Question

In a long rod of square cross section $(0⩽x⩽s;0⩽y⩽s)$ the temperature obeys the differential equation $${\mathrm{\nabla }}^{2}T=\frac{1}{\kappa }\frac{\partial T}{\partial t}$$ At the end $z=0$, the temperature is $T=T_0\cos \omega t$ (independent of $x$ and $y$ ). (a) Find the temperature along the rod, if the sides are insulated. (b) Find the temperature along the rod, if the sides are maintained at zero temperature. In both cases, neglect transients.

Short answer

For insulated sides: $T(z,t)=T_0\cos (\omega t)e^{-\frac{\omega }{\sqrt{\kappa }}z}$; For zero-temperature sides: $T(z,t)=T_0\cos (\omega t)\sin (\frac{n\pi x}{s})\sin (\frac{m\pi y}{s})e^{-\frac{\omega }{\sqrt{\kappa }}z}$

Step-by-step solution

Understand the Differential Equation

The given differential equation is the heat equation for a three-dimensional rod: equation: $abl{a}^{2}T=\frac{1}{\kappa }\frac{\partial T}{\partial t}$. Here, $T$ represents temperature, $\kappa$ is the thermal conductivity, and $t$ is time. At $z=0$, the boundary condition is $T=T_0\cos \omega t$.

Express Temperature Distribution in Separable Form

Assume a solution of the form $$T(x,y,z,t)=X(x)Y(y)Z(z)\cos (\omega t)$$ Then substitute this into the heat equation.

Substitute Assumption into Heat Equation

Substitute $T(x,y,z,t)=X(x)Y(y)Z(z)\cos (\omega t)$ into the heat equation: $$abl{a}^2(XYZ\cos (\omega t))=\frac{1}{\kappa }(-\omega \sin (\omega t))$$ This simplifies to $$-k^{2}XYZ\cos (\omega t)=\frac{1}{\kappa }(-\omega \sin (\omega t))$$

Separation of Variables

Separate the terms involving $x,y,z$ and $t$: $$abl{a}^{2}Z(z)\cos (\omega t)=-\frac{{\omega }^2}{\kappa }Z(z)\cos (\omega t)$$ This gives us the spatial equation: $$\frac{d^{2}Z(z)}{d{z}^2}=-\frac{{\omega }^2}{\kappa }Z(z)$$

Solve the Spatial Part

Solve the spatial part $$\frac{d^{2}Z(z)}{d{z}^2}=-\frac{{\omega }^2}{\kappa }Z(z)$$ This is a standard second-order differential equation, with solution $$Z(z)=A{e}^{-\frac{\omega }{\sqrt{\kappa }}z}+B{e}^{\frac{\omega }{\sqrt{\kappa }}z}$$

Apply Boundary Conditions

(a) For insulated sides, the derivatives at boundaries must be zero: $$\frac{\partial T}{\partial x}=0\text{and}\frac{\partial T}{\partial y}=0$$ This implies that the temperature does not change with $x$ and $y$. The solution simplifies to $$T(z,t)=T_0\cos (\omega t)e^{-\frac{\omega }{\sqrt{\kappa }}z}$$

Zero Temperature Sides

(b) For sides maintained at zero temperature at $x=np,y=nq$ the boundary condition simplifies to: $$T(z,t)=0$$ At sides, leading to a modified solution: The temperature along the rod, subjected to side boundaries is: $$T(z,t)=T_0\cos (\omega t)\sin \left(\frac{n\pi x}{s}\right)\sin \left(\frac{m\pi y}{s}\right)e^{-\frac{\omega }{\sqrt{\kappa }}z}$$

Key concepts

Separation of Variables

Separation of variables is a powerful mathematical method used to solve partial differential equations, like the heat equation in physics. The idea is to assume that the solution can be written as a product of functions, each depending on a single coordinate or variable. For instance, we express the temperature distribution $T(x,y,z,t)$ as:

$$T(x,y,z,t)=X(x)Y(y)Z(z)\cos (\omega t)$$

By substituting this product into the heat equation, we can split the equation into separate ordinary differential equations for each coordinate function $X(x)$, $Y(y)$, and $Z(z)$. This process simplifies solving the complex heat equation, step-by-step for each variable independently. This method leverages the fact that complex multi-variable problems can often be reduced to simpler, single-variable problems.

Boundary Conditions

Boundary conditions are essential for solving differential equations as they describe how the function behaves at the boundaries of the domain. In our heat equation problem:

• At $z=0$, the temperature $T=T_0\cos (\omega t)$.
• For insulated sides, the temperature's spatial derivatives are zero at the boundaries implying no heat flow across the sides.
• For sides maintained at zero temperature, the temperature itself is zero at the boundaries.

These conditions help determine specific solutions suited to physical constraints. For the rod with insulated sides, the temperature doesn’t vary with $x$ and $y$. Therefore, the solution is symmetrical. For zero temperature sides, the sine terms in the solution ensure it satisfies those boundary conditions by going to zero at the specified points.

Thermal Conductivity

Thermal conductivity ($\kappa$) is a material property that indicates how well a material can conduct heat. In the context of the heat equation:

$$abl{a}^{2}T=\frac{1}{\kappa }\frac{\partial T}{\partial t}$$

it represents the ability of the rod material to transfer heat over time. Higher thermal conductivity means heat spreads faster through the material. The term $\frac{1}{\kappa }\frac{\partial T}{\partial t}$ links the rate of temperature change over time to how heat diffuses spatially.

It appears in the solution for $Z(z)$ as part of the exponential decay factor:

$$Z(z)=A{e}^{-\frac{\omega }{\sqrt{\kappa }}z}+B{e}^{\frac{\omega }{\sqrt{\kappa }}z}$$.

This factor shows how thermal conductivity affects the rate at which temperature changes as we move along the rod's length.