Partial differential equation

[Jncik]

Homework Statement

solve u_xx = u_tt

The Attempt at a Solution

using the method of separation of variables I get

X''(x) - λ*X(x) = 0
T''(t) - λ*Τ(t) = 0

For λ = 0 I get X(x) = Ax + B where A and B are constants
T(t) = Dt + E where D and E are constants

hence u(x,t) = (Ax + B)*(Dt + E)

for λ>0 I get u(x,t) = [PLAIN]http://img690.imageshack.us/img690/8068/asdqu.gif [Broken] [Broken][/URL] where c1,c2,C1,C2 constants

for λ = -b^2<0 I get (E*cos(bx) + F*sin(bx))*(G*cos(bt) + H*sin(bt)) where E,F,G,H constants

hence the solution

is from the superposition principle

u(x,t) = (Ax + B)*(Dt + E) + [PLAIN]http://img690.imageshack.us/img690/8068/asdqu.gif [Broken] [Broken][/URL] + (E*cos(bx) + F*sin(bx))*(G*cos(bt) + H*sin(bt))


please is this correct? i think its wrong, because I am not sure about the last part

also our professor has only the [PLAIN]http://img690.imageshack.us/img690/8068/asdqu.gif [Broken] [Broken][/URL] as a reply, but i think its wrong, because we have to check for all λ to find all the possible solutions

 

[grey_earl]

λ is fixed. You either have λ > 0 or λ < 0, so you have either the solution
with exp[√(λ) x] or the one with the sines and cosines. Apart from that, you are correct but for one (I assume) typo: it's not exp[√(λx)] but exp[√(λ) x].

(PS: Of course you could also have λ = 0, but that appears seldom in reality.)

 

[phyzguy]

Note that given any function f, f(x-t) or f(x+t) satisfies this equation.

 

[hunt_mat]

Are you given any boundary conditions?

 

[Jncik]

thanks for your answers, no there are no boundary conditions

there is another exercise with boundary conditions and I get a simple result..

 

[HallsofIvy]

Without boundary or initial conditions, the most general possible solution is Af(x-t)+ Bf(x+t) where f is any twice differentiable function of a single variable, A and B constants, as phyzguy said.

That can be written in the form Jncik gives initially by summing over all possible values of $\lambda$.