Ross Ogilvie      16th September, 2024
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7.

Royale with Cheese

Recall Burgers’ equation from Example 1.5 of the lecture script:

$$\dot{u}+u{\partial }_{x}u=0,$$

for $u:ℝ\times ℝ\to ℝ$. In this question we will apply the method of characteristics to solve this equation for the initial condition $g(x)=2x$.

(a)

According to Theorem 1.4, there is a unique $C^1$ solution to this initial value problem, at least when $t$ is small. For how long does the theorem guarantee that the solution exists uniquely? (1 point)

(b)

Suppose that $u$ is a solution to this equation and suppose that $(x(s),t(s))$ is a path in the domain of $u$. What is the $s$ derivative of $u$ along this path? What constraints should we place on the derivatives of $x$ and $t$? (2 points)

(c)

On an $(x,t)$-plane draw the characteristics. (1 point)

(d)

Finally, derive the following solution to the initial value problem: (2 points)

$$u(x,t)=\frac{2x}{1+2t}.$$

8.

Linear Partial Differential Equations Consider a PDE of the form $F(\nabla <!--\; FUNCTION\; APPLICATION\; -->u(x),u(x),x)=0$. Suppose that $F$ is linear in the derivatives and has continuously differentiable coefficients. That is, it can be written in the form

$$F(p,z,x)=b(z,x)\cdot p+c(z,x)$$

with $b$ and $c$ continuously differentiable. Show that the characteristic curves $(x(s),z(s))$ for $z(s):=u(x(s))$ can be described by ODEs that are independent of $p(s):=\nabla <!--\; FUNCTION\; APPLICATION\; -->u(x(s))$. (4 points)

9.

Solving PDEs Solve the initial value problems of the following PDEs using the method of characteristics. You may assume that $g$ is continuously differentiable on the corresponding domain.

(a)

$x_2{\partial }_{1}u-x_1{\partial }_{2}u=u$ on the domain $x_1,x_2>0$, with initial condition $u(x_1,0)=g(x_1)$.
(3 points)

(b)

$x_1{\partial }_{1}u+3x_2{\partial }_{2}u+{\partial }_{3}u=2u$ on $x_1,x_2\in ℝ$, $x_3>0$, with initial condition $u(x_1,x_2,0)=g(x_1,x_2)$.
(3 points)

(c)

$u{\partial }_{1}u+{\partial }_{2}u=1$ on the domain $x_1,x_2>0$, with initial condition $u(x_1,x_1)=\frac{1}{2}{x}_1$.
(4 points)