M427K Handout: PDE’s: The Heat and Wave EquationsSalman ButtMay 4, 20061. Next week, I will continue to hold office hours. Regular discussion will also be office hours.The Heat EquationRecall the setup and solution for the heat equation:Let’s consider an example:uxx= ut, L = 1, f(x) = x(1 − x)1The Wave EquationOur next PDE of interest is the wave equation. Here is the setup: We have an elastic stringof length L tightly stretched between two supports along a horizontal line which we place on thex-axis. We pluck the string so that it vibrates in the vertical plane. We let u(x, t) be the verticaldisplacement of the string at some point 0 ≤ x ≤ L and at some time t. We ignore damping effectsand assume the amplitude of motion is not too large (this is to avoid pathologies in our solution).Then u(x, t) satisfies the differential equationThis is the 1-dimensional wave equation (see Appendix B for the derivation). The constant a isthe velocity of propagation of waves along the string. We need some initial and boundary conditions.We know the ends of the string are to remain fixed, so we have the boundary conditionsThe initial condition will be in terms of t and we have the second-derivative of u with respect to tin our e quation, so we want two initial conditions:Here f, g will be given functions. These functions must satisfy f (0) = f(L) = g(0) = g(L) = 0 inorder to be consistent with our initial condition. We nee d to now solve this problem. As before, itis an IVP with respect to t and a BVP with respect to x. But viewing the problem in the xt-plane,we see we have a BVP:Type I: g ≡ 0. We will be looking at problems where g ≡ 0. So we are looking at the BVPAs in the heat equation, we assume we can write u asPlugging in for our PDE, we find as beforeThus we have two ODEsThe first equation is the same as in the heat equation and we know the solution (the boundaryconditions w ill be homogeneous as in the heat e quation case):The second equation is new though:2We know how to solve this ODE, yieldingwhere k1, k2are arbitrary constants. Looking at the second condition (which tell us that T0(0) = 0),we find that k2must be 0. Thus (letting k1be normalized to 1), we find our fundamental solutionSticking all these solutions into a linear combination, we findFinally, we check our first initial condition u(x, 0) = f(x) to find thatBut this just means f has a Fourier sine series expansion, so our coefficients cnare given byThus once we determine these cn’s, we are done.Type II: f ≡ 0. We a re now looking at the BVPWe again use separation of variables, assuming that u(x, t) = X(x)T(t). The problem for X willbe as before, so we have the solution there. For T our initial conditions require that T (0) = 0. Asbefore, our solution for T isBut when we enforce that T (0) = 0, we find that k1is forced to be 0. Thus T (t) = sin(nπat/L)after normalizing k2to be 1. Thus our fundamental solution isPutting all these fundamental solutions together, we findTo determine the cn’s, we use the second initial condition which requires that we differentiate oursolution and se t t = 0:Thus (nπa/L)cnis the n-th coefficient in the Fourier sine series, so we have:3Determining these cn’s, we are done.Type III: f 6≡ 0, g 6≡ 0. We now have the BVPwhere f, g are the given initial position and velocity functions respectively. This problem can alsobe done using separation of variables, but it can also be done by just adding our two solutions fromabove together. To see this, let v(x, t) be a solution to Type I, w(x, t) be a solution to Type II, andlet u(x, t) = v(x, t) + w(x, t). We want u(x, t) to be a solution to a2uxx− utt= 0. Let’s plug in:Thus u satisfies the equation. Does it satisfy the initial and boundary conditions? Let’s check:Thus u = v + w is indeed the