To derive the heat conduction equation in cylindrical coordinates from Cartesian coordinates, we transform the Laplacian operator$\nabla^2 T = \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2}$ using the coordinate definitions $x = r \cos\theta$ and $y = r \sin\theta$.
Assuming a constant thermal conductivity $k$, the Cartesian equation is:
$$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} + \frac{\dot{q}}{k} = \frac{1}{\alpha} \frac{\partial T}{\partial t}$$
By transforming the $x$ and $y$ derivatives, the final cylindrical form is derived as:
$$\frac{\partial^2 T}{\partial r^2} + \frac{1}{r}\frac{\partial T}{\partial r} + \frac{1}{r^2}\frac{\partial^2 T}{\partial \theta^2} + \frac{\partial^2 T}{\partial z^2} + \frac{\dot{q}}{k} = \frac{1}{\alpha} \frac{\partial T}{\partial t}$$
1. Define Coordinate Relations
Establish the geometric links between Cartesian $(x, y)$ and polar $(r, \theta)$ systems:
$$r = \sqrt{x^2 + y^2}$$
$$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$
2. Calculate Partial Derivatives
Differentiate $r$ and $\theta$ with respect to $x$ and $y$ using standard calculus rules:
$$\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{r \cos\theta}{r} = \cos\theta$$
$$\frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} = \frac{r \sin\theta}{r} = \sin\theta$$
$$\frac{\partial \theta}{\partial x} = \frac{1}{1 + (y/x)^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2} = -\frac{\sin\theta}{r}$$
$$\frac{\partial \theta}{\partial y} = \frac{1}{1 + (y/x)^2} \cdot \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2} = \frac{\cos\theta}{r}$$
3. Expand the First Derivative
Apply the multi-variable chain rule to find $\frac{\partial T}{\partial x}$:
$$\frac{\partial T}{\partial x} = \frac{\partial T}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial T}{\partial \theta}\frac{\partial \theta}{\partial x}$$
$$\frac{\partial T}{\partial x} = \cos\theta \frac{\partial T}{\partial r} – \frac{\sin\theta}{r} \frac{\partial T}{\partial \theta}$$
This provides our spatial transformation operator for $x$:
$$\frac{\partial}{\partial x} = \cos\theta \frac{\partial}{\partial r} – \frac{\sin\theta}{r} \frac{\partial}{\partial \theta}$$
4. Expand the Second Derivative
Apply the transformation operator to itself to determine $\frac{\partial^2 T}{\partial x^2}$:
$$\frac{\partial^2 T}{\partial x^2} = \left(\cos\theta \frac{\partial}{\partial r} – \frac{\sin\theta}{r} \frac{\partial}{\partial \theta}\right) \left(\cos\theta \frac{\partial T}{\partial r} – \frac{\sin\theta}{r} \frac{\partial T}{\partial \theta}\right)$$
Distribute each term carefully, keeping in mind that $\frac{\partial}{\partial r}$ acts on $r$ and $\frac{\partial}{\partial \theta}$ acts on both the trigonometric functions and $T$:
$$\frac{\partial^2 T}{\partial x^2} = \cos^2\theta \frac{\partial^2 T}{\partial r^2} – \frac{2\sin\theta\cos\theta}{r}\frac{\partial^2 T}{\partial r \partial \theta} + \frac{2\sin\theta\cos\theta}{r^2}\frac{\partial T}{\partial \theta} + \frac{\sin^2\theta}{r}\frac{\partial T}{\partial r} + \frac{\sin^2\theta}{r^2}\frac{\partial^2 T}{\partial \theta^2}$$
5. Repeat for the $y$-Axis
Follow the identical chain rule workflow for the $y$ component:
$$\frac{\partial T}{\partial y} = \frac{\partial T}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial T}{\partial \theta}\frac{\partial \theta}{\partial y} = \sin\theta \frac{\partial T}{\partial r} + \frac{\cos\theta}{r} \frac{\partial T}{\partial \theta}$$
$$\frac{\partial^2 T}{\partial y^2} = \left(\sin\theta \frac{\partial}{\partial r} + \frac{\cos\theta}{r} \frac{\partial}{\partial \theta}\right) \left(\sin\theta \frac{\partial T}{\partial r} + \frac{\cos\theta}{r} \frac{\partial T}{\partial \theta}\right)$$
Expanding this expression yields:
$$\frac{\partial^2 T}{\partial y^2} = \sin^2\theta \frac{\partial^2 T}{\partial r^2} + \frac{2\sin\theta\cos\theta}{r}\frac{\partial^2 T}{\partial r \partial \theta} – \frac{2\sin\theta\cos\theta}{r^2}\frac{\partial T}{\partial \theta} + \frac{\cos^2\theta}{r}\frac{\partial T}{\partial r} + \frac{\cos^2\theta}{r^2}\frac{\partial^2 T}{\partial \theta^2}$$
6. Combine the Terms
Add the expansions for $\frac{\partial^2 T}{\partial x^2}$ and $\frac{\partial^2 T}{\partial y^2}$ together. Look for trigonometric identities ($\sin^2\theta + \cos^2\theta = 1$) to cancel out the complex cross-terms:
$$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = (\cos^2\theta + \sin^2\theta)\frac{\partial^2 T}{\partial r^2} + \frac{(\sin^2\theta + \cos^2\theta)}{r}\frac{\partial T}{\partial r} + \frac{(\sin^2\theta + \cos^2\theta)}{r^2}\frac{\partial^2 T}{\partial \theta^2}$$
This simplifies directly to the final Laplacian form:
$$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = \frac{\partial^2 T}{\partial r^2} + \frac{1}{r}\frac{\partial T}{\partial r} + \frac{1}{r^2}\frac{\partial^2 T}{\partial \theta^2}$$
✅ Derived Cylindrical Equation
Substituting this back into the core energy equation yields the comprehensive cylindrical coordinates heat conduction profile:
$$\frac{\partial^2 T}{\partial r^2} + \frac{1}{r}\frac{\partial T}{\partial r} + \frac{1}{r^2}\frac{\partial^2 T}{\partial \theta^2} + \frac{\partial^2 T}{\partial z^2} + \frac{\dot{q}}{k} = \frac{1}{\alpha}\frac{\partial T}{\partial t}$$
This can also be compactly written as:
$$\frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial T}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 T}{\partial \theta^2} + \frac{\partial^2 T}{\partial z^2} + \frac{\dot{q}}{k} = \frac{1}{\alpha}\frac{\partial T}{\partial t}$$