I have to solve $u_{tt}-u_{xx}=0$ with the given I.C.s
\begin{cases}
u_x(0,t)=u_x(\pi,t)=0\\
u(x,0)=\cos x \\
u_t(x,0)=-\cos x
\end{cases}
Solving the PDE with separation of variables :
\begin{equation}
\begin{array}
e-\frac{1}{X}u_{xx}=-k^2 \\
\frac{1}{T}u_{tt}=-k^2
\end{array}
\end{equation}
which give the following solutions:
\begin{gather}
u(x,t)=
\begin{cases}
Ae^{kx}\cos kt\\
Ae^{-kx}\cos kt \\
Ae^{kx}\sin kt\\
Ae^{-kx}\sin kt
\end{cases}
\end{gather}
Given the I.C. nr 2, I can see that the solution must have the cosine term, so I disregard the sine solutions. Then, since this is an oscillatory problem, I disregard the solution with $e^{kt}$ since $e^{kt}\rightarrow \infty$ (since this is not a perturbed system ). I am left with
\begin{equation}
u(x,t)= Ae^{-kx}\cos kt
\end{equation}
When I want to find A, I use I.C. 2:
\begin{equation}
cosx= Ae^{-kx}\cos 0t \rightarrow A=\cos xe^{-kx}
\end{equation}
This gives
\begin{equation}
u(x,t)= \cos x\cos kt
\end{equation}
But from here, I am not sure about $k$. If I use the other conditions, I just get $0=0$ or I use the third IC:
\begin{equation}
\begin{array}
fu_t(x,t)= -k\cos x\sin kt\\
IC: u_t(x,0)=-\cos x:\\
-\cos x= -k\cos x\sin 0\\
\frac{1}{k}=0\\
k=0?
\end{array}
\end{equation}
This gives
\begin{equation}
u(x,t)= cosx
\end{equation}
However, this is a strange result, since the PDE is for a function $u(x,t)$, having only one variable. Is there an apparent error here?
Thanks
UPDATE:
Separation of variables was not properly written:
Solving the PDE with separation of variables :
\begin{equation}
\begin{array}
e\frac{1}{F}F_{xx}=k^2 \\
\frac{1}{T}T_{tt}=k^2
\end{array}
\end{equation}
which gives the two ODEs:
\begin{equation}
\begin{array}
eF_{xx}-Fk^2=0 \\
T_{tt}-Tk^2=0
\end{array}
\end{equation}
with the solutions:
\begin{cases}
Ae^{kx} \\
Ae^{-kx}\\
Be^{kt}\\
Be^{-kt}
\end{cases}
Here we can disregard from the positive exponentials, and have:
$u(x,t)=Ae^{-kx}Be^{-kt}$ which we can write as:
$u(x,t)=Ae^{-k(x+t)}$
$u_t(x,t)=-Ake^{-k(x+t)}$
$u_x(x,t)=-Ake^{-k(x+t)}$
Using second IC:
\begin{equation}
\begin{array}
f u(x,t)=Ae^{-k(x+t)}\\
IC: u(x,0)=\cos x \\
Ae^{-kx}=\cos x \\
A=e^{kx}\cos x
\end{array}
\end{equation}
So we have now $u(x,t)=e^{kx}\cos x e^{-k(x+t)}\rightarrow \cos x e^{-kt}$
Therefore: $u(x,t)= \cos x e^{-kt}$
We find k by the 3d IC:
\begin{equation}
\begin{array}
f u(x,t)= \cos x e^{-kt}\\
IC: u_t(x,0)=-\cos x \\
-k\cos x e^{0}=-cosx\\
k= 1
\end{array}
\end{equation}
So the final answer is:
\begin{equation}
u(x,t)= \cos x e^{-t}
\end{equation}
A double check show that this is correct:
\begin{equation}
\begin{array}
fu_{xx}(x,t)= \cos x e^{-t} \\
u_{tt}(x,t)= \cos x e^{-t} \\
u_{tt}-u_{xx}=\cos x e^{-t}-\cos x e^{-t}=0
\end{array}
\end{equation}
So the original problem $u_{tt}-u_{xx}=0$ is satisfied by $u(x,t)= \cos x e^{-t} $.
A plot of $u(x,t)$ is given below
Best Answer
This problem can be solved using separation of variables. Setting $u=X(x)T(t)$ gives $$ \frac{T''}{T}=\lambda = \frac{X''}{X},\;\; X'(0)=X'(\pi)=0. $$ The solutions are $X_n(x)=\cos(n x),\;T_n(t)=A_n\cos(n t)+B_n\sin(n t)$, which correspond to $\lambda_n=-n^2$ for $n=0,1,2,3,\cdots$. So the general solution may be written as $$ u(t,x)=\sum_{n=0}^{\infty}(A_n\cos(n t)+B_n\sin(n t))\cos(n x). $$ The initial conditions are $$ u(0,x)=\cos(x),\;\; u_{t}(0,x)=-\cos(x). $$ The first condition determines the $A_n$'s and the second determines the $B_n$'s: $$ \cos(x)=\sum_{n=0}^{\infty}A_n\cos(n x),\;\; \cos(x)=\sum_{n=1}^{\infty}B_nn\cos(n x) $$