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The three distinguished limits are the outer expansion at $x=O(1)$, an intermediate asymptotic region at $x=O(ϵ)$, and an inner expansion at $x=O({ϵ}^2)$. The middle scale resolves the coefficient change from $x$ to $x+ϵ$; the narrower scale resolves the unmet left boundary condition. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second derivative at $x=0$ would be too strong for this degenerate ordinary differential equation.

In the outer expansion, set $y=y_0+ϵy_1+\cdots$. The leading equation is $x(x+1)y_0^{\prime }-x{y}_0=0$. Enforcing the right boundary condition gives $y_0=x+1$. Because $y_0^{\prime \prime }=0$, the next equation is particularly simple:An integrating factor, or the substitution $y_1=(x+1)u$, gives $u^{\prime }=-1/[x(x+1)]$. ThusThe error statement is for $x$ bounded away from zero. The left limiting form is $1+x-ϵ\log (2x)$ through the terms needed for matching. A logarithm in the correction makes a pure integer-power inner ansatz inadequate.

For the intermediate asymptotic region, put $x=ϵs$ and $y=V(s)$. The transformed equation isThe leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt{ϵ}$ forcing, so that homogeneous constant is zero after matching. At order $ϵ$, $(s+1)V_1^{\prime }=s$, giving $V_1=s-\log (1+s)+C(ϵ)$. Matching for $1\ll s\ll 1/ϵ$ with the outer expansion fixes $C=-\log (2ϵ)$. HenceThe switchback term $-ϵ\log ϵ$ is larger than an ordinary $O(ϵ)$ constant correction and must be retained. The highest derivative term first changes this middle solution at order ${ϵ}^{3/2}$, so it does not alter the displayed result.

To locate the inner expansion, compare the derivative terms for $x\ll ϵ$. Their ratio at a layer width $\delta$ is $ϵ/\sqrt{\delta }$, so the square-root-degenerate endpoint layer has $\delta ={ϵ}^2$. Put $x={ϵ}^{2}X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $ϵ$ isBoth orders therefore have the same homogeneous operator $\sqrt{X}{d}^2/d{X}^2+d/dX$. Its derivative mode is proportional to $e^{-2\sqrt{X}}$. Using $U(0)=0$ and the supplied elementary integral, defineThe intermediate asymptotic region gives the overlap value $1-ϵ\log (2ϵ)$ as $s\to 0$. ThereforeIt satisfies the left boundary condition exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll 1/ϵ$. The outer and middle limits match with $ϵ\ll x\ll 1$. These three formulae include all terms through $O(ϵ)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O({ϵ}^2)$.

For a check on that last point, continue the middle equation by one order. Its ${ϵ}^{3/2}$ term obeysThe integration constant is selected by the outer limit. Its value at $s=0$ is $\pi /2$, so the inner matching amplitude has a next correction $\pi {ϵ}^{3/2}/2$. This independently explains why an $o(ϵ)$ remainder is appropriate.

If a single additive composite expansion is useful, combine the three formulae and subtract their two common overlaps. Writing $A=1-ϵ\log (2ϵ)$ and interpreting $x\log x=0$ at zero givesIt is zero at $x=0$, gives $2+O({ϵ}^2)$ at $x=1$, and reproduces all three retained expansions. It does not assert an $O({ϵ}^2)$ error uniformly through the nested regions.

Finally, near zero $H(X)=2X-\frac{8}{3}{X}^{3/2}+O(X^2)$. Thus the solution has a finite nonzero first derivative but generally a second derivative diverging like $x^{-1/2}$. This is compatible with the equation for $x>0$: the vanishing coefficient of $y^{\prime \prime }$ multiplies that singular derivative to give a finite limit. If one instead required $y\in C^2[0,1]$ and the differential equation pointwise at $0$, the left condition would force $y^{\prime }(0)=0$. The normalized equation's other coefficients have only integrable $x^{-1/2}$ singularities, so uniqueness for the initial value problem with $y(0)=y^{\prime }(0)=0$ would give the zero solution, contradicting the right condition. Explicitly, write the normalized equation as a first-order system for $(y,y^{\prime })$ with an integrable coefficient matrix; its integral equation and the Gronwall inequality force a solution with zero initial vector to vanish. The continuous-endpoint interpretation is therefore essential.

A wavelet with $p$ vanishing moments satisfies $x^k\psi \in L^1$ and $\int x^k\psi (x)dx=0$ for $0\le k<p$. The moment differentiation of the Fourier transform theorem states that these weighted integrability hypotheses make $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported scaling function with a finite refinement filter. The wavelet identity givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ still gives $m(\pi +t)=o(|t{|}^{p-1})$ for the representative defined by the quotient: a Peano zero. This establishes the value at $\pi$, and for $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe Fourier coefficients of $\theta$ are absolutely summable. The Wiener algebra is closed under multiplication and the exponential series, so $a$ and $a^{-1}$ have absolutely summable Fourier coefficients. Thus $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves orthogonality and unit norm of its integer function translations; the inverse phase preserves their complete span. The same approximation spaces therefore give a multiresolution analysis.

The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ gives $a(2t)=a(t)a(t+\pi )$, so the periodic phase change of a scaling function produces $m(t)=a(t+\pi )m_0(t)$. In the high-pass formula the phases cancel:The associated wavelet is therefore exactly the original one, with unchanged vanishing moments.

