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The spatial shift in the PDF also requires a boundary value beyond the supplied right endpoint: if $m$ is the last interior index, the newest $m+2$ value is outside the interval. Thus the printed Dirichlet method has no specified boundary closure, and no unique stability range can be assigned to it as stated.

For completeness, the literal interior formula has an exact Von Neumann stability analysis on an infinite or periodic grid. Substitution of a Fourier mode $U_m^n={\xi }^n{e}^{im\theta }$ givesFor the root of a polynomial tending to one as $\theta \to 0$, expansion givesConsequently $\mu <3$ is unstable: sufficiently small nonzero frequencies have a root of a polynomial of modulus greater than one, and arbitrarily fine periodic grids contain such frequencies.

Conversely, if $\mu \ge 3$ and $s=\cos \theta$, thenFor $A=u+iv$ with $u\ge 3$,Since $|A|\ge 3$, the complex quadratic Schur criterion yields both roots of a polynomial in the unit disk, strictly inside except at $\theta =0$, where they are $1$ and $1/3$. Uniform stability follows directly from the BDF2 discrete energy identity: the modal recurrence is $3v^{n+2}-4v^{n+1}+v^n=-(A-3)v^{n+2}$, whose energy decreases because $\mathrm{Re}(A-3)\ge 0$. Summing the modal energies proves a bound independent of the grid and even of $\mu \ge 3$; no eigenvector-separation assumption is needed. This is the stability of a spatially shifted BDF2 stencil. Thus the literal periodic-grid result isThis cannot establish stability of an unspecified Dirichlet boundary condition closure. It also does not restore consistency of a numerical method.

For the intended unshifted backward differentiation formula, a spatial eigenvalue $-\beta \le 0$ givesThe Schur stability criterion gives the root condition for a multistep method for every $k\beta \ge 0$, with strict inequalities for $\beta >0$. A useful uniform energy method proof, including the repeated interior roots that a bare separation argument would miss, is the BDF2 discrete energy identity. With ${\delta }^{2}U=U^{n+2}-2U^{n+1}+U^n$ andthe corrected Dirichlet discrete Laplacian scheme satisfiesThis follows by taking the inner product of $3U^{n+2}-4U^{n+1}+U^n=2k{L}_h{U}^{n+2}$ with $U^{n+2}$ and using summation by parts. It controls both time levels uniformly, regardless of the mesh ratio. Hence the corrected method is stable for

For the test equation $y^{\prime }=\lambda y$, put $z=h\lambda$. The amplification roots satisfyThe quadratic Schur stability criterion shows that both roots lie in the closed unit disk whenTo check these inequalities on the closed left half-plane, write $z=-u+iv$ with $u\ge 0$. The right side of the second inequality isAfter squaring, the difference between its square and $16|5\overline{D}+1+z{|}^2$ is a polynomial in $u$ and $v^2$ with nonnegative coefficients:The first inequality follows from the same displayed positive expression. The unit-circle cases are semisimple, so the entire closed left half-plane is in the linear stability domain. Hence the method is A-stable.

Fourier transformation in $m$ turns the recurrence intoAn amplification factor $z$ therefore satisfiesFor $\mu >0$, the quadratic Schur stability criterion giveswhere $a_1=-2\mu \cos \theta /(1+\mu )$ and $a_2=(\mu -1)/(\mu +1)$. Thus both roots lie in the closed unit disk; roots on it at $\theta =0,\pi$ are simple. At $\mu =0$, the roots are $\pm 1$. Parseval's identity then gives stability precisely forwhich is the full stated parameter range.