Which of the following is a solution to ux + x2uy = 0 with u(x, 0) = ex?
Solving the PDE ux + x²uy = 0 using the Method of Characteristics
We are given a first-order linear partial differential equation (PDE):
\(u_x + x^2 u_y = 0\)
with the initial condition:
\(u(x, 0) = e^x\)
This PDE is in the form \(a(x, y) u_x + b(x, y) u_y = c(x, y)\), where \(a(x, y) = 1\), \(b(x, y) = x^2\), and \(c(x, y) = 0\). We can solve this using the method of characteristics.
Method of Characteristics
The characteristic equations are given by:
\(\frac{dx}{a} = \frac{dy}{b} = \frac{du}{c}\)
Substituting the coefficients from our PDE, we get:
\(\frac{dx}{1} = \frac{dy}{x^2} = \frac{du}{0}\)
Solving the Characteristic Equations
From the last part, \(\frac{du}{0}\), we have \(du = 0\). Integrating this gives us the first characteristic constant:
\(u = C_1\)
From the first two parts, \(\frac{dx}{1} = \frac{dy}{x^2}\), we can separate variables and integrate:
\(x^2 dx = dy\)
\(\int x^2 dx = \int dy\)
\(\frac{x^3}{3} = y + C_2\)
Rearranging to find the second characteristic constant \(C_2\):
\(C_2 = y - \frac{x^3}{3}\)
General Solution
The general solution of the PDE is of the form \(u(x, y) = F(C_2)\), where \(F\) is an arbitrary function. Substituting the expression for \(C_2\):
\(u(x, y) = F\left(y - \frac{x^3}{3}\right)\)
Applying the Initial Condition
We are given the initial condition \(u(x, 0) = e^x\). We substitute \(y = 0\) into the general solution:
\(u(x, 0) = F\left(0 - \frac{x^3}{3}\right) = F\left(-\frac{x^3}{3}\right)\)
Equating this to the given initial condition:
\(F\left(-\frac{x^3}{3}\right) = e^x\)
Determining the Function F
To find the form of the function \(F\), let \(v = -\frac{x^3}{3}\). Then \(x^3 = -3v\), which means \(x = (-3v)^{1/3}\).
Substitute \(x\) in terms of \(v\) into the equation \(F\left(-\frac{x^3}{3}\right) = e^x\):
\(F(v) = e^{(-3v)^{1/3}}\)
Now, replace \(v\) with \(y - \frac{x^3}{3}\) to get the solution \(u(x, y)\):
\(u(x, y) = F\left(y - \frac{x^3}{3}\right) = e^{\left(-3\left(y - \frac{x^3}{3}\right)\right)^{1/3}}\)
\(u(x, y) = e^{\left(-3y + 3\frac{x^3}{3}\right)^{1/3}}\)
\(u(x, y) = e^{\left(x^3 - 3y\right)^{1/3}}\)
Comparing with Options
Let's compare our derived solution with the given options:
- \(e^x\) - Incorrect.
- \(e^{\left(x^3+y\right)^{1/3}}\) - Incorrect.
- \(e^{\left(x^3-3y\right)^{1/3}}\) - Matches our derived solution.
- \((x^2 y + 1)e^x\) - Incorrect.
The solution that satisfies the PDE and the initial condition is \(u(x, y) = e^{\left(x^3 - 3y\right)^{1/3}}\).
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
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?
