Cauchy Problem Analysis

The given Cauchy problem involves a first-order linear partial differential equation:

\(\displaystyle \rm y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y}=0\)

This PDE is of the form \(\displaystyle \rm P\frac{\partial z}{\partial x} + Q\frac{\partial z}{\partial y} = R\), where \(\rm P = y\), \(\rm Q = -x\), and \(\rm R = 0\).

The initial data is given on the curve parameterized by \(\rm s\):

• \(\rm x_0(s) = \cos(s)\)
• \(\rm y_0(s) = \sin(s)\)
• \(\rm z_0(s) = 1\)

for \(\rm s > 0\).

We need to determine the nature of the solution (unique, no solution, finite solutions, infinitely many solutions) for this Cauchy problem.

Characteristic Equations

To solve this PDE, we use the method of characteristics. The characteristic equations associated with the PDE \(\displaystyle \rm P\frac{\partial z}{\partial x} + Q\frac{\partial z}{\partial y} = R\) are given by:

\(\displaystyle \frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}\)

Substituting the values of P, Q, and R from our PDE:

\(\displaystyle \frac{dx}{y} = \frac{dy}{-x} = \frac{dz}{0}\)

Solving Characteristic Equations

We can derive two independent first integrals from these characteristic equations.

From the equation \(\displaystyle \frac{dz}{0}\), we have \(\rm dz = 0\). Integrating this gives \(\rm z = c_1\), where \(\rm c_1\) is a constant.

From the equality of the first two terms, \(\displaystyle \frac{dx}{y} = \frac{dy}{-x}\), we can rearrange to get \(\rm -x \, dx = y \, dy\), or \(\rm x \, dx + y \, dy = 0\).

Integrating this equation:

\(\displaystyle \int x \, dx + \int y \, dy = C'\)

\(\displaystyle \frac{x^2}{2} + \frac{y^2}{2} = C'\)

\(\displaystyle x^2 + y^2 = 2C'\)

Let \(\rm c_2 = 2C'\). Then we have \(\rm x^2 + y^2 = c_2\), where \(\rm c_2\) is another constant.

General Solution Form

The general solution of the PDE is given by an arbitrary function relating the two constants of integration, i.e., \(\rm \Phi(c_1, c_2) = 0\). This can be expressed as \(\rm c_1 = f(c_2)\) for some arbitrary differentiable function \(\rm f\). Substituting the expressions for \(\rm c_1\) and \(\rm c_2\) in terms of x, y, and z, we get the general solution:

\(\rm z = f(x^2 + y^2)\)

Applying Cauchy Data

Now, we apply the initial data \(\rm x_0(s) = \cos(s)\), \(\rm y_0(s) = \sin(s)\), \(\rm z_0(s) = 1\) to the general solution form. The solution \(\rm z(x,y)\) must satisfy \(\rm z(x_0(s), y_0(s)) = z_0(s)\) along the initial curve.

\(\rm z_0(s) = f(x_0(s)^2 + y_0(s)^2)\)

Substituting the given initial data:

\(\rm 1 = f((\cos(s))^2 + (\sin(s))^2)\)

\(\rm 1 = f(\cos^2(s) + \sin^2(s))\)

Using the fundamental trigonometric identity \(\rm \cos^2(s) + \sin^2(s) = 1\), we get:

\(\rm 1 = f(1)\)

Interpreting the Condition for the Solution

The condition \(\rm f(1) = 1\) is a constraint on the arbitrary function \(\rm f\). It states that when the argument of \(\rm f\) is 1, its value must be 1. This condition specifies the behavior of \(\rm f\) at a single point, \(\rm u=1\), but it does not restrict the form of \(\rm f(u)\) for any other value of \(\rm u\).

Many different functions can satisfy the condition \(\rm f(1) = 1\). For example:

• \(\rm f(u) = u\) (since \(\rm f(1) = 1\))
• \(\rm f(u) = 1\) (since \(\rm f(1) = 1\))
• \(\rm f(u) = u^n\) for any integer n (since \(\rm f(1) = 1^n = 1\))
• \(\rm f(u) = u + (u-1)g(u)\) for any differentiable function \(\rm g(u)\) (since \(\rm f(1) = 1 + (1-1)g(1) = 1\))

Since there are infinitely many differentiable functions \(\rm f(u)\) that satisfy the condition \(\rm f(1) = 1\), there are infinitely many possible forms for the general solution \(\rm z = f(x^2 + y^2)\) that satisfy the initial data.

Characteristic Initial Curve

Let's examine the initial curve \(\rm x_0(s) = \cos(s)\), \(\rm y_0(s) = \sin(s)\). Squaring and adding these equations gives \(\rm x_0(s)^2 + y_0(s)^2 = \cos^2(s) + \sin^2(s) = 1\). This means the initial curve lies on the circle \(\rm x^2 + y^2 = 1\). The expression \(\rm x^2 + y^2 = c_2\) represents the characteristic curves of the PDE in the xy-plane. The initial curve \(\rm x^2 + y^2 = 1\) is therefore one of the characteristic curves itself.

When the initial data is given on a characteristic curve, the Cauchy problem typically does not have a unique solution. If the initial data is inconsistent along the characteristic curve, there is no solution. If the initial data is consistent, as it is here (\(\rm z\) is constant, which aligns with \(\rm dz=0\)), the problem has infinitely many solutions.

The condition \(\rm f(1)=1\) means that along the characteristic curve \(\rm x^2+y^2=1\) (where \(\rm c_2=1\)), the value of \(\rm z\) (which is \(\rm c_1\)) must be 1. This is precisely the initial data \(\rm z_0(s) = 1\). The initial data is consistent with the fact that the initial curve is a characteristic, but it does not provide enough information to uniquely determine the function \(\rm f\).

Therefore, the Cauchy problem has infinitely many solutions.