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**Unformatted text preview: **Math 124A November 04, 2010 Viktor Grigoryan 12 Heat conduction on the half-line In previous lectures we completely solved the initial value problem for the heat equation on the whole line, i.e. in the absence of boundaries. Next, we turn to problems with physically relevant boundary conditions. Let us first add a boundary consisting of a single endpoint, and consider the heat equation on the half-line D = (0 , ). The following initial/boundary value problem, or IBVP, contains a Dirichlet boundary condition at the endpoint x = 0. ( v t- kv xx = 0 , < x < , < t < , v ( x, 0) = ( x ) , x > , v (0 ,t ) = 0 , t > . (1) If the solution to the above mixed initial/boundary value problem exists, then we know that it must be unique from an application of the maximum principle. In terms of the heat conduction, one can think of v in (1) as the temperature in an infinite rod, one end of which is kept at a constant zero temperature. The initial temperature of the rod is then given by ( x ). Our goal is to solve the IBVP (1), and derive a solution formula, much like what we did for the heat IVP on the whole line. But instead of constructing the solution from scratch, it makes sense to try to reduce this problem to the IVP on the whole line, for which we already have a solution formula. This is achieved by extending the initial data ( x ) to the whole line. We have a choice of how exactly to extend the data to the negative half-line, and one should try to do this in such a fashion that the boundary condition of (1) is automatically satisfied by the solution to the IVP on the whole line that arises from the extended data. This is the case, if one chooses the odd extension of ( x ), which we describe next. By the definition a function ( x ) is odd, if (- x ) =- ( x ). But then plugging in x = 0 into this definition, one gets (0) = 0 for any odd function. Recall also that the solution u ( x,t ) to the heat IVP with odd initial data is itself odd in the x variable. This follows from the fact that the sum [ u ( x,t ) + u (- x,t )] solves the heat equation and has zero initial data, hence, it is the identically zero function by the uniqueness of solutions. Then, by our above observation for odd functions, we would have that u (0 ,t ) = 0 for any t > 0, which is exactly the boundary condition of (1)....

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