MA 30300: Differential Equations

9.5. Heat Equation on a Finite Rod

Goal: Understand solutions of the heat equation under different boundary conditions.

1. Zero Endpoint Temperatures

Heated Rod (Dirichlet Conditions): A rod of length \(L\) has both ends held at zero temperature. The temperature distribution evolves according to the heat equation and can be expressed as a Fourier sine series.

Let \(u(x,t)\) denote the temperature at position \(x\) and time \(t\). The constant \(k\) is the thermal diffusivity of the material, which measures how quickly heat spreads.

Some Thermal Diffusivity Constants

Material Thermal Diffusivity \(k\) (cm²/s)
Silver 1.70
Copper 1.15
Aluminum 0.85
Iron 0.15
Concrete 0.005

Heat Equation:

\[ u_t = k u_{xx}, \quad 0 < x < L,\; t > 0 \]

Endpoint Conditions:

\[ u(0,t)=0, \quad u(L,t)=0, \quad \text{ for all } t>0 \]

Initial Condition:

\[ u(x,0)=f(x), \quad 0 < x < L \]

Solution:

\[ u(x,t)=\sum_{n=1}^{\infty} b_n e^{-k (n\pi/L)^2 t}\sin\left(\frac{n\pi x}{L}\right) \]

Coefficients:

The coefficients \(b_n\) are the Fourier sine coefficients of the initial temperature distribution \(f(x)\):

\[ b_n=\frac{2}{L}\int_0^L f(x)\sin\left(\frac{n\pi x}{L}\right)\,dx \]

Note: As \(t \to \infty\), the temperature of the rod approaches 0:

\[ \displaystyle \lim_{t \to \infty} u(x,t) = 0 \]

Physically, the rod cools down completely because heat escapes through the ends.

2. Insulated Endpoint Temperatures

Heated Rod (Neumann Conditions): A rod of length \(L\) has both ends insulated, meaning no heat can flow in or out at the endpoints. The temperature distribution evolves according to the heat equation and can be expressed as a Fourier cosine series (plus a constant term for the average temperature).

Heat Equation:

\[ u_t = k u_{xx}, \quad 0 < x < L,\; t > 0 \]

Endpoint Conditions (Insulated):

\[ u_x(0,t)=0, \quad u_x(L,t)=0, \quad \text{ for all } t>0 \]

Initial Condition:

\[ u(x,0)=f(x), \quad 0 < x < L \]

Solution:

\[ u(x,t)= \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n e^{-k (n\pi/L)^2 t}\cos\left(\frac{n\pi x}{L}\right) \]

Coefficients:

The coefficients are the Fourier cosine coefficients of the initial temperature distribution \(f(x)\):

\[ a_0 = \frac{2}{L}\int_0^L f(x)\,dx, \quad a_n = \frac{2}{L}\int_0^L f(x)\cos\left(\frac{n\pi x}{L}\right)\,dx, \quad n \ge 1 \]

Note: As \(t \to \infty\), the temperature of the rod approaches the average of the initial temperature:

\[ \displaystyle \lim_{t \to \infty} u(x,t) = \frac{a_0}{2} = \frac{1}{L}\int_0^L f(x)\,dx \]

Physically, the heat redistributes evenly and the total energy is conserved because the ends are insulated.

3. Practice Questions

Zero Endpoint Temperatures

Example 1.1: Suppose a rod of length \(L = 50\ \text{cm}\) is immersed in steam until its temperature is \(u_0 = 100^\circ\text{C}\) throughout. At time \(t = 0\), its lateral surface is insulated, and its two ends are embedded in ice at \(0^\circ\text{C}\). Calculate the rod’s temperature at its midpoint after half an hour if it is made of iron.

Example 1.2: Suppose that a rod \(\text{40 cm}\) long with insulated lateral surface is heated to a uniform temperature of \(100^\circ\text{C}\), and that at time \(t = 0\) its two ends are embedded in ice at \(0^\circ\text{C}\).

  1. Find the formal series solution for the temperature \(u(x, t)\) of the rod.
  2. In the case the rod is made of copper, show that after \(5\) minutes the temperature at its midpoint is about \(15^\circ\text{C}\).
  3. In the case the rod is made of concrete, find the time required for its midpoint to cool to \(15^\circ\text{C}\).

