Lagrange-Charpit equations for specific expression

[gtfitzpatrick]

Homework Statement

write down the Lagrange-Charpit eqs for
$$\frac{ \partial u}{ \partial x} \frac{ \partial u}{ \partial y} - y \frac{ \partial u}{ \partial x} - x \frac{ \partial u}{ \partial y}= 0$$

and use them to show $$\frac{ d^2 p}{ d p^2} = P$$

assuming that u = x^2 when y=0 determine the characteristic curves (x(t),y(t))

The Attempt at a Solution

so out eq gives pq-yp-xq = 0 so F(x,y,u,p,q) = pq-yp-xq
so
F_x = -q
F_y = -p
F_p = q-y
F_q = p-x
F_u = 0

so the char. eqs are
$$\frac{ dx}{ dt}$$ = q-y
$$\frac{ dy}{ dt}$$ = p-x
$$\frac{ dx}{ dt}$$ = qp
$$\frac{ dx}{ dt}$$ = q
$$\frac{ dx}{ dt}$$ = p

so $$\frac{ dx}{ dt}$$ = q-p but then $${ d^2 p}{ d p^2} = 0$$? what am i doing wrong, any ideas anyone?

 

[hunt_mat]

I think there is a problem with notation here.

Mat

 

[gtfitzpatrick]

Homework Statement

write down the Lagrange-Charpit eqs for
$$\frac{ \partial u}{ \partial x} \frac{ \partial u}{ \partial y} - y \frac{ \partial u}{ \partial x} - x \frac{ \partial u}{ \partial y}= 0$$

and use them to show $$\frac{ d^2 p}{ d p^2} = P$$

assuming that u = x^2 when y=0 determine the characteristic curves (x(t),y(t))

The Attempt at a Solution

so out eq gives pq-yp-xq = 0 so F(x,y,u,p,q) = pq-yp-xq
so
F_x = -q
F_y = -p
F_p = q-y
F_q = p-x
F_u = 0

so the char. eqs are
$$\frac{ dx}{ dt}$$ = q-y
$$\frac{ dy}{ dt}$$ = p-x
$$\frac{ du}{ dt}$$ = qp
$$\frac{ dp}{ dt}$$ = q
$$\frac{ dq}{ dt}$$ = p

so $$\frac{ dx}{ dt}$$ = q-p but then $${ d^2 p}{ d p^2} = 0$$? what am i doing wrong, any ideas anyone?

sorry this should read-
so the char. eqs are
$$\frac{ dx}{ dt}$$ = q-y

$$\frac{ dy}{ dt}$$ = p-x

$$\frac{ du}{ dt}$$ = qp

$$\frac{ dp}{ dt}$$ = q

$$\frac{ dq}{ dt}$$ = p

so $$\frac{ dp}{ dt}$$ = q but then $$\frac{ d^2 p}{ d t^2} = 0$$

 

[fzero]

sorry this should read-
so the char. eqs are
$$\frac{ dx}{ dt}$$ = q-y

$$\frac{ dy}{ dt}$$ = p-x

$$\frac{ dp}{ dt}$$ = qp

$$\frac{ dq}{ dt}$$ = q

$$\frac{ du}{ dt}$$ = p

so $$\frac{ dp}{ dt}$$ = q but then $$\frac{ d^2 p}{ d t^2} = 0$$

I think you mean

$$\frac{ dq}{ dt} = p$$

$$\frac{ dp}{ dt} = q$$

Then

$$\frac{ d^2 p}{ d t^2} = \frac{ dq}{ dt}$$

which is not zero.

Note, I believe that you also have a mistake in $$du/dt$$, so you might want to double check that.

 

[gtfitzpatrick]

I think you mean

$$\frac{ dq}{ dt} = p$$

$$\frac{ dp}{ dt} = q$$

Then

$$\frac{ d^2 p}{ d t^2} = \frac{ dq}{ dt}$$

which is not zero.

Note, I believe that you also have a mistake in $$du/dt$$, so you might want to double check that.

thanks i had made a mistake in du/dt. I am not sure now
$$\frac{ d^2 p}{ d t^2} = \frac{ dq}{ dt}$$