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#[Chapter 6. Separation of variables](id:chapter-6)
In this Chapter we continue study separation of variables which we started in [Chapter 4](../Chapter4/S4.1.html) but interrupted to explore Fourier series and Fourier transform.
##[6.1. Separation of variables for heat equation](id:sect-6.1)
-----------------------------------------
> 1. [Dirichlet boundary condition](#sect-6.1.1)
> 2. [Corrolaries](#sect-6.1.2)
> 3. [Other boundary condition](#sect-6.1.3)
> 2. [Corrolaries](#sect-6.1.4)
###[Dirichlet boundary conditions](id:sect-6.1.1)
Consider problem
\begin{align}
& u\_t= ku\_{xx},&& t\>0,\ 0\< x\< l,\label{eq-6.1.1}\\\\[3pt]
& u|\_{x=0}=u|\_{x=l}=0.\label{eq-6.1.2}\\\\[3pt]
& u|\_{t=0}=g(x).
\label{eq-6.1.3}
\end{align}
Let us consider a simple solution $u(x,t)=X(x)T(t)$; then separating variables we arrive to $\frac{T'}{T}=k\frac{X''}{X}$ which implies $X''+\lambda X=0$,
\begin{equation}
T'=-k\lambda T \label{eq-6.1.4}
\end{equation}
(explain, how). We also get boundary conditions $X(0)=X(l)=0$ (explain, how).
So, we have eigenvalues $\lambda\_n=(\frac{\pi n}{l})^2$ and eigenfunctions $X\_n=\sin (\frac{\pi n x}{l})$ ($n=1,2,\ldots$) and equation (\ref{eq-6.1.4}) for $T$, which results in
\begin{equation}
T\_n=A\_ne^{-k\lambda\_n t}
\label{eq-6.1.5}
\end{equation}
and therefore a simple solution is
\begin{equation}
u\_n=A\_ne^{-k\lambda\_n t}\sin (\frac{\pi n x}{l})
\label{eq-6.1.6}
\end{equation}
and we look for a general solution in the form
\begin{equation}
u=\sum\_{n=1}^\infty A\_ne^{-k\lambda\_n t}\sin (\frac{\pi n x}{l}). \label{eq-6.1.7}
\end{equation}
Again, taking in account initial condition (\ref{eq-6.1.3}) we see that \begin{equation}
u=\sum\_{n=1}^\infty A\_n\sin (\frac{\pi n x}{l}).
\label{eq-6.1.8}
\end{equation}
and therefore
\begin{equation}
A\_n=\frac{2}{l}\int\_0^l g(x)\sin (\frac{\pi n x}{l})\,dx.
\label{eq-6.1.9}
\end{equation}
###[Corollaries](id:sect-6.1.2)
1. Formula (\ref{eq-6.1.6}) shows that the problem is really ill-posed for
$t\<0$.
2. Formula (\ref{eq-6.1.9}) shows that as $t\to +\infty$
\begin{equation}
u=O(e^{-k\lambda\_1 t});
\label{eq-6.1.10}
\end{equation}
3. Moreover we have as $t\to +\infty$
\begin{equation}
u=A\_1 e^{-k\lambda\_1t}X\_1(x)e^{-k\lambda\_1 t}+ O(e^{-k\lambda\_2 t}). \label{eq-6.1.11}
\end{equation}
Consider now inhomogeneous problem with the right-hand expression and boundary conditions independent on $t$:
\begin{align}
& u\_t= ku\_{xx}+f(x),&& t\>0,\\ 0\< x\< l,\label{eq-6.1.12}\\\\[3pt]
& u|\_{x=0}=\phi,\qquad u|\_{x=l}=\psi,\label{eq-6.1.13}\\\\[3pt]
& u|\_{t=0}=g(x).\label{eq-6.1.14}
\end{align}
Let us discard initial condition and find a *stationary solution* $u=v(x)$: \begin{align}
& v''=-\frac{1}{k}f(x),&& 0\< x \< l,\label{eq-6.1.15}\\\\[3pt]
& v(0)=\phi,\qquad v(l)=\psi.\label{eq-6.1.16}
\end{align}
Then (\ref{eq-6.1.15}) implies
\begin{equation\*}
v(x)=-\frac{1}{k}\int\_0^x\int\_0^{x'} f(x'')\,dx''dx'+A+Bx=
\int\_0^x (x-x')f(x')\,dx'+A+Bx
\end{equation\*}
where we used formula of $n$-th integral (you must know it from the 1st year calculus)
\begin{equation}
I\_n(x)= \frac{1}{(n-1)!}\int\_a^x (x-x')^{n-1}f(x')\,dx'\qquad n=1,2,\ldots \end{equation}
for $I\_n:=\int\_a^x I\_{n-1}(x')\,dx'$, $I\_0(x):=f(x)$.
