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I have linearised some equations and trying to solve them perturbatively in powers of small parameter $e$. Here is my script

Subscript[t, x] = 1 - (Subscript[B, x]^2 e^2)/2 + (7 Subscript[B, x]^4 e^4)/8; Subscript[t, y] = Subscript[B, x] e - Subscript[B, y]^3 e^3; Subscript[n, x] = -Subscript[t, y]; Subscript[n, y] = Subscript[t, x]; Subscript[\[Sigma], xx] = -d + 2 \[Mu] Subscript[u, x]; Subscript[\[Sigma], yy] = -d + 2 \[Mu] Subscript[v, y] e^-2; Subscript[\[Sigma], xy] = \[Mu] ( Subscript[v, x] e^-1 + Subscript[u, y] e^-1); Subscript[u, x] = Subscript[u, x0] + Subscript[u, x2] e^2 + Subscript[u, x4] e^4; Subscript[u, y] = Subscript[u, y0] + Subscript[u, y2] e^2 + Subscript[u, y4] e^4; Subscript[v, x] = Subscript[v, x0] + Subscript[v, x2] e^2 + Subscript[v, x4] e^4; Subscript[v, y] = Subscript[v, y0] + Subscript[v, y2] e^2 + Subscript[v, y4] e^4; d = d0 + d2 e^2 + d4 e^4; Series[FullSimplify[ e (Subscript[\[Sigma], xx] Subscript[n, x] + Subscript[\[Sigma], xy] Subscript[n, y])]==0, {e, 0, 4}] Series[FullSimplify[ e^2 (Subscript[\[Sigma], yy] Subscript[n, y] + Subscript[\[Sigma], xy] Subscript[n, x])]==0, {e, 0, 4}] 

This outputs two equations that I need to solve. In the first order $e^0$, these equations simplify to

(Subscript[u, y0] + Subscript[v, x0])=0 

and

`2 \[Mu] Subscript[v, y0]=0` 

Which mean I can use $2 \mu v_{y0} =0$ and $(u_{y0}=- v_{x0})$ to simplify the higher order equations $e^2$ and $e^4$. But how can I do this without manually subbing lower order equations to higher order equations?

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  • $\begingroup$ Calling the equations e1 and e2, one can do SolveAlways[{e1, e2}, e]. $\endgroup$ Commented Apr 24, 2024 at 14:25
  • $\begingroup$ could you please post your complete answer? @DanielLichtblau $\endgroup$ Commented Apr 24, 2024 at 16:26

1 Answer 1

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t[x] = 1 - (b[x]^2 e^2)/2 + (7 b[x]^4 e^4)/8; t[y] = b[x] e - b[y]^3 e^3; n[x] = -t[y]; n[y] = t[x]; sig[xx] = -d + 2 mu u[x]; sig[yy] = -d + 2 mu v[y] e^-2; sig[xy] = mu (v[x] e^-1 + u[y] e^-1); u[x] = u[x0] + u[x2] e^2 + u[x4] e^4; u[y] = u[y0] + u[y2] e^2 + u[y4] e^4; v[x] = v[x0] + v[x2] e^2 + v[x4] e^4; v[y] = v[y0] + v[y2] e^2 + v[y4] e^4; d = d0 + d2 e^2 + d4 e^4; e1 = Series[e (sig[xx] n[x] + sig[xy] n[y]) == 0, {e, 0, 4}]; e2 = Series[e^2 (sig[yy] n[y] + sig[xy] n[x]) == 0, {e, 0, 4}]; SolveAlways[{e1, e2}, e] (* Out[129]= {{d2 -> 0, d0 -> 0, mu -> 0}, {u[y4] -> (d0 b[y]^3 - 2 mu b[y]^3 u[x0] - mu v[x4])/mu, v[y4] -> d2/(2 mu), v[y2] -> d0/(2 mu), u[y2] -> -v[x2], v[y0] -> 0, u[y0] -> -v[x0], b[x] -> 0}, {u[y4] -> (1/( 2 mu b[x]))(-2 d2 b[x]^2 + 4 mu b[x]^2 u[x2] + mu b[x]^3 u[y2] - 2 mu b[y]^3 u[y2] + mu b[x]^3 v[x2] - 2 mu b[y]^3 v[x2] - 2 mu b[x] v[x4]), v[y4] -> (d2 + mu b[x] u[y2] + mu b[x] v[x2])/(2 mu), v[y2] -> d0/(2 mu), u[x0] -> (d0 b[x] + mu u[y2] + mu v[x2])/(2 mu b[x]), v[y0] -> 0, u[y0] -> -v[x0]}} *) 
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