Chapter 2 #

Notation: $\Delta u = \sum_{i=1}^{n} u_{x_i x_i}$

$\nabla f(x) = \sum_{i=1}^{n} \frac{\partial f}{\partial x_i} e_i = (\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \ldots, \frac{\partial f}{\partial x_n})$

$x \in \mathbb{R}^n$

$x = (x_1, x_2, \ldots, x_n)$

First, recall that the directional derivative of a function $f$ at a point $x$ in the direction of a vector $v$ is defined as $D_v f(x) = \nabla f(x) \cdot v$. This means that the directional derivative measures how much the function $f$ changes at the point $x$ when we move in the direction of the vector $v$.

Second, Recall the chain rule for partial derivatives: if $f$ is a function of $y_1, y_2, \ldots, y_m$, and each $y_j$ is a function of $x_1, x_2, \ldots, x_n$, then the partial derivative of $f$ with respect to $x_i$ can be expressed as: $$\frac{\partial f}{\partial x_i} = \sum_{j=1}^{m} \frac{\partial f}{\partial y_j} \frac{\partial y_j}{\partial x_i}$$

Let b be given.

So, $$ u_t + b \cdot D u = 0 $$ $$ u_t + b \cdot \nabla u = 0 $$ $$ \Rightarrow \frac{\partial u}{\partial t} + b \cdot \sum_{i=1}^{n} u_{x_i} e_i = 0 $$

Note that we lose a dimension when we take the derivative with respect to $t$. The function $u$ is defined on $\mathbb{R}^n \times [0, \infty)$, but after taking the derivative with respect to $t$, we get a function defined on $\mathbb{R}^n$. This is because the derivative with respect to $t$ only depends on the values of $u$ at a fixed time, and does not depend on the values of $u$ at different times.

We have $z(s) := u(x + sb,t + s)$ where $s\in \mathbb{R}$

Now, using the chain rule, we can compute the derivative of $z(s)$ with respect to $s$: $$ \frac{d z}{ d s} = \partial u/\partial t \cdot \frac{d (t + s)}{d s} + \frac{\partial u}{\partial x} \cdot \frac{d (x + sb)}{d s} $$

$$\frac{dz}{ds} = \frac{\partial u}{\partial t} \frac{dt}{ds} + \sum_{i=1}^{n} \frac{\partial u}{\partial x_i} \frac{dx_i}{ds}$$

$$\frac{dz}{ds} = \nabla u(x + sb, t + s) \cdot b + u_t(x + sb, t + s)$$

Notice that what we have done is taken $u$ at $(x,t)$ and translated it by $s(1,b)$. Hence, along this translation, the function $u$ is constant. This means that the derivative of $z(s)$ with respect to $s$ is zero as we had from our derivation.

2.1.1 #

Okay, now we assume some things about the function $u$. That is, we know a value. I think if we know any value we know all the values, but they say we need to know the value at $t=0$. This is on the hyperplane

${t=0}$, which is the initial condition.

Call the hyperplane ${t=0}$: $\Gamma$.

Yeah, so suppose you have $(x,t)$, you can make a line ${(x + sb, t + s) : s \in \mathbb{R}}$

We have to traverse the line until we hit the hyperplane $\Gamma$. This happens when $t + s = 0$, which means $s = -t$. So, we have to traverse the line until we hit the point $(x - tb, 0)$ on the hyperplane $\Gamma$.

That is, once we have the value of $u$ at $(x - tb, 0)$, we can use the fact that $u$ is constant along the line ${(x + sb, t + s) : s \in \mathbb{R}}$ to conclude that $u(x,t) = u(x - tb, 0)$.

It is weird that we would have u = g at t=0 as an initial condition. Why not have u = g at some (x,t) or whatever? You could equally traverse the line.

Okay, well, that kind of gets at the heart of the matter. We want something defined on the whole space and knowing the value at one point is not enough. We need to know the value at the hyperplane $\Gamma$ to be able to traverse the line and get the value at any point in the space.

What I just said is not completely true. We can have any

Non-Characteristic Initial Condition which here means anything not parallel to the line ${(x + sb, t + s) : s \in \mathbb{R}}$, which means anything not parallel to the vector $(b,1)$. Sorry if I have the things flipped.

For example, if we had initial conditions on a line parallel, chose a point not on the line, and then traverse the line, we would never hit an initial condition and thus, we are out of luck.

I was told weak solutions are later.

2.1.2 NonHomogeneous Transport Equation #

$$u_t + b \cdot \nabla u = f \quad \text{in } \mathbb{R}^n \times (0, \infty)$$

$$\text{with initial condition:}$$

$$u = g \quad \text{on } \mathbb{R}^n \times {t = 0}$$

I think this all in the book pretty good.

We need to do the integral change though.

$$ \int_{-t}^{0} f(x + sb, t+s) ds $$

Let $\tau = t+ s$ so that $s = \tau - t$. This is kind of stupid, but we are going to change in terms of $\tau$, understand it was chosen as a dummy variable, and then replace $\tau$ with s.

From $\tau = t + s$, we have $d\tau = ds$.

When $s = -t$, we have $\tau = 0$. When $s = 0$, we have $\tau = t$.

We are integrating with a change of variable, so we have to change the limits of integration and the integrand. $$ \int_{-t}^{0} f(x + sb, t+s) ds = \int_{0}^{t} f(x + (\tau - t)b, \tau) d\tau $$

The limits are “when $s = -t$” and “when $s = 0$”. The rest is self explanatory.

I need help with why $ds = d\tau$. I know it is true, but I don’t understand why. TODO

We note notes from the lecture that I don’t understand.: Rmk 1: The solution has no regularizing effect. That is, u is C1 and so is $\bar{u}$. They have the same smoothness or something.

Rmk 2: We have used that $(1,b)$ is transverse to the hyperplane $\Gamma$. This is the non-characteristic condition. This comes from the 1 in (1,b). Okay, what I wrote down is from the lecture, but I don’t know why it can’t be (1,0). TODO

Rmk 3: we are transporting the equation at initial time

Rmk 4: If $\bar{u}(x-tb) = u(t,x)$ is non-smooth, then we call it a weak solution.

Rmk 5: The solution also solves the system: $X = (X_0, \dots, X_n)$ $$ \dot{X}(s) = (1,b) \in \mathbb{R}^{n+1}, \quad X(0) = (0,\bar{x}) $$ $X_0(s) = s$

$X_i(s) = \bar{x}_i + sb_i$ for $i = 1, \dots, n$

We call the system a characteristic system and the solution is called a characteristic curve.

okay, the rest is for later about 30 minutes left in the lecture.