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Solve Uxx3Uxt4Utt=0 (hyperbolic) 
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#1
Oct404, 09:48 AM

P: 11

solve Uxx3Uxt4Utt=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 (Uttc^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 UbbUaa=0. This is where I am stuck.. I know the general solution is u(a,b)=f(a+b)+g(ab) also the explicit solution is u(a,b)=(1/2)*[φ(a+b)+φ(ab)]*(1/2c)*(integral ψ(s)ds from ab 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^(xt)]+x^2+(1/4)*t^2 but how to obtain it? 


#2
Oct404, 12:11 PM

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P: 39,552

"I know the general solution is u(a,b)=f(a+b)+g(ab)
also the explicit solution is u(a,b)=(1/2)*[φ(a+b)+φ(ab)]*(1/2c)*(integral ψ(s)ds from ab 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)?? 


#3
Oct404, 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. 


#4
Oct504, 01:02 PM

P: 11

Solve Uxx3Uxt4Utt=0 (hyperbolic)
Problem solved, thanks for taking the time to look at it



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