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Assignment3-Solutions-HandIn

# Assignment3-Solutions-HandIn - 1 Consider the principal...

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1. Consider the principal part of the equation, u xx - 2 u xy - 3 u yy . Based on this, we construct the polynomial α 2 - 2 α - 3 and we note that it has roots r 1 = 3 and r 2 = - 1 (the equation is hyperbolic). Consequently, we will suggest a linear coordinate transformation ξ = r 1 x + y = 3 x + y, η = r 2 x + y = - x + y. If we define ˜ u ( ξ, η ) = u ( x, y ), then u x = u ξ - ˜ u η u y = ˜ u ξ + ˜ u η u x + u y = u ξ and u xx = 9 u ξξ - 6 u ξη + ˜ u ηη - 2 u xy = - u ξξ - u ηξ + 2˜ u ηη - 3 u yy = - u ξξ - u ξη - u ηη 0 = - 16˜ u ξη We can conclude that u xx - 2 u xy - 3 u yy + u x + u y = 0 is equivalent to ˜ u ξη - 1 4 ˜ u ξ = 0 . If we set v = ˜ u ξ the PDE above becomes, v η = 1 4 v , and so v = f ( ξ ) e η/ 4 = ˜ u ξ . Integrating with respect to ξ we get ˜ u = e η/ 4 F ( ξ ) + G ( η ) or u ( x, y ) = e ( y - x ) / 4 F (3 x + y ) + G ( y - x ) X = e y/ 4 - x/ 4 - 3 x/ 4+3 x/ 4 F (3 x + y ) + G ( y - x ) = e - x H (3 x + y ) + G ( y - x ) Page 1

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3. Exercise 4, page 89. We look for a solution of the form u ( x, t ) = X ( x ) T ( t ). If the solution has this form, then XT 00 + rXT 0 = c 2 X 00 T. Dividing by - c 2 XT we get - 1 c 2 T 00 T - r c 2 T 0 T = - X 00 X = λ. The homogeneous Dirichlet BCs tell us that the eigenfunctions in x are X n ( x ) = sin nπx n = 1 , 2 , 3 , . . . . Turning to T we see that for a given n , T 00 n + r T 0 n + c 2 λ n T n = 0 where λ n = ( nπ/‘ ) 2 . Since 0 < r < 2 πc/‘ , we can say that r 2 < 4( πc/‘ ) 2 and so the discriminant r 2 - 4( nπc/‘ ) 2 < 0 . It follows that the roots of the characteristic equation are the complex conjugate pair, - r 2 ± i 1 2 r 4 nπc 2 - r 2 .
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