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?
Cauchy Problem Analysis
We are given two Cauchy problems for the same first-order linear partial differential equation (PDE):
\(\partial_x w + 2 \partial_y w = 0\)
This is a homogeneous first-order linear PDE. The characteristic curves are given by the equations:
\(\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 2\)
Integrating these equations, we get \(x = t + c_1\) and \(y = 2t + c_2\), where \(c_1\) and \(c_2\) are constants. Eliminating \(t\), we find the equation for the characteristic curves:
\(y - 2x = c_2 - 2c_1 = c\)
where \(c\) is a constant along each characteristic curve. The general solution to the PDE is of the form \(w(x, y) = f(y - 2x)\) for some arbitrary function \(f\).
Solution for u
The first Cauchy problem is for \(u(x, y)\) with the initial condition \(u(x, y) = \sin(x)\) on the line \(y = 3x + 1\). Using the general solution form \(u(x, y) = f_u(y - 2x)\), we apply the initial condition:
\(u(x, 3x + 1) = f_u((3x + 1) - 2x) = f_u(x + 1)\)
From the initial condition, \(u(x, 3x + 1) = \sin(x)\). So, we have:
\(f_u(x + 1) = \sin(x)\)
Let \(z = x + 1\). Then \(x = z - 1\). Substituting this into the equation for \(f_u\):
\(f_u(z) = \sin(z - 1)\)
Replacing \(z\) with the characteristic variable \(y - 2x\), the solution for \(u(x, y)\) is:
\(u(x, y) = \sin((y - 2x) - 1) = \sin(y - 2x - 1)\)
Solution for v
The second Cauchy problem is for \(v(x, y)\) with the initial condition \(v(x, 0) = \sin(x)\) on the line \(y = 0\). Using the general solution form \(v(x, y) = f_v(y - 2x)\), we apply the initial condition:
\(v(x, 0) = f_v(0 - 2x) = f_v(-2x)\)
From the initial condition, \(v(x, 0) = \sin(x)\). So, we have:
\(f_v(-2x) = \sin(x)\)
Let \(z = -2x\). Then \(x = -z/2\). Substituting this into the equation for \(f_v\):
\(f_v(z) = \sin(-z/2) = -\sin(z/2)\)
Replacing \(z\) with the characteristic variable \(y - 2x\), the solution for \(v(x, y)\) is:
\(v(x, y) = -\sin((y - 2x)/2)\)
Statement Analysis
Let \(\delta = [0, 1] \times [0, 1]\). We analyze the given statements:
u changes sign in interior of \(\delta\)
The interior of \(\delta\) is \((0, 1) \times (0, 1)\). The solution is \(u(x, y) = \sin(y - 2x - 1)\). Let \(Z_u = y - 2x - 1\). For \((x, y) \in (0, 1) \times (0, 1)\):
- The minimum value of \(Z_u\) approaches \(0 - 2(1) - 1 = -3\).
- The maximum value of \(Z_u\) approaches \(1 - 2(0) - 1 = 0\).
So, for \((x, y) \in (0, 1) \times (0, 1)\), the range of \(Z_u\) is \((-3, 0)\). The sine function \(\sin(z)\) is negative for \(z \in (-\pi, 0)\). Since \((-3, 0) \subset (-\pi, 0)\) (because \(-\pi \approx -3.14\)), \(\sin(Z_u)\) is negative for all \((x, y) \in (0, 1) \times (0, 1)\). Thus, \(u(x, y)\) is negative in the interior of \(\delta\) and does not change sign. Statement 1 is false.
v changes sign in interior of \(\delta\)
The interior of \(\delta\) is \((0, 1) \times (0, 1)\). The solution is \(v(x, y) = -\sin((y - 2x)/2)\). Let \(Z_v = (y - 2x)/2\). For \((x, y) \in (0, 1) \times (0, 1)\):
- The minimum value of \(Z_v\) approaches \((0 - 2(1))/2 = -1\).
- The maximum value of \(Z_v\) approaches \((1 - 2(0))/2 = 0.5\).
