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显然特征线:x+t=c1,t-x=c2.
1.
设α=x+t,β=t-x,
设w(α,β)=u(x,t)
==>
Әu/Әx=Әw/Әα-Әw/Әβ,
Әu/Әt=Әw/Әα+Әw/Әβ,
Ә^2u/Әx^2=Ә^2w/Әα^2-2Ә^2w/(ӘαӘβ)+Ә^2w/Әβ^2,
Ә^2u/Әt^2=Ә^2w/Әα^2+2Ә^2w/(ӘαӘβ)+Ә^2w/Әβ^2.
==>
4Ә^2w/(ӘαӘβ)+2Әw/Әα=0.

2.
==>
w(α,β)=e^(-β/2)f(α)+g(β),
其中f在(0,+∞)中二阶连续可导,
g(t)在(-∞,0)(0,+∞)中二阶连续可导.
==>
u(x,t)=e^[(x-t)/2]f(x+t)+g(t-x),
Әu/Әt=e^[(x-t)/2]f'(x+t)+g'(t-x)-e^[(x-t)/2]f(x+t)/2,

3.
e^(-t/2)f(t)+g(t)=0
e^(x/2)f(x)+g(-x)=φ(x)==>φ(0)=g(0-)-g(0+)
e^[x/2]f'(x)+g'(-x)-e^[x/2]f(x)/2=ψ(x)
==>

f(t)=-e^(t/2)g(t),
f'(t)=-e^(t/2)g'(t)-e^(t/2)g(t)/2
==>

-e^xg(x)+g(-x)=φ(x),
-e^xg'(x)+g'(-x)=ψ(x),
==>

-2e^xg'(x)-e^xg(x)=φ'(x)+ψ(x),

==>
解出g(x),当x>0,
g(-x)=φ(x)+e^xg(x),解出g(x),当0>x.
f(t)=-e^(t/2)g(t),解出f(t),当t≥0.

4.
若φ(0)=0,则在0<x,t上有解.
φ(0)≠0,则在x=t时无解.

若对称开拓:
(1)偶开拓
e^xg(x)+g(-x)=φ(x)=φ(-x)=
e^(-x)g(-x)+g(x)==>e^x=1矛盾.

(2)同理奇开拓也不行.