Solve the wave equation $u_{tt} = c^2 u_{xx}$ for $u(x,0) = sin(pi x)$ and $u_t(x,0) = 0$ on the domain $[0, 1]$ with fixed endpoints.

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start by using the method of separation of variables. Assume a solution of the form $u(x,t) = X(x)T(t)$. ‖ ‖ step 2 ⋮ Substitute $u(x,t) = X(x)T(t)$ into the wave equation $u_{tt} = c^2 u_{xx}$ to get $X(x)T''(t) = c^2 X''(x)T(t)$. ‖ ‖ step 3 ⋮ Divide both sides by $X(x)T(t)$ to separate the variables: $\frac{T''(t)}{c^2 T(t)} = \frac{X''(x)}{X(x)} = -\lambda$. ‖ ‖ step 4 ⋮ This gives us two ordinary differential equations: $T''(t) + \lambda c^2 T(t) = 0$ and $X''(x) + \lambda X(x) = 0$. ‖ ‖ step 5 ⋮ Solve the spatial part $X''(x) + \lambda X(x) = 0$ with boundary conditions $X(0) = 0$ and $X(1) = 0$. ‖ ‖ step 6 ⋮ The general solution for $X(x)$ is $X(x) = A \sin(\sqrt{\lambda} x) + B \cos(\sqrt{\lambda} x)$. ‖ ‖ step 7 ⋮ Applying the boundary conditions, we get $B = 0$ and $\sqrt{\lambda} = n\pi$ for $n = 1, 2, 3, \ldots$. Thus, $X_n(x) = A_n \sin(n\pi x)$. ‖ ‖ step 8 ⋮ Solve the temporal part $T''(t) + (n\pi c)^2 T(t) = 0$. The general solution is $T_n(t) = C_n \cos(n\pi c t) + D_n \sin(n\pi c t)$. ‖ ‖ step 9 ⋮ Combine the solutions: $u_n(x,t) = (A_n \sin(n\pi x))(C_n \cos(n\pi c t) + D_n \sin(n\pi c t))$. ‖ ‖ step 10 ⋮ Apply the initial conditions $u(x,0) = \sin(\pi x)$ and $u_t(x,0) = 0$. ‖ ‖ step 11 ⋮ From $u(x,0) = \sin(\pi x)$, we get $A_1 = 1$ and $A_n = 0$ for $n eq 1$. ‖ ‖ step 12 ⋮ From $u_t(x,0) = 0$, we get $D_1 = 0$. ‖ ‖ step 13 ⋮ Thus, the solution is $u(x,t) = \sin(\pi x) \cos(\pi c t)$. ‖ ∻Answer∻ ⚹ $u(x,t) = \sin(\pi x) \cos(\pi c t)$ ⚹ ∻Key Concept∻ ⚹Separation of Variables⚹ ∻Explanation∻ ⚹The method of separation of variables allows us to transform a partial differential equation into two ordinary differential equations, which can be solved independently.⚹◊What are the boundary conditions for this wave equation problem?⍭ Generate me a similar question◊

Sia