Question

Consider the Cauchy problem \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = u; \] \[ u = f(t) \text{ on the initial curve } \Gamma = (t, t); t>0. \] Consider the following statements:
P: If \(f(t) = 2t + 1\), then there exists a unique solution to the Cauchy problem in a neighbourhood of \(\Gamma\).
Q: If \(f(t) = 2t - 1\), then there exist infinitely many solutions to the Cauchy problem in a neighbourhood of \(\Gamma\).
Then

• A. both P and Q are TRUE
• B. P is FALSE and Q is TRUE
• C. P is TRUE and Q is FALSE
• D. both P and Q are FALSE

Hint:

For a first-order PDE, always start by checking the transversality condition. If the determinant is non-zero, a unique solution exists. If it is zero, the initial curve is a characteristic. You must then check if the initial data is compatible with the characteristic equations to decide between no solution and infinitely many solutions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem deals with the existence and uniqueness of solutions to a first-order quasi-linear partial differential equation (PDE) using the method of characteristics. The existence and uniqueness depend on a transversality condition at the initial curve.
Step 2: Key Formula or Approach:
The PDE is of the form \(a(x,y,u)u_x + b(x,y,u)u_y = c(x,y,u)\). Here, \(a=x, b=y, c=u\). The initial curve is parametrized as \(x_0(s) = s\), \(y_0(s) = s\), and the initial condition is \(u_0(s) = f(s)\) (let's use s instead of t for the parameter). The characteristic equations are: \(\frac{dx}{dt} = x\), \(\frac{dy}{dt} = y\), \(\frac{du}{dt} = u\). For existence and uniqueness, the Jacobian determinant (transversality condition) must be non-zero along the initial curve: \[ J = \begin{vmatrix} a(x_0, y_0) & b(x_0, y_0) \\ \frac{dx_0}{ds} & \frac{dy_0}{ds} \end{vmatrix} = a(x_0, y_0)\frac{dy_0}{ds} - b(x_0, y_0)\frac{dx_0}{ds} \neq 0 \] If \(J=0\), we must check a compatibility condition. We solve the characteristic equations along the curve. If the initial data is consistent with the characteristics, there are infinitely many solutions. If it is inconsistent, there is no solution.
Step 3: Detailed Calculation:
The initial curve is \(\Gamma\), parametrized by \(x_0(s) = s, y_0(s) = s\) for \(s>0\).
Step 3.1: Check the Transversality Condition. Here \(a=x, b=y\). On \(\Gamma\), \(a(x_0, y_0) = s\) and \(b(x_0, y_0) = s\). We have \(\frac{dx_0}{ds} = 1\) and \(\frac{dy_0}{ds} = 1\). The Jacobian is: \[ J = (s)(1) - (s)(1) = s - s = 0 \] Since \(J=0\) for all \(s\), the initial curve \(\Gamma\) is a characteristic curve itself. This means we do not have a unique solution. The problem will either have no solution or infinitely many solutions.
Step 3.2: Check the Compatibility Condition.
To determine which case it is, we solve the characteristic equations.
From \(\frac{dx}{dt} = x\), we get \(x(t) = C_1 e^t\).
From \(\frac{dy}{dt} = y\), we get \(y(t) = C_2 e^t\).
From \(\frac{du}{dt} = u\), we get \(u(t) = C_3 e^t\).
From the first two, we see that \(y/x = C_2/C_1\) is constant along characteristics. The characteristic curves are lines through the origin. Our initial curve \(y=x\) is one such line.
Also, \(u/x = C_3/C_1\) is constant. This means the general solution must have the form \(u = x \cdot \phi(y/x)\) for some function \(\phi\).
Let's check the given initial condition \(u = f(t)\) on \(x=t, y=t\).
Substitute this into the general solution: \(f(t) = t \cdot \phi(t/t) = t \cdot \phi(1)\).
This implies \(f(t)\) must be a linear function of \(t\), specifically \(f(t) = C t\) where \(C = \phi(1)\) is a constant.
Analysis of Statement P:
Here \(f(t) = 2t + 1\). This is not of the form \(Ct\). Therefore, the initial condition is incompatible with the PDE along the characteristic curve. There is NO solution. So, statement P, which claims a unique solution exists, is FALSE.
Analysis of Statement Q:
Here \(f(t) = 2t - 1\). This is also not of the form \(Ct\). Therefore, the initial condition is incompatible. There is NO solution.
So, statement Q, which claims infinitely many solutions exist, is FALSE.
(Note: if the condition had been, for example, \(f(t) = 2t\), it would satisfy the compatibility condition with \(C=2\), and there would be infinitely many solutions).
Step 4: Final Answer:
both P and Q are FALSE. Step 5: Why This is Correct:
The transversality condition determinant is zero, indicating the initial curve is a characteristic. This rules out a unique solution. The compatibility condition requires the initial data to be of the form \(u=Ct\) on the curve \(x=t, y=t\). Neither of the given functions \(f(t)\) satisfies this form, meaning the initial data is inconsistent with the PDE's behavior along the characteristic, resulting in no solution in both cases.