Partial Differential Equations/Answers to the exercises

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

Chapter 1[edit]

Exercise 1[edit]

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[edit]

Using the one-dimensional chain rule, we directly calculate

\partial_t u(t, x) = \partial_t g(x + ct) = c g'(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) = c g'(x + ct) - c g'(x +ct) = 0

Exercise 3[edit]

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[edit]

Exercise 1[edit]

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| < b

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[edit]

\begin{align}
\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{align}
Partial Differential Equations
 ← The Malgrange-Ehrenpreis theorem Answers to the exercises