Uniqueness of Solutions for DiffEq’s

Let {V} be a normed finite-dimensional real vector space and let {U \subseteq V} be an open set. A vector field on {U} is a function {\xi : U \rightarrow V}. (In the words of Gaitsgory: “you should imagine a vector field as a domain, and at every point there is a little vector growing out of it.”)

The idea of a differential equation is as follows. Imagine your vector field specifies a velocity at each point. So you initially place a particle somewhere in {U}, and then let it move freely, guided by the arrows in the vector field. (There are plenty of good pictures online.) Intuitively, for nice {\xi} it should be the case that the trajectory resulting is unique. This is the main take-away; the proof itself is just for completeness.

This is a so-called differential equation:

Definition 1

Let {\gamma : (-\varepsilon, \varepsilon) \rightarrow U} be a continuous path. We say {\gamma} is a solution to the differential equation defined by {\xi} if for each {t \in (-\varepsilon, \varepsilon)} we have

\displaystyle  \gamma'(t) = \xi(\gamma(t)).

Example 2 (Examples of DE’s)

Let {U = V = \mathbb R}.

  1. Consider the vector field {\xi(x) = 1}. Then the solutions {\gamma} are just {\gamma(t) = t+c}.
  2. Consider the vector field {\xi(x) = x}. Then {\gamma} is a solution exactly when {\gamma'(t) = \gamma(t)}. It’s well-known that {\gamma(t) = c\exp(t)}.

Of course, you may be used to seeing differential equations which are time-dependent: i.e. something like {\gamma'(t) = t}, for example. In fact, you can hack this to fit in the current model using the idea that time is itself just a dimension. Suppose we want to model {\gamma'(t) = F(\gamma(t), t)}. Then we instead consider

\displaystyle  \xi : V \times \mathbb R \rightarrow V \times \mathbb R \qquad\text{by}\qquad \xi(v, t) = (F(v,t), 1)

and solve the resulting differential equation over {V \times \mathbb R}. This does exactly what we want. Geometrically, this means making time into another dimension and imagining that our particle moves at a “constant speed through time”.

The task is then mainly about finding which conditions guarantee that our differential equation behaves nicely. The answer turns out to be:

Definition 3

The vector field {\xi : U \rightarrow V} satisfies the Lipschitz condition if

\displaystyle  \left\lVert \xi(x')-\xi(x'') \right\rVert \le \Lambda \left\lVert x'-x'' \right\rVert

holds identically for some fixed constant {\Lambda}.

Note that continuously differentiable implies Lipschitz.

Theorem 4 (Picard-Lindelöf)

Let {V} be a finite-dimensional real vector space, and let {\xi} be a vector field on a domain {U \subseteq V} which satisfies the Lipschitz condition.

Then for every {x_0 \in U} there exists {(-\varepsilon,\varepsilon)} and {\gamma : (-\varepsilon,\varepsilon) \rightarrow U} such that {\gamma'(t) = \xi(\gamma(t))} and {\gamma(0) = x_0}. Moreover, if {\gamma_1} and {\gamma_2} are two solutions and {\gamma_1(t) = \gamma_2(t)} for some {t}, then {\gamma_1 = \gamma_2}.

In fact, Peano’s existence theorem says that if we replace Lipschitz continuity with just continuity, then {\gamma} exists but need not be unique. For example:

Example 5 (Counterexample if {\xi} is not differentiable)

Let {U = V = \mathbb R} and consider {\xi(x) = x^{\frac23}}, with {x_0 = 0}. Then {\gamma(t) = 0} and {\gamma(t) = \left( t/3 \right)^3} are both solutions to the differential equation

\displaystyle  \gamma'(t) = \gamma(t)^{\frac 23}.

Now, for the proof of the main theorem. The main idea is the following result (sometimes called the contraction principle).

Lemma 6 (Banach Fixed-Point Theorem)

Let {(X,d)} be a complete metric space. Let {f : X \rightarrow X} be a map such that {d(f(x_1), f(x_2)) < \frac{1}{2} d(x_1, x_2)} for any {x_1, x_2 \in X}. Then {f} has a unique fixed point.

