Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Dear all, I am studying symmetry method for ODEs. The book written by George W.bluman (can be found in google books) discusses the procedure of determining ODE admitting a given group in details. However, as to the converse way, it only gives some theories, but no examples.

I am trying to solve the exercise 3.2 - 9, P120:

given ODE


find a nontrivial Lie group of transformations admitted by the ODE; and find the general solution of the ODE.

Please help me out here. Thank you.

Cheers, Jan

share|cite|improve this question
up vote 5 down vote accepted

Before I say anything, let me say that I very much think you should try to get a hold of Peter Olver's Applications of Lie Groups to Differential Equations, published by Springer in its GTM series. It is literally packed with examples.

As for your equation: notice that you can rewrite it as $$\frac{\mathrm d y}{\mathrm d x}=\frac{-y}{y-\tfrac32x-3}.$$ There is a point $(x_0,y_0)$ such that if we change variables to $X=x-x_0$ and $Y=y-y_0$, the equation becomes $$\frac{\mathrm d Y}{\mathrm d X}=\frac{-Y}{Y-\tfrac32X}.$$ Now this equation is homogeneous, i.e., the right hand side depends only on $Y/X$. This observation can be used to solve it explicitly, by changing variables once more so that the new dependent variable is $T(X)=Y(X)/X$. But it also tells us that for each $\lambda\in\mathbb R\setminus\{0\}$ the transformation

\begin{align} X &\leadsto \lambda X \\ Y &\leadsto \lambda Y \end{align}

leaves the equation invariant, so we have a $1$-dimensional Lie group of symmetries for the equation for $X$ and $Y$. Undoing the change of variables that got us there will give us a group of symmetries for the original equation.

share|cite|improve this answer
"Before I say anything, let me say ..." :) Cheers, – Matt E Mar 12 '11 at 6:54

If you are interested I have developed a symbolic package in Mathematica for obtaining the symmetries of a DE. You can learn more about it here. If you are interested I can send you an updated version of the package via email.

This specific ODE is a first order one so don't expect to find a finite Lie group without making an ansantz in the form of the symmetries. For example, let your symmetries be represented by the infinitesimal generator $$\mathcal X=\xi(x,y)\partial_x+\eta(x,y)\partial_y$$ as a Lie algebra and assume that the coefficients $\xi,\eta$ are polynomials of order up to three with respect to $x,y$ then we have the symmetries: \begin{array}{l} (-2-x+2 y)\partial _x \\ (2+x)\partial _x+y\partial _y \\ 2 x y\partial _y+x (6+3 x-2 y)\partial _x \\ 2 x y^2\partial _y+x (6+3 x-2 y) y\partial _x \\ 2 y^3\partial _y+(6+3 x-2 y) y^2\partial _x \\ 2 x y (-6+3 x+2 y)\partial _y+x \left(9 x^2-4 (-3+y)^2\right)\partial _x \\ 2 y^2 (3+y)\partial _y+\left(3 x \left(3+3 y+y^2\right)-2 \left(-9+y^3\right)\right)\partial _x \\ \end{array}

You can see that the symmetry found by Mariano Suárez-Alvarez is the second one from the above list.

From that point by exponentiation you can obtain the respective local Lie group trasformation. For example, for the second symmetry we have the ode's \begin{array}{l} \bar x^\prime(\epsilon) = \bar x(\epsilon)+2,\, \bar x(0)=x\\ \bar y^\prime(\epsilon) = \bar y,\, \bar y(0)=y \end{array} by solving this system of ODEs you get \begin{array}{l} \bar x(\epsilon) =e^\epsilon(x+2)-2=\lambda (x+2)-2\\ \bar y(\epsilon) =e^\epsilon y=\lambda y \end{array}

Now, to obtain a solution for the ODE you can you use one of the symmetries to get a canonical set of coordinates. For example, from the second one the change of coordinates $$ X=\frac{y}{x+2},\, Y=\ln\left|y\right| $$ will turn the symmetry to the canonical form $\partial_Y$. By applying this change of variables to the ODE you get the ODE: $$ Y^\prime = \frac{1}{X(X-\frac{1}{2})}. $$ After solving it and returning back to the original variables you have: $$ y^3=c(x+2-2y)^2. $$ Bottomline: symmetries encompass most of the empirical methods we know for solving ODEs and PDEs, but the advantage of Lie's method is that they can be obtained in a systematic and algorithmic way.

share|cite|improve this answer
This would be better as a comment as opposed to an answer. Regards – Amzoti Mar 12 '13 at 0:15
@Amzoti I have expanded and clarify further my answer. I hope now things are clearer :) – Spawn1701D Apr 4 '13 at 22:05
Nice update +1 (upvote). Regards – Amzoti Apr 4 '13 at 22:58
Amazing package! May I ask how those symmetries are obtained via Mathematica in detail? – davyjones Jan 12 at 8:33
@davyjones In essence I "taught" Mathematica the Lie theory. For more details about the package itself see the article, A new symmetry finding package for Mathematica – Spawn1701D Jan 12 at 14:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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