# Partial Differential Equations/Answers to the exercises

## Chapter 1

[edit | edit source]### Exercise 1

[edit | edit source]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 2

[edit | edit source]Using the one-dimensional chain rule, we directly calculate

and

Therefore,

### Exercise 3

[edit | edit source]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 2

[edit | edit source]### Exercise 1

[edit | edit source]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.