# Partial Differential Equations/Answers to the exercises

Partial Differential Equations
 ← The Malgrange-Ehrenpreis theorem Answers to the exercises

## Chapter 1

### 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.

## Chapter 2

### 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}
Partial Differential Equations
 ← The Malgrange-Ehrenpreis theorem Answers to the exercises