• Support PF! Buy your school textbooks, materials and every day products Here!

Lagrange-Charpit equations for specific expression

  • #1

Homework Statement



write down the Lagrange-Charpit eqs for
[tex] \frac{ \partial u}{ \partial x} \frac{ \partial u}{ \partial y} - y \frac{ \partial u}{ \partial x} - x \frac{ \partial u}{ \partial y}= 0 [/tex]

and use them to show [tex] \frac{ d^2 p}{ d p^2} = P [/tex]

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
[tex] \frac{ dx}{ dt} [/tex] = q-y
[tex] \frac{ dy}{ dt} [/tex] = p-x
[tex] \frac{ dx}{ dt} [/tex] = qp
[tex] \frac{ dx}{ dt} [/tex] = q
[tex] \frac{ dx}{ dt} [/tex] = p

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

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,741
25


I think there is a problem with notation here.

Mat
 
  • #3


Homework Statement



write down the Lagrange-Charpit eqs for
[tex] \frac{ \partial u}{ \partial x} \frac{ \partial u}{ \partial y} - y \frac{ \partial u}{ \partial x} - x \frac{ \partial u}{ \partial y}= 0 [/tex]

and use them to show [tex] \frac{ d^2 p}{ d p^2} = P [/tex]

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
[tex] \frac{ dx}{ dt} [/tex] = q-y
[tex] \frac{ dy}{ dt} [/tex] = p-x
[tex] \frac{ du}{ dt} [/tex] = qp
[tex] \frac{ dp}{ dt} [/tex] = q
[tex] \frac{ dq}{ dt} [/tex] = p

so [tex] \frac{ dx}{ dt} [/tex] = q-p but then [tex]{ d^2 p}{ d p^2} = 0 [/tex]??? what am i doing wrong, any ideas anyone?
sorry this should read-
so the char. eqs are
[tex] \frac{ dx}{ dt} [/tex] = q-y

[tex] \frac{ dy}{ dt} [/tex] = p-x

[tex] \frac{ du}{ dt} [/tex] = qp

[tex] \frac{ dp}{ dt} [/tex] = q

[tex] \frac{ dq}{ dt} [/tex] = p

so [tex] \frac{ dp}{ dt} [/tex] = q but then [tex]\frac{ d^2 p}{ d t^2} = 0 [/tex]
 
Last edited:
  • #4
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289


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

[tex] \frac{ dy}{ dt} [/tex] = p-x

[tex] \frac{ dp}{ dt} [/tex] = qp

[tex] \frac{ dq}{ dt} [/tex] = q

[tex] \frac{ du}{ dt} [/tex] = p

so [tex] \frac{ dp}{ dt} [/tex] = q but then [tex]\frac{ d^2 p}{ d t^2} = 0 [/tex]
I think you mean

[tex] \frac{ dq}{ dt} = p[/tex]

[tex] \frac{ dp}{ dt} = q[/tex]

Then

[tex]\frac{ d^2 p}{ d t^2} = \frac{ dq}{ dt} [/tex]

which is not zero.

Note, I believe that you also have a mistake in [tex]du/dt[/tex], so you might want to double check that.
 
  • #5


I think you mean

[tex] \frac{ dq}{ dt} = p[/tex]

[tex] \frac{ dp}{ dt} = q[/tex]

Then

[tex]\frac{ d^2 p}{ d t^2} = \frac{ dq}{ dt} [/tex]

which is not zero.

Note, I believe that you also have a mistake in [tex]du/dt[/tex], so you might want to double check that.
thanks i had made a mistake in du/dt. im not sure now
[tex]\frac{ d^2 p}{ d t^2} = \frac{ dq}{ dt} [/tex]
 

Related Threads on Lagrange-Charpit equations for specific expression

  • Last Post
Replies
1
Views
2K
Replies
0
Views
2K
Replies
0
Views
865
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
0
Views
999
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
1K
Replies
2
Views
640
Replies
3
Views
730
Top