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Sorted by: 2. utt(x, t) = 4uxx(x, t), x ∈ (0, π), t> 0. ux(0, t) = ux(π, t) = 0. u(x, 0) = 0, ut(x, 0) = sin(x) The general solution has the form u(x, t) ≡ f(x − 2t) + g(x + 2t) where f and g are functions to be determined. First, we'll find the solution for a general ut(x, 0) ≡ ϕ(x). Later on, ϕ(x) is chosen to agree with the ...
To solve the Wave Equation. utt − c2uxx = 0. One method is to start with operator factorization. utt − c2uxx = (∂ ∂t − c ∂ ∂x)(∂ ∂t + c ∂ ∂x)u = 0. Then it is claimed the solution is. u(x, t) = f(x + ct) + g(x − ct) where f ve g are arbitrary functions of single variable. The proof goes by letting v = ut + cux. then.
Wave equation, initial conditions. The displacement of an infinite string obeys the wave equation: ∂2u ∂t2 =c2∂2u ∂x2 ∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2. Find the solution in the form: u(x, t) = f(x − ct) + g(x + ct) u (x, t) = f (x − c t) + g (x + c t) u(x, 0) = 0 u (x, 0) = 0. ∂ ∂tu(x, 0) = x (1 +x2)2 ∂ ∂ t u (x, 0) = x ...
$\begingroup$ Check the laplace method for solving the wave equation. It is well-suited for problems with different boundary conditions and more robust than the d'alembert solution $\endgroup$ – user32882
To work according to the hint, substitute v(x, t) = u(x, t) − / phi(t) to receive a non homogeneous pde of the function v with homogeneous boundary conditions. Since this problem also contains the conditions u(x, 0) = 0 and ut(x, 0) = 0 , we do not necessarily have to consider the issues about the conditions u(0, t) = ϕ(t) and ux(π, t) = 0 .
So it would be possible to construct (if it does not already exist) a frame work that will have a physically tangible meaning for a complex speed. Another way to view this is to transform into the frequency domain. The "usual" wave equation transforms to the Helmholtz equation ∇2u^(r , ω) +k2u^(r , ω) = f^(r , ω) ∇ 2 u ^ (r →, ω) + k ...
After I read some books dedicated to variational calculus, I came up with the following answer, which is perfectly consistent with the accepted answer by @user10354138.
The term Ae − α (ct − x) is the homogeneous solution to the Robin boundary condition of the 1D wave problem. This term vanishes (A = 0) when the problem is read as an Initial Value Problem. Other boundary conditions (if the domain of interest has finite length) will most likely imply A = 0.
Limitations of D'Alembert solution of wave equation. 1. d'Alembert solution to wave equation. 2.
It is very intuitive that any function of the form y = f(x + vt) y = f (x + v t) would describe a wave in two spatial dimensions and time. From that it is easy to use the chain rule, letting w = x + vt w = x + v t and doing: ∂y ∂x = dy dw ∂w ∂x = dy dw ∂ y ∂ x = d y d w ∂ w ∂ x = d y d w.