For the proof of the main theorem, we are given {x_0 \in V}. Let {X} be the metric space of continuous functions from {(-\varepsilon, \varepsilon)} to the complete metric space {\overline{B}(x_0, r)} which is the closed ball of radius {r} centered at {x_0}. (Here {r > 0} can be arbitrary, so long as it stays in {U}.) It turns out that {X} is itself a complete metric space when equipped with the sup norm

\displaystyle  d(f, g) = \sup_{t \in (-\varepsilon, \varepsilon)} \left\lVert f(t)-g(t) \right\rVert.

This is well-defined since {\overline{B}(x_0, r)} is compact.

We wish to use the Banach theorem on {X}, so we’ll rig a function {\Phi : X \rightarrow X} with the property that its fixed points are solutions to the differential equation. Define it by, for every {\gamma \in X},

\displaystyle  \Phi(\gamma) : t \mapsto x_0 + \int_0^t \xi(\gamma(s)) \; ds.

This function is contrived so that {(\Phi\gamma)(0) = x_0} and {\Phi\gamma} is both continuous and differentiable. By the Fundamental Theorem of Calculus, the derivative is exhibited by

\displaystyle  (\Phi\gamma)'(t) = \left( \int_0^t \xi(\gamma(s)) \; ds \right)' = \xi(\gamma(t)).

In particular, fixed points correspond exactly to solutions to our differential equation.

A priori this output has signature {\Phi\gamma : (-\varepsilon,\varepsilon) \rightarrow V}, so we need to check that {\Phi\gamma(t) \in \overline{B}(x_0, r)}. We can check that

\displaystyle  \begin{aligned} \left\lVert (\Phi\gamma)(t) - x_0 \right\rVert &=\left\lVert \int_0^t \xi(\gamma(s)) \; ds \right\rVert \\ &\le \int_0^t \left\lVert \xi(\gamma(s)) \; ds \right\rVert \\ &\le t \max_{s \in [0,t]} \left\lVert \xi\gamma(s) \right\rVert \\ &< \varepsilon \cdot A \end{aligned}

where {A = \max_{x \in \overline{B}(x_0,r)} \left\lVert \xi(x) \right\rVert}; we have {A < \infty} since {\overline{B}(x_0,r)} is compact. Hence by selecting {\varepsilon < r/A}, the above is bounded by {r}, so {\Phi\gamma} indeed maps into {\overline{B}(x_0, r)}. (Note that at this point we have not used the Lipschitz condition, only that {\xi} is continuous.)

It remains to show that {\Phi} is contracting. Write

\displaystyle  \begin{aligned} \left\lVert (\Phi\gamma_1)(t) - (\Phi\gamma_2)(t) \right\rVert &= \left\lVert \int_{s \in [0,t]} \left( \xi(\gamma_1(s))-\xi(\gamma_2(s)) \right) \right\rVert \\ &= \int_{s \in [0,t]} \left\lVert \xi(\gamma_1(s))-\xi(\gamma_2(s)) \right\rVert \\ &\le t\Lambda \sup_{s \in [0,t]} \left\lVert \gamma_1(s)-\gamma_2(s) \right\rVert \\ &< \varepsilon\Lambda \sup_{s \in [0,t]} \left\lVert \gamma_1(s)-\gamma_2(s) \right\rVert \\ &= \varepsilon\Lambda d(\gamma_1, \gamma_2) . \end{aligned}

Hence once again for {\varepsilon} sufficiently small we get {\varepsilon\Lambda \le \frac{1}{2}}. Since the above holds identically for {t}, this implies

\displaystyle  d(\Phi\gamma_1, \Phi\gamma_2) \le \frac{1}{2} d(\gamma_1, \gamma_2)

as needed.

This is a cleaned-up version of a portion of a lecture from Math 55b in Spring 2015, instructed by Dennis Gaitsgory.

Some Notes on Valuations

There are some notes on valuations from the first lecture of Math 223a at Harvard.

1. Valuations

Let {k} be a field.

Definition 1

A valuation

\displaystyle  \left\lvert - \right\rvert : k \rightarrow \mathbb R_{\ge 0}

is a function obeying the axioms

  • {\left\lvert \alpha \right\rvert = 0 \iff \alpha = 0}.
  • {\left\lvert \alpha\beta \right\rvert = \left\lvert \alpha \right\rvert \left\lvert \beta \right\rvert}.
  • Most importantly: there should exist a real constant {C}, such that {\left\lvert 1+\alpha \right\rvert < C} whenever {\left\lvert \alpha \right\rvert \le 1}.

