The general solution of the equation \(x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0\) is
The given partial differential equation (PDE) is:
\(x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0\)
This is a first-order linear homogeneous partial differential equation of the form \(P \frac{\partial z}{\partial x} + Q \frac{\partial z}{\partial y} = R\), where \(P=x\), \(Q=y\), and \(R=0\).
Method of Characteristics
To find the general solution of this type of PDE, we use the method of characteristics. The characteristic equations are given by:
\(\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}\)
Substituting the values of \(P\), \(Q\), and \(R\) from the given equation, we get:
\(\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{0}\)
Solving Characteristic Equations
We can obtain independent first integrals (characteristics) from these equations.
Consider the first pair of equations:
\(\frac{dx}{x} = \frac{dy}{y}\)
Integrating both sides:
\(\int \frac{dx}{x} = \int \frac{dy}{y}\)
\(\ln|x| = \ln|y| + \ln C_1\)
\(\ln|x| - \ln|y| = \ln C_1\)
\(\ln\left|\frac{x}{y}\right| = \ln C_1\)
\(\left|\frac{x}{y}\right| = C_1\)
where \(C_1\) is an arbitrary positive constant. Let \(u = \left|\frac{x}{y}\right|\).
Now consider the equation involving \(dz\):
\(\frac{dz}{0}\)
This implies \(dz = 0\). Integrating this gives:
\(\int dz = \int 0\)
\(z = C_2\)
where \(C_2\) is another arbitrary constant.
General Solution Form
For a homogeneous PDE (\(R=0\)), the general solution is given by expressing one characteristic constant as an arbitrary function of the other characteristic constants. In this case, the second constant \(z\) is a function of the first constant \(u\).
So, the general solution is of the form:
\(z = \phi(u)\)
Substituting the expression for \(u\), we get:
\(z = \phi\left(\left|\frac{x}{y}\right|\right)\)
This can also be written as \(z = \phi\left(\frac{|x|}{|y|}\right)\).
The problem statement specifies that \(\phi \in C^1(\mathbb{R})\), which means \(\phi\) is a continuously differentiable function of its argument.
Comparing with Options
Let's compare our derived general solution with the given options:
- Option 1: \(z=\phi\left(\frac{|x|}{|y|}\right), \phi \in C^1(\mathbb{R})\)
- Option 2: \(z=\varphi\left(\frac{x-1}{y}\right), ϕ ∈ C^1(\mathbb{R})\)
- Option 3: \(z=\phi\left(\frac{x+1}{y}\right), \phi \in C^1(\mathbb{R})\)
- Option 4: \(z = ϕ(|x| + |y|), ϕ ∈ C^1(\mathbb{R})\)
Our derived general solution matches Option 1.
Therefore, the general solution of the equation \(x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0\) is \(z=\phi\left(\frac{|x|}{|y|}\right)\), where \(\phi\) is an arbitrary continuously differentiable function.
Important Questions from Lagrange and Charpit Methods
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)?
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?
