Getting Canonical Coordinates in differential equations from Lie Group
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I've been trying to understand how Lie Groups can help solve differential equations (this 12-page pdf is the most straightforward explanation I've seen).
My understanding is that when a first-order differential equation $frac{dy}{dx}=h(x,y)$ has a continuous "translation symmetry" $(x,y)mapsto(x,y+lambda)$, it becomes very easy to solve the equation using basic integration of one variable.
The problem is that most differential equations in standard $(x,y)$ coordinates don't have this nice translation symmetry, and so we need to change to a set of "canonical coordinates" $(X,Y)$ which do have the desired translation symmetry $(X,Y)mapsto(X,Y+lambda)$.
The main condition that these new canonical coordinates need to satisfy, is that the $Y$ coordinate should be aligned with the $lambda$ variable ($frac{dY}{dlambda}=1,frac{dX}{dlambda}=0$), so that we get the nice translation symmetry we want:
$$frac{dY}{dlambda}=frac{dY}{dx}frac{dx}{dlambda}+frac{dY}{dy}frac{dy}{dlambda}=Y_xxi+Y_yeta=1$$
$$frac{dX}{dlambda}=frac{dX}{dx}frac{dx}{dlambda}+frac{dX}{dy}frac{dy}{dlambda}=X_xxi+X_yeta=0$$
We also require that the symmetry condition be satisfied:
$$frac{dY}{dX}=frac{D_xY}{D_xX}=frac{Y_x+Y_y y'}{X_x+X_y y'}=h(X,Y)$$
Here's where I'm stuck. Basically all the terms in the above equations are unknowns, and it seems incredibly difficult to solve for the new coordinates $(X,Y)$. I've also seen the "linearized symmetry condition" below, but it still results in a difficult-to-solve equation:
$$eta_x-xi_y h^2 +(eta_y -xi_x)h - (xi h_x + eta h_y) = 0$$
Is there a nice, algorithmic approach for deducing a set of canonical coordinates? Or does it ultimately come down to "guess and check" with the above equation?
ordinary-differential-equations lie-groups
asked Nov 24 '18 at 22:52
eigenchriseigenchris
1,530616
$endgroup$
add a comment |
$begingroup$
I've been trying to understand how Lie Groups can help solve differential equations (this 12-page pdf is the most straightforward explanation I've seen).
My understanding is that when a first-order differential equation $frac{dy}{dx}=h(x,y)$ has a continuous "translation symmetry" $(x,y)mapsto(x,y+lambda)$, it becomes very easy to solve the equation using basic integration of one variable.
The problem is that most differential equations in standard $(x,y)$ coordinates don't have this nice translation symmetry, and so we need to change to a set of "canonical coordinates" $(X,Y)$ which do have the desired translation symmetry $(X,Y)mapsto(X,Y+lambda)$.
The main condition that these new canonical coordinates need to satisfy, is that the $Y$ coordinate should be aligned with the $lambda$ variable ($frac{dY}{dlambda}=1,frac{dX}{dlambda}=0$), so that we get the nice translation symmetry we want:
$$frac{dY}{dlambda}=frac{dY}{dx}frac{dx}{dlambda}+frac{dY}{dy}frac{dy}{dlambda}=Y_xxi+Y_yeta=1$$
$$frac{dX}{dlambda}=frac{dX}{dx}frac{dx}{dlambda}+frac{dX}{dy}frac{dy}{dlambda}=X_xxi+X_yeta=0$$
We also require that the symmetry condition be satisfied:
$$frac{dY}{dX}=frac{D_xY}{D_xX}=frac{Y_x+Y_y y'}{X_x+X_y y'}=h(X,Y)$$
Here's where I'm stuck. Basically all the terms in the above equations are unknowns, and it seems incredibly difficult to solve for the new coordinates $(X,Y)$. I've also seen the "linearized symmetry condition" below, but it still results in a difficult-to-solve equation:
$$eta_x-xi_y h^2 +(eta_y -xi_x)h - (xi h_x + eta h_y) = 0$$
Is there a nice, algorithmic approach for deducing a set of canonical coordinates? Or does it ultimately come down to "guess and check" with the above equation?
ordinary-differential-equations lie-groups
asked Nov 24 '18 at 22:52
eigenchriseigenchris
1,530616
$endgroup$
add a comment |
$begingroup$
I've been trying to understand how Lie Groups can help solve differential equations (this 12-page pdf is the most straightforward explanation I've seen).
