Summary of Separation of Variables — Question 3

Question 3

Consider the separated spatial problem $$-X''=\lambda X,\qquad 0<x<L,\quad X(0)=0,\quad X'(L)+hX(L)=0,$$ where $L>0$ and $h\in\mathbb R$. The same boundary conditions are imposed on the heat equation $u_t=\kappa u_{xx}$ and the wave equation $v_{tt}=c^2v_{xx}$, with $\kappa,c>0$.

Tasks

1. Derive the energy identity for $\lambda$. Use $X(L)^2\le L\int_0^LX'^2\,dx$ to identify when every eigenvalue is positive and when a zero eigenvalue occurs.

2. For $h<-1/L$, derive the negative-eigenvalue equation and prove there is exactly one negative eigenvalue.

3. Choose $L=1$ and $h=-\coth 1$. Construct a heat solution and a wave solution with initial profile $\sinh x$, taking zero initial wave velocity.

4. Explain why these solutions invalidate a rule that separated heat modes always decay and separated wave modes always oscillate. Identify the boundary term responsible.

Question 3 – Solution

Strategy. The admissible spatial spectrum depends on the boundary operator, including the sign of a Robin coefficient.

Step 1: Include the Robin boundary term. Multiplication by $X$ and integration by parts give $$\boxed{\lambda\int_0^LX^2\,dx=\int_0^LX'^2\,dx+hX(L)^2.}$$ For $h\ge 0$ the right side is positive for nonzero $X$ with $X(0)=0$. For $-1/L<h<0$, the supplied trace inequality bounds it below by $(1+hL)\int X'^2>0$. At $h=-1/L$ it is nonnegative, with equality precisely for an affine function through zero; $X=x$ is the zero mode. Conversely, solving $X''=0$ shows a nonzero zero mode occurs only at $h=-1/L$.

Step 2: Find the negative mode. For $\lambda=-\sigma^2$, $\sigma>0$, the left condition gives $X=A\sinh(\sigma x)$. The right condition is $$\boxed{\sigma\coth(\sigma L)=-h.}$$ The left side increases strictly from $1/L$ to infinity: with $z=\sigma L$, the derivative of $z\coth z$ is $(\sinh z\cosh z-z)/\sinh^2z>0$. Thus there is exactly one positive $\sigma$ when $-h>1/L$, and none otherwise. This counts all negative eigenvalues because every such eigenfunction must have the displayed hyperbolic form.

Step 3: Build the two growing examples. For the specified $L,h$, $\sigma=1$ and $\lambda=-1$. Hence $$\boxed{u(x,t)=e^{\kappa t}\sinh x,\qquad v(x,t)=\cosh(ct)\sinh x.}$$ Their spatial second derivatives equal the fields; their time derivatives satisfy the respective PDEs. The left traces are zero, and at $x=1$ the Robin expression is proportional to $\cosh 1-\coth 1\,\sinh 1=0$. The initial profiles are $\sinh x$ and the initial wave velocity is zero.

Step 4: Locate the failure of the rule. Heat factors satisfy $a'=-\kappa\lambda a$; wave factors satisfy $b''+c^2\lambda b=0$. A negative eigenvalue therefore gives growth rather than decay or pure oscillation. In the spatial energy identity, $hX(L)^2$ is negative here. The usual nonnegative energy form for dissipative or fixed boundaries cannot be assumed for this Robin condition. The counterexample changes the spectrum, not the separation calculation.