Consider the Cauchy problem$u \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1$, (x, y) ∈ &reals; × (0, ∞),u(x, 0) = kx, x ∈ &reals;with a given real parameter k. For which of the following values of k does the above problem have a solution defined on R × (0, ∞)?

Question

Consider the Cauchy problem$u \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1$, (x, y) &isin; &reals; &times; (0, &infin;),u(x, 0) = kx, x &isin; &reals;with a given real parameter k. For which of the following values of k does the above problem have a solution defined on R &times; (0, &infin;)?

Prepp Team · Accepted Answer

Cauchy Problem Description

We are given a first-order quasilinear partial differential equation (PDE):

$u \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1$

This PDE is defined for $(x, y) \in \mathbb{R} \times (0, \infty)$. We are also given an initial condition on the line $y=0$:

$u(x, 0) = kx$

where $k$ is a real parameter. We need to find for which values of $k$ from the given options the solution to this Cauchy problem exists and is defined on the entire domain $\mathbb{R} \times (0, \infty)$.

Characteristics Method

To solve this non-linear first-order PDE, we can use the method of characteristics. The characteristic equations are given by:

$\frac{dx}{dt} = u$
$\frac{dy}{dt} = 1$
$\frac{du}{dt} = 1$

Characteristic Equations Solution

We can solve these ordinary differential equations. From the second equation, integrating with respect to $t$, we get $y = t + C_1$. Let's set the initial time $t=0$ on the initial curve $y=0$. So, if $y(0) = 0$, then $C_1 = 0$, which gives $y(t) = t$.

From the third equation, integrating with respect to $t$, we get $u = t + C_2$. At $t=0$, let the initial value of $u$ be $u_0 = u(x(0), y(0))$. So, $u_0 = 0 + C_2$, which gives $C_2 = u_0$. Thus, $u(t) = t + u_0$.

From the first equation, we have $\frac{dx}{dt} = u$. Substituting $u = t + u_0$, we get $\frac{dx}{dt} = t + u_0$. Integrating with respect to $t$, we get $x = \frac{t^2}{2} + u_0 t + C_3$. At $t=0$, let the initial position be $x(0) = x_0$. So, $x_0 = 0 + 0 + C_3$, which gives $C_3 = x_0$. Thus, $x(t) = x_0 + u_0 t + \frac{t^2}{2}$.

Initial Condition Application

The initial condition is given on the curve $y=0$. Let's parameterize the initial curve $y=0$ by $s$, so $x_0 = s$ and $y_0 = 0$. The initial condition is $u(x, 0) = kx$. So, at $t=0$ on the initial curve $y=0$, we have $u_0 = u(x_0, y_0) = u(s, 0) = ks$.

Now, substitute $t=y$, $x_0=s$, and $u_0=ks$ into the characteristic equations:

$y = t$
$u = t + u_0 = y + ks$
$x = x_0 + u_0 t + \frac{t^2}{2} = s + ks y + \frac{y^2}{2}$

We now have a system of equations relating $x, y, u$ to the initial parameter $s$ and the parameter along the characteristics $y$:

$x = s + ks y + \frac{y^2}{2}$
$u = y + ks$

Solution Derivation

From the second equation, we can express $ks$ in terms of $u$ and $y$: $ks = u - y$.

Substitute this into the first equation:

$x = s + (u - y) y + \frac{y^2}{2}$

$x = s + uy - y^2 + \frac{y^2}{2}$

$x = s + uy - \frac{y^2}{2}$

To find the solution $u(x, y)$, we need to express $s$ in terms of $x$ and $y$ from the equation for $x$ and substitute it back into the equation for $u$.

From $x = s + ks y + \frac{y^2}{2}$, we can rearrange to solve for $s$:

$x - \frac{y^2}{2} = s(1 + ky)$

Provided that $1 + ky \neq 0$, we can solve for $s$:

$s = \frac{x - y^2/2}{1 + ky}$

Substitute this expression for $s$ back into $u = y + ks$:

$u = y + k \left( \frac{x - y^2/2}{1 + ky} \right)$

$u = \frac{y(1 + ky) + k(x - y^2/2)}{1 + ky}$

$u = \frac{y + ky^2 + kx - ky^2/2}{1 + ky}$

$u(x, y) = \frac{y + kx + ky^2/2}{1 + ky}$

Existence Domain Analysis

The solution $u(x, y)$ is defined as long as the denominator $1 + ky$ is not zero. The domain of interest is $(x, y) \in \mathbb{R} \times (0, \infty)$, which means $y > 0$.

The denominator is zero if $1 + ky = 0$.

If $k = 0$, the denominator is $1 + 0 \cdot y = 1$, which is never zero. In this case, the PDE becomes $\frac{\partial u}{\partial y} = 1$ with initial condition $u(x, 0) = 0$. Integrating the PDE gives $u(x, y) = y + C(x)$. Applying the initial condition gives $u(x, 0) = 0 + C(x) = 0$, so $C(x) = 0$. The solution is $u(x, y) = y$, which is defined on the entire domain $\mathbb{R} \times (0, \infty)$.

If $k \neq 0$, the denominator is zero when $y = -\frac{1}{k}$. For the solution to be defined on the entire domain $\mathbb{R} \times (0, \infty)$ (where $y > 0$), the point where the denominator is zero, $y = -\frac{1}{k}$, must not be in the domain $y > 0$. This means we must have $-\frac{1}{k} \le 0$.

If $k > 0$, then $-\frac{1}{k} < 0$. This value of $y$ is not in the domain $y > 0$. So, $1 + ky \neq 0$ for all $y > 0$. The solution is defined on the entire domain.
If $k < 0$, then $-\frac{1}{k} > 0$. This value of $y$ is in the domain $y > 0$. At $y = -\frac{1}{k}$, the denominator becomes zero, and the solution blows up. This indicates that characteristics originating from different points on the initial curve intersect, leading to a breakdown of the classical solution. Thus, the solution is not defined on the entire domain $\mathbb{R} \times (0, \infty)$ if $k < 0$.

Therefore, the solution is defined on the entire domain $\mathbb{R} \times (0, \infty)$ if and only if $k \ge 0$.

Parameter k Values

We check the given options for the value of $k$ based on the condition $k \ge 0$:

$k = 0$: Since $0 \ge 0$, the solution exists on the entire domain.
$k = -2$: Since $-2 < 0$, the solution does not exist on the entire domain.
$k = 4$: Since $4 > 0$, the solution exists on the entire domain.
$k = 1$: Since $1 > 0$, the solution exists on the entire domain.

Thus, the values of $k$ for which the problem has a solution defined on $\mathbb{R} \times (0, \infty)$ are $k=0$, $k=4$, and $k=1$.

Prepp Team · Answer

k = -2