Lie Theory-Symmetry of Differential Equations

Consider the differential equation
\displaystyle \frac{dy}{dx} =F(x,y)

\displaystyle \frac{dy}{dx} =f(x)
\displaystyle y(x)=\int_{0}^{x} f(t) d t+C

This equation can be solved in integration (quadrature) and every solution is obtained by shifting one particular solution.(Constants of integration)
Is it possible to change coordinates in the first equation and reduce to the second case where we can solve equation explicitly.
One way to view the reason for this solvability is the symmetry- {f(x)} is independent of {y}, so the vector field is invariant under vertical shifts.
So given one solution, all the solutions are obtained by shifting it along {y} axis.

Can we always change coordinates to solve the first equation? Assume that we have a symmetry to the vector field, does that help you to solve the equation?

Assume a symmetry group acts continuously on our system:
{g_{s}.(x,y) =(x_s, y_s)}
\displaystyle \frac{dy_s}{dx_s} =F(x_s, y_s)

Let {a(x, y) \partial_{x}+b(x, y) \partial_{y}} be the the infinitesimal generator of a one-parameter subgroup of these symmetries.

\displaystyle \frac{d y}{d x}=F(x,y)=\frac{Y(x, y)}{X(x, y)}

Lie (1874): Assume {g_{s}} is a 1-parameter group leaving the system invariant. There exists a function {H(x, y)} such that

\displaystyle \frac{\partial H}{\partial x} =\frac{-Y}{X b-Y a}
\displaystyle \frac{\partial H}{\partial y}=\frac{X}{X b-Y a}

and {H(x, y)=C} is the solution to {\frac{d y}{d x}=F(x,y)=\frac{Y(x, y)}{X(x, y)}.}

Proof: If {H(x,y)} is a solution then,

\displaystyle H\left(x_{s}, y_{s}\right)=c(s)
\displaystyle \frac{\partial H}{\partial x} \frac{d x_{s}}{d s}+\frac{\partial H}{\partial y} \frac{d y_{s}}{d s}=c^{\prime}(s)
\displaystyle \frac{\partial H}{\partial x} a+\frac{\partial H}{\partial y} b=c^{\prime}(0)

\displaystyle dH(x,y)=0
\displaystyle \frac{\partial H}{\partial x} X+\frac{\partial H}{\partial y} Y=0
\displaystyle \frac{\partial H}{\partial x} dx+\frac{\partial H}{\partial y} dy=0

Thus we see that
\displaystyle \frac{\partial H}{\partial x} =\frac{-Y}{X b-Y a}
\displaystyle \frac{\partial H}{\partial y}=\frac{X}{X b-Y a}

So {\mu(x,y)=\displaystyle \frac{1}{X b-Y a}} is the integrating factor for our problem.

Examples:

1. \quad\displaystyle \frac{d y}{d x}=g(x)

The vertical translations {(x,y) \rightarrow (x,y+s)} leave the equation invariant.
Hence {a=0, b=1.}
{\mu(x,y)=1} works and we just get
\displaystyle y(x) =\int_{x_0}^{x} g(x) dx +C

2. \quad\displaystyle \frac{d y}{d x}=\frac{y+x\left(x^{2}+y^{2}\right)}{x-y\left(x^{2}+y^{2}\right)}

Rotations leave this equation invariant: Infact, we see that the equation says that the vector field is rotationally symmetric, but the angle between the vectors and the radial direction changes with the radius (at a fixed radius this angle is contant)

\quad{\displaystyle -y \frac{\partial}{\partial x}+x \frac{\partial}{\partial y} } is the generator for rotations.
{a =-y, ~b=x}
\displaystyle \mu(x,y) = \frac{1}{X b-Y a}=\frac{1}{(x^{2}+y^{2})}

3. \quad\displaystyle \frac{d y}{d x}=\frac{1-y^{2}}{x}

The symmetry is the transformation: {(x,y) \rightarrow (e^sx,y).}
The generator is {x\frac{\partial}{\partial x}}.
\displaystyle a =x, ~b=0
\displaystyle \mu(x,y) = \frac{1}{X b-Y a}= \frac{1}{x(y^2-1)}

4. \quad\displaystyle \frac{d y}{d x}=F(x) y+G(x)

The symmetry is the transformation: {(x,y) \rightarrow (x, y+se^{\int F(x) d x}).}
The generator is {e^{\int F(x) d x}\frac{\partial}{\partial y}}.
\displaystyle a=0, ~b=e^{\int F(x) d x}
\displaystyle \mu(x,y) =\frac{1}{X b-Y a}=\frac{1}{e^{\int F(x) d x}}

5. Riccati equation:
\quad\displaystyle \frac{d y}{d x}=x y^{2}-\frac{2 y}{x}-\frac{1}{x^{3}}

The symmetry is the transformation: {(x,y) \rightarrow \left(e^{s} x, e^{-2s} y\right)}, with generator
{ x \frac{\partial}{\partial x}-2y \frac{\partial}{\partial y}}

6. \quad\displaystyle \frac{d y}{d x}=f\left(\frac{y}{x}\right)

The symmetry is the transformation: {(x,y) \rightarrow \left(e^{s} x, e^{s} y\right)} with generator { x \frac{\partial}{\partial x}+y \frac{\partial}{\partial y}}
\displaystyle \mu(x,y) =\frac{1}{X b-Y a}= \frac{1}{y-f\left(\frac{y}{x}\right) x}

7. \quad\displaystyle \frac{d y}{d x}=\frac{y(x-y)}{x^{2}}

The symmetry is the transformation: {(x,y) \rightarrow \left(e^{s} x, e^{s} y\right)} with generator { x \frac{\partial}{\partial x}+y \frac{\partial}{\partial y}}
\displaystyle \mu(x,y) =\frac{1}{X b-Y a}= \frac{1}{x^2y -y(x-y)x} =\frac{1}{y^2x}

Differential Galois Theory: If there is solvable {r}-dimensional stability group for \displaystyle \frac{d y_{j}}{dx}=f^{j}\left(x, y_{1}, \dots, y_{r}\right)
Then the solution can be found by repeated integration.

Posted in $.

Leave a comment