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Then <spanclass="arithmatex">\(u_I\)</span> matches all prescribed initial derivatives.</p>
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\]</div>
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<p>Then <spanclass="arithmatex">\(u_I\)</span> matches all prescribed initial derivatives.</p>
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<p><strong>Proposition A.4 (exact initial derivatives).</strong> Assume <spanclass="arithmatex">\(u\)</span> is <spanclass="arithmatex">\(K+1\)</span> times differentiable in <spanclass="arithmatex">\(t\)</span>.
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Then <spanclass="arithmatex">\(\partial_t^k u_I(\cdot,t_0)=g_k(\cdot)\)</span> for <spanclass="arithmatex">\(k=0,\dots,K\)</span>.</p>
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<p><em>Proof.</em> Write <spanclass="arithmatex">\(u_I=P_K+g(u-P_K)\)</span>. By Leibniz, every derivative of <spanclass="arithmatex">\(g(u-P_K)\)</span> of order <spanclass="arithmatex">\(\le K\)</span> contains a factor
@@ -2584,16 +2584,10 @@ <h2 id="a5-higher-order-initial-constraints-via-a-gated-taylor-ansatz">A.5. High
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<p>In implementation, two gate families are supported:</p>
where <spanclass="arithmatex">\(\varepsilon=\texttt{gate\_eps}>0\)</span>, and <spanclass="arithmatex">\(\Delta t\)</span> is the oriented (or absolute) distance from the initial slice
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<spanclass="arithmatex">\(q=K+1\)</span>, <spanclass="arithmatex">\(\tau=\frac{\Delta t}{L\,\varepsilon^{1/q}}\)</span>, and
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<spanclass="arithmatex">\(g_{\text{rat}}(t)=\frac{\tau^q}{1+\tau^q}\)</span>, where <spanclass="arithmatex">\(\varepsilon=\texttt{gate\_eps}>0\)</span>, and <spanclass="arithmatex">\(\Delta t\)</span> is the oriented (or absolute) distance from the initial slice
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depending on whether the initial component is <code>FixedStart</code>, <code>FixedEnd</code>, or <code>Fixed</code>.</li>
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</ul>
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<p>Near <spanclass="arithmatex">\(t_0\)</span>, <spanclass="arithmatex">\(g_{\text{rat}}(t)=\mathcal O((t-t_0)^{K+1})\)</span>, so it preserves the same exact-derivative property while
than a prescribed threshold <spanclass="arithmatex">\(\varepsilon_{\text{snap}}\)</span> to an anchor, the weight becomes one-hot.</p>
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<p><strong>Proposition A.6 (exactness at anchors under snapping).</strong> Suppose <spanclass="arithmatex">\(M(z_k)>0\)</span> and <spanclass="arithmatex">\(w_k(z_k)=1\)</span>. Then <spanclass="arithmatex">\(\tilde u(z_k)=y_k\)</span>.</p>
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<p><em>Proof.</em> Evaluate the overlay at <spanclass="arithmatex">\(z=z_k\)</span>:</p>
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<p>[
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\begin{aligned}
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\tilde u(z_k) &= u(z_k) + M(z_k)\,r_k \
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&= u(z_k) + M(z_k)\frac{y_k-u(z_k)}{M(z_k)} \
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&= y_k.
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\end{aligned}
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]
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<p>$$
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\tilde u(z_k)
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= u(z_k) + M(z_k)\,r_k
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= u(z_k) + M(z_k)\frac{y_k-u(z_k)}{M(z_k)}
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= y_k.
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$$
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<spanclass="arithmatex">\(\square\)</span></p>
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<h3id="a75-preservation-of-boundary-and-initial-constraints">A.7.5. Preservation of boundary and initial constraints<aclass="headerlink" href="#a75-preservation-of-boundary-and-initial-constraints" title="Permanent link">¤</a></h3>
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<p>The interior correction has the form <spanclass="arithmatex">\(\Delta u(z)=M(z)\,h(z)\)</span> where <spanclass="arithmatex">\(h(z)=\sum_i w_i(z)r_i\)</span>.</p>
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