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4 - Second Order Constant Coefficient PDE

# 4 - Second Order Constant Coefficient PDE - MATH 220 SECOND...

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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 =1 Ξ ∂x k ( ξ u ) X. 1

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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 , are corresponding (unit length) eigenvectors e j . (Thus, the orthogonality of O is a consequence of the orthogonality of the e j .) Choosing B = O , we deduce that BAB = OO Λ OO = Λ is indeed diagonal.
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• Fall '10
• Vasy
• Partial differential equation, Diagonal matrix, initial conditions, Hyperbolic partial differential equation, diagonal entries

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