$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ #Chapter 6. Separation of variables 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 ----------------------------------------- > 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 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 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 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 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} --------------- [$\Leftarrow$](../Chapter5/S5.3.html)  [$\Uparrow$](../contents.html)  [$\downarrow$](./S6.A.html)  [$\Rightarrow$](./S6.2.html)