Problem statement
Separation of variables
General solution
Boundary value problems:
plucked string
standing waves
superposition
String of length L with initial shape f(x), initial velocity g(x), pinned ends
Denote vertical displacement at time t, position x by u(x,t)
Wave equation
Initial conditions
Boundary conditions
pinned ends: u(x=0,t)=0, u(x=L,t)=0, (e.g., u0=u1=0)
ut⁡t=a2⁡ux⁡x, u(x,0)=f(x), ut(x,0)=g(x), u(0,t)=0, u(L,t)=0. (Dirichlet problem)
u(x,t)=X(x)⁡T(t)⇒ X⁡T''=a2⁡X''⁡T
λ=0⇒X''=0⇒X(x)=c1+c2x, T''=0⇒T(t)=c3+c4t.
u(x,t)=X(x)T(t)=(c1+c2x)(c3+c4t)=(1+b1x)(b2+b3t)
u(0,t)=(b2+b3t)=0⇒b2=b3=0. Only solution u(x,t)=0 ✠
λ=-α2<0⇒X''-α2X=0⇒X(x)=c4⁡cosh(α⁡x)+c5⁡sinh(α⁡x)
T''-a2⁡α2⁡T=0⇒T(t)=c6⁡cosh(a⁡α⁡t)+c7⁡sinh(a⁡α⁡t) ✠
λ=α2>0⇒X''+α2X=0⇒X(x)=c7⁡cos(α⁡x)+c8⁡sin(α⁡x),
T''+a2⁡α2⁡T=0⇒T(t)=c9⁡cos(a⁡α⁡t)+c10⁡sin(a⁡α⁡t)
Only possible choice of λ=α2>0⇒
u(0,t)=0⇒u(0,t)=b1[⁡b3⁡cos(a⁡α⁡t)+⁡sin(a⁡α⁡t)]=0⇒b1=0
u(L,t)=0⇒u(L,t)=b2⁡sin(α⁡L)[⁡b3⁡cos(a⁡α⁡t)+⁡sin(a⁡α⁡t)]=0
b2=0⇒u(x,t)=0⁡⁡✠
sin(α⁡L)=0⇒α⁡L=n⁡π⇒αn=n⁡πL all lead to a possible solution
u(x,t=0)=f(x)=∑n=1∞⁡An⁡sin(αn⁡x)⁡,
ut(x,t=0)=g(x)=⁡∑n=1∞⁡a⁡αn⁡Bn⁡sin(αn⁡x)⁡
Solution is
Damped harmonic oscillator y''+f⁡y'+k⁡y=0
Damped wave equation ut⁡t+f⁡ut=a2⁡ux⁡x