[proofplan]
We use the positivity of $u$ to define the smooth map $f:M\times I\to\mathbb{R}$ by $f=\log u$, so that $u=e^f$. Differentiating this identity in time gives $\partial_t u=e^f\partial_t f$. At each fixed time, applying the Riemannian product rule and chain rule to the spatial function $e^f$ gives $\Delta_g u=e^f(\Delta_g f+|\nabla f|_g^2)$. Substituting these two identities into the [heat equation](/page/Heat%20Equation) and dividing by the positive factor $e^f=u$ gives the claimed logarithmic equation.
[/proofplan]
[step:Use positivity to define a smooth logarithm]
Let $\ell \in C^\infty((0,\infty);\mathbb{R})$ be the logarithm map from $(0,\infty)$ to $\mathbb{R}$ defined by $\ell(s)=\log s$ for each $s\in(0,\infty)$. Since $u:M\times I\to(0,\infty)$ is smooth and $\ell$ is smooth on $(0,\infty)$, the composition
\begin{align*}
f=\ell\circ u:M\times I&\to\mathbb{R}
\end{align*}
is smooth. By definition of $f$, for every $(p,t)\in M\times I$,
\begin{align*}
u(p,t)=e^{f(p,t)}.
\end{align*}
[/step]
[step:Differentiate $u=e^f$ in time]
Fix $p\in M$. The map from $I$ to $(0,\infty)$ sending $t$ to $u(p,t)$ equals the composition of the map from $I$ to $\mathbb{R}$ sending $t$ to $f(p,t)$ with the exponential map. Applying the one-variable chain rule in the time variable gives, for every $(p,t)\in M\times I$,
\begin{align*}
\partial_t u(p,t)=e^{f(p,t)}\partial_t f(p,t).
\end{align*}
Equivalently,
\begin{align*}
\partial_t u=e^f\partial_t f
\end{align*}
on $M\times I$.
[/step]
[step:Compute the Riemannian Laplacian of $e^f$ at a fixed time]
Fix $t\in I$. Define the time-slice function $u_t:M\to(0,\infty)$ by $u_t(p)=u(p,t)$ for each $p\in M$. Define the time-slice function $f_t:M\to\mathbb{R}$ by $f_t(p)=f(p,t)$ for each $p\in M$. Then $u_t=e^{f_t}$. Let $p\in M$, and let $n:=\dim M$ denote the dimension of the Riemannian manifold. Choose a coordinate neighbourhood $W\subset M$ of $p$ with a smooth coordinate frame consisting of $n$ vector fields. Applying the pointwise Gram-Schmidt procedure to this coordinate frame and shrinking to an open neighbourhood $V\subset W$ of $p$ if necessary gives a smooth local $g$-orthonormal frame $E_1,\dots,E_n$ of vector fields on $V$. For any smooth function $h:V\to\mathbb{R}$, the Laplace-Beltrami operator is locally
\begin{align*}
\Delta_g h=\sum_{i=1}^n \left(E_i(E_i h)-(\nabla^g_{E_i}E_i)h\right),
\end{align*}
where $\nabla^g$ denotes the Levi-Civita connection.
For each $i\in\{1,\dots,n\}$, the chain rule for the smooth one-variable exponential map composed with the directional derivative $E_i$ gives
\begin{align*}
E_i(e^{f_t})=e^{f_t}E_i f_t.
\end{align*}
Applying $E_i$ once more and using the product rule for directional derivatives gives
\begin{align*}
E_i(E_i(e^{f_t}))=E_i(e^{f_t}E_i f_t).
\end{align*}
Using the already computed identity $E_i(e^{f_t})=e^{f_t}E_i f_t$, this becomes
\begin{align*}
E_i(E_i(e^{f_t}))=e^{f_t}(E_i f_t)^2+e^{f_t}E_i(E_i f_t).
\end{align*}
Also, applying the first chain rule to the vector field $\nabla^g_{E_i}E_i$,
\begin{align*}
(\nabla^g_{E_i}E_i)(e^{f_t})=e^{f_t}(\nabla^g_{E_i}E_i)f_t.
\end{align*}
Substituting these identities into the local formula for $\Delta_g$ gives
First,
\begin{align*}
\Delta_g u_t=\sum_{i=1}^n \left(E_i(E_i(e^{f_t}))-(\nabla^g_{E_i}E_i)(e^{f_t})\right).
\end{align*}
Substituting the two derivative identities gives
\begin{align*}
\Delta_g u_t=e^{f_t}\sum_{i=1}^n \left(E_i(E_i f_t)-(\nabla^g_{E_i}E_i)f_t\right)+e^{f_t}\sum_{i=1}^n (E_i f_t)^2.
\end{align*}
The first sum is $\Delta_g f_t$, and the second sum is the Riemannian gradient norm $|\nabla f_t|_g^2$ because $E_1,\dots,E_n$ is a $g$-orthonormal frame. Therefore
\begin{align*}
\Delta_g u_t=e^{f_t}\Delta_g f_t+e^{f_t}|\nabla f_t|_g^2.
\end{align*}
Factoring $e^{f_t}$ yields
\begin{align*}
\Delta_g u_t=e^{f_t}\left(\Delta_g f_t+|\nabla f_t|_g^2\right).