My understanding is that when a first-order differential equation $frac{dy}{dx}=h(x,y)$ has a continuous "translation symmetry" $(x,y)mapsto(x,y+lambda)$, it becomes very easy to solve the equation using basic integration of one variable.
The problem is that most differential equations in standard $(x,y)$ coordinates don't have this nice translation symmetry, and so we need to change to a set of "canonical coordinates" $(X,Y)$ which do have the desired translation symmetry $(X,Y)mapsto(X,Y+lambda)$.
The main condition that these new canonical coordinates need to satisfy, is that the $Y$ coordinate should be aligned with the $lambda$ variable ($frac{dY}{dlambda}=1,frac{dX}{dlambda}=0$), so that we get the nice translation symmetry we want:
$$frac{dY}{dlambda}=frac{dY}{dx}frac{dx}{dlambda}+frac{dY}{dy}frac{dy}{dlambda}=Y_xxi+Y_yeta=1$$
$$frac{dX}{dlambda}=frac{dX}{dx}frac{dx}{dlambda}+frac{dX}{dy}frac{dy}{dlambda}=X_xxi+X_yeta=0$$
We also require that the symmetry condition be satisfied:
$$frac{dY}{dX}=frac{D_xY}{D_xX}=frac{Y_x+Y_y y'}{X_x+X_y y'}=h(X,Y)$$
Here's where I'm stuck. Basically all the terms in the above equations are unknowns, and it seems incredibly difficult to solve for the new coordinates $(X,Y)$. I've also seen the "linearized symmetry condition" below, but it still results in a difficult-to-solve equation:
$$eta_x-xi_y h^2 +(eta_y -xi_x)h - (xi h_x + eta h_y) = 0$$
Is there a nice, algorithmic approach for deducing a set of canonical coordinates? Or does it ultimately come down to "guess and check" with the above equation?
ordinary-differential-equations lie-groups
asked Nov 24 '18 at 22:52
eigenchriseigenchris
1,530616
$endgroup$
I've been trying to understand how Lie Groups can help solve differential equations (this 12-page pdf is the most straightforward explanation I've seen).
My understanding is that when a first-order differential equation $frac{dy}{dx}=h(x,y)$ has a continuous "translation symmetry" $(x,y)mapsto(x,y+lambda)$, it becomes very easy to solve the equation using basic integration of one variable.
The problem is that most differential equations in standard $(x,y)$ coordinates don't have this nice translation symmetry, and so we need to change to a set of "canonical coordinates" $(X,Y)$ which do have the desired translation symmetry $(X,Y)mapsto(X,Y+lambda)$.
The main condition that these new canonical coordinates need to satisfy, is that the $Y$ coordinate should be aligned with the $lambda$ variable ($frac{dY}{dlambda}=1,frac{dX}{dlambda}=0$), so that we get the nice translation symmetry we want:
$$frac{dY}{dlambda}=frac{dY}{dx}frac{dx}{dlambda}+frac{dY}{dy}frac{dy}{dlambda}=Y_xxi+Y_yeta=1$$
$$frac{dX}{dlambda}=frac{dX}{dx}frac{dx}{dlambda}+frac{dX}{dy}frac{dy}{dlambda}=X_xxi+X_yeta=0$$
We also require that the symmetry condition be satisfied:
$$frac{dY}{dX}=frac{D_xY}{D_xX}=frac{Y_x+Y_y y'}{X_x+X_y y'}=h(X,Y)$$
Here's where I'm stuck. Basically all the terms in the above equations are unknowns, and it seems incredibly difficult to solve for the new coordinates $(X,Y)$. I've also seen the "linearized symmetry condition" below, but it still results in a difficult-to-solve equation:
$$eta_x-xi_y h^2 +(eta_y -xi_x)h - (xi h_x + eta h_y) = 0$$
Is there a nice, algorithmic approach for deducing a set of canonical coordinates? Or does it ultimately come down to "guess and check" with the above equation?
ordinary-differential-equations lie-groups
ordinary-differential-equations lie-groups
asked Nov 24 '18 at 22:52
eigenchriseigenchris
1,530616
asked Nov 24 '18 at 22:52
eigenchriseigenchris
1,530616
asked Nov 24 '18 at 22:52
eigenchriseigenchris
1,530616
asked Nov 24 '18 at 22:52
eigenchriseigenchris
1,530616
asked Nov 24 '18 at 22:52
eigenchriseigenchris
1,530616
1,530616
add a comment |
add a comment |
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