The third property is the interesting one. Note in particular it can be rewritten as {\left\lvert a+b \right\rvert < C\max\{ \left\lvert a \right\rvert, \left\lvert b \right\rvert \}}.

Note that we can recover {\left\lvert 1 \right\rvert = \left\lvert 1 \right\rvert \left\lvert 1 \right\rvert \implies \left\lvert 1 \right\rvert = 1} immediately.

Example 2 (Examples of Valuations)

If {k = \mathbb Q}, we can take the standard absolute value. (Take {C=2}.)

Similarly, the usual {p}-adic evaluation, {\nu_p}, which sends {p^a t} to {p^{-a}}. Here {C = 1} is a valid constant.

These are the two examples one should always keep in mind: with number fields, all valuations look like one of these too. In fact, over {\mathbb Q} it turns out that every valuation “is” one of these two valuations (for a suitable definition of equality). To make this precise:

Definition 3

We say {\left\lvert - \right\rvert_1 \sim \left\lvert - \right\rvert_2} (i.e. two valuations on a field {k} are equivalent) if there exists a constant {k > 0} so that {\left\lvert \alpha \right\rvert_1 = \left\lvert \alpha \right\rvert_2^k} for every {\alpha \in k}.

In particular, for any valuation we can force {C = 2} to hold by taking an equivalent valuation to a sufficient power.

In that case, we obtain the following:

Lemma 4

In a valuation with {C = 2}, the triangle inequality holds.

Proof: First, observe that we can get

\displaystyle  \left\lvert \alpha + \beta \right\rvert \le 2 \max \left\{ \left\lvert \alpha \right\rvert, \left\lvert \beta \right\rvert \right\}.

Applying this inductively, we obtain

\displaystyle  \left\lvert \sum_{i=1}^{2^r} a_i \right\rvert \le 2^r \max_i \left\lvert a_i \right\rvert

or zero-padding,

\displaystyle  \sum_{i=1}^{n} a_i \le 2n\max_i \left\lvert a_i \right\rvert.

From this, one can obtain

\displaystyle  \left\lvert \alpha+\beta \right\rvert^n \le \left\lvert \sum_{j=0}^n \binom nj \alpha^j \beta^{n-j} \right\rvert \le 2(n+1) \sum_{j=0}^n \left\lvert \binom nj \right\rvert \left\lvert \alpha \right\rvert^j \left\lvert \beta \right\rvert^{n-j} \le 4(n+1)\left( \left\lvert \alpha \right\rvert+\left\lvert \beta \right\rvert \right)^n.

Letting {n \rightarrow \infty} completes the proof. \Box

Next, we prove that

Lemma 5

If {\omega^n=1} for some {n}, then{\left\lvert \omega \right\rvert = 1}. In particular, on any finite field the only valuation is the trivial one which sends {0} to {0} and all elements to {1}.

Proof: Immediate, since {\left\lvert \omega \right\rvert^n = 1}. \Box

2. Topological field induced by valuations

Let {k} be a field. Given a valuation on it, we can define a basis of open sets

\displaystyle  \left\{ \alpha \mid \left\lvert \alpha - a \right\rvert < d \right\}

across all {a \in K}, {d \in \mathbb R_{> 0}}. One can check that the same valuation gives rise to the same topological spaces, so it is fine to assume {C = 2} as discussed earlier; thus, in fact we can make {k} into a metric space, with the valuation as the metric.

In what follows, we’ll always assume our valuation satisfies the triangle inequality. Then:

Lemma 6

Let {k} be a field with a valuation. Viewing {k} as a metric space, it is in fact a topological field, meaning addition and multiplication are continuous.

Proof: Trivial; let’s just check that multiplication is continuous. Observe that

\displaystyle  \begin{aligned} \left\lvert (a+\varepsilon_1)(b+\varepsilon_2) - ab \right\rvert & \le \left\lvert \varepsilon_1\varepsilon_2 \right\rvert + \left\lvert a\varepsilon_2 \right\rvert + \left\lvert b\varepsilon_1 \right\rvert \\ &\rightarrow 0. \end{aligned}

\Box

Now, earlier we saw that two valuations which are equivalent induce the same topology. We now prove the following converse:

Proposition 7

If two valuations {\left\lvert - \right\rvert_1} and {\left\lvert - \right\rvert_2} give the same topology, then they are in fact equivalent.

