Partial Differential Equations/Answers to the exercises

Partial Differential Equations
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Chapter 1Edit

Exercise 1Edit

The general ordinary differential equation is given by

 

for a set  . Noticing that

 

, we observe that the general ordinary differential equation is just the general one-dimensional partial differential equation.

Exercise 2Edit

Using the one-dimensional chain rule, we directly calculate

 

and

 

Therefore,

 

Exercise 3Edit

By choosing

 

, we see that in our function   first order derivatives suffice to depict the partial differential equation. On the other hand, if   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  

 

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

 

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 2Edit

Exercise 1Edit

Let   and   be arbitrary. We choose   and   for an arbitrary   and apply Leibniz' integral rule. We first check that all the three conditions for Leibniz' integral rule are satisfied.

1.

Since  ,   is continuous in all variables (so therefore in particular in the first), and the same is true for the composition of   with any continuous function. Thus, as all continuous functions of one variable are integrable on an interval,

 

exists for all   such that   (in fact, it exists everywhere, but this is the requirement of Leibniz' rule in our situation).

2.

  was supposed to be in  , which is why

 

exists for all   and all   such that   (in fact, it exists everywhere, but this is the requirement of Leibniz' rule in our situation).

3.

We note that  . Therefore, also  , where   is compact. Furthermore, as   was supposed to be continuous (by definition of   and

 

is continuous as well as composition of continuous functions, also the function   is continuous in   and  . Thus, due to the extreme value theorem, it is bounded for  , i. e. there is a  ,   such that for all

 

for all   such that  , provided that   are fixed. Therefore, we might choose   and obtain that

 

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

 

for all   such that  . Setting   gives the result for  .

Since   and   were arbitrary, this completes the exercise.

Exercise 2Edit

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