Example 1.3: Solve the boundary value problems:

  1. \[ u_t = 3 u_{xx}, \quad 0 < x < \pi, \ t>0; \quad u(0,t)=u(\pi,t)=0, \quad u(x,0)=4\sin 2x \]
  2. \[ u_t = 2 u_{xx}, \quad 0 < x < 1, \ t>0; \quad u(0,t)=u(1,t)=0, \quad u(x,0)=5\sin \pi x - 5\sin 3\pi x \]
  3. \[ u_t = u_{xx}, \quad 0 < x < \pi, \ t>0; \quad u(0,t)=u(\pi,t)=0, \quad u(x,0)=4\sin 4x \cos 2x \]
  4. \[ 2 u_t = u_{xx}, \quad 0 < x < 1, \ t>0; \quad u(0,t)=u(1,t)=0, \quad u(x,0)=4\sin \pi x \cos^3\pi x \]
  5. \[ u_t = 2 u_{xx}, \quad 0 < x < 3, \ t>0; \quad u(0,t)=u(3,t)=0, \quad u(x,0)=x(3-x) \]

Insulated Ends

Example 2.1: We consider a rod of length \(L = 50\ \text{cm}\) whose initial temperature forms a triangular profile along its length: the temperature starts at 0°C at the left end (\(x = 0\)), rises linearly to 100°C at the midpoint (\(x = 25\ \text{cm}\)), and then decreases linearly back to 0°C at the right end (\(x = 50\ \text{cm}\)). At time \(t = 0\), the rod’s lateral surface and its two ends are insulated. Assume the thermal diffusivity of the material is \(k\). Find the temperature distribution \(u(x,t)\) along the rod for \(t > 0\).

\[ u_t = k\, u_{xx}, \quad 0 < x < 50, \quad t > 0 \] \[ u_x(0,t) = u_x(50,t) = 0, \quad t > 0 \] \[ u(x,0) = f(x), \quad 0 < x < 50 \]

Example 2.2: Consider a copper rod of length \(\text{50 cm}\) with its sides insulated. Initially, the rod’s temperature varies along its length according to \(u(x, 0) = 2x\), and the two ends are insulated from the start.

  1. Determine the temperature distribution \(u(x, t)\) along the rod over time.
  2. Calculate the temperature at the point \(x = 10\) after \(1\) minute has passed.
  3. Estimate the time it will take for the temperature at \(x = 10\) to drop to \(45^\circ\text{C}\).

Example 2.3: Solve the boundary value problems:

  1. \[ u_t = 10 u_{xx}, \quad 0 < x < 5, \ t>0; \quad u_x(0,t)=u_x(5,t)=0, \quad u(x,0)=7 \]
  2. \[ 3 u_t = u_{xx}, \quad 0 < x < 2, \ t>0; \quad u_x(0,t)=u_x(2,t)=0, \quad u(x,0)=\cos^2 2 \pi x \]
  3. \[ u_t = u_{xx}, \quad 0 < x < 2, \ t>0; \quad u_x(0,t)=u_x(2,t)=0, \quad u(x,0)=10 \cos \pi x \cos 3 \pi x \]
  4. \[ 5 u_t = u_{xx}, \quad 0 < x < 10, \ t>0; \quad u_x(0,t)=u_x(10,t)=0, \quad u(x,0)=4x \]
  5. \[ u_t = u_{xx}, \quad 0 < x < 100, \ t>0; \quad u_x(0,t)=u_x(100,t)=0, \quad u(x,0)=x(100-x) \]

Worked examples

Example 3.1: Consider the linear, homogeneous Convection-Diffusion Equation in one dimension:

\[ \frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial x^2} - v \frac{\partial u}{\partial x} \]
Assume a separated solution \(u(x,t) = X(x)T(t)\) and separation constant \(-\lambda\). The resulting ODE system is:
  1. \[ X'' - \frac{v}{D}X' - \frac{\lambda}{D}X = 0, \quad T' - \lambda T = 0 \]
  2. \[ D X'' - v X' + \lambda X = 0, \quad T' - \lambda T = 0 \]
  3. \[ D X'' - v X' - \lambda X = 0, \quad T' + \lambda T = 0 \]
  4. \[ \frac{1}{D}X'' - \frac{v}{D}X' + \lambda X = 0, \quad T' + \lambda T = 0 \]
  5. \[ D X'' - v X' + \lambda X = 0, \quad T' + \lambda T = 0 \]

Solution: Assume a separated solution \(u(x,t) = X(x)T(t)\) and substitute into the PDE:

\[ u_t = XT', \quad u_x = X'T, \quad u_{xx} = X''T \]
\[ XT' = D X''T - v X'T \]

Divide through by \(XT\):

\[ \frac{T'}{T} = D\frac{X''}{X} - v\frac{X'}{X} \]

Set both sides equal to a separation constant \(-\lambda\):

\[ \frac{T'}{T} = -\lambda, \quad D\frac{X''}{X} - v\frac{X'}{X} = -\lambda \]

