Partial Differential Equations: Exercises

04 General PDEs and boundary conditions

Exercise 1: Visualizing spatial slices

Consider the wave equation for a vibrating rectangular membrane of dimensions \(a\) by \(b\), where the displacement is given by \(u(x,y,t)\) with \((x,y) \in [0,a] \times [0,b]\) and \(t \geq 0\).

  1. What is the dimension of the domain \(D\) for this PDE?

  2. Sketch or describe what a spatial slice at time \(t = t_0\) looks like geometrically.

  3. If we track the displacement at a single point \((x_0, y_0)\) over all time, what geometric object does this trace out in the domain \(D\)?

  4. Suppose instead we consider the steady-state version where \(u\) solves Laplace’s equation (so \(u = u(x,y)\) with no time dependence). How does the domain change?

  1. The domain is \[ D =\{ (x,y,t) : 0 \leq x \leq a, \, 0 \leq y \leq b, \, t \geq 0\}, \] which is 3-dimensional (two spatial dimensions plus time).

  2. A spatial slice at time \(t = t_0\) is the rectangle \([0,a] \times [0,b]\) in the \(xy\)-plane. Geometrically, this represents the entire membrane at a single frozen moment in time. The value \(u(x,y,t_0)\) at each point gives the vertical displacement of the membrane at that location at time \(t_0\).

  3. Tracking a single point \((x_0, y_0)\) over all time traces out a vertical line (or ray) in the \(t\)-direction: the set \(\{(x_0, y_0, t) : t \geq 0\}\). This is a 1-dimensional curve in the 3-dimensional domain.

  4. For Laplace’s equation, there is no time dependence, so the domain becomes \[ D = \{(x,y) : 0 \leq x \leq a, \, 0 \leq y \leq b\}, \] which is 2-dimensional. This is just the rectangle representing the membrane itself.

Exercise 2: Recognizing linear vs. nonlinear operators

For each differential operator below, determine whether it is linear or nonlinear. If it is linear, state whether the PDE \(L(u) = 0\) is homogeneous or non-homogeneous.

  1. \(L(u) = \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} + e^x\)

  2. \(L(u) = \frac{\partial^2 u}{\partial x^2} + \left(\frac{\partial u}{\partial x}\right)^2\)

  3. \(L(u) = x^2 \frac{\partial^2 u}{\partial x^2} + y \frac{\partial u}{\partial y} - 3u\)

  4. \(L(u) = \nabla^2 u + e^u\)

  1. This operator is not linear because it includes a term \(e^x\). In particular, notice that we don’t have \(L(0) = 0\) due to this \(e^x\) term. (Every linear transformation must map the zero function to zero.)

  2. This operator is not linear because of the \(\left(\frac{\partial u}{\partial x}\right)^2\) term. To verify, notice, in particular, that \[ L(2x^2) = \frac{\partial^2 (2x^2)}{\partial x^2} + \left(\frac{\partial (2x^2)}{\partial x}\right)^2 = 4 + 16x^2. \] However, we have \[ 2L(x^2) = 2\left(\frac{\partial^2 (x^2)}{\partial x^2} + \left(\frac{\partial (x^2)}{\partial x}\right)^2\right) = 2(2 + 4x^2) = 4 + 8x^2. \] Thus, \(L(2x^2) \neq 2L(x^2)\), so \(L\) is nonlinear.

  3. This operator is linear. The coefficients \(x^2\), \(y\), and \(-3\) are all functions of the independent variables (not of \(u\)), and \(u\) and its derivatives appear linearly (i.e., no products or nonlinear functions of \(u\) or its derivatives).

  4. This operator is not linear because of the \(e^u\) term, which depends nonlinearly on \(u\) itself. To verify, notice, in particular, that \[ L(0) = \nabla^2 0 + e^0 = 1 \neq 0, \]

while for any linear operator we must have \(L(0) = 0\).

Exercise 3: Boundary condition detective work

For each physical scenario below, determine the appropriate type of boundary condition (Dirichlet or Neumann) and write it down mathematically. Explain your reasoning.

  1. A metal wire of length \(L\) has its left end held at a constant temperature of \(100°C\) and its right end held at \(0°C\). We model the temperature as \(u(x,t)\) where \(x \in [0,L]\).

  2. A metal rod of length \(L\) is perfectly insulated at both ends, so no heat can flow in or out. The temperature is \(u(x,t)\) where \(x \in [0,L]\).

  3. A vibrating guitar string of length \(L\) is pinned down at both ends (displacement is zero there). The displacement is \(u(x,t)\) where \(x \in [0,L]\).

  4. A hot metal plate occupies the square \([0,1] \times [0,1]\), and you hold the temperature constant at \(50°C\) along the entire boundary. The temperature is \(u(x,y,t)\).

  1. This requires Dirichlet boundary conditions because the temperature values are specified directly at the endpoints: \[ u(0,t) = 100, \quad u(L,t) = 0 \quad \text{for all } t > 0. \]

  2. This requires Neumann boundary conditions. Since the rod is insulated, no heat flows through the endpoints, meaning that \[ \frac{\partial u}{\partial x}(0,t) = 0, \quad \frac{\partial u}{\partial x}(L,t) = 0 \quad \text{for all } t > 0. \]

  3. This requires Dirichlet boundary conditions because the displacement is specified (as zero) at the endpoints: \[ u(0,t) = 0, \quad u(L,t) = 0 \quad \text{for all } t > 0. \]

  4. This requires Dirichlet boundary conditions on the entire boundary of the square. If we denote the boundary as \(\partial D\), then: \[ u(x,y,t) = 50 \quad \text{for all } (x,y) \in \partial D, \, t > 0. \] More explicitly, this means: \[ \begin{align*} u(0,y,t) &= 50 \quad \text{for } y \in [0,1], \\ u(1,y,t) &= 50 \quad \text{for } y \in [0,1], \\ u(x,0,t) &= 50 \quad \text{for } x \in [0,1], \\ u(x,1,t) &= 50 \quad \text{for } x \in [0,1]. \end{align*} \]

Exercise 4: Initial vs. boundary value problems

Consider the wave equation \(\frac{\partial^2 u}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2}\) on the interval \([0,L]\), modeling a string of length \(L\) held fixed at both ends. At \(t=0\), the string has the shape \(u(x,0) = x(L-x)\) and is released from rest. Write down all the boundary and initial conditions for this problem.

Boundary conditions (Dirichlet, since the string is fixed at both ends): \[ u(0,t) = 0, \quad u(L,t) = 0 \quad \text{for all } t > 0. \] Initial conditions: \[ u(x,0) = x(L-x) \quad \text{for all } x \in [0,L], \] \[ \frac{\partial u}{\partial t}(x,0) = 0 \quad \text{for all } x \in [0,L]. \]