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Function: require_once
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Function: require_once
Theorem 1 (Uniqueness). Let
u_(1)(x,t)and
u_(2)(x,t)be
C^(2)solutions of the following problem D.E.
u(0,t)=A(t),u(L,t)=B(t)u(x,0)=f(x)u_(t)(x,0)=g(x)u_(1)(x,t)=u_(2)(x,t)0<=x<=L,-\infty .u_(tt)=a^(2)u_(\times ),0<=x<=L,-\infty
B.C. u(0,t)=A(t),u(L,t)=B(t)
I.C. u(x,0)=f(x)u_(t)(x,0)=g(x).
Then u_(1)(x,t)=u_(2)(x,t) for all 0<=x<=L,-\infty .Let
v(x,t)and
w(x,t)be two
C^(2)solutions of the problem D.E.
u(0,t)=0,u(L,t)=0(d)/(dt)\int_0^L [a^(2)v_(x)(x,t)w_(x)(x,t)+v_(t)(x,t)w_(t)(x,t)]dx=0v_(xt)w_(x)+v_(x)w_(xt)+v_(\times )w_(t)+v_(t)w_(\times )=(v_(t)w_(x)+v_(x)w_(t))_(x)u_(tt)=a^(2)u_(\times ),0<=x<=L,-\infty
B.C. u(0,t)=0,u(L,t)=0.
(a) Use the technique in the proof of Theorem 1 to show that
(d)/(dt)\int_0^L [a^(2)v_(x)(x,t)w_(x)(x,t)+v_(t)(x,t)w_(t)(x,t)]dx=0
Hint. v_(xt)w_(x)+v_(x)w_(xt)+v_(\times )w_(t)+v_(t)w_(\times )=(v_(t)w_(x)+v_(x)w_(t))_(x)