Differential Equations Lecture Work Solutions 187

Differential Equations Lecture Work Solutions 187 - u = 1(u...

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u ξη = 1 6( η ξ ) ( u η u ξ ) The parabolic case is easier, the only characteristic is y = 1 2 x + K and so the transformation is ξ = y 1 2 x η = x The last equation is an arbitrary function and one should check the Jacobian. The details are left to the reader. One can easily show that A = B =0 Also C =1 and the rest of the coefficients are zero. Therefore the equation is u ηη =0 In the elliptic case, one can use the transformation
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