# lec10 - Math 124A Viktor Grigoryan 10 Heat equation...

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Math 124A – October 28, 2010 Viktor Grigoryan 10 Heat equation: interpretation of the solution Last time we considered the IVP for the heat equation on the whole line u t - ku xx = 0 ( -∞ < x < , 0 < t < ) , u ( x, 0) = φ ( x ) , (1) and derived the solution formula u ( x, t ) = Z -∞ S ( x - y, t ) φ ( y ) dy, for t > 0 , (2) where S ( x, t ) is the heat kernel, S ( x, t ) = 1 4 πkt e - x 2 / 4 kt . (3) Substituting this expression into (2), we can rewrite the solution as u ( x, t ) = 1 4 πkt Z -∞ e - ( x - y ) 2 / 4 kt φ ( y ) dy, for t > 0 . (4) Recall that to derive the solution formula we first considered the heat IVP with the following particular initial data Q ( x, 0) = H ( x ) = 1 , x > 0 , 0 , x < 0 . (5) Then using dilation invariance of the Heaviside step function H ( x ), and the uniqueness of solutions to the heat IVP on the whole line, we deduced that Q depends only on the ratio x/ t , which lead to a reduction of the heat equation to an ODE. Solving the ODE and checking the initial condition (5), we arrived at the following explicit solution Q ( x, t ) = 1 2 + 1 π Z x/ 4 kt 0 e - p 2 dp, for t > 0 . (6) The heat kernel S ( x, t ) was then defined as the spatial derivative of this particular solution Q ( x, t ), i.e. S ( x, t ) = ∂Q ∂x ( x, t ) , (7) and hence it also solves the heat equation by the differentiation property. The key to understanding the solution formula (2) is to understand the behavior of the heat kernel S ( x, t ). To this end some technical machinery is needed, which we develop next. 10.1 Dirac delta function Notice that, due to the discontinuity in the initial data of Q , the derivative Q x ( x, t ), which we used in the definition of the function S in (7), is not defined in the traditional sense when t = 0. So how can one make sense of this derivative, and what is the initial data for S ( x, t )? It is not difficult to see that the problem is at the point x = 0. Indeed, using that Q ( x, 0) = H ( x ) is constant for any x 6 = 0, we will have S ( x, 0) = 0 for all x different from zero. However, H ( x ) has a jump discontinuity at x = 0, as is seen in Figure 1, and one can imagine that at this point the rate of growth of H is infinite. Then the “derivative” δ ( x ) = H 0 ( x ) (8) is zero everywhere, except at x = 0, where it has a spike of zero width and infinite height. Refer to Figure 2 below for an intuitive sketch of the graph of δ . Of course, δ is not a function in the traditional sense, 1

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1 x H Figure 1: The graph of the Heaviside step function. x Δ Figure 2: The sketch of the Dirac δ function. but is rather a generalized function , or distribution . Unlike regular functions, which are characterized by their finite values at every point in their domains, distributions are characterized by how they act on regular functions. To make this rigorous, we define the set of test functions D = C c , the elements of which are smooth functions with compact support. So φ ∈ D , if and only if φ has continuous derivatives of any order k N , and the closure of the support of φ , supp( φ ) = { x R | φ ( x ) 6 = 0 } , is compact. Recall that compact sets in R are those that are closed and bounded. In particular for any test function φ there is a rectangle [ - R, R ], outside of which φ vanishes. Notice that derivatives of test
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