Proof: Again, we may safely assume that both satisfy the triangle inequality. Next, observe that {\left\lvert a \right\rvert < 1 \iff a^n \rightarrow 0} (according to the metric) and by taking reciprocals, {\left\lvert a \right\rvert > 1 \iff a^{-n} \rightarrow 0}.

Thus, given any {\beta}, {\gamma} and integers {m}, {n} we derive that

\displaystyle  \left\lvert \beta^n\gamma^m \right\rvert_1 < 1 \iff \left\lvert \beta^n\gamma^m \right\rvert < 1

with similar statements holding with “{<}” replaced by “{=}”, “{>}”. Taking logs, we derive that

\displaystyle  n \log\left\lvert \beta \right\rvert_1 + m \log \left\lvert \gamma \right\rvert_1 < 0 \iff n \log\left\lvert \beta \right\rvert_2 + m \log \left\lvert \gamma \right\rvert_1 < 0

and the analogous statements for “{=}”, “{>}”. Now just choose an appropriate sequence of {m}, {n} and we can deduce that

\displaystyle  \frac{\log \left\lvert \beta_1 \right\rvert}{\log \left\lvert \beta_2 \right\rvert} = \frac{\log \left\lvert \gamma_1 \right\rvert}{\log \left\lvert \gamma_2 \right\rvert}

so it equals a fixed constant {c} as desired. \Box

3. Discrete Valuations

Definition 8

We say a valuation {\left\lvert - \right\rvert} is discrete if its image around {1} is discrete, meaning that if {\left\lvert a \right\rvert \in [1-\delta,1+\delta] \implies \left\lvert a \right\rvert = 1} for some real {\delta}. This is equivalent to requiring that {\{\log\left\lvert a \right\rvert\}} is a discrete subgroup of the real numbers.

Thus, the real valuation (absolute value) isn’t discrete, while the {p}-adic one is.

4. Non-Archimedian Valuations

Most importantly:

Definition 9

A valuation {\left\lvert - \right\rvert} is non-Archimedian if we can take {C = 1} in our requirement that {\left\lvert a \right\rvert \le 1 \implies \left\lvert 1+a \right\rvert \le C}. Otherwise we say the valuation is Archimedian.

Thus the real valuation is Archimedian while the {p}-adic valuation is non-Archimedian.

Lemma 10

Given a non-Archimedian valuation {\left\lvert - \right\rvert}, we have {\left\lvert b \right\rvert < \left\lvert a \right\rvert \implies \left\lvert a+b \right\rvert = \left\lvert a \right\rvert}.

Proof: We have that

\displaystyle  \left\lvert a \right\rvert = \left\lvert (a+b)-b \right\rvert \le \max\left\{ \left\lvert a+b \right\rvert, \left\lvert b \right\rvert \right\}.

On the other hand, {\left\lvert a+b \right\rvert \le \max \{ \left\lvert a \right\rvert, \left\lvert b \right\rvert\}}. \Box

Given a field {k} and a non-Archimedian valuation on it, we can now consider the set

\displaystyle  \mathcal O = \left\{ a \in k \mid \left\lvert a \right\rvert \le 1 \right\}

and by the previous lemma, this turns out to be a ring. (This is the point we use the fact that the valuation is non-Archimedian; without that {\mathcal O} need not be closed under addition). Next, we define

\displaystyle  \mathcal P = \left\{ a \in k \mid \left\lvert a \right\rvert < 1 \right\} \subset \mathcal O

which is an ideal. In fact it is maximal, because {\mathcal O/\mathcal P} is the set of units in {\mathcal O}, and is thus necessarily a field.

Lemma 11

Two valuations are equal if they give the same ring {\mathcal O} (as sets, not just up to isomorphism).

Proof: If the valuations are equivalent it’s trivial.

For the interesting converse direction (they have the same ring), the datum of the ring {\mathcal O} lets us detect whether {\left\lvert a \right\rvert < \left\lvert b \right\rvert} by simply checking whether {\left\lvert ab^{-1} \right\rvert < 1}. Hence same topology, hence same valuation. \Box

We will really only work with valuations which are obviously discrete. On the other hand, to detect non-Archimedian valuations, we have

Lemma 12

{\left\lvert - \right\rvert} is Archimedian if {\left\lvert n \right\rvert \le 1} for every {n = 1 + \dots + 1 \in k}.

