More for Chapter 5

Author: ctkelley

SIAMFANLCh5.ipynb

@@ -44,14 +44,13 @@
     "\\newcommand{\\ball}{{\\cal B}}\n",
     "\\newcommand{\\ptc}{{\\Psi TC}}\n",
     "\\newcommand{\\diag}{\\mbox{diag}}\n",
-    "\\newcommand{\\begeq}{{\\begin{equation}}}\n",
-    "\\newcommand{\\endeq}{{\\end{equation}}}\n",
+    "\\newcommand{\\frechet}{{Fr\\'echet\\ }}\n",
     "$"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 4,
+   "execution_count": 8,
    "metadata": {},
    "outputs": [],
    "source": [
@@ -508,7 +507,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 5,
+   "execution_count": 9,
    "metadata": {},
    "outputs": [
     {
@@ -562,7 +561,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 6,
+   "execution_count": 10,
    "metadata": {},
    "outputs": [
     {
@@ -610,6 +609,157 @@
     "vary_xferheat_parms();"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Section 5.2: A Continuation Problem for the $H$-equation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Section 5.2.1: Properties of $H$ as a function of $c$\n",
+    "\n",
+    "In this section we will explore some properties of the solution of\n",
+    "$H$-equation as a function of $c$. As before, we will do the the analysis\n",
+    "for the continuous problem, but the results are valid for any discretization\n",
+    "with a quadrature rule with positive weights that integrates constants\n",
+    "exactly.\n",
+    "\n",
+    "Recall that the equation in $C[0,1]$ is\n",
+    "$$\n",
+    "\\calf(H)(\\mu) =\n",
+    "H(\\mu) - \\frac{1}{1 - \\int_0^1 \\frac{\\mu H(\\mu)}{\\mu+\\nu} \\dnu} = 0.\n",
+    "$$\n",
+    "We will also refer to the fixed point form of the equation\n",
+    "$$\n",
+    "H(\\mu) = \\calg(H)(\\mu) =\n",
+    "\\int_0^1 \\frac{\\mu H(\\mu)}{\\mu+\\nu} \\dnu.\n",
+    "$$\n",
+    "\n",
+    "We will use an alternative form of the $H$-equation to compute\n",
+    "the $L^1$ norm of $H$.\n",
+    "$$\n",
+    "H(\\mu) = 1 + \\frac{c}{2} H(\\mu) \\int_0^1 \\frac{\\mu}{\\mu + \\nu} H(\\nu) \\dnu.\n",
+    "$$\n",
+    "\n",
+    "Integrating both sides of the alternative formulation yields\n",
+    "$$\n",
+    "\\| H \\|_1 =\n",
+    "1 + \\frac{c}{2} \\int_0^1 \\int_0^1 H(\\mu) H(\\nu) \\frac{\\mu}{\\mu + \\nu} \\dnu.\n",
+    "$$\n",
+    "Since $\\mu$ and $\\nu$ are variables of integration, we may interchange\n",
+    "them to obtain\n",
+    "$$\n",
+    "\\| H \\|_1 =\n",
+    "1 + \\frac{c}{2} \\int_0^1 \\int_0^1 H(\\mu) H(\\nu) \\frac{\\nu}{\\mu + \\nu} \\dnu.\n",
+    "$$\n",
+    "When we average the two equations we obtain\n",
+    "$$\n",
+    "\\| H \\|_1 =\n",
+    "1 + \\frac{c}{4} \\| H \\|_1^2.\n",
+    "$$\n",
+    "So\n",
+    "$$\n",
+    "\\| H \\|_1  = \\frac{1 \\pm \\sqrt{1 - c}}{c/2}.\n",
+    "$$\n",
+    "\n",
+    "As we said in Chapter __2__, the H-equation has two solutions\n",
+    "for $0 < c < 1$.  The\n",
+    "__lower branch__, where $\\| H \\|_1  = \\frac{1 - \\sqrt{1 - c}}{c/2}$\n",
+    "is the solution of physical interest and the one you're likely to find\n",
+    "with any iteration. In this section we will explore computation of the\n",
+    "__upper branch__, where $\\| H \\|_1  = \\frac{1 + \\sqrt{1 - c}}{c/2}$.\n",
+    "\n",
+    "The formula for the $L^1$ norm implies that the $H$-equation has no real\n",
+    "solutions for $c > 1$. This implies that the Frechet\n",
+    "derivative\n",
+    "$\\calf'$ of $\\calf$ must be singular. The solution path cannot stop\n",
+    "abruptly <cite data-cite=\"crandall\"><a href=\"siamfa.html#crandall\">(CR71)</cite>,\n",
+    "so there must be an upper\n",
+    "branch. The formula for the $L^1$ norm implies that $\\| H \\|_1 \\to \\infty$\n",
+    "on the upper branch.\n",
