# solve Uxx-3Uxt-4Utt=0 (hyperbolic)

by forget_f1
Tags: hyperbolic, solve, uxx3uxt4utt0
 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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,899 "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)??
 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.
P: 11

## solve Uxx-3Uxt-4Utt=0 (hyperbolic)

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

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