fix(ch12): replace invalid \pd[2] with explicit second partial notation
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@@ -45,12 +45,12 @@ Taking the limit $\Delta x \to 0$, the right side becomes $-\pd{J}{x}$:
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\paragraph{Derivation of the heat equation.} Substitute Fourier's law \cref{eq:fouriers_law_1d} into the energy conservation equation:
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\[
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\rho\,c\,\pd{u}{t} = -\pd{}{x}\bigl(-\kappa\,\pd{u}{x}\bigr)
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= \kappa\,\pd[2]{u}{x}.
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= \kappa\,\frac{\partial^2 u}{\partial x^2}.
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\]
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Assuming the material is homogeneous ($\kappa$, $\rho$, and $c$ are constant), we divide by $\rho\,c$ to obtain the \textbf{one-dimensional heat equation}:
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\begin{equation}
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\label{eq:heat_equation_1d}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x}, \qquad 0 < x < L, \;\; t > 0,
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, \qquad 0 < x < L, \;\; t > 0,
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\end{equation}
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where
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\begin{equation}
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@@ -65,7 +65,7 @@ is the \textbf{thermal diffusivity} (units: $\mathrm{m^2/s}$).
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\paragraph{Higher dimensions.} In three dimensions, Fourier's law becomes $\bm{J} = -\kappa\nabla u$ and energy conservation gives
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\[
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\pd{u}{t} = \alpha\,\nabla^2 u = \alpha\left(\pd[2]{u}{x} + \pd[2]{u}{y} + \pd[2]{u}{z}\right).
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\pd{u}{t} = \alpha\,\nabla^2 u = \alpha\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right).
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\]
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Throughout this chapter, we focus on the one-dimensional case \cref{eq:heat_equation_1d}, which captures the essential mathematical structure.
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@@ -134,7 +134,7 @@ we have the initial-boundary value problem:
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\begin{equation}
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\label{eq:heat_dirichlet_problem}
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\begin{cases}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < L, \;\; t > 0, \\[6pt]
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < L, \;\; t > 0, \\[6pt]
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u(0,t) = 0, \;\; u(L,t) = 0, & t > 0, \\[6pt]
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u(x,0) = f(x), & 0 < x < L.
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\end{cases}
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@@ -210,7 +210,7 @@ Higher modes ($n \geq 2$) decay much faster because their decay rates scale as $
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Solve the heat equation on a rod of length $L = \pi$ with zero-temperature ends and initial temperature $f(x) = \sin(2x)$:
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\[
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\begin{cases}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt]
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u(x,0) = \sin(2x), & 0 < x < \pi.
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\end{cases}
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@@ -244,7 +244,7 @@ Higher modes ($n \geq 2$) decay much faster because their decay rates scale as $
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Solve the heat equation on a rod of length $L = \pi$ with zero-temperature ends and a triangular initial temperature distribution:
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\[
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\begin{cases}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt]
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u(x,0) = x(\pi - x), & 0 < x < \pi.
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\end{cases}
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@@ -418,7 +418,7 @@ This means the $n=0$ mode is a \textbf{steady-state component} that does not dec
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\begin{keyresult}
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\textbf{Heat equation with homogeneous Neumann BCs.} For the problem
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\[
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\pd{u}{t} = \alpha\,\pd[2]{u}{x}, \qquad
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, \qquad
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u_x(0,t) = 0,\; u_x(L,t) = 0, \qquad
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u(x,0) = f(x),
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\]
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@@ -444,7 +444,7 @@ This means the $n=0$ mode is a \textbf{steady-state component} that does not dec
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Solve the heat equation on a rod of length $L = \pi$ with insulated ends and initial temperature $f(x) = \cos(2x) + 3$:
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\[
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\begin{cases}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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\pd{u}{x}(0,t) = 0, \;\; \pd{u}{x}(\pi,t) = 0, & t > 0, \\[6pt]
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u(x,0) = \cos(2x) + 3, & 0 < x < \pi.
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\end{cases}
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@@ -491,7 +491,7 @@ A \textbf{steady-state solution} is a solution that does not change in time: $\p
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\paragraph{Derivation.} Setting $\pd{u}{t} = 0$ in \cref{eq:heat_equation_1d} gives
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\[
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0 = \alpha\,\pd[2]{u}{x} \quad\Longrightarrow\quad \pd[2]{u}{x} = 0.