\end{align*}
Since $p\in M$ and $t\in I$ were arbitrary, this is the pointwise identity
\begin{align*}
\Delta_g u=e^f\left(\Delta_g f+|\nabla f|_g^2\right)
\end{align*}
on $M\times I$.
[guided]
We want to compute the spatial Laplacian of $u=e^f$. The time variable is held fixed during this computation, because $\Delta_g$ acts only in the $M$ variable. Fix $t\in I$. Define $u_t:M\to(0,\infty)$ by $u_t(p)=u(p,t)$ for each $p\in M$. Also define $f_t:M\to\mathbb{R}$ by $f_t(p)=f(p,t)$ for each $p\in M$. Then $u_t=e^{f_t}$ as functions on $M$.
Now fix $p\in M$, and let $n:=\dim M$ denote the dimension of the Riemannian manifold. Choose a coordinate neighbourhood $W\subset M$ of $p$ with a smooth coordinate frame consisting of $n$ vector fields. Applying the pointwise Gram-Schmidt procedure to this coordinate frame and shrinking to an open neighbourhood $V\subset W$ of $p$ if necessary gives a smooth local $g$-orthonormal frame $E_1,\dots,E_n$ on $V$. In such a frame, the Laplace-Beltrami operator applied to a smooth function $h:V\to\mathbb{R}$ is
\begin{align*}
\Delta_g h=\sum_{i=1}^n \left(E_i(E_i h)-(\nabla^g_{E_i}E_i)h\right),
\end{align*}
where $\nabla^g$ is the Levi-Civita connection. This formula reduces the computation to repeated directional derivatives along the vector fields $E_i$.
For each $i\in\{1,\dots,n\}$, the ordinary chain rule along the vector field $E_i$, applied to the smooth exponential map composed with $f_t$, gives
\begin{align*}
E_i(e^{f_t})=e^{f_t}E_i f_t.
\end{align*}
Applying $E_i$ again requires the product rule for directional derivatives:
\begin{align*}
E_i(E_i(e^{f_t}))=E_i(e^{f_t}E_i f_t).
\end{align*}
The product rule expands the right-hand side as
\begin{align*}
E_i(E_i(e^{f_t}))=E_i(e^{f_t})E_i f_t+e^{f_t}E_i(E_i f_t).
\end{align*}
Using $E_i(e^{f_t})=e^{f_t}E_i f_t$, we obtain
\begin{align*}
E_i(E_i(e^{f_t}))=e^{f_t}(E_i f_t)^2+e^{f_t}E_i(E_i f_t).
\end{align*}
The connection term is also a directional derivative, now along the vector field $\nabla^g_{E_i}E_i$. Applying the same chain rule gives
\begin{align*}
(\nabla^g_{E_i}E_i)(e^{f_t})=e^{f_t}(\nabla^g_{E_i}E_i)f_t.
\end{align*}
Substitute these two identities into the local expression for the Laplacian:
\begin{align*}
\Delta_g u_t=\sum_{i=1}^n \left(E_i(E_i(e^{f_t}))-(\nabla^g_{E_i}E_i)(e^{f_t})\right).
\end{align*}
Substituting the formulas for $E_i(E_i(e^{f_t}))$ and $(\nabla^g_{E_i}E_i)(e^{f_t})$ gives
\begin{align*}
\Delta_g u_t=\sum_{i=1}^n \left(e^{f_t}(E_i f_t)^2+e^{f_t}E_i(E_i f_t)-e^{f_t}(\nabla^g_{E_i}E_i)f_t\right).
\end{align*}
Factoring the common term $e^{f_t}$ and grouping the second-order terms gives
\begin{align*}
\Delta_g u_t=e^{f_t}\sum_{i=1}^n \left(E_i(E_i f_t)-(\nabla^g_{E_i}E_i)f_t\right)+e^{f_t}\sum_{i=1}^n (E_i f_t)^2.
\end{align*}
The first sum is exactly $\Delta_g f_t$ by the same local formula for the Laplacian. The second sum is the Riemannian gradient norm $|\nabla f_t|_g^2$ because $E_1,\dots,E_n$ is a $g$-orthonormal frame. Hence
\begin{align*}
\Delta_g u_t=e^{f_t}\Delta_g f_t+e^{f_t}|\nabla f_t|_g^2.
\end{align*}
Factoring $e^{f_t}$ gives
\begin{align*}
\Delta_g u_t=e^{f_t}\left(\Delta_g f_t+|\nabla f_t|_g^2\right).
\end{align*} Since the point $p$ and the time $t$ were arbitrary, the identity holds on all of $M\times I$:
\begin{align*}
\Delta_g u=e^f\left(\Delta_g f+|\nabla f|_g^2\right).
\end{align*}
[/guided]
[/step]
[step:Substitute the derivative identities into the heat equation]
Since $u$ solves the [heat equation](/page/Heat%20Equation) on $M\times I$,
\begin{align*}
\partial_t u=\Delta_g u.
\end{align*}
Using the time derivative identity and the Laplacian identity, we obtain
\begin{align*}
e^f\partial_t f=e^f\left(\Delta_g f+|\nabla f|_g^2\right).
\end{align*}
For every $(p,t)\in M\times I$, the factor $e^{f(p,t)}=u(p,t)$ is positive. Dividing pointwise by $e^f$ gives
\begin{align*}
\partial_t f=\Delta_g f+|\nabla f|_g^2.
\end{align*}
This is the claimed logarithmic form of the heat equation.
[/step]
