we have a certain region of either two dimensional or three dimensional space where thermal energy is stored. It could be a metal plate (two dimensions) or a solid body (three dimensions). At a generic point P of this region we consider a very small portion of this region surrounding P ; let s call this small portion. The total thermal energy in is E(x, y, t) da (2 dims.) E(x, y, z, t) dv (3 dims.) To keep the treatment unified, I will write merely x to stand for either (x, y) or (x, y, z), dx stands for either da or dv and the total energy in the tiny portion is E(x, t) dx c(x)ρ(x)u(x, t) dx. The rate at which this energy changes is thus given by d c(x)ρ(x)u(x, t) dx = dt c(x)ρ(x) u (x, t) dx. t This rate must equal the total amount of energy that has entered or left the portion per unit time. Over a small patch S of the boundary of this rate equals the flux φ(x, t) which, according to Fourier s law equals K 0 u ν, for a constant K. Adding all this up and equating one gets that c(x)ρ(x) u t (x, t) dx = K 0 u(x) ν(x) ds x. Here is where auss divergence theorem comes in. Applying it, one gets, because the divergence of a gradient is the Laplacian ( u = u): c(x)ρ(x) u t (x, t) dx = K 0 u(x) x, or ( c(x)ρ(x) u ) t (x, t) K 0 u(x) x = 0. This is true for any portion of our big region, and a function cannot have integral 0 over all little regions of a big region without being itself 0, so this shows that u must satisfy u t = k u. where k = k(x) = K 0 /(c(x)ρ(x)). This is how one shows that the temperature satisfies the heat equation.

6 5 LAPLACE S EQUATION 6 5 Laplace s equation As time goes on, if there are no outside sources of heat, the temperature becomes stable; it does not change any more in time, u t = 0. The temperature distribution, which satisfied the heat equation now satisfies Laplace s equation u = 0. This equation appears in a lot of situations, not only for stable temperature. For example it is satisfied by the electric field generated by a charged body. It is also satisfied by functions u = u(x, y) whose graph has minimal area for the given boundary. The graph of such a function can be visualized by shaping a piece of wire into the boundary, then dipping it into a soap solution. The soap film is the graph of a harmonic function, as solutions to Laplace s equation are called. Laplace s equation is usually associated with a boundary value, similar to the boundary value for the heat equation. 6 The Wave Equation This equation has the form 2 u t 2 = c2 u. it is a model for wave motion; for example in 3 dimensions u(x, y, z, t) can measure a disturbance propagating through space; for example it could be pressure at point (x, y, z) at time t due to a sound source. Or, in two dimensions, we could have thrown a stone (or several) into a swimming pool, and u(x, y, t) is how high above (or below) the horizontal is the water at time t, at point (x, y). In other words, it measures the ripple at (x, y), time t. Or, in one dimension, u(x, t) could measure the displacement from rest position at time t at x, of a taut string that has been plucked. In any case, the interpretation of c is the speed at which the disturbance propagates.