8.2.2 Example B Equation (5.119) rewritten with the assumed form z(k, l) = Ch De is Ck+3 De – 3Ck+2De+1+ 3Ck+1De+2 - Ck De43 = 0. (8.30) Dividing all terms by Cr Dk gives ´Ck+3 ´Ck+2 -3 De+1 ´Ck+1 +3...

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Extracted text: 8.2.2 Example B Equation (5.119) rewritten with the assumed form z(k, l) = Ch De is Ck+3 De – 3Ck+2De+1+ 3Ck+1De+2 - Ck De43 = 0. (8.30) Dividing all terms by Cr Dk gives ´Ck+3 ´Ck+2 -3 De+1 ´Ck+1 +3 De+2 De+3 = 0. (8.31) Ck De Ck De De We can start with either one of the following choices Ck+1 De+1 = a Ck (8.32) or = a. De Using the second expression gives De = Aa", A = arbitrary function of a. (8.33) Substitution of this result in equation (8.31) yields a () () ´Ck+1 Ck+3 Ck Ck+2 За - a³ = 0, + 3a? (8.34) or Ck+3 – (3a)Ck+2 + 3a²Ck+1 – a³Ck = 0. (8.35) The corresponding characteristic equation is p3 – (3a)r2 + (3a²)r – a³ = (r – a)³ = 0. (8.36) Therefore, C (a) is Ch(a) = A1(a)a* + A2(@)ka* + A3(a)k²a* (8.37) and z(k, l, a) = Cr(@)De(a) = á(a)a*+l + Ã2(a)ka*+e + Ã3(a)k²a*+e. (8.38) If we sum/integrate over a, we obtain the following solution to equation (5.119) C2(k, l) = f(k + l) + kg(k + l) + k²h(k+ l), (8.39) where (f, g, h) are arbitrary functions of (k + e).


Extracted text: ull stc ksa 9:28 PM @ 9 59% The equation z(k +3, l) – 32(k + 2, l + 1) + 3z(k + 1, l + 2) – z(k, l + 3) = 0 (5.119) cannot be solved by the method of separation of variables. However, La- Cancel Actual Size (393 KB) Choose