# Partial Differential Equations/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
    $\partial _{t}u(t,x)=\partial _{t}g(x+ct)=cg'(x+ct)$
and
    $\partial _{x}u(t,x)=\partial _{x}g(x+ct)=g'(x+ct)$
Therefore,
    $\partial _{t}u(t,x)-c\partial _{x}u(t,x)=cg'(x+ct)-cg'(x+ct)=0$

### 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

    ${\begin{aligned}\partial _{t}g(x+t\mathbf {v} )&={\begin{pmatrix}\partial _{x_{1}}g(x+t\mathbf {v} )&\cdots &\partial _{x_{d}}g(x+t\mathbf {v} )\end{pmatrix}}\mathbf {v} &{\text{by the chain rule}}\\&=\mathbf {v} \cdot \nabla _{x}g(x+t\mathbf {v} )&\end{aligned}}$