Robin BC in the 1D wave equation - Mathematics Stack Exchange
2019年3月14日 · $\begingroup$ @pluton, to be precise, the boundary condition you are considering, strictly speaking, is not the standard Robin condition: this has the following form $$\left[\partial_x u(x,t)+\alpha u(x,t)\right]_{x=0}=0 \iff \partial_x u(0,t)=\color{red}{-\alpha} u(0,t).$$ In this form, the condition $\alpha>0$ implies existence and uniqueness of a solution for the …
Second order 1-D wave equation, Implicit scheme
I am trying to solve the second order wave equation in 1 dimension from the implicit method by finite difference. $$ \frac{\partial^2U}{\partial t^2} = c^2\frac{\partial^2U}{\partial x^2} , \quad t>0 $$ where initially $$ U(x,0)=f(x), \quad \frac{\partial U(x,0)}{\partial t} = g(x) $$ The discretized form of the above equation can be written as:
Is the wave equation a hyperbolic, parabolic, or elliptic PDE?
2022年9月9日 · 0. Generally speaking, wave equations are hyperbolic. They have the similar form that. ∂2u ∂t2 =a2Δu, ∂ 2 u ∂ t 2 = a 2 Δ u, where Δ Δ is the Laplacian and u u is the displacement of the wave. Typical examples are acoustic wave, elastic wave, and electromagnetic. In one dimensional, the equation is written as.
1D wave equation with Boundary Conditions: Fourier Transform …
2019年4月12日 · $\begingroup$ The problem is a bit further back. Your $\hat{u}$ is not strictly correct; the point is that the boundary conditions can only be satisfied by a non-identically-zero function if $\omega=\omega_k$.
1D Wave equation mixed boundary conditions and I.C.
2019年2月2日 · 1. You need to subtract off the boundary conditions before you can apply separation of variables. Try a solution of the form. u(x, t) = Axcos(Ωt) + v(x, t) where the boundary function was obtained from f(x)Acos(Ωt) such that f(0) = 0 and f ′ (L) = 1. Then v(x, t) satisfies. {vtt − c2vxx = Ω2Axcos(Ωt) v(0, t) = vx(L, t) = 0 v(x, 0) = − ...
calculus - Greens function of 1-d forced wave equation
Fundamental solution for 1D nonhomogeneous wave equation. 3. Fourier Transform of Heaviside Step Function. 3.
Help visualizing solutions to the (1D) wave equation.
2016年6月5日 · 3. I know that the one-dimensional wave equation can be written as $$ \frac {\partial^2 u} {\partial t^2} = c^2 \frac {\partial ^2 u} {\partial x^2}$$ and has solutions of the form $$ u = F (x+ct) + G (x-ct)$$. I'm having trouble developing a proper intuition about the meaning of the solution, though.
Applying Green's function for one dimensional wave equation
2020年9月22日 · 3. The Green's function of the one dimensional wave equation. (∂2t −∂2z)ϕ = 0. fulfills. (∂2t −∂2z)G(z, t) = δ(z)δ(t) I calculated that its retarded part is given by: G+(z, t) = Θ(t)Θ(t −|z|). In Wikipedia I find a very similar expression without the first Θ(t). I think this has to do with the fact that in Wikipedia the full ...
Deduce 1d wave equation from 2d by method of descent
Deduce 1d wave equation from 2d by method of descent. Ask Question Asked 10 years, 10 months ago.
Laplace transform of the wave equation - Mathematics Stack …
2015年5月28日 · Your equation $$\dfrac{d^2 v}{dx^2} - s^2 v = - s \sin(\pi x)$$ is a second-order linear differential equation. Like any linear differential equation, its general solution is of the form (particular solution) + (general solution of the homogeneous equation).