- #1
JasonJo
- 429
- 2
Consider a DE
dy/dx = f(x,y)
Suppose that all the scaling transformations Tt, t not equal to 0, are symmetries of the DE. Show that the value of f(x,y) only depends on the ratio y/x, by showing that:
f(x,y) = f(1, y/x)
I think I have a rough idea of how to do this, my professor did it in class, but every step doesn't seem to be justified, or it's one of those solutions where the solution seems to come out of thin air, to me.
I tried:
Let T be a symmetry of the DE:
T takes (x,y) to (u = tx, v = ty)
du = t*dx, dv = t*dy
dv/du = (t/t)*(dy/dx) = dy/dx = f(y/x)
but how do you deduce that?
another problem:
Consider the 2nd order nonlinear DE:
m*(dv/dt) = mg - kv^2
at time t=0, the initial velocity is v(0) = 0.
(a) Solve for velocity v as a function of t be rewriting the above DE as a seperable first-order DE in v and t.
I don't really see how to do this, any hints?
dy/dx = f(x,y)
Suppose that all the scaling transformations Tt, t not equal to 0, are symmetries of the DE. Show that the value of f(x,y) only depends on the ratio y/x, by showing that:
f(x,y) = f(1, y/x)
I think I have a rough idea of how to do this, my professor did it in class, but every step doesn't seem to be justified, or it's one of those solutions where the solution seems to come out of thin air, to me.
I tried:
Let T be a symmetry of the DE:
T takes (x,y) to (u = tx, v = ty)
du = t*dx, dv = t*dy
dv/du = (t/t)*(dy/dx) = dy/dx = f(y/x)
but how do you deduce that?
another problem:
Consider the 2nd order nonlinear DE:
m*(dv/dt) = mg - kv^2
at time t=0, the initial velocity is v(0) = 0.
(a) Solve for velocity v as a function of t be rewriting the above DE as a seperable first-order DE in v and t.
I don't really see how to do this, any hints?