1. **Problem statement:** Derive the one-dimensional heat equation, which models how heat diffuses through a rod over time. 2. **Physical setup:** Consider a thin rod along the $x$-axis with temperature distribution $u(x,t)$, where $x$ is position and $t$ is time. 3. **Assumptions:** - Heat flow is only along the $x$-direction. - The rod is homogeneous and isotropic. - No internal heat sources. 4. **Fourier's law of heat conduction:** The heat flux $q$ (heat flow per unit area per unit time) is proportional to the negative temperature gradient: $$q = -k \frac{\partial u}{\partial x}$$ where $k$ is the thermal conductivity. 5. **Conservation of energy:** The change in heat energy in a small segment $[x, x+\Delta x]$ over time $\Delta t$ equals the net heat flux into the segment: $$\rho c \Delta x \frac{\partial u}{\partial t} = q(x) - q(x + \Delta x)$$ where $\rho$ is density and $c$ is specific heat capacity. 6. **Express net heat flux difference:** $$q(x) - q(x + \Delta x) = -\left(q(x + \Delta x) - q(x)\right) = -\Delta x \frac{\partial q}{\partial x}$$ 7. **Substitute Fourier's law into the above:** $$\frac{\partial q}{\partial x} = \frac{\partial}{\partial x} \left(-k \frac{\partial u}{\partial x}\right) = -k \frac{\partial^2 u}{\partial x^2}$$ 8. **Combine and simplify:** $$\rho c \Delta x \frac{\partial u}{\partial t} = -\Delta x \left(-k \frac{\partial^2 u}{\partial x^2}\right) = k \Delta x \frac{\partial^2 u}{\partial x^2}$$ 9. **Cancel $\Delta x$ from both sides:** $$\cancel{\rho c \Delta x} \frac{\partial u}{\partial t} = \cancel{\Delta x} k \frac{\partial^2 u}{\partial x^2}$$ 10. **Divide both sides by $\rho c$:** $$\frac{\partial u}{\partial t} = \frac{k}{\rho c} \frac{\partial^2 u}{\partial x^2}$$ 11. **Define thermal diffusivity:** $$\alpha = \frac{k}{\rho c}$$ 12. **Final heat equation in 1D:** $$\boxed{\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}}$$ This equation describes how temperature $u(x,t)$ evolves over time due to heat conduction in one dimension.