a bit more on finite difference

_build/bvps/finite-difference.html

@@ -9,9 +9,9 @@
 prev_page:
   url: /bvps/shooting-method.html
 next_page:
-  url: /quizzes/quiz2-IVPs.html
+  url: /bvps/eigenvalue.html
 suffix: .ipynb
-search: x y delta f prime equation t frac begin end align right boundary left b derivative dx difference xi fin equations solve yi l infty finite c conditions where n heat transfer m lets our solution ac point system condition text k p differences mathcal o example ode points linear matlab h exact using order second consider nonlinear gives domain through formula q dt theta well rightarrow approx into recursion get bmatrix term guess temperature d value also accurate five above set mathbf implement hand fixed octave e control volume approximations derivatives function approximate forward backward taylor series discrete us
+search: x y delta f t prime equation frac begin end right align boundary left b derivative dx infty m difference xi fin solve equations yi l finite c conditions where solution n heat transfer lets our ac point system condition nonlinear text k p ti differences mathcal o example ode points linear matlab h exact using order second consider gives domain through formula term q dt theta well rightarrow approx into above recursion get bmatrix hand guess temperature d value also accurate five set mathbf implement fixed octave because e control volume method approximations derivatives function approximate forward backward taylor
 
 comment: "***PROGRAMMATICALLY GENERATED, DO NOT EDIT. SEE ORIGINAL FILES IN /content***"
 ---
@@ -699,6 +699,12 @@ <h3 id="Heat-transfer-with-radiation">Heat transfer with radiation<a class="anch
 \begin{equation}
 \frac{d^2 T}{dx^2} - \frac{h P}{k A_c} \left(T - T_{\infty}\right) - \frac{\sigma \epsilon P}{h A_c} \left(T^4 - T_{\infty}^4 \right) = 0
 \end{equation}</p>
+<p>This is a bit trickier to solve because of the nonlinear term involving $T^4$. But, we can handle it via the iterative solution method discussed above, moving the nonlinear parts to the right-hand side:
+\begin{align}
+\frac{T_{i-1} - 2T_i + T_{i+1}}{\Delta x^2} - m^2 \left( T_i - T_{\infty} \right) - M^2 \left( T_i^4 - T_{\infty}^4 \right) &amp;= 0 \\
+\frac{T_{i-1} - 2T_i + T_{i+1}}{\Delta x^2} - m^2 T_i &amp;= M^2 \left( T_i^4 - T_{\infty}^4 \right) - m^2 T_{\infty} \\
+T_{i-1} + T_i (-2 - \Delta x^2 m^2
+\end{align}</p>
 
 </div>
 </div>

content/bvps/finite-difference.ipynb

@@ -629,7 +629,14 @@
     "Let's now consider a more-complicated case, where we also have radiation heat transfer occuring along the length of the fin. Now, our governing ODE is\n",
     "\\begin{equation}\n",
     "\\frac{d^2 T}{dx^2} - \\frac{h P}{k A_c} \\left(T - T_{\\infty}\\right) - \\frac{\\sigma \\epsilon P}{h A_c} \\left(T^4 - T_{\\infty}^4 \\right) = 0\n",
-    "\\end{equation}"
+    "\\end{equation}\n",
+    "\n",
+    "This is a bit trickier to solve because of the nonlinear term involving $T^4$. But, we can handle it via the iterative solution method discussed above, moving the nonlinear parts to the right-hand side:\n",
+    "\\begin{align}\n",
+    "\\frac{T_{i-1} - 2T_i + T_{i+1}}{\\Delta x^2} - m^2 \\left( T_i - T_{\\infty} \\right) - M^2 \\left( T_i^4 - T_{\\infty}^4 \\right) &= 0 \\\\\n",
+    "\\frac{T_{i-1} - 2T_i + T_{i+1}}{\\Delta x^2} - m^2 T_i &= M^2 \\left( T_i^4 - T_{\\infty}^4 \\right) - m^2 T_{\\infty} \\\\\n",
+    "T_{i-1} + T_i (-2 - \\Delta x^2 m^2\n",
+    "\\end{align}"
    ]
   },
   {