Substitution of parts of a PDE

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Question:Substitution of parts of a PDE

Posted: salim-barzani 1410 Product: Maple 2024

May 07 2025

I'm trying to extract the real and imaginary parts of a given PDE. However, during the substitution step, something unexpected occurs. Specifically, my replacement does not yield the expected result, and an extra term appears: D(k)(t*w + x) in the expression P11. I'm unsure why this term arises and would appreciate any insight into what might be going wrong during the substitution process.

(1)

(2)

pde := I*(diff(u(x, t), `$`(t, 2))-s^2*(diff(u(x, t), `$`(x, 2))))+(1/24)*c[1]*(diff(u(x, t), t, `$`(x, 4)))-alpha*s*c[1]*(diff(u(x, t), `$`(x, 5)))+diff(c[2]*u(x, t)*U(-t*v+x)^2+c[3]*u(x, t)*U(-t*v+x)^4+c[4]*u(x, t)*(diff(U(-t*v+x)^2, `$`(x, 2))), t)-beta*s*(diff(u(x, t), `$`(x, 5)))+diff(c[2]*u(x, t)*U(-t*v+x)^2+c[3]*u(x, t)*U(-t*v+x)^4+c[4]*u(x, t)*(diff(U(-t*v+x)^2, `$`(x, 2))), x)

(3)

G1 := U(-t*v+x) = U(xi); G2 := (D(U))(-t*v+x) = diff(U(xi), xi); G3 := ((D@@2)(U))(-t*v+x) = diff(U(xi), `$`(xi, 2)); G4 := ((D@@3)(U))(-t*v+x) = diff(U(xi), `$`(xi, 3)); G5 := ((D@@4)(U))(-t*v+x) = diff(U(xi), `$`(xi, 4)); G6 := ((D@@5)(U))(-t*v+x) = diff(U(xi), `$`(xi, 5))

(4)

(5)

P1 := I*(diff(u(x, t), `$`(t, 2))-s^2*(diff(u(x, t), `$`(x, 2))))+(1/24)*c[1]*(diff(u(x, t), t, `$`(x, 4)))-alpha*s*c[1]*(diff(u(x, t), `$`(x, 5)))+diff(c[2]*u(x, t)*U(-t*v+x)^2+c[3]*u(x, t)*U(-t*v+x)^4+c[4]*u(x, t)*(diff(U(-t*v+x)^2, `$`(x, 2))), t)-beta*s*(diff(u(x, t), `$`(x, 5)))+diff(c[2]*u(x, t)*U(-t*v+x)^2+c[3]*u(x, t)*U(-t*v+x)^4+c[4]*u(x, t)*(diff(U(-t*v+x)^2, `$`(x, 2))), x)