
Lecture 21. Separation of variables: Misc equations

In the previous Lecture 20 we considered heat equation.

Schrödinger equation

Consider problem \begin{align} & u_t= iku_{xx},&& t>0,\ 0<x<l,\label{eq-1}\\[3pt] & (\alpha_0u_x-\alpha u)|_{x=0}=(\beta_0u_x+\beta u)|_{x=l}=0.\label{eq-2}\\[3pt] & u|_{t=0}=g(x)\label{eq-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 $X''+\lambda X=0$, $$T'=-i k\lambda T \label{eq-4}$$ (explain, how). We also get boundary conditions $(\alpha_0 X'-\alpha X)=(\beta_0 X'+\beta X)(l)=0$ (Explain, how).

So, we have eigenvalues $\lambda_n$ and eigenfunctions $X_n$ ($n=1,2,\ldots$) and equation (\ref{eq-4}) for $T$, which results in $$T_n=A_ne^{-ik\lambda_n t} \label{eq-5}$$ and therefore a simple solution is $$u_n=A_ne^{-ik\lambda_n t}X_n(x) \label{eq-6}$$ and we look for a general solution in the form $$u=\sum_{n=1}^\infty A_ne^{-ik\lambda_n t}X_n(x). \label{eq-7}$$ Again, taking in account initial condition (\ref{eq-3}) we see that $$u=\sum_{n=1}^\infty A_nX_n(x). \label{eq-8}$$ and therefore $$A_n=\frac{1}{\|X_n\|^2}\int_0^l g(x)X_n(x)\,dx. \label{eq-9}$$

Remark.

1. Formula (\ref{eq-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 ofanalysis of Lecture 19.

1D wave equation

Consider problem \begin{align} & u_{tt}= c^2u_{xx},&& -\infty<x<\infty,\label{eq-10}\\[3pt] & (\alpha_0u_x-\alpha u)|_{x=0}=(\beta_0u_x+\beta u)|_{x=l}=0.\label{eq-11}\\[3pt] & u|_{t=0}=g(x),\qquad u_t|_{t=0}=h(x).\label{eq-12} \end{align} Actually we started from this equation in Lecture 12 but now we consider more general boundary conditions. Now we have $$T''= -k\lambda T \label{eq-13}$$ and we have 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-14} and respectively we get $$u =\sum _n T_n(t)X_n(x) \label{eq-15}$$ 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-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-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-18}\\[3pt] & (\alpha_0u_x-\alpha u)|_{x=0}=(\beta_0u_x+\beta u)|_{x=l}=0, \label{eq-19}\\[3pt] & u|_{y=0}=g(x).\label{eq-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 $$Y''= \lambda Y \label{eq-21}$$ and therefore assuming that $\lambda>0$ we get $$Y=Ae^{-\sqrt{\lambda}y}+Be^{\sqrt{\lambda}y} \label{eq-22}$$ We discard the second term in the right-hand expression of (\ref{eq-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-20}) and $$u_n=A_ne^{-\sqrt{\lambda_n}y}X_n(x) \label{eq-23}$$ and assuming that all $\lambda_n>0$ we get $$u=\sum_n A_ne^{-\sqrt{\lambda_n}y}X_n(x) \label{eq-24}$$ and (\ref{eq-19}) yields $$A_n=\frac{1}{\|X_n\|^2}\int_0^l g(x)X_n(x)\,dx. \label{eq-25}$$

Remark.

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-26}\\[3pt] & (\alpha_0u_x-\alpha u)|_{x=0}=\phi(y), \label{eq-27}\\[3pt] & (\beta_0u_x+\beta u)|_{x=l}=\psi(y), \label{eq-28}\\[3pt] & u|_{y=0}=g(x)\label{eq-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 Lecture 19).

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. everywher--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-30}\\[3pt] & (\alpha_0u_x-\alpha u)|_{x=0}=(\beta_0u_x+\beta u)|_{x=a}=0, \label{eq-31}\\[3pt] & u|_{y=0}=g(x), \label{eq-32}\\[3pt] & u|_{y=b}=h(x).\label{eq-33} \end{align}

Then we get (\ref{eq-21}) and (\ref{eq-22}) again but with two b.c. we cannot diacard anything; we get instead \begin{align} & A_n && + B_n&& =g_n,\label{eq-34}\\[3pt] & A_n e^{-\sqrt{\lambda_n}b} && + B_n e^{\sqrt{\lambda_n}b}&& =h_n\label{eq-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 $$Y_n(y)=\frac{\sinh (\sqrt{\lambda_n}(b-y))}{\sinh (\sqrt{\lambda_n}b)} g_n+ \frac{\sinh (\sqrt{\lambda_n}y)}{\sinh (\sqrt{\lambda_n}b)} h_n. \label{eq-36}$$ 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. Investigate other boundary conditions (Robin, Neumann, mixed).

Remark. a. 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 $\int_0^a h(x)X_0(x)\,dx -\int_0^a g(x)X_0(x)\,dx=0$.

1. 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.

2. Assume that we have Neumann b.c. everywhere--as $x=0$, $x=a$, $y=0$, $y=b$. Then solution does not exist unless $$\int_0^a h(x)\,dx -\int_0^a g(x)\,dx+\int_0^b \psi(y)\,dy -\int_0^b \phi(y)\,dy=0 \label{eq-37}$$ 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.