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I have a little Problem in understanding how to derive invariants form Lie Symmetries (or their infinitesimals).

As one can show the heat equation $u_t=u_{xx}$ has the following symmetries (infinitesimal transformations): $$ v_1 = \partial_x \quad v_2 = \partial_t \quad v_3 = u\partial_u \quad v_4 = x\partial_x +2t\partial_t$$ $$ v_5 = 2t\partial_x-xu\partial_u \quad v_6 = 4tx\partial_x+4t^2\partial_t-(x^2+2t)u\partial_u$$ $$ v_{\alpha} = \alpha(x,t)\partial_u$$

Where $\alpha(x,t)$ is a solution of the heat equation.

Now I read in the Book Symmetry Analysis of Differential Equations with Mathematica (Baumann, 1999) that one can find the invariants by applying the infinitesimal transformations $v_i$ onto a scalar function $F(x,t,u)$ and equate this to 0.

$$v_iF(x,t,u)=0$$

Now in the text the author picks $v=x\partial_x+2t\partial_t+cu\partial_u$ as infinitesimal. I actually dont understand why one can choose this this way. Because it seems to me that this is some kind of linear combination of $v=v_4+cv_3$. Solving the resulting PDE fesults in $F(x,t,u)=H(\frac{t}{x^2},ux^{-c})$. Where $H$ is an arbitrary function and $t/x^2$ and $ux^{-c}$ are the invariants.

Here is my question. Why did the author use this specific $v$? Do I have to check all possible combinations to get all possible invariants?

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The answer to your query is very simple in fact, the element $v_{4}+c\,v_{3}$ the author has taken belong to optimal algebra of Lie algebra $\{v_{1},...,v_{6}\}$. The optimal algebra of heat equation consist of following vectors

$\{v_{4}+c\,v_{3}, v_{1}+v_{6}+c\,v_{4}, v_{1}, v_{1}+v_{5}, v_{3}, v_{2}+c v_{3}\}$

proof of which can be found in several text and research articles but I recommend you reading Peter Olver book (See for e.g. Example 3.13).

The question you have raised about possible combinations of vectors for group invariants has answer lies with optimal Lie algebra. You have to find group invariants corresponds to every vector in 1$-$ dimensional optimal sub$-$algebra for complete and mutually in equivalent classification of invariants. If you don't do this process through optimal algebra then in equivalency among group invariants would not be guaranteed.

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  • $\begingroup$ Thank you alot. Just a small question. Does it mean that only vector combinations from the optimal algebra can reduce the number of independent variables? Or to ask a little broader, is it necessary and sufficient for the reduction of independent variables, if the combination of vectors is from the optimal algebra? $\endgroup$ – MrYouMath May 6 '16 at 7:12
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    $\begingroup$ Any vector combination can reduce PDE, but vectors from 1-dimensional optimal algebra will give all possible inequivalent reductions. $\endgroup$ – IgotiT May 6 '16 at 7:57
  • $\begingroup$ So it is enough to check them? $\endgroup$ – MrYouMath May 6 '16 at 10:29
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    $\begingroup$ Yes, once you have constructed your optimal sub-algebra then you can go for inequivalent reductions. $\endgroup$ – IgotiT May 6 '16 at 11:38
  • $\begingroup$ I have read the example and as always Mr Olver keeps all the important ideas well hidden. Could you explain the process in more detail (math.stackexchange.com/questions/1778100/…)? $\endgroup$ – MrYouMath May 9 '16 at 13:16

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