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Article Contents

# Conditional symmetries of nonlinear third-order ordinary differential equations

• * Corresponding author: Aeeman Fatima.
• In this work, we take as our base scalar second-order ordinary differential equations (ODEs) which have seven equivalence classes with each class possessing three Lie point symmetries. We show how one can calculate the conditional symmetries of third-order non-linear ODEs subject to root second-order nonlinear ODEs which admit three point symmetries. Moreover, we show when scalar second-order ODEs taken as first integrals or conditional first integrals are inherited as Lie point symmetries and as conditional symmetries of the derived third-order ODE. Furthermore, the derived scalar nonlinear third-order ODEs without substitution are considered for their conditional symmetries subject to root second-order ODEs having three symmetries.

Mathematics Subject Classification: 34A05, 22E05, 22E45, 22E60, 22E70.

 Citation:

• Table Ⅰ.  Lie group classification of scalar second-order equations in the real plane

 $p=\partial /\partial x$ and $q=\partial /\partial y$ Algebra Canonical forms of generators Representative equations $L_{1}$ $X_1=p$ $y''=g(y,y')$ $L_{2;1}^I$ $X_1=p,X_2=q$ $y''=g(y')$ $L_{2;1}^{II}$ $X_1=q,X_2=xp+yq$ $xy''=g(y')$ $L_{3;3}^I$ $X_1=p, X_2=q, X_3=xp+(x+y)q$ $y''=Ae^{-y'}$ $L_{3;6}^I$ $X_1=p, X_2=q, X_3=xp+ayq$ $y''=Ay'^{(a-2)/(a-1)}$ $L_{3;7}^I$ $X_1=p, X_2=q, X_3=(bx+y)p+(by-x)q$ $y''=A(1+y'^{2})^{\frac{3}{2}}e^{b\arctan y'}$ $L_{3;8}^I$ $X_1=q, X_2=xp+yq, X_3=2xyp+y^2q$ $xy''=Ay'^{3}-\frac{1}{2}y'$ $L_{3;8}^{II}$ $X_1=q, X_2=xp+yq, X_3=2xyp+(y^2+x^2)q$ $xy''=y'+y'^{3}+A(1+y'^{2})^{\frac{3}{2}}$ $L_{3;8}^{III}$ $X_1=q, X_2=xp+yq, X_3=2xyp+(y^2-x^2)q$ $xy''=y'-y'^{3}+A(1-y'^{2})^{\frac{3}{2}}$ $L_{3;9}$ $X_1=(1+x^2)p+xyq, X_2=xyp+(1+y^2)q$, $X_3=yp-xq$ $y''=A[\displaystyle{1+y'+(y-xy')^2\over 1+x^2+y^2}]^{3/2}$

Table Ⅱ.  Inherited symmetries of derived scalar third-order equations

 Representative 2nd-order ODE $A\ne0$ Derived 3rd-order ODE Inherited algebra $y''=Ae^{-y'}$ $y'''+y''^{2}=0$ $L_{3;3}^{I}$ $y''=Ay'^{(a-2)/(a-1)}$ $y'y'''-\frac{a-2}{a-1}y''^2=0$ $L_{3;6}^I$ $y''=A(1+y'^{2})^{\frac{3}{2}}e^{b\arctan y'}$ $y'''-\frac{3y'+b}{1+y'^2}y''^2=0$ $L_{3;7}^I$ $xy''=Ay'^{3}-\frac{1}{2}y'$ $y'y'''-3y''^2=0$ $L_{3;8}^I$ $xy''=y'+y'^{3}+A(1+y'^{2})^{\frac{3}{2}}$ $y'''+y'''y'^2-3y'y''^2=0$ $L_{3;8}^{II}$ $xy''=y'-y'^{3}+A(1-y'^{2})^{\frac{3}{2}}$ $y'''-y'''y'^2-3y'y''^2=0$ $L_{3;8}^{III}$ $y''=AK^{3/2}$ $y'''=\frac32y''K^{-1}D_x K$ $L_{3;9}$ where $K=\displaystyle{1+y'+(y-xy')^2\over 1+x^2+y^2}$
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