anonymous
  • anonymous
show that the expression u = sin (x-vt) describing a sinusoidal wave, satisfies the wave equation. Show that in general u = f(x - vt) and u = f(x + vt) satisfies the wave equations, where f is any function with a second derivative
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@mukushla
anonymous
  • anonymous
slow down buddy . you wrote u=f(x-vt) twice
anonymous
  • anonymous
i wrote once, u = f(x - vt) and u = f(x + vt)

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anonymous
  • anonymous
The wave equation is: \[\frac{\partial^2 u}{\partial t^2} = v^2\frac{\partial^2 u}{\partial x^2}\] In your case, \(u = \sin(x - vt)\) And \[u_{tt} = -\sin(x - vt) * v^2 \] \[u_{xx} = -\sin(x - vt)\] So since \(u_{tt} = v^2 u_{xx}\), u satifies the wave equation
anonymous
  • anonymous
good job @Meepi i appreciate
anonymous
  • anonymous
wait a sec
anonymous
  • anonymous
isn't it 1/v2 ??
anonymous
  • anonymous
No, \(u_{tt} = -\sin(x - vt) * v^2\) and \(u_{xx} = -\sin(x - vt)\) the v^2 is from the chain rule when you derive with respect to t in \(u_{tt}\) since \(u_{xx} = -\sin(x - vt)\), \(u_{tt} = u_{xx} * v^2\)
anonymous
  • anonymous
isn't it \[\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\] ??? you typed \(\frac{ ∂^{2}u }{ ∂t^{2} } = v^{2} \frac{ ∂^{2}u }{ ∂x^{2} } \)
anonymous
  • anonymous
i thought \(\frac{ ∂^{2}u }{ ∂t^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂x^{2} }\)
anonymous
  • anonymous
ohh my bad.., sorry
anonymous
  • anonymous
\[\frac{ ∂^{2}u }{ ∂x^{2} } = \frac{ 1 }{ v^{2} } \frac{ ∂^{2}u }{ ∂t^{2} }\]
anonymous
  • anonymous
It's alright your text book probably has the partial derivatives swapped
anonymous
  • anonymous
yeah
anonymous
  • anonymous
ur allright :), cool thanks :)
anonymous
  • anonymous
Also, for the general case, just use the chain rule as well to get \[u_{xx} = f''(x + vt)\] \[u_{tt} = f''(x + vt) * v^2\] Then \[u_{xx} = \frac{u_{tt}}{v^2}\]
anonymous
  • anonymous
Same can be done for u = f(x - vt)
anonymous
  • anonymous
ok.., i got it now :)
anonymous
  • anonymous
awesome :D
anonymous
  • anonymous
wanna help me with Legendre series ??
anonymous
  • anonymous
Haven't really done much of that but I can give it a shot :)

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