Htykuut

The question is as follows:

Solve the following initial boundary value problem for the wave equation in one spatial dimension:
$$
begin{cases}
displaystyle frac{partial^2 u}{partial t^2} - mu^2 frac{partial^2 u}{partial x^2_1} = 0, ~~ (x_1,t) in mathbb{R}_+ times (0,infty),\
\
displaystyle frac{partial^ell u}{partial t^ell}(x_1,0) = Q_ell(x_1), ~~ x_1 in mathbb{R}_+,\
\
displaystylefrac{partial u}{partial x_1} (0,t) = 0, ~~ t in (0,infty).
end{cases}
$$

For reference, I've included the way I was taught to solve IBVPs in general (my specific question is listed below this):

Find $u:Omega:[0,T]rightarrowmathbb{R}$ that is governed by
$$begin{cases}frac{partial^ku}{partial t^k}=-rhoDelta u=f(x,t), (x,t)inOmegatimes(0,T], ~~ leftarrowmbox{PDE}\
Bu(x,t)=g(x,t), ~~ (x,t)inpartialOmegatimes(0,T], ~~ leftarrowmbox{Boundary Condition}\
frac{partial^lu}{partial t^l}(x,0)=Q_l(x), ~~ xinOmega, l=0,cdots,k-1. ~~ leftarrowmbox{Initial Condition}end{cases}$$
For $k=1$, this is the heat equation and for $k=2$, this is the wave equation. Here, $f:Omegatimes(0,T]rightarrowmathbb{R}$, $g:partialOmegatimes(0,T]rightarrowmathbb{R}$, and $Q_l:Omegarightarrowmathbb{R}$ are given data, $B$ is the boundary operator, and $rho>0$.

The construction of such a solution relies upon a basis of $L^2(Omega)$, denoted by ${phi_alpha}_alpha$ and ${lambda_alpha}subsetmathbb{R}$, governed by
$$begin{cases}-Deltaphi_alpha=lambda_alphaphi_alpha & mbox{in }Omega,\Bphi_alpha=0 & mbox{on }partialOmega.end{cases}$$
The goal is to find
begin{equation}
tag{$1$}
u(x,t)=sum_alpha U_alpha(t)phi_alpha(x),
end{equation}
where
$$U_alpha(t)=A_alphaint_Omega u(x,t)phi_alpha(x)dx$$
and
$$A_alpha=|phi_alpha|_{L^2(Omega)}^{-2}.$$
(Note that $alpha=(alpha_1,cdots,alpha_n)$ is a multiindex.)
The following assumption is made: for all $tin[0,T]$, $frac{partial^ku}{partial t^k},-Delta u,fin L^2(Omega)$. This allows for a representation of each of these functions as a linear combination of ${phi_alpha}_alpha$. Then,
begin{equation}
tag{$2$}
f(x,t)=sum_alpha f_alpha(t)phi_alpha(x), ~~ mbox{where }f_alpha(t)=A_alphaint_Omega f(x,t)phi_alpha(x)dx,
end{equation}
and
begin{equation}
tag{$3$}
frac{partial^k u}{partial t^k}(x,t)=sum_alpha T_alpha(t)phi_alpha(x), ~~ mbox{where } T_alpha(t)=A_alphaint_Omegafrac{partial^k u}{partial t^k}(x,t)cdotphi_alpha(x)dx.
end{equation}
The goal is now to relate $T_alpha(t)$ to $U_alpha(t)$. Hence,
begin{equation}
tag{$4$}
frac{d^kU_alpha}{dt^k}=A_alphafrac{d^k}{dt^k}int_Omega u(x,t)phi_alpha(x)dx=A_alphaint_Omegafrac{partial^ku}{partial t^k}(x,t)phi_alpha(x)dx=T_alpha,
end{equation}
and
begin{equation}
tag{$5$}
-Delta u(x,t)=sum_alpha L_alpha(t)phi_alpha(x), ~~ mbox{where }L_alpha(t)=A_alphaint_Omega -Delta u(x,t)phi_alpha(x)dx.
end{equation}
Similarly, the goal is now to relate $L_alpha(t)$ to $U_alpha(t)$. By Green's formula,
begin{align*}
tag{$6$}
L_alpha(t)&=A_alphaleft(int_Omeganabla ucdotnablaphi_alpha dx-int_{partialOmega}phi_alphanabla ucdotnu dSright)\
&=A_alphaint_Omega(-Deltaphi_alpha)udx+A_alphaleft(int_{partialOmega}unablaphi_alphacdotnu dS-int_{partialOmega}phi_alphanabla ucdotnu dSright)\
&=A_alphaint_Omegalambda_alphaphi_alpha udx+Bdry\
&=lambda_alpha U_alpha(t)+Bdry.
end{align*}
Similarly,
$$frac{partial^lu}{partial t^l}(x,0)=sum_alpha T_alpha^lphi_alpha(x), ~~ mbox{where }T_alpha^l=A_alphaint_Omegafrac{partial^lu}{partial t^l}(x,0)phi_alpha(x)dx=A_alphaint_Omega Q_l(x)phi_alpha(x)dx,$$
so by the same process as in $(4)$,
begin{equation}
tag{$7$}
frac{d^lU_alpha}{dt^l}(0)=T_alpha^l=A_alphaint_Omega Q_l(x)phi_alpha(x)dx.
end{equation}
Substituting $(2),(3),$ and $(5)$ pack into the PDE:
$$sum_alphaleft(frac{d^kU_alpha}{dt^k}+rholambda_alpha U_alpha+rho Bdryright)phi_alpha=sum_alpha f_alpha(t)phi_alpha.$$
By the linear independence of ${phi_alpha}_alpha$, it follows that
$$frac{d^kU_alpha}{dt^k}+rholambda_alpha U_alpha+rho Bdry=f_alpha(t), ~~ forallalpha.$$
This is a nonhomogeneous $k^{mbox{th}}$ order linear ODE with initial condition given by $(7)$. It needs to be made precise what $Bdry$ is. The expression depends on the boundary condition $Bu=g$ on $partialOmegatimes(0,T]$.

1. When $Bu=u$ (i.e., Dirichlet),
   $$Bdry=A_alphaint_{partialOmega}g(x,t)nablaphi_alphacdotnu dS(x).$$
   


2. when $Bu=-nabla ucdotnu$ (i.e., Neumann),
   $$Bdry=A_alphaint_{partialOmega}g(x,t)phi_alpha dS(x).$$
   


3. Similarly for when $Bu=-nabla ucdotnu+bu$ (i.e., Robin).


4. The same can be derived for when $Bu$ is a mixed boundary condition.

I'm struggling to even get started on this problem. This is a homework question, so I'm asking for guidance, not a full solution. My struggle stems from the fact that all of the examples I can find in textbooks and online use vastly different notation (which I suppose is another way to say that I don't truly understand the method being applied). Sure, it's simple to see things like $Omega=mathbb{R}_+$ and $T=infty$, but I'm not sure how to apply any of the info from the method beyond initial setup.