There is a short cut to obtaining \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) if the first order ode type is known or can be determined. (of course, if we know the ode type, then a direct method for solving the ode can be used, since the type is known and there is no need to use Lie symmetry), but still Lie symmetry can be useful in this case, and also it allows us to find the integrating factor quickly, which provides one more method to solve the ode. An example of a first order ode which does not have known type is
The above can be solved using Lie symmetry but with functional form of anstaz \(\xi =f\left ( x\right ) g\left ( y\right ) ,\eta =0\). which gives \(\xi =e^{-\sin y},\eta =0\).
I am in the process of building table for ready to use infinitesimal based on the first ode type. The following small list is the current ones determined. For some first order ode such as linear \(y^{\prime }=f\left ( x\right ) y\left ( x\right ) +g\left ( x\right ) \) or separable \(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) the infinitesimals can be written directly (but again, for these simple ode’s Lie method is not really needed but it provides good illustration on how to use it. Lie method is meant to be used for ode’s which have no known type or difficult to solve otherwise). For an ode type not given in this list, an anstaz have to be used to solve the similarity PDE.
|
ode type |
form |
\(\xi \) |
\(\eta \) |
notes |
|
linear ode |
\(y'=f(x) y(x) +g(x)\) |
\(0\) |
\(e^{\int fdx}\) |
Notice that \(g\left ( x\right ) \) does not affect the result |
|
separable ode |
\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) |
\(\frac {1}{f}\) |
\(0\) |
This works for any \(g\) function that depends on \(y\) only |
|
quadrature ode |
\(y^{\prime }=f\left ( x\right ) \) |
\(0\) |
\(1\) |
of course for quadrature we do not need Lie symmetry as ode is already quadrature |
|
quadrature ode |
\(y^{\prime }=g\left ( y\right ) \) |
\(1\) |
\(0\) |
For example \(y^{\prime }=\frac {x+y}{-x+y}\) or \(y^{\prime }=\frac {y+2\sqrt {yx}}{x}\) |
|
homogeneous ODEs of Class A |
\(y^{\prime }=f\left ( \frac {y}{x}\right ) \) |
\(x\) |
\(y\) |
|
|
homogeneous ODEs of Class C |
\(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\) |
\(1\) |
\(-\frac {b}{c}\) |
Also \(\xi =0,\eta =c\left ( bx+cy+a\right ) ^{\frac {n}{m}}\) are possible. For example, for \(y^{\prime }=\left ( 1+2x+3y\right ) ^{\frac {1}{2}}\) then use the first option as simpler which is \(\xi =1,\eta =-\frac {2}{3}\). Notice that \(\xi =1,\eta =-\frac {b}{c}\) does not depend on \(a\) and not on \(n,m\). Hence these odes \(y^{\prime }=\left ( 1+x+y\right ) ^{\frac {1}{3}}\),\(y^{\prime }=\left ( 10+x+y\right ) ^{\frac {1}{3}}\) and \(y^{\prime }=\left ( 10+x+y\right ) ^{\frac {2}{3}}\) all have the same infinitesimals \(\xi =1,\eta =-\frac {b}{c}=-1\) |
|
homogeneous class D |
\(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \) |
\(x^{2}\) |
\(xy\) |
example \(y^{\prime }=\frac {y}{x}+\frac {1}{x}e^{-\frac {y}{x}}\). Where here \(g\left ( x\right ) =\frac {1}{x},F\left ( \frac {y}{x}\right ) =e^{-\frac {y}{x}}\). |
|
First order special form ID 1 |
\(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \) |
\(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
\(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
For an example, for the ode \(y^{\prime }=5e^{x^{2}+20y}+\sin x\), here \(g\left ( x\right ) =5,h\left ( x\right ) =x^{2},b=20,f\left ( x\right ) =\sin x\), hence \(\xi =\frac {e^{-\int 20\sin dx-x^{2}}}{5},\eta =\frac {\sin xe^{-\int 20\sin xdx-x^{2}}}{5}\) or \(\xi =\frac {1}{5}\sin x\left ( e^{20\cos \left ( x\right ) -x^{2}}\right ) ,\eta =\frac {\sin \left ( x\right ) }{5}\left ( e^{20\cos \left ( x\right ) -x^{2}}\right ) \). In this form, \(b\) must be constant. |
|
polynomial type ode |
\(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) |
\(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
\(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
For example for \(y^{\prime }=\frac {x+y+3}{2x+y}\) then \(a_{1}=1,b_{1}=1,c_{1}=3,a_{2}=2,b_{2}=1,c_{2}=0\). Hence \(\xi =x-3,\eta =y+6\). |
|
Bernoulli ode |
\(y' = f(x) y+ g(x)y^{n}\) |
\(0\) |
\( y^n e^{\int (1-n) f(x) \,dx} \) |
\(n\) is integer \(n\neq 1,n\neq 0\). For example, for \(y^{\prime }=-\sin \left ( x\right ) y+x^{2}y^{2}\) then \(f\left ( x\right ) =-\sin x,g\left ( x\right ) =x^{2},n=2\) and \(\xi =0,\eta =e^{\int \sin xdx}y^{2}\) or \(\xi =0,\eta =e^{-\cos x}y^{2}\). Notice that \(g\left ( x\right ) \) does not show up in the infinitesimals Another example is \(y^{\prime }=2\frac {y}{x}+\frac {y^{3}}{x^{2}}\) where here \(f\left ( x\right ) =\frac {2}{x}\). Hence \(\xi =0,\eta =e^{-\int \left ( 3-1\right ) \frac {2}{x}dx}y^{3}\) or \(\xi =0,\eta =\eta =\frac {y^{3}}{x^{4}}\) |
|
Reduced Riccati |
\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) |
\(0\) |
\(e^{-\int f_{1}dx}\) |
For example, for \(y^{\prime }=xy+\sin \left ( x\right ) y^{2}\) then \(f_{1}=x,f_{2}=\sin x\) and hence \(\xi =0,\eta =e^{-\int xdx}\) or \(\xi =0,\eta =e^{\frac {1}{2}x^{2}}\). Notice that \(f_{2}\left ( x\right ) \) does not show up in the infinitesimals. I could not find infinitesimals for the full Riccati ode \(y^{\prime }=f_{0}\left ( x\right ) +f_{1}\left ( x\right ) y+f_{2}\left ( x\right ) y^{2}\). Notice that \(f_{1},f_{2}\) can not be both constants, else this becomes separable |
|
Abel first kind |
\(y^{\prime }=f_{0}\left ( x\right ) +f_{1}\left ( x\right ) y+f_{2}\left ( x\right ) y^{2}+f_{3}\left ( x\right ) y^{3}\) |
|
|
No infinitesimals found |
Currently the above are the ones I am able to determine for known first order ode’s. If I find more, will add them. The table lookup is much faster to use than having to solve the similarity PDE each time using anstaz in order to find \(\xi ,\eta \).