This gives the resulting system of ordinary differential equations:

\[ D X'' - v X' + \lambda X = 0, \quad T' + \lambda T = 0 \]

Example 3.2: The solution to the heat conduction problem:

\[ \begin{array}{lll} u_{t}=9u_{xx}, && 0 < x < 5 \\ u(0,t)=0, \quad u(5,t)=0,&&t >0\\ u(x,0)=25, &&0 \leq x \leq 5 \end{array} \]

is given by

\[ u(x,t)=\sum_{n=1}^{\infty} c_{n}e^{-\frac{9 n^2 \pi^2 }{25}t} \sin\left( \frac{n\pi x}{5}\right) \]

for some constants \(c_n\), \(n \geq 1\). What is \(c_5\)?

Solution: Since \(u(x,0)=25\),

\[ 25 = \sum_{n=1}^{\infty} c_n \sin\left(\frac{n\pi x}{5}\right) \]

This is the Fourier sine series of \(25\) on \(0 < x < 5\). The coefficients are given by:

\[ c_n = \frac{2}{5}\int_{0}^{5} 25 \sin\left(\frac{n\pi x}{5}\right)\,dx \]

Compute the integral:

\[ c_n = 10 \int_{0}^{5} \sin\left(\frac{n\pi x}{5}\right)\,dx \]
\[ = 10 \cdot \frac{5}{n\pi}\left(1 - \cos(n\pi)\right) \]
\[ = \frac{50}{n\pi}\left(1 - (-1)^n\right) \]

Now substitute \(n = 5\):

\[ (-1)^5 = -1 \quad \Rightarrow \quad 1 - (-1) = 2 \]
\[ c_5 = \frac{50}{5\pi} \cdot 2 = \frac{20}{\pi} \]

Final Answer:

\[c_5= \frac{20}{\pi}. \]

Example 3.3: The heat conduction problem:

\[ \begin{array}{lll} u_{t}=u_{xx}, &&0 < x < 30\\ u_x(0,t)=0, \quad u_x(30,t)=0,&&t >0\\ \\ u(x,0)= \left\{ \begin{array}{ll} 40 & 0 \leq x <10\\ 20 & 10 \leq x < 20\\ 30 & 20 \leq x \leq 30 \end{array} \right. \end{array} \]

has solution:

\[ u(x,t)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n e^{-\frac{n^2 \pi^2 t}{900}} \cos \left(\frac{n\pi x}{30}\right) \]
Determine the temperature at \(x = 15\) as \(t \to \infty\), when the temperature distribution along the rod has stabilized.

Solution: As \(t \to \infty\), all exponential terms decay to zero:

\[ e^{-\frac{n^2 \pi^2 t}{900}} \to 0 \]

So

\[u(x,t) \to \frac{a_0}{2} \]

Compute \(a_0\):

\[ a_0 = \frac{2}{30}\int_0^{30} u(x,0)\,dx \]

Split the integral:

\[ a_0 = \frac{1}{15}\left( \int_0^{10} 40\,dx + \int_{10}^{20} 20\,dx + \int_{20}^{30} 30\,dx \right) \]
\[\implies a_0 = \frac{1}{15}(400 + 200 + 300) = \frac{900}{15} = 60 \]

Therefore as \(t \to \infty\), for all \(x\in[0, 30]\)

\[ u(x,t) \to \frac{a_0}{2} = 30 \]

In particular at \(x = 15\):

\[ u(15,\infty) = 30 \]

Example 3.4: Consider the heat conduction problem:

\[ \begin{array}{lll} u_{xx}=u_{t}, &&0 < x < 30, t>0\\ u(0,t)=20, \quad u(30,t)=50,&&t >0\\ u(x,0)=60-2x, &&0 \leq x \leq 30 \end{array} \]

It is observed empirically that as \(t \to \infty\), \(u(x,t)\) approaches a steady-state temperature \(u_{ss}(x)\) that corresponds to setting \(u_t = 0\) in the boundary value problem. What is the temperature at \(x=5\) as \(t \to \infty\)?

Solution: As \(t \to \infty\), the solution approaches a steady-state profile, so:

\[ u_t = 0 \;\Rightarrow\; u_{xx} = 0 \]

Solve:

\[ u_{xx} = 0 \Rightarrow u(x) = Ax + B \]

Apply boundary conditions:

\[ u(0)=20 \Rightarrow B = 20 \]
\[ u(30)=50 \Rightarrow 30A + 20 = 50 \]
\[ 30A = 30 \Rightarrow A = 1 \]

So the steady-state solution is:

\[ u(x) = x + 20 \]

Evaluate at \(x=5\):

\[ u(5) = 5 + 20 = 25 \]

Final Answer: 25