Then satisfying b.c. $A=\phi$ and
$B=\frac{1}{l}(\psi-\phi+\frac{1}{k}\int\_0^l (l-x') f(x')\,dx')$ and \begin{equation}
v(x) =\int\_0^x G(x,x') f(x')\,dx'+\phi (1-\frac{x}{l})+\psi \frac{x}{l} \label{eq-6.1.18}
\end{equation}
with
\begin{equation}
G(x,x') =\frac{1}{k}\left\\{ \begin{aligned}
& x'(1-\frac{x}{l})&& 0\< x'\< x,\\\\[3pt]
& x(1-\frac{x'}{l}) && x\< x'\< l.
\end{aligned}\right.
\label{eq-6.1.19}
\end{equation}
Returning to the original problem we note that $u-v$ satisfies (\ref{eq-6.1.1})--(\ref{eq-6.1.3}) with $g(x)$ replaced by $g(x)-v(x)$ and therefore $u-v=O(e^{-k\lambda\_1t})$. So
\begin{equation}
u=v+O(e^{-k\lambda\_1t}).
\label{eq-6.1.20}
\end{equation}
In other words, solution stabilizes to the stationary solution. For more detailed analysis of BVP for ODEs see [Section 6.A](S6.A.html).
###[Other boundary conditions](id:sect-6.1.3)
Similar approach works in the cases of boundary conditions we considered
before:
a. Dirichlet on one and and Neumann on the other
$u|\_{x=0}=u\_x|\_{x=l}=0$;
b. Neumann on both ends $u\_x|\_{x=0}=u\_x|\_{x=l}=0$;
c. Periodic $u|\_{x=l}=u|\_{x=0}$, $u\_x|\_{x=l}=u\_x|\_{x=0}$;
4. Dirichlet on one and and Robin on the other
$u|\_{x=0}=(u\_x+\beta u)|\_{x=l}=0$;
d. Robin on both ends
$(u\_x-\alpha u)|\_{x=0}=(u\_x+\beta u)|\_{x=l}=0$
but in (d), (e) we cannot find eigenvalues explicitly.
###[Corollaries](id:sect-6.1.4)
All corollaries remain valid as long as $\lambda\_1\>0$ which happens in cases (a), (d) with $\beta\ge 0$, (e) with $\alpha \ge 0,\beta\ge 0$ except $\alpha=\beta=0$.
Let us consider what happens when $\lambda\_1=0$ (cases (b) and (c)).
First, solution of the problem with r.h.e. and b.c. equal to $0$ does
not decay as $t\to +\infty$, instead
\begin{equation}
u=A\_1 +O(e^{-k\lambda\_2 t})
\label{eq-6.1.21}
\end{equation}
because in (b) and (c) $X\_1(x)=1$.
Second, solution of stationary problem exists *only conditionally*: iff
\begin{equation}
\frac{1}{k}\int\_0^l f(x)\,dx-\phi+\psi =0
\label{eq-6.1.22}
\end{equation}
in the case of Neumann b.c. on both ends $u\_x|\_{x=0}=\phi$, $u\_x|\_{x=l}=\psi$ and
\begin{equation}
\frac{1}{k}\int\_0^l f(x)\,dx =0
\label{eq-6.1.23}
\end{equation}
in the case of periodic b.c.
To cover the case when (\ref{eq-6.1.22}) or (\ref{eq-6.1.23}) fails (i.e. total heat flow is not $0$ so there is no balance) it is sufficient to consider the case $f=p$, $\phi=\psi=0$; then $u=pt$ with
\begin{equation}
p=\frac{1}{l} \int\_0^l f(x)\,dx
\label{eq-6.1.24}
\end{equation}
and in the general case
\begin{equation}
u = pt +A\_1 +O(e^{-k\lambda\_2 t}).
\label{eq-6.1.25}
\end{equation}
---------------
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