+    "\n",
+    "We will briefly consider the details of the singularity before\n",
+    "following the path. We begin by showing that $\\calf'(H)$ is nonsingular\n",
+    "for $0 \\le c < 1$ on the lower branch. To simplify the analysis we define\n",
+    "the linear operator $\\call$ on $C[0,1]$ by\n",
+    "$$\n",
+    "(\\call u)(\\mu) = \\int_0^1 \\frac{\\mu u(\\nu)}{\\mu + \\nu} \\dnu\n",
+    "$$\n",
+    "and express the $H$-equation as\n",
+    "$$\n",
+    "H = \\frac{1}{1 - (c/2) \\call H}\n",
+    "$$\n",
+    "where the division is understood to be pointwise. Now\n",
+    "note that for $u \\in C[0,1]$,\n",
+    "$$\n",
+    "\\calf'(H) u = u - \\frac{(c/2) \\call u}{(1 - (c/2) \\call H)^2}\n",
+    " = u - H^2 (c/2) \\call u.\n",
+    "$$\n",
+    "Here we use, as we have before, the simple trick to compute a Frechet\n",
+    "derivative\n",
+    "$$\n",
+    "\\calf'(H)u = \\frac{d}{d \\epsilon} \\calf(H + \\epsilon u)\n",
+    "\\bigg|_{\\epsilon = 0 }.\n",
+    "$$\n",
+    "Since $\\calf'$ is the sum of a compact operator and the identity, it is\n",
+    "singular only if $0$ is an eigenvector. So singularity of $\\calf'$ is\n",
+    "equivalent to $\\calg'(H)$ having $\\lambda = 1$ as an eigenvalue.\n",
+    "    \n",
+    "When $c = 1$, the Perron theorem\n",
+    "<cite data-cite=\"karlin\"><a href=\"siamfa.html#karlin\">(Kar59)</cite>\n",
+    "and the positivity of the\n",
+    "$H$-function imply that the largest eigenvalue in absolute value\n",
+    "of $\\calg'(H)$ is positive and the corresponding eigenfunction\n",
+    "does not change sign, and hence can be taken as\n",
+    "non-negative. That eigenvalue is $\\lambda = 1$ and the eigenfunction\n",
+    "is $u(\\mu) = \\mu H(\\mu)$. To see this use the formula for the $L^1$ norm and compute,\n",
+    "with $\\| H \\|_1 = 2$,\n",
+    "$$\n",
+    "\\begin{array}{ll}\n",
+    "\\calg'(H)(u)(\\mu) & = H^2(\\mu) (1/2) \\int_0^1 \\frac{\\mu \\nu H(\\nu)}{\\mu + \\nu}\n",
+    "\\dnu \\\\\n",
+    "\\\\\n",
+    "& = (\\mu H(\\mu)) H(\\mu) (1/2) \\int_0^1 \\frac{\\nu H(\\nu)}{\\mu + \\nu} \\dnu \\\\\n",
+    "\\\\\n",
+    "& = u(\\mu) H(\\mu) (1/2) \\int_0^1 H(\\nu) (1 - \\frac{\\mu}{\\mu + \\nu} ) \\dnu \\\\\n",
+    "\\\\\n",
+    "& = u(\\mu) \\left( H(\\mu) - (1/2) (H \\call H)(\\mu) \\right) = u(\\mu)\n",
+    "\\end{array}\n",
+    "$$\n",
+    "Since $u \\ge 0$, $\\lambda = 1$ is the Perron eigenvalue. Therefore\n",
+    "the eigenvalue has multiplicity one.\n",
+    "\n",
+    "The Perron theory is also applicable if $0 < c < 1$. Let\n",
+    "$\\lambda > 0$ be the Perron eigenvalue of\n",
+    "$\\calg'(H)$ with eigenfunction $u$. Set $u(\\mu) = \\mu H(\\mu) p(\\mu)$.\n",
+    "Then\n",
+    "$$\n",
+    "\\begin{array}{ll}\n",
+    "\\lambda p(\\mu)  & =  H(\\mu) (c/2) \\int_0^1\n",
+    "\\frac{H(\\nu) p(\\nu) \\nu}{\\mu + \\nu} \\dnu\n",
+    "\\le \\| p \\|_\\infty H(\\mu) (c/2) \\int_0^1 \\frac{H(\\nu) \\nu}{\\mu + \\nu} \\dnu\\\\\n",
+    "\\\\\n",
+    "& \\le \\| p \\|_\\infty H(\\mu) (c/2)\n",
+    "\\int_0^1 H(\\nu) \\left(1 - \\frac{\\mu}{\\mu + \\nu} \\right) \\dnu\n",
+    "= \\| p \\|_\\infty H(\\mu) (1 - \\sqrt{1-c} ) - (H(\\mu) - 1)\\\\\n",
+    "\\\\\n",
+    "& = \\| p \\|_\\infty (1  - H(\\mu) \\sqrt{1-c} )\n",
+    "\\le \\| p \\|_\\infty (1 - \\sqrt{1-c})\n",
+    "\\end{array}\n",
+    "$$\n",
+    "Hence, taking the $L^\\infty$ norm of the left side of \\eqnok{lambdaest},\n",
+    "we have\n",
+    "$$\n",
+    "\\lambda \\le 1 - \\sqrt{1-c} < 1.\n",
+    "$$\n",
+    "Hence $\\calf'(H)$ is nonsingular for $0 < c < 1$ on the lower branch.\n",
+    "This also proves the norm estimate from Chapter __4__.\n"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,