# 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

$\partial _{t}u(t,x)=\partial _{t}g(x+ct)=cg'(x+ct)$

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|

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