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0 = \alpha\,\frac{\partial^2 u}{\partial x^2} \quad\Longrightarrow\quad \frac{\partial^2 u}{\partial x^2} = 0.
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\]
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Integrating twice:
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\begin{equation}
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@@ -540,7 +540,7 @@ When the boundary conditions are nonhomogeneous --- for example, one end held at
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\paragraph{The shifting technique.} Consider the problem
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\[
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\begin{cases}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < L, \;\; t > 0, \\[6pt]
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < L, \;\; t > 0, \\[6pt]
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u(0,t) = T_1, \;\; u(L,t) = T_2, & t > 0, \\[6pt]
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u(x,0) = f(x), & 0 < x < L.
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\end{cases}
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@@ -559,11 +559,11 @@ The simplest choice is the linear steady-state profile \cref{eq:steady_state_dir
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\]
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Substituting $u = v + \phi$ into the heat equation:
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\[
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\pd{v}{t} = \alpha\,\pd[2]{v}{x} + \alpha\,\phi''(x).
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\pd{v}{t} = \alpha\,\frac{\partial^2 v}{\partial x^2} + \alpha\,\phi''(x).
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\]
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Since $\phi(x)$ is linear, $\phi''(x) = 0$, so $v$ satisfies the same homogeneous heat equation:
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\[
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\pd{v}{t} = \alpha\,\pd[2]{v}{x}.
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\pd{v}{t} = \alpha\,\frac{\partial^2 v}{\partial x^2}.
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\]
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The boundary conditions for $v$ are homogeneous:
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\[
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@@ -592,7 +592,7 @@ The initial condition for $v$ is
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\textbf{Solution.} The problem is
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\[
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\begin{cases}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2}, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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u(0,t) = 0, \;\; u(\pi,t) = 100, & t > 0, \\[6pt]
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u(x,0) = 50, & 0 < x < \pi.
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\end{cases}
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@@ -605,7 +605,7 @@ The initial condition for $v$ is
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\textit{Step 2: Define $v(x,t) = u(x,t) - \phi(x)$.} Then $v$ satisfies
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\[
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\begin{cases}
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\pd{v}{t} = \alpha\,\pd[2]{v}{x}, & 0 < x < \pi, \\[6pt]
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\pd{v}{t} = \alpha\,\frac{\partial^2 v}{\partial x^2}, & 0 < x < \pi, \\[6pt]
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v(0,t) = 0, \;\; v(\pi,t) = 0, \\[6pt]
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v(x,0) = 50 - \frac{100}{\pi}\,x.
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\end{cases}
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@@ -679,7 +679,7 @@ The initial condition for $v$ is
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The basic heat equation \cref{eq:heat_equation_1d} assumes no internal heat generation. In many practical situations, there is a \textbf{source term} $Q(x,t)$ representing internal heat production (e.g., electrical heating, chemical reactions, nuclear decay):
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\begin{equation}
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\label{eq:heat_with_source}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x} + Q(x,t), \qquad 0 < x < L, \;\; t > 0.
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2} + Q(x,t), \qquad 0 < x < L, \;\; t > 0.
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\end{equation}
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\paragraph{Eigenfunction expansion method.} We assume the solution can be expanded in the eigenfunctions of the spatial problem:
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@@ -722,7 +722,7 @@ This is a first-order linear ODE for each mode $T_n(t)$, solvable by the integra
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Solve the heat equation with a spatially uniform source on a rod of length $L = \pi$ with zero-temperature ends and initially zero temperature:
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\[
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\begin{cases}
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\pd{u}{t} = \alpha\,\pd[2]{u}{x} + Q_0, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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\pd{u}{t} = \alpha\,\frac{\partial^2 u}{\partial x^2} + Q_0, & 0 < x < \pi, \;\; t > 0, \\[6pt]
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u(0,t) = 0, \;\; u(\pi,t) = 0, & t > 0, \\[6pt]
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u(x,0) = 0, & 0 < x < \pi,
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\end{cases}
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