Question

Let \(u(x, t)\) be the solution of \[ \frac{\partial^2 u}{\partial x^2} - \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = 0, \quad x \in (-\infty, \infty), t>0, \] \[ u(x, 0) = \sin x, \quad x \in (-\infty, \infty), \] \[ \frac{\partial u}{\partial t}(x, 0) = \cos x, \quad x \in (-\infty, \infty), \] for some positive real number c.
Let the domain of dependence of the solution u at the point P(3,2) be the line segment on the x-axis with end points Q and R.
If the area of the triangle PQR is 8 square units, then the value of \(c^2\) is ................

Hint:

Remember that the solution to the wave equation propagates outwards from the initial data at speed \(c\). The "cone of influence" determines which future points are affected, while the "domain of dependence" determines which past data affects a given point. The boundary of this region is defined by the characteristic lines \(x \pm ct = \text{constant}\).

Answer 1

Step 1: Understanding the Concept:
This problem involves the 1D wave equation and the concept of the "domain of dependence". The value of the solution \(u(x_0, t_0)\) at a specific point in spacetime depends only on the initial conditions (at \(t=0\)) within a certain interval on the x-axis. This interval is called the domain of dependence.
Step 2: Key Formula or Approach:
For the wave equation \(u_{tt} = c^2 u_{xx}\) (note the rearrangement), the value of the solution at a point \((x_0, t_0)\) is determined by the initial data on the interval \([x_0 - ct_0, x_0 + ct_0]\). This interval on the x-axis is the domain of dependence.
Step 3: Detailed Calculation:
The given wave equation is \(\frac{\partial^2 u}{\partial x^2} - \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = 0\), which is equivalent to \(u_{tt} = c^2 u_{xx}\).
We are interested in the solution at the point \(P(x_0, t_0) = (3, 2)\).
The domain of dependence is the interval \([x_0 - ct_0, x_0 + ct_0]\) on the x-axis (\(t=0\)). Substituting the values \(x_0=3\) and \(t_0=2\), the interval is \([3 - c(2), 3 + c(2)] = [3 - 2c, 3 + 2c]\).
The endpoints of this segment are Q and R. Let's assign their coordinates: \(Q = (3 - 2c, 0)\) and \(R = (3 + 2c, 0)\).
We are given a triangle PQR with vertices \(P(3, 2)\), \(Q(3 - 2c, 0)\), and \(R(3 + 2c, 0)\).
The base of the triangle is the line segment QR, which lies on the x-axis. The length of the base is the distance between R and Q: \[ \text{Base} = (3 + 2c) - (3 - 2c) = 4c \] The height of the triangle is the perpendicular distance from the point P to the x-axis, which is the y-coordinate of P. \[ \text{Height} = 2 \] The area of the triangle PQR is given by: \[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times (4c) \times 2 = 4c \] We are given that the area of the triangle is 8 square units. \[ 4c = 8 \implies c = 2 \] The question asks for the value of \(c^2\). \[ c^2 = 2^2 = 4 \] Step 4: Final Answer:
The value of \(c^2\) is 4.
Step 5: Why This is Correct:
The domain of dependence for the wave equation was correctly identified. The geometry of the resulting triangle was used to set up an equation relating the wave speed \(c\) to the given area. Solving this equation yields \(c=2\), and therefore \(c^2=4\).