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Unformatted text preview: MATH 220: SECOND ORDER CONSTANT COEFFICIENT PDE ANDRAS VASY We consider second order constant coefficient scalar linear PDEs on R n . These have the form L u = f, L = n summationdisplay i,j =1 a ij ∂ x i ∂ x j + n summationdisplay i =1 b i ∂ x i + c where a ij , b i and c are (complex) constants, and f is given. As a general principle, the leading, i.e. second order, terms are the most important in analyzing the PDE. The second order terms then take the form L = n summationdisplay i,j =1 a ij ∂ x i ∂ x j ; here we are interested in the case when the a ij are real constants. Since ∂ x i ∂ x j = ∂ x j ∂ x i , we might as well assume that a ij = a ji (otherwise replace both by 1 2 ( a ij + a ji ), which does not change L ), i.e. that A is symmetric. We rewrite L in a different form. With ∇ x = ∂ x 1 ∂ x 2 ... ∂ x n and A = a 11 ... a 1 n ... a n 1 ... a nn , we can write L = ∇ † x A ∇ x , where ∇ † x is the transpose of ∇ x : ∇ † x = bracketleftbig ∂ x 1 ... ∂ x n bracketrightbig . Technically, there is no need for the x in the subscripts, and one should write e.g. ∇ † = bracketleftbig ∂ 1 ... ∂ n bracketrightbig . However, the subscripts are very useful below as they enable us to compress (and abuse) the notation in the change of variables formula. We would like to bring it to a simpler form; the simpler form we wish for may depend on the intended use. As a first step we would like to have L brought into a diagonal form, i.e. A made diagonal. We can achieve this by changing coordinates. Namely, if we let ξ = Ξ( x ), i.e. ξ j = Ξ j ( x 1 ,...,x n ), j = 1 ,...,n , with inverse function X , i.e. x = X ( ξ ), meaning that X ◦ Ξ and Ξ ◦ X are both the identity maps (both X and Ξ are assumed to be C 1 ), then by the chain rule for any C 1 function u we have ∂ x k ( u ◦ X ) = n summationdisplay ℓ =1 ∂ Ξ ℓ ∂x k ( ∂ ξ ℓ u ) ◦ X. 1 2 ANDRAS VASY We can rewrite this in a matrix form: (1) ∂ x 1 ... ∂ x n = ∂ Ξ 1 ∂ x 1 ... ∂ Ξ n ∂x 1 ... ∂ Ξ 1 ∂ xn ... ∂ Ξ n ∂x n ∂ ξ 1 ... ∂ ξ n , so in particular if ξ is a linear function of x , i.e. ξ = Bx , B = bracketleftbig b ij bracketrightbig n i,j =1 , so ξ i = ∑ n j =1 B ij x j , then (2) ∇ x = B † ∇ ξ . Thus, L = ∇ † x A ∇ x = ∇ † ξ BAB † ∇ ξ . Hence, in order to make L be given by a diagonal matrix in the ξ coordinates, we need that BAB † is diagonal. Recalling that A is symmetric, we proceed as follows. Any diagonal matrix, such as A , can be diagonalized by conjugating it by an orthogonal matrix, i.e. there is an orthogonal matrix O (recall that orthogonal means that OO † = O † O = Id, i.e. O † = O − 1 ) and a diagonal matrix Λ such that A = O † Λ O, where the diagonal entries λ j of Λ are the eigenvalues of A and the rows of O , i.e. the columns of O †...
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This note was uploaded on 03/16/2010 for the course CME 303 taught by Professor Vasy during the Fall '10 term at Stanford.
 Fall '10
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