So, for \((x, y) \in (0, 1) \times (0, 1)\), the range of \(Z_v\) is \((-1, 0.5)\). The sine function \(\sin(z)\) changes sign in \((-1, 0.5)\) (e.g., \(\sin(-0.5) < 0\), \(\sin(0.5) > 0\), and \(\sin(0) = 0\)). Since \(v(x, y) = -\sin(Z_v)\), \(v\) will be positive when \(\sin(Z_v)\) is negative, negative when \(\sin(Z_v)\) is positive, and zero when \(\sin(Z_v)\) is zero. Therefore, \(v(x, y)\) changes sign in the interior of \(\delta\). Statement 3 is true.
v vanishes along a line in \(\delta\)
The solution is \(v(x, y) = -\sin((y - 2x)/2)\). \(v(x, y)\) vanishes when \(\sin((y - 2x)/2) = 0\). This occurs when \((y - 2x)/2 = k\pi\) for some integer \(k\). This simplifies to \(y - 2x = 2k\pi\). This is the equation of a line with slope 2.
For \(k = 0\), the line is \(y - 2x = 0\), or \(y = 2x\). We check if this line intersects the interior of \(\delta\), which is \((0, 1) \times (0, 1)\). The line \(y=2x\) passes through the origin \((0,0)\) and \((0.5, 1)\). For \(x \in (0, 0.5)\), the points \((x, 2x)\) lie in \((0, 0.5) \times (0, 1)\), which is contained within \((0, 1) \times (0, 1)\). For example, the point \((0.1, 0.2)\) is in the interior of \(\delta\), and \(v(0.1, 0.2) = -\sin((0.2 - 2(0.1))/2) = -\sin(0/2) = -\sin(0) = 0\). Therefore, \(v\) vanishes along the line \(y = 2x\), which intersects the interior of \(\delta\). Statement 4 is true.
u(x, y) = v(x, y) along a line in \(\delta\)
We need to find if there is a line where \(u(x, y) = v(x, y)\). The line must be a characteristic line, where \(y - 2x = c\) for some constant \(c\). On such a line, the condition becomes:
\(\sin(c - 1) = -\sin(c/2)\)
\(\sin(c - 1) + \sin(c/2) = 0\)
Using the sum-to-product formula, \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\), we get:
\(2 \sin\left(\frac{(c - 1) + c/2}{2}\right) \cos\left(\frac{(c - 1) - c/2}{2}\right) = 0\)
\(2 \sin\left(\frac{3c/2 - 1}{2}\right) \cos\left(\frac{c/2 - 1}{2}\right) = 0\)
\(2 \sin\left(\frac{3c - 2}{4}\right) \cos\left(\frac{c - 2}{4}\right) = 0\)
This equation holds if \(\sin\left(\frac{3c - 2}{4}\right) = 0\) or \(\cos\left(\frac{c - 2}{4}\right) = 0\).
- Case 1: \(\sin\left(\frac{3c - 2}{4}\right) = 0\). This means \(\frac{3c - 2}{4} = k\pi\) for some integer \(k\). \(3c - 2 = 4k\pi\), so \(c = \frac{2 + 4k\pi}{3}\).
- Case 2: \(\cos\left(\frac{c - 2}{4}\right) = 0\). This means \(\frac{c - 2}{4} = (k + 1/2)\pi\) for some integer \(k\). \(c - 2 = (4k + 2)\pi\), so \(c = 2 + (4k + 2)\pi\).
The range of \(y - 2x\) for \((x, y) \in (0, 1) \times (0, 1)\) is \((0 - 2(1), 1 - 2(0)) = (-2, 1)\). We need to find a value of \(c\) in \((-2, 1)\) from the cases above.
From Case 1: If \(k = 0\), \(c = \frac{2}{3}\). \(2/3 \in (-2, 1)\). The line is \(y - 2x = 2/3\), or \(y = 2x + 2/3\). This line intersects the interior of \(\delta\). For example, when \(x = 0.1\), \(y = 0.2 + 2/3 \approx 0.867\). The point \((0.1, 0.867)\) is in \((0, 1) \times (0, 1)\). Thus, \(u(x, y) = v(x, y)\) along the line \(y = 2x + 2/3\), which intersects the interior of \(\delta\). Statement 2 is true.
Conclusion
Based on the analysis:
- Statement 1: u changes sign in the interior of \(\delta\). (False)
- Statement 2: u(x, y) = v(x, y) along a line in \(\delta\). (True)
- Statement 3: v changes sign in the interior of \(\delta\). (True)
- Statement 4: v vanishes along a line in \(\delta\). (True)
Statements 2, 3, and 4 are 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(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)?
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?
