I recently attempted to derive a differential equation equivalent to the heat equation except for a three-dimensional curve instead of one-dimensional. I ended up with the following equation;
$$k \frac{\partial^2 u}{\partial\theta^2}|r'(\theta)|^2-k\frac{\partial u}{\partial \theta}(r'(\theta)\cdot r''(\theta))=c\rho u_t |r'(\theta)|^4$$.
where $c$ is the specific heat (a constant), $\rho$ is the density (a constant), $r(\theta)=x(\theta)\hat i + y(\theta) \hat j + z(\theta) \hat k$ is the parametric vector equation of the curve, $u(x,y,z,t)$ is the temperature at a point $(x,y,z)$ at time $t$ and $k$ is the proportionality constant.
How I got here was I assumed the heat in a segment of the curve was the line integral;
$$H(t)=\int_C (c\rho u) ds$$
based on how the one dimensional heat equation is derived. Then I expressed the line integral wrt $\theta$ and used the Leibniz rule to find the derivative of the integral, $dH/dt$. Then I used similar reasoning to what's used to derive the diffusion equation to come up with another expression for $dH/dt$.
In particular I assumed that
$$dH/dt=\text{heat flow in - heat flow out}$$ $$=k\Bigg [\nabla u \cdot \frac{r'(b)}{|r'(b)|} - \nabla u \cdot \frac{r'(a)}{|r'(a)|} \Bigg ]$$
I assumed that the flow of heat at the boundary is equal to the directional derivative of $u$ in the direction of the unit tangent vector and that heat only flows at the end of the curve i.e. the rest of the curve is laterally insulated.
Equating these two expressions and differentiating wrt b and assuming $c$ and $\rho$ are constant, you get;
$$\frac{d}{d\theta} \Bigg [ k \nabla u (r(\theta)) \cdot \frac{r'(\theta)}{|r'(\theta)|}\Bigg ] = c \rho u_t (r(\theta)) |r'(\theta)|$$
With the chain rule and some more algebra I end up with the equation;
$$k \frac{\partial^2 u}{\partial\theta^2}|r'(\theta)|^2-k\frac{\partial u}{\partial \theta}(r'(\theta)\cdot r''(\theta))=c\rho u_t |r'(\theta)|^4$$.
Is this derivation correct?
Is it possible to solve it?