082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

25.5 Partial differential equations 833
wherecisthespeedofthewave.Thisp.d.e.iscalledtheone-dimensionalwaveequation.
Underprescribedconditionsthesubsequentdisplacementofthewavecanbecalculated
as a function of position and time.
Example25.8 Verifythat
u(x,t) = sin(x +2t)
isasolution ofthe one-dimensional wave equation
∂ 2 u
∂t = u
2 4∂2 ∂x 2
Solution The first partialderivatives arecalculated.
∂u
∂u
= cos(x +2t)
∂x
∂t =2cos(x+2t)
The second derivatives, ∂2 u
∂x and ∂2 u
, arenow found.
2 ∂t2 ∂ 2 u
∂x = −sin(x+2t) ∂ 2 u
= −4sin(x +2t)
2 ∂t2 Now
∂ 2 u
∂t = −4sin(x +2t) = 4[−sin(x +2t)] = u
2 4∂2 ∂x 2
Henceu(x,t) = sin(x +2t) isasolution ofthe given wave equation.
Another equally important p.d.e. is Laplace’s equation. This equation is used extensivelyinelectrostatics.Undercertainconditionstheelectrostaticpotentialinaregionis
describedby a function φ(x,y) which satisfiesLaplace’s equation intwo dimensions.
∂ 2 φ
∂x + ∂2 φ
2 ∂y = 0 2
Thisequationissoimportantthatawholeareaofappliedmathematics,calledpotential
theory, isdevoted tothe studyof its solution.
Example25.9 Verifythat
φ(x,y,z) =
1
√
x2 +y 2 +z 2
satisfies the three-dimensional Laplace’s equation
∂ 2 φ
∂x + ∂2 φ
2 ∂y + ∂2 φ
2 ∂z = 0 2
Solution We begin by calculating the first partialderivative, ∂φ . We aregiven
∂x
φ = (x 2 +y 2 +z 2 ) −1/2
and so
∂φ
∂x = −1 2 (x2 +y 2 +z 2 ) −3/2 (2x) = −x(x 2 +y 2 +z 2 ) −3/2