Recall from Formula (6) of Section 13.3 that under appropriate conditions a pluc

Question

Recall from Formula (6) of Section 13.3 that under appropriate conditions a plucked string satisfies the wave equation partial differential^2 u/partial differential t^2 = c^2 partial differential^2 u/partial differential x^2 where c is a positive constant. (a) Show that a function of the form u(x, t) = f(x + ct) satisfies the wave equation. (b) Show that a function of the form u(x, t) = g(x - ct) satisfies the wave equation. (c) Show that a function of the form u(x, t) = f(x + ct) + g(x - ct) satisfies the wave equation. (d) It can be proved that every solution of the wave equation is expressible in the form stated in part (c). Confirm that u(x, t) = sin t sin x satisfies the wave equation in which c = 1, and then use appropriate trigonometric identities to express this function in the form f(x + t) + g(x - t).

Explanation / Answer

a . the given PDE is : Utt = c2 Uxx

to verify U = f( x+ct ) is a solution , find Ut = cf ' (x+ct) , U tt = c2 f ''(x+ct) ----(1)

Ux =f ' (x+ct) , Uxx = f '' (x+ct) ----(2)

from 1 ,2 eliminating f '' we get Utt = c2  Uxx

Hence f(x+ct ) is the solution of the given PDE

b . similarly Utt = c2 g '' (x-ct ) and Uxx= g '' (x+ct )

eliminating g '' we get Utt = c2 Uxx and hence U= g(x-ct ) is also the solution of the given PDE

c . from a,b we can verify that U = f(x+ct ) + g(x-ct) also satisfies the given PDE .

hence it is also the solution

d. sint sin x= 1/2[ cos ( x-t) - cos (x+t)] = 1/2 [ cos (x-t) ] - 1/2 [ cos (x+t) ]

= f (x-t) - g(x+t)

here c=1 if f(x- t ) satisfies the PDE and g)x+ t) satisfies the PDE

then f(x-t) - g(x+t) also satisfies the Pde . hence sin t sinx is a solution

  

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