Partial Differential Equations/Answers to the exercises

Chapter 1

Exercise 1

The general ordinary differential equation is given by

$\mathrm{\forall }x\in B:F(x,u(x),\stackrel{\text{arbitrarily but finitely high derivatives}}{\overbrace{u^{\prime }(x),u^{″}(x),\dots }})=0$

for a set $B\subseteq ℝ$. Noticing that

$u^{\prime }(x)={\partial }_{x}u(x),u^{″}(x)=\underset{x}{\overset{2}{\partial }}u(x),\dots$

, 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

${\partial }_{t}u(t,x)={\partial }_{t}g(x+ct)=c{g}^{\prime }(x+ct)$

and

${\partial }_{x}u(t,x)={\partial }_{x}g(x+ct)=g^{\prime }(x+ct)$

Therefore,

${\partial }_{t}u(t,x)-c{\partial }_{x}u(t,x)=c{g}^{\prime }(x+ct)-c{g}^{\prime }(x+ct)=0$

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 $g:ℝ\to ℝ$

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

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 $n\in \{1,\dots ,d\}$ and $(t,x)\in ℝ\times {ℝ}^d$ 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 $f\in {𝒞}^1(ℝ\times {ℝ}^d)$, $f$ 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,

$\underset{0}{\overset{t}{\int }}f(s,y+𝐯(t-s))ds$

exists for all $y=(y_1,\dots ,y_n,\dots ,y_d)$ 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 ${𝒞}^1({ℝ}^2)$, which is why

$\frac{d}{d{y}_n}f(s,y+𝐯(t-s))\stackrel{\text{chain rule}}{=}{\partial }_{x_n}f(s,y+𝐯(t-s))$

exists for all $s\in B$ and all $y=(y_1,\dots ,y_n,\dots ,y_d)$ 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 ${𝒞}^1(ℝ\times {ℝ}^d)$ and

$\frac{d}{d{y}_n}f(s,y+𝐯(t-s))\stackrel{\text{chain rule}}{=}{\partial }_{x_n}f(s,y+𝐯(t-s))$

is continuous as well as composition of continuous functions, also the function ${\partial }_{x_n}f(s,y+𝐯(t-s))$ 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 $b\in ℝ$, $b>0$ such that for all

$|{\partial }_{x_n}f(s,y+𝐯(t-s))|<b$

for all $y=(y_1,\dots ,y_n,\dots ,y_d)$ such that $y_n\in O$, provided that $y_1,\dots ,y_{n-1},y_{n+1},\dots ,y_d$ are fixed. Therefore, we might choose $g(s)=b$ and obtain that

$\mathrm{\forall }|{\partial }_{x_n}f(s,y+𝐯(t-s))|<|g(s)|\text{ and }\underset{B}{\int }|g(s)|ds=tb<\mathrm{\infty }$

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

${\partial }_{x_n}\underset{0}{\overset{t}{\int }}f(s,y+𝐯(t-s))ds=\underset{0}{\overset{t}{\int }}{\partial }_{x_n}f(s,y+𝐯(t-s))ds$

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

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

Exercise 2

$\begin{array}{ccc}{\partial }_{t}g(x+t𝐯)& =\left(\begin{array}{ccc}{\partial }_{x_1}g(x+t𝐯)& \cdots & {\partial }_{x_d}g(x+t𝐯)\end{array}\right)𝐯& \text{by the chain rule}\\ & =𝐯\cdot {\mathrm{\nabla }}_{x}g(x+t𝐯)& \end{array}$