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Solve Uxx-3Uxt-4Utt=0 (hyperbolic)

by forget_f1
Tags: hyperbolic, solve, uxx3uxt4utt0
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forget_f1
#1
Oct4-04, 09:48 AM
P: 11
solve Uxx-3Uxt-4Utt=0 with u(x,0)=x^2 and Ut(x,0)=e^x

I know that this is hyperbolic since D=(-1.5)^2+4 >0 so I have to transform the variables x and t linearly to obtain the wave equation of the form
(Utt-c^2Uxx=0). The above equation is equivalent to:

(d/dx - 1.5 d/dt)*(d/dx - 1.5 d/dt)u - 6.25 d^2u/dt^2 = 0

let x=b
let t=-1.5b + 2.5a
Thus,
Ub=Ux - (1.5) Ut
Ua=2.5 Ut

thus Ubb-Uaa=0. This is where I am stuck..

I know the general solution is u(a,b)=f(a+b)+g(a-b)
also the explicit solution is u(a,b)=(1/2)*[φ(a+b)+φ(a-b)]*(1/2c)*(integral
ψ(s)ds from a-b to a+b).
where u(a,0)=φ(a) and Ub(a,0)=ψ(a).

The solution is (4/5)*[e^(x+t/4)-e^(x-t)]+x^2+(1/4)*t^2
but how to obtain it?
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HallsofIvy
#2
Oct4-04, 12:11 PM
Math
Emeritus
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Thanks
PF Gold
P: 39,353
"I know the general solution is u(a,b)=f(a+b)+g(a-b)
also the explicit solution is u(a,b)=(1/2)*[φ(a+b)+φ(a-b)]*(1/2c)*(integral
ψ(s)ds from a-b to a+b).
where u(a,0)=φ(a) and Ub(a,0)=ψ(a)."

Excuse me? That the first time φ has appeared. What is φ(x)??
forget_f1
#3
Oct4-04, 04:06 PM
P: 11
u(a,0)=φ(a) and Ub(a,0)=ψ(a)

These are the initial conditions that would satisfy the explicit solution, in terms of a and b. φ and ψ ar functions.

Now what functions they are, that is where I need help, if I need them at all that is.

forget_f1
#4
Oct5-04, 01:02 PM
P: 11
Solve Uxx-3Uxt-4Utt=0 (hyperbolic)

Problem solved, thanks for taking the time to look at it


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