0
$\begingroup$

I need to solve $y''=0$ for $y=y(x)$ using Symmetry methods.

Here is my work:

TOTAL DERIVATIVE OPERATOR: $ D_x = \frac{\partial }{\partial x} + y'\frac{\partial}{\partial y}$

LINEARIZED SYMMETRY CONDITION: $Q_{xx}+2y'Q_{xy}+y'^2Q_{yy}=0$

From the function: $Q(x,y,y')=\eta(x,y) - \xi(xy)y'$

I have :

$Q_{xx}=\eta_{xx}-\xi_{xx}y'$

$Q_{xy}=\eta_{xy}-\xi_{xy}y'$

$Q_{yy}=\eta_{yy}-\xi_{xy}y'$

I sub these into the LSC and get:

$\eta_{xx}+(2\eta_{xy}-\xi_{xx})y'+(\eta_{yy}-2\xi_{xy})y'^{2}-\xi_{yy}y'^{3}=0$

Therefore:

$\eta_{xx}=0$

$\eta_{yy}-2\xi_{xy}=0$

$2\eta_{xy}-\xi_{xx}=0$

$\xi_{yy}=0 $

from $\xi_{yy}=0$ I get the equation

$\xi(x,y)=a_1xy+a_2y+a_3x^2+a_4x+a_5$ - call this eq1

from $\eta_{xx}=0$ I get the equation:

$\eta(x,y)=c_1xy+c_2x+c_3y^2+c_4y+c_5$ - call this eq2

From eq2 we get $\eta_{xy}=c_1 \text{ and } \eta_{yy}=2c_3$

From eq1 we get $\xi_{xx}=2a_3 \text{ and } \xi_{xy}=a_1$ when we sub this into $2\eta_{xy}-\xi_{xx}=0 \text{ and } \eta_{yy}-2\xi_{xy}=0$ we get $c_1=a_3 \text{ and } c_3=a_1$

This lets me rewrite equations 1 and 2 as :

$\xi(x,y)=c_1x^2+c_2xy+c_3x+c_4y+c_5$

$\eta(x,y)=c_1xy+c_2y^2+c_6x+c_7y+c_8$

The infinitismal Creator of Lie Symmetries is :

$X = \xi(xy)\frac{\partial}{\partial x} + \eta(x,y)\frac{\partial}{\partial y}$

this yields the following symmetries:

$X_1=x^2\frac{\partial}{\partial x} + xy\frac{\partial}{\partial y}$

$X_2=xy\frac{\partial}{\partial x} + y^2\frac{\partial}{\partial y}$

$X_3=x\frac{\partial}{\partial x}$

$X_4=y\frac{\partial}{\partial x}$

$X_5=\frac{\partial}{\partial x}$

$X_6=x\frac{\partial}{\partial y}$

$X_7=y\frac{\partial}{\partial y}$

$X_8=\frac{\partial}{\partial y}$

I solved $X_1$ as follows:

$\frac{dx}{x^2}=\frac{dy}{xy}$

$\frac{dy}{dx}=\frac{xy}{x^2}$

$\frac{dy}{dx}=\frac{y}{x}$

therefore $y=xe^{k_1}$

I get the same solution when solving $X_2$

For $X_3-X_5$ I am getting $y=k_2$

However I don't know if any of what I have done is correct and I don't know how to solve the other symmetries.

Please help.

$\endgroup$

1 Answer 1

3
$\begingroup$

You have solved the equations to find the generators of these lie point symmetries. To find the exact transformations, you should use the property that the lie point symmetry is the unique solution to: $$\frac{d}{ds}(u(s))=X(u(s))\\u(0)=(x,y)\\\implies u(s)=(\tilde x,\tilde y)$$where $u(s)$ is the symmetry you wish to solve for, and $X$ is the generator.

To illustrate this, take the generator $X_5=\frac{\partial}{\partial x}$. As a vector field, this is written as $X_5(x,y)=\left(\begin{matrix}1\\0\end{matrix}\right)$. So $$\frac{d}{ds}\left(\begin{matrix}\tilde x\\\tilde y\end{matrix}\right)=\left(\begin{matrix}1\\0\end{matrix}\right)$$ So $\tilde x=x+s$ and $\tilde y=y$.

A more tricky example is $X_1=x^2\frac{\partial}{\partial x} + xy\frac{\partial}{\partial y}$. Solving the differential equations gives: $$\tilde x(s)=\frac x{1-sx}\\\tilde y(s)=\frac y{1-sx}$$

As a check, we see that $$\frac{\partial}{\partial \tilde x}=\frac{\partial x}{\partial \tilde x}\frac{\partial}{\partial x}=\frac{1}{(1+s\tilde x)^2}\frac{\partial}{\partial x}=(1-sx)^2\frac{\partial}{\partial x}$$So $$\begin{align}\tilde y_{\tilde x\tilde x}&=(1-sx)^2\frac{\partial}{\partial x}\left((1-sx)^2\frac{\partial}{\partial x}\left(\frac y{1-sx}\right)\right)\\&=(1-sx)^2\frac{\partial}{\partial x}\left((1-sx)^2\left(\frac {y_x(1-sx)+sy}{(1-sx)^2}\right)\right)\\&=(1-sx)^2\frac{\partial}{\partial x}\left({y_x(1-sx)+sy}\right)\\&=(1-sx)^2((1-sx)y_{xx}-sy_x+sy_x)\\&=(1-sx)^3 y_{xx}\end{align}$$So $\tilde y_{\tilde x\tilde x}=0\iff y_{xx}=0$, which is the definition of a symmetry of a p.d.e.

$\endgroup$
2
  • $\begingroup$ would this apply for y=y(x)? Please excuse my ignorance. I have not studied symmetry methods for solving ODEs. I am essentially referring to examples I find online and applying them to solve y''=0 for y=y(x). Thanks again for your help. :) $\endgroup$ Jan 3, 2018 at 6:17
  • $\begingroup$ Yep, it is for y=y(x) $\endgroup$
    – John Doe
    Jan 4, 2018 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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