Let u = u(x, y) be the solution to the following Cauchy problem ux + uy = eu for (x, y) ∈ ℝ × \(\rm\left(0, \frac{1}{e}\right)\) and u(x, 0) = 1 for x ∈ ℝ. Which of the following statements are true?
Cauchy Problem Statement
The given problem is a first-order nonlinear partial differential equation (PDE) with an initial condition, also known as a Cauchy problem:
\(\qquad u_x + u_y = e^u \quad \text{for } (x, y) \in \mathbb{R} \times \left(0, \frac{1}{e}\right)\)
\(\qquad u(x, 0) = 1 \quad \text{for } x \in \mathbb{R}\)
Characteristics Method
We can solve this type of PDE using the method of characteristics. This method involves converting the PDE into a system of ordinary differential equations (ODEs) along specific curves in the domain called characteristic curves.
Solving Characteristic Equations
For a PDE of the form \(a u_x + b u_y = c\), the characteristic equations are given by:
- \(\frac{dx}{dt} = a\)
- \(\frac{dy}{dt} = b\)
- \(\frac{du}{dt} = c\)
In our specific problem, we have \(a=1\), \(b=1\), and \(c=e^u\). So, the characteristic equations are:
- \(\frac{dx}{dt} = 1\)
- \(\frac{dy}{dt} = 1\)
- \(\frac{du}{dt} = e^u\)
Integrating these equations with respect to \(t\), we get:
- \(\int dx = \int 1 \, dt \implies x = t + c_1\)
- \(\int dy = \int 1 \, dt \implies y = t + c_2\)
- \(\int e^{-u} du = \int 1 \, dt \implies -e^{-u} = t + c_3\)
Here \(c_1, c_2, c_3\) are constants of integration along the characteristic curves.
Applying Initial Condition
The initial condition is given on the line \(y=0\) as \(u(x, 0) = 1\). We can parameterize the initial curve at \(t=0\) as \((x_0(s), y_0(s)) = (s, 0)\), and the initial value of \(u\) is \(u_0(s) = u(s, 0) = 1\).
We use the initial condition at \(t=0\) to find the constants \(c_1, c_2, c_3\):
- At \(t=0\), \(x=s\): \(s = 0 + c_1 \implies c_1 = s\)
- At \(t=0\), \(y=0\): \(0 = 0 + c_2 \implies c_2 = 0\)
- At \(t=0\), \(u=1\): \(-e^{-1} = 0 + c_3 \implies c_3 = -e^{-1}\)
Solution \(u(x, y)\)
Substitute the values of the constants back into the general solutions of the characteristic equations:
- \(x = t + s\)
- \(y = t\)
- \(-e^{-u} = t - e^{-1}\)
From the equation \(y=t\), we have \(t=y\).
Substitute \(t=y\) into the third equation:
\(\qquad -e^{-u} = y - e^{-1}\)
Rearranging this equation, we get:
\(\qquad e^{-u} = e^{-1} - y\)
Taking the natural logarithm of both sides gives:
\(\qquad -u = \log(e^{-1} - y)\)
Thus, the solution to the Cauchy problem is:
\(\qquad u(x, y) = -\log(e^{-1} - y) = \log\left(\frac{1}{e^{-1} - y}\right)\)
This solution is valid for \(e^{-1} - y > 0\), which means \(y < e^{-1}\), matching the given domain \(\left(0, \frac{1}{e}\right)\).
Partial Derivatives
Now, let's calculate the partial derivatives of the solution \(u(x, y) = -\log(e^{-1} - y)\).
- Partial derivative with respect to \(x\):
\(u_x = \frac{\partial}{\partial x} (-\log(e^{-1} - y)) = 0\) - Partial derivative with respect to \(y\):
\(u_y = \frac{\partial}{\partial y} (-\log(e^{-1} - y)) = - \frac{1}{e^{-1} - y} \cdot (-1) = \frac{1}{e^{-1} - y}\)
So, \(u_x = 0\) and \(u_y = \frac{1}{e^{-1} - y}\).
Verifying Options
We will now evaluate each given statement using the derived solution \(u(x, y)\) and its partial derivatives \(u_x\) and \(u_y\).
