6
$\begingroup$

I am using perturbation theory to solve a problem of the following form: $$ R(h,\theta)f(\theta) = h g(\theta) = 0 $$ where $h$ is small, and I assume $$ \theta = \sum_{i=0}^\infty \theta_i h^i $$ Further, I have expanded $R(h,\theta)$ as $$ R(h,\theta) = \sum_{i=0}^\infty R_i(\theta)h^i $$ and I expand $f$, $g$, and $R_i(\theta)$ as Taylor series \begin{align} f(\theta) &= f(\theta_0) + f'(\theta_0)(\theta-\theta_0) + \frac{1}{2}f''(\theta_0)(\theta-\theta_0)^2 + ...\\ g(\theta) &= g(\theta_0) + g'(\theta_0)(\theta-\theta_0) + \frac{1}{2}g''(\theta_0)(\theta-\theta_0)^2 + ...\\ R_i(\theta) &= R_i(\theta_0) + R_i'(\theta_0)(\theta-\theta_0) + \frac{1}{2}R_i''(\theta_0)(\theta-\theta_0)^2 + ... \end{align}

I would like Mathematica to list the terms of the equation I'm trying to solve at the various orders in $h$. For instance, at $O(1)$, we have just $$R_{00}f_0=0$$, where $R_{00}$ stands for $R_0(\theta_0)$ and $f_0$ stands for $f(\theta_0)$. Similarly, at $O(h)$ the term is $$(R_{00}'\theta_1 +R_{10})f_0 + R_{00}f_0'\theta_1+g_0 = 0$$ and at $O(h^2)$ we get $$ (R_{00}'\theta_1 + R_{10})f_0'\theta_1 + R_{00}(f_0'\theta_2 + \frac{1}{2}f_0''\theta_1^2) + g_0'\theta_1 = 0 $$ These terms are obviously becoming more complicated at the higher orders, and the chances of me making a mistake in finding these terms increases probably exponentially with the order of $h$. So I'm wondering, how can I get Mathematica to display these terms for me? I have tried already: here is what I'm using:

    n = 5;
    θh := θ0 + Sum[θ[i] h^i, {i, 1, n}] + O[h]^(n + 1);
    rSymb[θ_] = 
    Sum[rrExp[i, θ] h^i, {i, 0, n}] + O[h]^(n + 1);
    rrExp[i_, θ_] := Series[rr[i][θ], {h, 0, n}];
    fSymb[θ_] := Series[ff[θ], {h, 0, n}];
    gSymb[θ_] := Series[gg[θ], {h, 0, n}];
    ySymb[θ_] := 
    rSymb[θ] fSymb[θ] + h*gSymb[θ]

rrExp is the Taylor expansion of the individual $R_i(\theta)$, fSymb and gSymb are the Taylor series expansions of $f$ and $g$, and ySymb is supposed to be the series consisting of the terms I want Mathematica to compute/display for me. However, there is a problem with rSymb. I don't know how to get it to properly regroup the rrExp terms by powers of $h$. I'm looking for rSymb to give me a series of the form \begin{multline} R(h,\theta) = R_{00} + h(R_{00}'\theta_1 + R_{10})\\ + h^2(R_{00}'\theta_2 + \frac{1}{2}R_{00}''\theta_1^2 + R_{10}'\theta_1 + R_{20})\\ +h^3(R_{00}'\theta_3 + R_{00}''\theta_1\theta_2 +\frac{1}{6} R_{00}'''\theta_1^3+R_{10}'\theta_2 + \frac{1}{2}R_{10}''\theta_1^2+R_{20}'\theta_1 + R_{30})\\ + O(h^4) \end{multline}

How can I modify my code to get Mathematica to properly find rSymb and then use that in ySymb to automatically determine the terms at arbitrary order in h?

$\endgroup$
3
  • $\begingroup$ Have you tried applying Collect to the result of rSymb? $\endgroup$
    – MarcoB
    Commented Jun 18, 2016 at 19:12
  • $\begingroup$ I had not tried that, but now I have based on your suggestion. I think it's working properly now---thank you! I'm sorry I missed such a simple solution to the problem. $\endgroup$
    – Mike Bell
    Commented Jun 18, 2016 at 19:31
  • $\begingroup$ No worries! I'm glad it worked. Perhaps you could pen a quick self-answer if you'd like. $\endgroup$
    – MarcoB
    Commented Jun 18, 2016 at 19:38

2 Answers 2

1
$\begingroup$

I would write

R[h, θ[h]] f[θ[h]] == h g[θ[h]] == 0
Series[%, {h, 0, 3}]
LogicalExpand[%]
Solve[%]

This returns a list of possible solutions, mostly degenerate, from which you would need to select the appropriate one.

$\endgroup$
0
$\begingroup$

Does this solve the problem?

ExtractTerm[n_] := Block[{T, R, b},
  T = Sum[Subscript[t, i] h^i, {i, 0, n}];
  R = Sum[Subscript[r, i][T] h^i, {i, 0, n}];
  b = SeriesCoefficient[R f[T] + h g[T], {h, 0, n}]
  ]

In the original post I think you meant $$R(h,\theta)f(\theta) + h\, g(\theta)=0$$ but if you meant something else I think you can edit the above code.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.