Partial Differential Equations/Answers to the exercises

Exercise 1
The general ordinary differential equation is given by
 * $$\forall x \in B : F(x, u(x), \overbrace{u'(x), u''(x), \ldots}^\text{arbitrarily but finitely high derivatives}) = 0$$

for a set $$B \subseteq \mathbb R$$. Noticing that
 * $$u'(x) = \partial_x u(x), u''(x) = \partial_x^2 u(x), \ldots$$

, we observe that the general ordinary differential equation is just the general one-dimensional partial differential equation.

Exercise 2
Using the one-dimensional chain rule, we directly calculate
 * $$\partial_t u(t, x) = \partial_t g(x + ct) = c g'(x + ct)$$

and
 * $$\partial_x u(t, x) = \partial_x g(x + ct) = g'(x + ct)$$

Therefore,
 * $$\partial_t u(t, x) - c \partial_x u(t, x) = c g'(x + ct) - c g'(x +ct) = 0$$

Exercise 3
By choosing
 * $$h(t, x, z, p, q) = p - cq$$

, we see that in our function $$h$$ first order derivatives suffice to depict the partial differential equation. On the other hand, if $$h$$ needs no derivatives as arguments, we have due to theorem 1.4 (which you may have just proven in exercise 2) that for all continuously differentiable functions $$g : \mathbb R \to \mathbb R$$
 * $$\forall (t, x) \in \mathbb R^2 : h(t, x, g(x + ct)) = 0$$

Since we can choose $$g$$ as a constant function with an arbitrary real value, $$h$$ does not depend on $$z$$, since otherwise it would be nonzero somewhere for some constant function $$g$$, as dependence on $$z$$ means that a different $$z$$ changes the value at least in one point. Therefore, the one-dimensional homogenous transport equation would be given by
 * $$h(x, t) = 0$$

and there would be many more solutions to the initial value problem of the homogenous one-dimensional transport equation than those given by theorem and definition 1.5.

Exercise 1
Let $$n \in \{1, \ldots, d\}$$ and $$(t, x) \in \mathbb R \times \mathbb R^d$$ be arbitrary. We choose $$B = [0, t]$$ and $$O = (-R, R)$$ for an arbitrary $$R > |x_n|$$ and apply Leibniz' integral rule. We first check that all the three conditions for Leibniz' integral rule are satisfied.

1.

Since $$f \in \mathcal C^1(\mathbb R \times \mathbb R^d)$$, $$f$$ is continuous in all variables (so therefore in particular in the first), and the same is true for the composition of $$f$$ with any continuous function. Thus, as all continuous functions of one variable are integrable on an interval,
 * $$\int_0^t f(s, y + \mathbf v (t - s)) ds$$

exists for all $$y = (y_1, \ldots, y_n, \ldots, y_d)$$ such that $$y_n \in O$$ (in fact, it exists everywhere, but this is the requirement of Leibniz' rule in our situation).

2.

$$f$$ was supposed to be in $$\mathcal C^1(\mathbb R^2)$$, which is why
 * $$\frac{d}{dy_n} f(s, y + \mathbf v (t - s)) \overset{\text{chain rule}}{=} \partial_{x_n} f(s, y + \mathbf v (t - s))$$

exists for all $$s \in B$$ and all $$y = (y_1, \ldots, y_n, \ldots, y_d)$$ such that $$y_n \in O$$ (in fact, it exists everywhere, but this is the requirement of Leibniz' rule in our situation).

3.

We note that $$O = (-R, R) \subset [-R, R]$$. Therefore, also $$O \times B \subset [-R, R] \times B =: K$$, where $$K$$ is compact. Furthermore, as $$\partial_{x_n} f$$ was supposed to be continuous (by definition of $$\mathcal C^1(\mathbb R \times \mathbb R^d)$$ and
 * $$\frac{d}{dy_n} f(s, y + \mathbf v (t - s)) \overset{\text{chain rule}}{=} \partial_{x_n} f(s, y + \mathbf v (t - s))$$

is continuous as well as composition of continuous functions, also the function $$\partial_{x_n} f(s, y + \mathbf v (t - s))$$ is continuous in $$s$$ and $$y_n$$. Thus, due to the extreme value theorem, it is bounded for $$(s, y_n) \in K$$, i. e. there is a $$b \in \mathbb R$$, $$b > 0$$ such that for all
 * $$\left| \partial_{x_n} f(s, y + \mathbf v (t - s)) \right| < b$$

for all $$y = (y_1, \ldots, y_n, \ldots, y_d)$$ such that $$y_n \in O$$, provided that $$y_1, \ldots, y_{n-1}, y_{n+1}, \ldots, y_d$$ are fixed. Therefore, we might choose $$g(s) = b$$ and obtain that
 * $$\forall \left| \partial_{x_n} f(s, y + \mathbf v (t - s)) \right| < |g(s)| \text{ and } \int_B |g(s)| ds = tb < \infty$$

Now we have checked all three ingredients for Leibniz' integral rule and thus obtain by it:
 * $$\partial_{x_n} \int_0^t f(s, y + \mathbf v(t - s)) ds = \int_0^t \partial_{x_n} f(s, y + \mathbf v(t - s)) ds$$

for all $$y = (y_1, \ldots, y_n, \ldots, y_d)$$ such that $$y_n \in O$$. Setting $$y = x$$ gives the result for $$x$$.

Since $$n \in \{1, \ldots, d\}$$ and $$(t, x) \in \mathbb R \times \mathbb R^d$$ were arbitrary, this completes the exercise.

Exercise 2

 * $$\begin{align}

\partial_t g(x + t \mathbf v) &= \begin{pmatrix}\partial_{x_1} g (x + t \mathbf v) & \cdots & \partial_{x_d} g (x + t \mathbf v) \end{pmatrix} \mathbf v & \text{by the chain rule} \\ &= \mathbf v \cdot \nabla_x g (x + t \mathbf v) & \end{align}$$