# 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

${\displaystyle \forall x\in B:F(x,u(x),\overbrace {u'(x),u''(x),\ldots } ^{\text{arbitrarily but finitely high derivatives}})=0}$

for a set ${\displaystyle B\subseteq \mathbb {R} }$ . Noticing that

${\displaystyle u'(x)=\partial _{x}u(x),u''(x)=\partial _{x}^{2}u(x),\ldots }$

, 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

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

and

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

Therefore,

${\displaystyle \partial _{t}u(t,x)-c\partial _{x}u(t,x)=cg'(x+ct)-cg'(x+ct)=0}$

### Exercise 3

By choosing

${\displaystyle h(t,x,z,p,q)=p-cq}$

, we see that in our function ${\displaystyle h}$  first order derivatives suffice to depict the partial differential equation. On the other hand, if ${\displaystyle h}$  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 ${\displaystyle g:\mathbb {R} \to \mathbb {R} }$

${\displaystyle \forall (t,x)\in \mathbb {R} ^{2}:h(t,x,g(x+ct))=0}$

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

${\displaystyle h(x,t)=0}$

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

1.

Since ${\displaystyle f\in {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})}$ , ${\displaystyle f}$  is continuous in all variables (so therefore in particular in the first), and the same is true for the composition of ${\displaystyle f}$  with any continuous function. Thus, as all continuous functions of one variable are integrable on an interval,

${\displaystyle \int _{0}^{t}f(s,y+\mathbf {v} (t-s))ds}$

exists for all ${\displaystyle y=(y_{1},\ldots ,y_{n},\ldots ,y_{d})}$  such that ${\displaystyle y_{n}\in O}$  (in fact, it exists everywhere, but this is the requirement of Leibniz' rule in our situation).

2.

${\displaystyle f}$  was supposed to be in ${\displaystyle {\mathcal {C}}^{1}(\mathbb {R} ^{2})}$ , which is why

${\displaystyle {\frac {d}{dy_{n}}}f(s,y+\mathbf {v} (t-s)){\overset {\text{chain rule}}{=}}\partial _{x_{n}}f(s,y+\mathbf {v} (t-s))}$

exists for all ${\displaystyle s\in B}$  and all ${\displaystyle y=(y_{1},\ldots ,y_{n},\ldots ,y_{d})}$  such that ${\displaystyle y_{n}\in O}$  (in fact, it exists everywhere, but this is the requirement of Leibniz' rule in our situation).

3.

We note that ${\displaystyle O=(-R,R)\subset [-R,R]}$ . Therefore, also ${\displaystyle O\times B\subset [-R,R]\times B=:K}$ , where ${\displaystyle K}$  is compact. Furthermore, as ${\displaystyle \partial _{x_{n}}f}$  was supposed to be continuous (by definition of ${\displaystyle {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})}$  and

${\displaystyle {\frac {d}{dy_{n}}}f(s,y+\mathbf {v} (t-s)){\overset {\text{chain rule}}{=}}\partial _{x_{n}}f(s,y+\mathbf {v} (t-s))}$

is continuous as well as composition of continuous functions, also the function ${\displaystyle \partial _{x_{n}}f(s,y+\mathbf {v} (t-s))}$  is continuous in ${\displaystyle s}$  and ${\displaystyle y_{n}}$ . Thus, due to the extreme value theorem, it is bounded for ${\displaystyle (s,y_{n})\in K}$ , i. e. there is a ${\displaystyle b\in \mathbb {R} }$ , ${\displaystyle b>0}$  such that for all

${\displaystyle \left|\partial _{x_{n}}f(s,y+\mathbf {v} (t-s))\right|

for all ${\displaystyle y=(y_{1},\ldots ,y_{n},\ldots ,y_{d})}$  such that ${\displaystyle y_{n}\in O}$ , provided that ${\displaystyle y_{1},\ldots ,y_{n-1},y_{n+1},\ldots ,y_{d}}$  are fixed. Therefore, we might choose ${\displaystyle g(s)=b}$  and obtain that

${\displaystyle \forall \left|\partial _{x_{n}}f(s,y+\mathbf {v} (t-s))\right|<|g(s)|{\text{ and }}\int _{B}|g(s)|ds=tb<\infty }$

Now we have checked all three ingredients for Leibniz' integral rule and thus obtain by it:

${\displaystyle \partial _{x_{n}}\int _{0}^{t}f(s,y+\mathbf {v} (t-s))ds=\int _{0}^{t}\partial _{x_{n}}f(s,y+\mathbf {v} (t-s))ds}$

for all ${\displaystyle y=(y_{1},\ldots ,y_{n},\ldots ,y_{d})}$  such that ${\displaystyle y_{n}\in O}$ . Setting ${\displaystyle y=x}$  gives the result for ${\displaystyle x}$ .

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

### Exercise 2

{\displaystyle {\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