Cauchy Problem Description

The problem asks us to solve a Cauchy problem for a first-order partial differential equation (PDE) and then evaluate the solution at a specific point. The given PDE is:

\(\frac{{\partial u}}{{\partial y}} - x\frac{{\partial u}}{{\partial x}} + u - 1 = 0\)

This equation holds for \(-\infin < x < \infin\) and \(y \ge 0\).

The initial condition is given on the line \(y=0\):

\(u(x, 0) = \sin x\)

We need to find the value of the solution \(u(x,y)\) at the point \((0,1)\), i.e., \(u(0,1)\).

Characteristic Equations

We can rewrite the PDE in the standard form for the method of characteristics: \(a \frac{\partial u}{\partial x} + b \frac{\partial u}{\partial y} = c\).

Rearranging the given PDE:

\(\frac{{\partial u}}{{\partial y}} - x\frac{{\partial u}}{{\partial x}} = 1 - u\)

Comparing this with the standard form \(a \frac{\partial u}{\partial x} + b \frac{\partial u}{\partial y} = c\), we identify the coefficients:

• \(a = -x\)
• \(b = 1\)
• \(c = 1 - u\)

The system of characteristic equations is given by:

\(\frac{dx}{a} = \frac{dy}{b} = \frac{du}{c}\)

Substituting the coefficients:

\(\frac{dx}{-x} = \frac{dy}{1} = \frac{du}{1-u}\)

Solving Characteristics

We solve these ordinary differential equations (ODEs) to find the characteristic curves.

From the first equality, \(\frac{dx}{-x} = \frac{dy}{1}\):

Integrate both sides:

\(\int \frac{dx}{-x} = \int dy\)

\(-\ln|x| = y + C'\)

\(\ln|x| + y = -C'\)

Exponentiating, we get \(|x|e^y = e^{-C'}\). This gives us the first invariant:

\(xe^y = C_1\)

From the second equality, \(\frac{dy}{1} = \frac{du}{1-u}\):

Integrate both sides:

\(\int dy = \int \frac{du}{1-u}\)

\(y = -\ln|1-u| + C''\)

\(\ln|1-u| + y = C''\)

Exponentiating, we get \(|1-u|e^y = e^{C''}\). This gives us the second invariant:

\((1-u)e^y = C_2\)

General Solution Derivation

The general solution of the PDE is expressed as a relationship between the two invariants \(C_1\) and \(C_2\). We can write \(C_2 = F(C_1)\), where \(F\) is an arbitrary differentiable function determined by the initial condition.

Substituting the expressions for \(C_1\) and \(C_2\):

\((1-u)e^y = F(xe^y)\)

Now, we solve for \(u(x,y)\):

\(1-u(x,y) = e^{-y} F(xe^y)\)

\(u(x,y) = 1 - e^{-y} F(xe^y)\)

Applying Initial Condition

We use the initial condition \(u(x, 0) = \sin x\) to find the specific form of the function \(F\).

Substitute \(y=0\) into the general solution:

\(u(x, 0) = 1 - e^{-0} F(xe^0)\)

\(\sin x = 1 - 1 \cdot F(x)\)

\(\sin x = 1 - F(x)\)

Solving for \(F(x)\):

\(F(x) = 1 - \sin x\)

Specific Solution Formula

Now substitute the expression for \(F(x)\) back into the general solution formula \(u(x,y) = 1 - e^{-y} F(xe^y)\). Since \(F(z) = 1 - \sin z\), replacing \(z\) with \(xe^y\) gives \(F(xe^y) = 1 - \sin(xe^y)\).

The specific solution to the Cauchy problem is:

\(u(x,y) = 1 - e^{-y} (1 - \sin(xe^y))\)

This can also be written as:

\(u(x,y) = 1 - e^{-y} + e^{-y} \sin(xe^y)\)

Evaluate u(0,1)

Finally, we need to evaluate the solution at the point \((0,1)\). Substitute \(x=0\) and \(y=1\) into the specific solution formula:

\(u(0,1) = 1 - e^{-1} + e^{-1} \sin(0 \cdot e^1)\)

\(u(0,1) = 1 - e^{-1} + e^{-1} \sin(0)\)

Since \(\sin(0) = 0\):

\(u(0,1) = 1 - e^{-1} + e^{-1} \cdot 0\)

\(u(0,1) = 1 - e^{-1}\)

The value of \(u(0,1)\) is \(1 - \frac{1}{e}\).