$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ##6.2. Separation of variables: Misc equations --------------------------------------------------- > 1. [Schrödinger equation](#sect-6.2.1) > 2. [1D wave equation](#sect-6.2.2) > 3. [Laplace equation in half-strip](#sect-6.2.3) > 4. [Laplace equation in rectangle](#sect-6.2.4) In the previous [Section 6.1](./S6.1.html) we considered heat equation. ###Schrödinger equation Consider problem \begin{align} & u\_t= iku\_{xx},&& t\>0,\ 0\< x\< l,\label{eq-6.2.1}\\\\[3pt] & (\alpha\_0u\_x-\alpha u)|\_{x=0}=(\beta\_0u\_x+\beta u)|\_{x=l}=0. \label{eq-6.2.2}\\\\[3pt] & u|\_{t=0}=g(x)\label{eq-6.2.3} \end{align} where *either* $\alpha\_0=0$, $\alpha=1$ and we have a Dirichlet boundary condition *or* $\alpha\_0=1$ and we have either Neumann or Robin boundary condition and the same at $x=l$. Let us consider a simple solution $u(x,t)=X(x)T(t)$; then separating variables we arrive to $\frac{T'}{T}=i k\frac{X''}{X}$ which implies $X''+\lambda X=0$, \begin{equation} T'=-i k\lambda T \label{eq-6.2.4} \end{equation} (explain, how). We also get boundary conditions $(\alpha\_0 X'-\alpha X)(0)=(\beta\_0 X'+\beta X)(l)=0$ (explain, how). So, we have eigenvalues $\lambda\_n$ and eigenfunctions $X\_n$ ($n=1,2,\ldots$) and equation (\ref{eq-6.2.4}) for $T$, which results in \begin{equation} T\_n=A\_ne^{-ik\lambda\_n t} \label{eq-6.2.5} \end{equation} and therefore a simple solution is \begin{equation} u\_n=A\_ne^{-ik\lambda\_n t}X\_n(x) \label{eq-6.2.6} \end{equation} and we look for a general solution in the form \begin{equation} u=\sum\_{n=1}^\infty A\_ne^{-ik\lambda\_n t}X\_n(x). \label{eq-6.2.7} \end{equation} Again, taking in account initial condition (\ref{eq-6.2.3}) we see that \begin{equation} u=\sum\_{n=1}^\infty A\_nX\_n(x). \label{eq-6.2.8} \end{equation} and therefore \begin{equation} A\_n=\frac{1}{\\|X\_n\\|^2}\int\_0^l g(x)X\_n(x)\,dx. \label{eq-6.2.9} \end{equation} ###Remark 1.** 1. Formula (\ref{eq-6.2.6}) shows that the problem is well-posed for $t\<0$ and $t\>0$. 2. However there is no stabilization. 3. What we got is a finite interval version of analysis of [Subsection 5.3.2](../Chapter5/S5.3.html#sect-5.3.2). ###1D wave equation Consider problem \begin{align} & u\_{tt}= c^2u\_{xx},&& 0\< x\< l,\label{eq-6.2.10}\\\\[3pt] & (\alpha\_0u\_x-\alpha u)|\_{x=0}=(\beta\_0u\_x+\beta u)|\_{x=l}=0. \label{eq-6.2.11}\\\\[3pt] & u|\_{t=0}=g(x),\qquad u\_t|\_{t=0}=h(x).\label{eq-6.2.12} \end{align} Actually we started from this equation in [Section 4.1](../Chapter4/S4.1.html) but now we consider more general boundary conditions. Now we have \begin{equation} T''= -k\lambda T \label{eq-6.2.13} \end{equation} and we have \begin{equation} T\_n=\left\\{\begin{aligned} & A\_n \cos (\omega\_n t)+ B\_n \sin (\omega\_n t) && \omega\_n = c\lambda\_n^{\frac{1}{2}}\text{as } \lambda\_n\>0,\qquad\\\\ & A\_n + B\_n t && \text{as } \lambda\_n=0,\\\\ & A\_n \cosh (\eta\_n t)+ B\_n \sinh (\eta\_n t) && \eta\_n = c(-\lambda\_n)^{\frac{1}{2}} \text{as } \lambda\_n\>0, \end{aligned}\right. \label{eq-6.2.14} \end{equation} and respectively we get \begin{equation} u =\sum \_n T\_n(t)X\_n(x) \label{eq-6.2.15} \end{equation} and we find from initial conditions \begin{align} & A\_n=\frac{1}{\\|X\_n\\|^2}\int\_0^l g(x)X\_n(x)\,dx, \label{eq-6.2.16}\\\\ & B\_n = \frac{1}{\\|X\_n\\|^2}\int\_0^l h(x)X\_n(x)\,dx \times\left\\{\begin{aligned} & \frac{1}{\omega\_n} && \text{as } \lambda\_n\>0,\\\\ & 1 && \text{as } \lambda\_n=0,\\\\ & \frac{1}{\eta\_n} && \text{as } \lambda\_n\<0. \label{eq-6.2.17} \end{aligned}\right. \end{align} ###Laplace equation in half-strip Consider problem \begin{align} & \Delta u:=u\_{xx}+u\_{yy}=0,&& y\>0,\\ 0\< x\< l,\label{eq-6.2.18}\\\\[3pt] & (\alpha\_0u\_x-\alpha u)|\_{x=0}=(\beta\_0u\_x+\beta u)|\_{x=l}=0, \label{eq-6.2.19}\\\\[3pt] & u|\_{y=0}=g(x).\label{eq-6.2.20} \end{align} To make it uniquely solvable we need to add condition $|u|\le M$. Again separating variables $u(x,y)=X(x)Y(y)$ we get \begin{equation} Y''= \lambda Y \label{eq-6.2.21} \end{equation} and therefore *assuming that $\lambda\>0$* we get \begin{equation} Y=Ae^{-\sqrt{\lambda}y}+Be^{\sqrt{\lambda}y} \label{eq-6.2.22} \end{equation} We discard the second term in the right-hand expression of (\ref{eq-6.2.22}) because it is unbounded. However if we had Cauchy problem (i.e. $u|\_{y=0}=g(x)$, $u\_y|\_{y=0}=h(x)$) we would not be able to do this and this problem will be ill-posed. So, $u=Ae^{-\sqrt{\lambda}y}$ and (\ref{eq-6.2.20}) and \begin{equation} u\_n=A\_ne^{-\sqrt{\lambda\_n}y}X\_n(x) \label{eq-6.2.23} \end{equation} and *assuming that all $\lambda\_n\>0$* we get \begin{equation} u=\sum\_n A\_n e^{-\sqrt{\lambda\_n}y}X\_n(x) \label{eq-6.2.24} \end{equation} and (\ref{eq-6.2.19}) yields \begin{equation} A\_n=\frac{1}{\\|X\_n\\|^2}\int\_0^l g(x)X\_n(x)\,dx. \label{eq-6.2.25} \end{equation} **Remark 2.** 1. If there is eigenvalue $\lambda=0$ we have $Y= A+By$ and as we are looking for a bounded solution we discard the second term again; so *our analysis remain valid as all $\lambda\_n\ge 0$*. 2. If we have a problem \begin{align} & \Delta u:=u\_{xx}+u\_{yy}=0,&& y\>0,\ 0\< x\< l,\label{eq-6.2.26}\\\\[3pt] & (\alpha\_0u\_x-\alpha u)|\_{x=0}=\phi(y), \label{eq-6.2.27}\\\\[3pt] & (\beta\_0u\_x+\beta u)|\_{x=l}=\psi(y), \label{eq-6.2.28}\\\\[3pt] & u|\_{y=0}=g(x)\label{eq-6.2.29} \end{align} with $g(x)=0$ we could reduce it by the method of continuation to the problem in the whole strip and solve it by Fourier transform (see [Subsecion 5.3.5](../Chapter5/S5.3.html#sect-5.3.5)). 3. In the general case we can find $u= u\_{(1)} +u\_{(2)}$ where $u\_{(1)}$ solves problem with $g=0$ and $u\_{(2)}$ solves problem with $\phi=\psi=0$ (explain how it follows from the linearity). 4. One can replace Dirichlet boundary condition $u|\_{y=0}$ by Robin boundary condition $(u\_y-\gamma u)|\_{y=0}=g(x)$ ($\gamma\ge 0$) but there is an exceptional case: there is an eigenvalue $\lambda\_0=0$ and as $y=0$ we have Neumann boundary condition. 