At any dyadic point $t_0=2\pi k/2^j$, the difference quotient of the terms with $j\le n\le {\log }_2(1/|h|)$ is $1+O((2^{n}h{)}^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of differentiability. Dyadic points are dense, and $m_0(\pi +t)$ is nonzero for sufficiently small $t\ne 0$. Thus $m$ is not differentiable on any neighborhood of $\pi$, so an ordinary second derivative at $\pi$, understood as the derivative of a locally defined first derivative, need not exist. The Peano zero remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.

Under the smoothness hypotheses in the smooth-mask vanishing-moment criterion, write any nonzero integer $j$ as $j=2^r\ell$, where $r\ge 0$ and $\ell$ is odd, possibly negative. Iterating the scaling refinement equation exactly $r+1$ times givesThe last MRA low-pass filter factor is $m(\pi \ell +t/2^{r+1})$. Its derivatives of all orders less than $p$ vanish at $t=0$, because $\ell$ is odd and the MRA low-pass filter is $2\pi$-periodic. If the remaining factors are $C^{p-1}$ at the displayed arguments, the product rule forces every derivative of order less than $p$ of the product to vanish. In particular compact support and a finite MRA low-pass filter supply all these hypotheses. Therefore the integer-frequency zeros of a scaling function are

Under only the printed assumptions, the exact same finite refinement argument gives the Peano zero $\hat{\phi }(2\pi j+t)=o(|t{|}^{p-1})$. Indeed, the last MRA low-pass filter factor has that estimate, the other MRA low-pass filter factors are bounded by one almost everywhere by the quadrature mirror filter identity, and the last Fourier transform factor is bounded because $\phi \in L^1$. The estimate initially holds almost everywhere and extends to every $t$ by continuity of $\hat{\phi }$. It does not imply arbitrary ordinary higher derivatives. In the lacunary scaling-phase regularity counterexample, $\hat{\phi }=a{\hat{\phi }}_0$ is nondifferentiable at dense dyadic points next to the isolated integer-frequency zeros of the smooth base transform. Thus the same regularity qualification is necessary here.

Take the indicator function $u(x_1,x_2)={\chi }_{\{x_1>1/2\}}$ inside $(0,1{)}^2$. For a compactly supported test vector field, integration in $x_1$ and then $x_2$ givesThe vertical derivative term integrates to zero. The absolute pairing is at most one, and test vector fields with ${\xi }_1=-1$ along all but arbitrarily short endpoint pieces approach one. Thus $|Du|(\Omega )=1$. More explicitly, the distributional derivative is $Du=e_1{\mathcal{H}}^1\upharpoonright \{(1/2,x_2):0<x_2<1\}$, where ${\mathcal{H}}^1$ is Hausdorff measure along the segment. Also $\Vert u{\Vert }_1=1/2$.

A Sobolev space function in $W^{1,1}(\Omega )$ has each weak derivative represented by an $L^1$ density with respect to planar Lebesgue measure. The nonzero segment measure here is singular with respect to that measure, so this requirement fails. Equivalently, the Sobolev fundamental theorem of calculus on lines would require an absolutely continuous representative on almost every horizontal slice, whereas each slice has a jump. ThereforeThis is the bounded-variation step outside W11 example.

On the Dahlquist test equation, $g(y)={\lambda }^{2}y$, so the amplification polynomial of a multistep method becomesBoth roots of a polynomial, including the parasitic one, must satisfy the root condition for a multistep method.

For $\alpha =1/7$, multiply by seven and set $a=7-6z+2z^2$. The polynomial is $a{\xi }^2-8\xi +1$. The complex quadratic Schur criterion says that its roots of a polynomial are strictly in the unit disk ifHere is a direct verification on the entire left half-plane. Put $z=-r+iy$, $r\ge 0$, $t=y^2$, and $a_0=7+6r+2r^2$. Thenand expansion of the squared inequality givesEvery term is nonnegative, and at least one is positive unless $z=0$. Since $q-1>0$, taking square roots proves the strict complex quadratic Schur criterion for $z\ne 0$. At $z=0$, the roots of a polynomial are the simple values $1$ and $1/7$. The leading coefficient cannot vanish because $|a{|}^2\ge 49$. Thus the method at $\alpha =1/7$ is A-stable. Its order three is compatible with the Second Dahlquist barrier: that barrier concerns first-derivative linear multistep methods, whereas this method uses a second derivative.

For $\alpha =1/2$, choose $z=-4$, strictly in the left half-plane. The amplification polynomial of a multistep method reduces toOne root of a polynomial has modulus $(3+\sqrt{17})/4>1$. Thus the method at $\alpha =1/2$ is not A-stable.

Insert an exact smooth solution of a differential equation and expand about $t=t_{n+2}$. The unscaled local truncation error, with every term moved to the left, isHere the constant, first-derivative, and second-derivative terms all cancel, using the chain rule identity $y^{\prime \prime }=f^{\prime }(y)f(y)$. Hence the formal defect convention $d_h=O(h^{p+1})$ givesAt $\alpha =1/7$, the coefficient of $h^4{y}^{(4)}$ is $1/21$, so there is no further cancellation. At any other $\alpha$, the cubic coefficient is nonzero.

There is a degeneracy at $\alpha =1$: both the first-derivative coefficient and ${\rho }^{\prime }(1)$ vanish. The raw defect still has the displayed formal order two, but it cannot be normalized by $(1-\alpha )h$ into an ordinary first-order consistency of a numerical method test. Thus a convention requiring this nondegenerate normalization would leave the ODE order undefined at $\alpha =1$. In either convention, it is not a convergent approximation to arbitrary initial-value data; the next part explains why.