Partial Differentials Notes_Part_5

# Partial Differentials Notes_Part_5 - so there are no...

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Unformatted text preview: so there are no non-separable eigenfunctions † . As a consequence, the general solution to the initial-boundary value problem can be expressed as a linear combination u ( t, x, y ) = ∞ summationdisplay m,n = 1 c m,n u m,n ( t, x, y ) = ∞ summationdisplay m,n = 1 c m,n e − λ m,n t v m,n ( x, y ) (12 . 41) of the eigenfunction modes. The coefficients c m,n are prescribed by the initial conditions, which take the form of a double Fourier sine series f ( x, y ) = u (0 , x, y ) = ∞ summationdisplay m,n = 1 c m,n v m,n ( x, y ) = ∞ summationdisplay m,n = 1 c m,n sin mπx a sin nπy b . Self-adjointness of the Laplacian coupled with the boundary conditions implies that † the eigenfunctions v m,n ( x, y ) are orthogonal with respect to the L 2 inner product on the rectangle: ( v k,l ; v m,n ) = integraldisplay b integraldisplay a v k,l ( x, y ) v m,n ( x, y ) dx dy = 0 unless k = m and l = n. (The skeptical reader can verify the orthogonality relations directly from the eigenfunction formulae (12.38).) Thus, we can appeal to our usual orthogonality formula (12.25) to evaluate the coefficients c m,n = ( f ; v m,n ) bardbl v m,n bardbl 2 = 4 ab integraldisplay b integraldisplay a f ( x, y ) sin mπx a sin nπy b dxdy, (12 . 42) where the formula for the norms of the eigenfunctions bardbl v m,n bardbl 2 = integraldisplay b integraldisplay a v m,n ( x, y ) 2 dx dy = integraldisplay b integraldisplay a sin 2 mπx a sin 2 nπy b dx dy = 1 4 ab. (12 . 43) follows from a direct evaluation of the double integral. Unfortunately, while the orthogo-follows from a direct evaluation of the double integral....
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## This note was uploaded on 02/10/2012 for the course MATH 5587 taught by Professor Olver during the Fall '10 term at University of Central Florida.

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Partial Differentials Notes_Part_5 - so there are no...

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