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)?
1/2
This problem asks us to find the value of the solution \(u(x, y)\) at a specific point \((2, 3)\), given a first-order partial differential equation (PDE) and an initial condition. The given PDE is a quasilinear equation.
Cauchy Problem Analysis
The given Cauchy problem is:
\(u u_x + u_y = 0\), for \(x \in \reals, y > 0\)
\(u(x, 0) = x\), for \(x \in \reals\)
This is a first-order quasilinear PDE. We can solve this type of PDE using the method of characteristics.
Method of Characteristics
For a quasilinear PDE of the form \(a(x, y, u) u_x + b(x, y, u) u_y = c(x, y, u)\), the characteristic equations are given by:
- \(\frac{dx}{dt} = a(x, y, u)\)
- \(\frac{dy}{dt} = b(x, y, u)\)
- \(\frac{du}{dt} = c(x, y, u)\)
In our problem, the PDE is \(u u_x + 1 u_y = 0\). Comparing this with the general form, we have \(a = u\), \(b = 1\), and \(c = 0\).
So, the characteristic equations are:
- \(\frac{dx}{dt} = u\)
- \(\frac{dy}{dt} = 1\)
- \(\frac{du}{dt} = 0\)
Solving Characteristic Equations
Let's solve these ordinary differential equations:
1. From \(\frac{dy}{dt} = 1\), integrating with respect to \(t\) gives \(y(t) = t + C_1\). We can set \(t=0\) at the initial curve \(y=0\), so \(y(0) = 0 = 0 + C_1\), which means \(C_1 = 0\). Thus, \(y(t) = t\).
2. From \(\frac{du}{dt} = 0\), integrating with respect to \(t\) gives \(u(t) = C_2\). This means that \(u\) is constant along each characteristic curve.
3. From \(\frac{dx}{dt} = u\), since \(u\) is constant along a characteristic (\(u = C_2\)), we have \(\frac{dx}{dt} = C_2\). Integrating with respect to \(t\) gives \(x(t) = C_2 t + C_3\). At \(t=0\), let the initial point on the curve \(y=0\) be \(x_0\). So \(x(0) = x_0 = C_2(0) + C_3\), which means \(C_3 = x_0\). Thus, \(x(t) = C_2 t + x_0\).
Using Initial Condition
The initial condition is \(u(x, 0) = x\). This means that at \(y=0\) (which corresponds to \(t=0\)), the value of \(u\) is equal to the initial \(x\)-coordinate, \(x_0\). Since \(u\) is constant along a characteristic and \(u(0) = x_0\), we have \(u = x_0\) for all \(t\) along that characteristic. Therefore, the constant \(C_2\) is equal to \(x_0\).
Finding the Solution u(x, y)
Substitute \(u = x_0\) (i.e., \(C_2 = x_0\)) and \(t = y\) back into the equation for \(x(t)\):
\(x = x_0 y + x_0\)
\(x = x_0(y + 1)\)
Since \(u = x_0\), we can replace \(x_0\) with \(u\):
\(x = u(y + 1)\)
Now, solve for \(u\):
\(u(x, y) = \frac{x}{y + 1}\)
This is the solution to the Cauchy problem.
Evaluating u(2, 3)
We need to find the value of \(u(x, y)\) at the point \((2, 3)\). Substitute \(x = 2\) and \(y = 3\) into the solution we found:
\(u(2, 3) = \frac{2}{3 + 1}\)
\(u(2, 3) = \frac{2}{4}\)
\(u(2, 3) = \frac{1}{2}\)
The value of \(u(2, 3)\) is \(1/2\).
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 = 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?
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?