Option 1: \(u\left(\frac{1}{2 e}, \frac{1}{2 e}\right) = 1\)
Substitute \(y = \frac{1}{2e}\) into the solution \(u(x, y) = \log\left(\frac{1}{e^{-1} - y}\right)\):
\(\qquad u\left(x, \frac{1}{2e}\right) = \log\left(\frac{1}{e^{-1} - \frac{1}{2e}}\right) = \log\left(\frac{1}{\frac{1}{e} - \frac{1}{2e}}\right) = \log\left(\frac{1}{\frac{2-1}{2e}}\right) = \log\left(\frac{1}{\frac{1}{2e}}\right) = \log(2e)\)
\(\qquad u\left(x, \frac{1}{2e}\right) = \log(2) + \log(e) = \log(2) + 1\)
Since \(\log(2) \ne 0\), \(u\left(\frac{1}{2 e}, \frac{1}{2 e}\right) = \log(2) + 1 \ne 1\). Thus, statement 1 is false.
Option 2: \(u_x\left(\frac{1}{2 e}, \frac{1}{2 e}\right) = 0\)
We found that \(u_x = 0\) for all \((x, y)\) in the domain. Therefore, \(u_x\left(\frac{1}{2 e}, \frac{1}{2 e}\right) = 0\). Thus, statement 2 is true.
Option 3: \(u_x \left( \frac{1}{4e}, \frac{1}{4e}\right) = \log 4\)
We found that \(u_x = 0\) for all \((x, y)\) in the domain. Therefore, \(u_x \left( \frac{1}{4e}, \frac{1}{4e}\right) = 0\).
Since \(\log 4 = 2\log 2 \ne 0\), \(u_x \left( \frac{1}{4e}, \frac{1}{4e}\right) \ne \log 4\). Thus, statement 3 is false.
Option 4: \(u_y\left(0, \frac{1}{4 e}\right) = \frac{4 e}{3}\)
Substitute \(y = \frac{1}{4e}\) into the expression for \(u_y = \frac{1}{e^{-1} - y}\):
\(\qquad u_y\left(0, \frac{1}{4 e}\right) = \frac{1}{e^{-1} - \frac{1}{4e}} = \frac{1}{\frac{1}{e} - \frac{1}{4e}} = \frac{1}{\frac{4-1}{4e}} = \frac{1}{\frac{3}{4e}} = \frac{4e}{3}\)
Thus, statement 4 is true.
Summary
Based on the verification of each statement:
- Statement 1: False
- Statement 2: True
- Statement 3: False
- Statement 4: True
Important Questions from Lagrange and Charpit Methods
The general solution of the equation
\(x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0\) is
Which of the following is a solution to ux + x2uy = 0 with u(x, 0) = ex?
Let u ∶ ℝ2 → ℝ be the solution to the Cauchy problem:
\(\left\{\begin{aligned} \rm \partial_x u+2 \partial_y u & =0 & & \text { for } \rm (x, y) \in ℝ ^2, \\ \rm u(x, y) & =\rm \sin (x) & & \text { for } \rm y=3 x+1, x \in ℝ.\end{aligned}\right.\)
Let v ∶ ℝ2 → ℝ be the solution to the Cauchy problem:
\(\left\{\begin{aligned} \rm \partial_x v+2 \partial_y v & =0 & & \text { for } \rm (x, y) \in ℝ ^2, \\ \rm v(x, 0) & =\rm\sin (x) & & \text { for } \rm x \in ℝ.\end{aligned}\right.\)
Let δ = [0, 1] × [0, 1].
Which of the following statements are true?
Let u(x, y) be the solution of the Cauchy problem
uux + uy = 0, x ∈ ℝ, y > 0,
u(x, 0) = x, x ∈ ℝ.
Which of the following is the value of u(2, 3)?
Consider the partial differential equation (PDE)
\(x\left(\frac{\partial u}{\partial x}\right)^2+y\left(\frac{\partial u}{\partial y}\right)^2+(x+y) \frac{\partial u}{\partial x} \frac{\partial u}{\partial y}-u\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\right)+1=0\)
Which of the following statements are true?
