- #1
Stalker_VT
- 10
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I think i found the solution to my problem but i was hoping to have someone check to make sure i did not make a mistake.
[itex]\xi[/itex] = x - ct...... (1)
u(t,x) = v(t,[itex]\xi[/itex])......(2)
Taking the derivative
d[u(t,x) = v(t,[itex]\xi[/itex])]
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex] d[itex]\xi[/itex].....(3)
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex][ [itex]\frac{\partial \xi}{\partial x}[/itex]dx + [itex]\frac{\partial \xi}{\partial t}[/itex]dt]......(4)
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex] [itex]\frac{\partial \xi}{\partial x}[/itex]dx + [itex]\frac{\partial v}{\partial \xi}[/itex][itex]\frac{\partial \xi}{\partial t}[/itex]dt.....(5)
Question 1
Is it legal to cancel out partial fractions as such
[itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex] [itex]\frac{\partial \xi}{\partial x}[/itex]dx + [itex]\frac{\partial v}{\partial \xi}[/itex][itex]\frac{\partial \xi}{\partial t}[/itex]dt = [itex]\frac{\partial v}{\partial x}[/itex]dx + [itex]\frac{\partial v}{\partial t}[/itex] dt
Question 2
Is it legal to group like terms to get two separate equations as such:
Using (1) to get [itex]\frac{\partial \xi}{\partial x}[/itex] = 1 and [itex]\frac{\partial \xi}{\partial t}[/itex] = -c we obtain
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial\xi}[/itex]dx - c[itex]\frac{\partial v}{\partial \xi}[/itex] dt
simplifying slightly
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [[itex]\frac{\partial v}{\partial t}[/itex] - c[itex]\frac{\partial v}{\partial \xi}[/itex]]dt + [itex]\frac{\partial v}{\partial\xi}[/itex]dx
From this can we conclude the following:
A) [itex]\frac{\partial u}{\partial t}[/itex] = [itex]\frac{\partial v}{\partial t}[/itex] - c[itex]\frac{\partial v}{\partial \xi}[/itex]
B) [itex]\frac{\partial u}{\partial x}[/itex] = [itex]\frac{\partial v}{\partial\xi}[/itex]
by matching parameters in front of dt and dx?
Thank you for any insight
[itex]\xi[/itex] = x - ct...... (1)
u(t,x) = v(t,[itex]\xi[/itex])......(2)
Taking the derivative
d[u(t,x) = v(t,[itex]\xi[/itex])]
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex] d[itex]\xi[/itex].....(3)
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex][ [itex]\frac{\partial \xi}{\partial x}[/itex]dx + [itex]\frac{\partial \xi}{\partial t}[/itex]dt]......(4)
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex] [itex]\frac{\partial \xi}{\partial x}[/itex]dx + [itex]\frac{\partial v}{\partial \xi}[/itex][itex]\frac{\partial \xi}{\partial t}[/itex]dt.....(5)
Question 1
Is it legal to cancel out partial fractions as such
[itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex] [itex]\frac{\partial \xi}{\partial x}[/itex]dx + [itex]\frac{\partial v}{\partial \xi}[/itex][itex]\frac{\partial \xi}{\partial t}[/itex]dt = [itex]\frac{\partial v}{\partial x}[/itex]dx + [itex]\frac{\partial v}{\partial t}[/itex] dt
Question 2
Is it legal to group like terms to get two separate equations as such:
Using (1) to get [itex]\frac{\partial \xi}{\partial x}[/itex] = 1 and [itex]\frac{\partial \xi}{\partial t}[/itex] = -c we obtain
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial\xi}[/itex]dx - c[itex]\frac{\partial v}{\partial \xi}[/itex] dt
simplifying slightly
[itex]\frac{\partial u}{\partial t}[/itex]dt + [itex]\frac{\partial u}{\partial x}[/itex]dx = [[itex]\frac{\partial v}{\partial t}[/itex] - c[itex]\frac{\partial v}{\partial \xi}[/itex]]dt + [itex]\frac{\partial v}{\partial\xi}[/itex]dx
From this can we conclude the following:
A) [itex]\frac{\partial u}{\partial t}[/itex] = [itex]\frac{\partial v}{\partial t}[/itex] - c[itex]\frac{\partial v}{\partial \xi}[/itex]
B) [itex]\frac{\partial u}{\partial x}[/itex] = [itex]\frac{\partial v}{\partial\xi}[/itex]
by matching parameters in front of dt and dx?
Thank you for any insight