5. In this exceptional case (usually as we have Neumann b.c. everywhere--as $x=0$, $x=l$, $y=0$) a required solution simply does not exists unless $\int\_0^l g(x)X\_0(x)\,dx=0$. ###Laplace equation in rectangle Consider problem \begin{align} &\Delta u:=u\_{xx}+u\_{yy}=0,&& 0\< y\< b,\ 0\< x\< a,\label{eq-6.2.30}\\\\[3pt] &(\alpha\_0u\_x-\alpha u)|\_{x=0}=(\beta\_0u\_x+\beta u)|\_{x=a}=0, \label{eq-6.2.31}\\\\[3pt] & u|\_{y=0}=g(x), \label{eq-6.2.32}\\\\[3pt] & u|\_{y=b}=h(x).\label{eq-6.2.33} \end{align} Then we get (\ref{eq-6.2.21}) and (\ref{eq-6.2.22}) again but with two b.c. we cannot diacard anything; we get instead \begin{align} & A\_n && + B\_n&& =g\_n,\label{eq-6.2.34}\\\\[3pt] & A\_n e^{-\sqrt{\lambda\_n}b} && + B\_n e^{\sqrt{\lambda\_n}b} && =h\_n\label{eq-6.2.35} \end{align} where $g\_n$ and $h\_n$ are Fourier coefficients of $g$ and $h$ respectively, which implies \begin{align\*} & A\_n= \frac{e^{\sqrt{\lambda\_n}b}}{2\sinh (\sqrt{\lambda\_n}b)}g\_n -\frac{1}{2\sinh (\sqrt{\lambda\_n}b)}h\_n) , \\\\[3pt] & B\_n= -\frac{e^{-\sqrt{\lambda\_n}b}}{2\sinh (\sqrt{\lambda\_n}b)}g\_n +\frac{1}{2\sinh (\sqrt{\lambda\_n}b)}h\_n \end{align\*} and therefore \begin{equation} Y\_n(y)=\frac{\sinh (\sqrt{\lambda\_n}(b-y))}{\sinh (\sqrt{\lambda\_n}b)} g\_n+ \frac{\sinh (\sqrt{\lambda\_n}y)}{\sinh (\sqrt{\lambda\_n}b)} h\_n. \label{eq-6.2.36} \end{equation} One can see easily that $\frac{\sinh (\sqrt{\lambda\_n}(b-y))}{\sinh (\sqrt{\lambda\_n}b)}$ and $\frac{\sinh (\sqrt{\lambda\_n}y)}{\sinh (\sqrt{\lambda\_n}b)}$ are bounded as $0\le y\le b$. **Problem 1.** Investigate other boundary conditions (Robin, Neumann, mixed). **Remark 3.** 1. There is an exeptional case: there is an eigenvalue $\lambda\_0=0$ and as $y=0$ and $y=b$ we have Neumann boundary conditions. Then solution does not exist unless $\int\_0^a h(x)X\_0(x)\\,dx -\int\_0^a g(x)X\_0(x)\\,dx=0$. 2. We can consider general b.c. with $(\alpha\_0u\_x-\alpha u)|\_{x=0}=\phi(y)$, $(\beta\_0u\_x+\beta u)|\_{x=a}=\psi(y)$. Then we can find $u= u\_{(1)} +u\_{(2)}$ where $u\_{(1)}$ solves problem with $g=h=0$ and $u\_{(2)}$ solves problem with $\phi=\psi=0$ (explain how it follows from the linearity). The second problem is also "our" problem with $x$ and $y$ permutted. 3. Assume that we have Neumann b.c. everywhere--as $x=0$, $x=a$, $y=0$, $y=b$. Then solution does not exist unless \begin{equation} \int\_0^a h(x)\,dx -\int\_0^a g(x)\,dx + \int\_0^b \psi(y)\,dy - \int\_0^b \phi(y)\,dy=0 \label{eq-6.2.37} \end{equation} which means that the total heat flow is $0$. How from two assumptions we can get one? Well, we just need to consider $\phi=g=0$, $\psi=\frac{y}{b}$, $h=-\frac{x}{a}$ (explain why) but there is a solution $u=\frac{y}{b}-\frac{x}{a}$ for that. -------------- [$\Leftarrow$](./S6.1.html)  [$\Uparrow$](../contents.html)  [$\Rightarrow$](./S6.3.html)