 Earlier we found that the one-dimensional heat equation has a solution that can be written as a series where our g functions are exponentials and our f functions are trigonometric or exponential functions depending on the value of lambda. For every lambda in some set, capital lambda. But this raises the important question, which value lambda will work for a given problem? To answer this question, we have to deal with the real universe. Oh, stop that. There's nothing wrong with applied mathematics. So suppose we try to solve a realistic problem. Let L be the length of the rod. We'll make the following assumptions. First, we'll assume the endpoints are maintained at a constant temperature, 0. So u of t0 is 0, and u of tL is equal to 0. Next, we'll assume the rod itself has an initial temperature distribution. So u of 0x is i of x for some function i. Now our goal is to try and find a function that gives the temperature at any point along the rod at any time t. And so this leads to the problem, solve the one-dimensional heat equation, where we have a set of boundary conditions. To begin with, our theorem tells us that our solutions will be the sum of products of functions. Now our boundary conditions tell us that u of t0 is 0, and u of tL is 0 as well. So remember, this sum comes from adding up solutions to the heat equation. So let's consider one of those solutions, u of tx equals c e to the power minus lambda alpha squared t f of x, where we'll determine f of x later. So incorporating those boundary conditions tells us, and since we have an exponential function here, we can divide by it, and so we know that f of 0 is 0, and f of L is 0. How about those lambdas? Well, lambda has two possibilities. Suppose lambda is less than 0. Remember, the value of lambda determines the form of f of x. If lambda is less than 0, then f of x will just be an exponential function, where alpha is square root of minus lambda. And we have our initial conditions, f of 0 equals 0, so we can substitute those into our function, and we have a system of equations that we can solve for c1 and c2. Then you can solve this algebraically, but a little cleverness never hurts. Notice that this is a homogeneous system of equations, and a homogeneous system of linear equations always has the trivial solution. So we know that c1 is equal to 0, and c2 is equal to 0. Moreover, because these exponential coefficients are going to be different, we know this is the unique solution, and so that tells us that f of x must be 0 for all x. But that means u of t of x must also be 0 for all t and x. And that's a trivial solution, and we're not looking for those. So we know that lambda cannot be less than 0. That means we know that lambda is greater than 0. Incidentally, this is why we made that separation constant minus lambda, because that way, lambda itself would be positive. Which reminds me, I have to go back in time and remind myself to make it minus lambda. So I'll open up a portal of the spacetime continuum. Wow, did my hair really look like that? Well, whatever. Hey, you! What's that? I'm the future you. I've opened up this portal of the spacetime continuum to tell you to make that separation constant minus lambda. Things will work out better. Oh, wait. Sorry. A portal in the spacetime. All right, so where were we? So if lambda is greater than 0, then our eigenvalues will turn out to be complex, and our solutions will have the form of a trigonometric function. Substituting in our boundary conditions. So c1 has to be 0. There's no other possibility. Now if c1 is 0, then in order for our second boundary condition to be met, c2 sine square root lambda l must also be equal to 0. This means that square root lambda l equals n pi, where n is any integer. And so our function looks like c2 sine of n pi over lx. And if we want to know the value of lambda, we can take our equation and find that lambda is equal to n squared pi squared over l squared. And so the general solution, absorbing the constants, is going to be... So what about those constants? Again, we can rely on our boundary value since u of 0x is i of x, our initial temperature distribution we have. And so our initial temperature distribution gives us a trigonometric series. So now i of x is a trigonometric series. To find the coefficients bn, we'll make a few observations. First, i of x is a sum of odd functions, so it will also be an odd function. Since sine of n pi over lx will have period 2l, we'll multiply by sine k pi over lx, and then integrate over the interval between minus l and l. So assuming convergence, we can interchange the integral and the summation. As before, since we're integrating over one pole period, if k is not equal to n, the integrals will vanish, leaving only the integral of bk sine k pi l sine k pi l. And that's going to be where the cosine component will vanish on integration, and so we get bk l. Now since i of x is odd and sine k pi over lx is also odd, their product is an even function. And if f of x is even, then the integral over any interval symmetric about the origin is going to be twice the integral from zero to the endpoint. And that allows us to simplify this expression, and we can find b of k. And so putting everything together, we now have an approach to solving the one-dimensional heat equation with boundary conditions. Remember, these correspond to the idea that the endpoints are at temperature zero, and that we have some initial temperature distribution along the rod. Our solution is going to be a trigonometric series multiplied by an exponential function, where the coefficients of our series are going to be found by an integration.