Proof: Clearly Archimedian {\implies} {\left\lvert n \right\rvert \le 1}. The converse direction is more interesting; the proof is similar to the analytic trick we used earlier. Given {\left\lvert a \right\rvert \le 1}, we wish to prove {\left\lvert 1+a \right\rvert \le 1}. To do this, first assume the triangle inequality as usual, then

\displaystyle  \left\lvert 1+a \right\rvert^n < \sum_j \left\lvert \binom nj \right\rvert\left\lvert a \right\rvert^j \le \sum_{j=0}^n \left\lvert a \right\rvert^j \le \sum_{j=0}^n 1 = n+1.

Finally, let {n \rightarrow \infty} again. \Box
In particular, any field of finite characteristic in fact has {\left\lvert n \right\rvert = 1} and thus all valuations are non-Archimedian.

5. Completions

We say that a field {k} is complete with respect to a valuation {\left\lvert - \right\rvert} if it is complete in the topological sense.

Theorem 13

Every field {k} is with a valuation {\left\lvert - \right\rvert} can be embedded into a complete field {\overline{k}} in a way which respects the valuation.

For example, the completion of {\mathbb Q} with the Euclidean valuation is {\mathbb R}. Proof: Define {\overline{k}} to be the topological completion of {k}; then extend by continuity; \Box
Given {k} and its completion {\overline{k}} we use the same notation for the valuations of both.

Proposition 14

A valuation {\left\lvert - \right\rvert} on {\overline{k}} is non-Archimedian if and only if the valuation is non-Archimedian on {k}.

Proof: We saw non-Archimedian {\iff} {\left\lvert n \right\rvert \le 1} for every {n = 1 + \dots + 1}. \Box

Proposition 15

Assume {\left\lvert - \right\rvert} is non-Archimedian on {k} and hence {\overline{k}}. Then the set of values achieved by {\left\lvert - \right\rvert} coincides for {k} and {\overline{k}}, i.e. {\{ \left\lvert k \right\rvert \} = \{ \left\lvert \overline{k} \right\rvert \}}.

Not true for Archimedian valuations; consider {\left\lvert \sqrt2 \right\rvert = \sqrt2 \notin \mathbb Q}. Proof: Assume {0 \neq b \in \overline{k}}; then there is an {a \in k} such that {\left\lvert b-a \right\rvert < \left\lvert b \right\rvert} since {k} is dense in {\overline{k}}. Then, {\left\lvert b \right\rvert \le \max \{ \left\lvert b-a \right\rvert, \left\lvert a \right\rvert \}} which implies {\left\lvert b \right\rvert = \left\lvert a \right\rvert}. \Box

6. Weak Approximation Theorem

Proposition 16 (Weak Approximation Theorem)

Let {\left\lvert-\right\rvert_i} be distinct nontrivial valuations of {k} for {i=1,\dots,n}. Let {k_i} denote the completion of {k} with respect to {\left\lvert-\right\rvert_i}. Then the image

\displaystyle  k \hookrightarrow \prod_{i=1}^n k_i

is dense.

This means that distinct valuations are as different as possible; for example, if {\left\lvert-\right\rvert _1 = \left\lvert-\right\rvert _2} then we might get, say, a diagonal in {\mathbb R \times \mathbb R} which is as far from dense as one can imagine. Another way to think of this is that this is an analogue of the Chinese Remainder Theorem.

Proof: We claim it suffices to exhibit {\theta_i \in k} such that

\displaystyle  \left\lvert \theta_i \right\rvert_j \begin{cases} > 1 & i = j \\ < 1 & \text{otherwise}. \end{cases}

Then

\displaystyle  \frac{\theta_i^r}{1+\theta_i^r} \rightarrow \begin{cases} 1 & \text{ in } \left\lvert-\right\rvert_i \\ 0 & \text{ otherwise}. \end{cases}

Hence for any point {(a_1, \dots, a_n)} we can take the image of {\sum \frac{\theta_i^r}{1+\theta_i^r} a_i \in k}. So it would follow that the image is dense.

Now, to construct the {\theta_i} we proceed inductively. We first prove the result for {n=2}. Since the topologies are different, we exhibit {\alpha}, {\beta} such that {\left\lvert \alpha_1 \right\rvert < \left\lvert \alpha_2 \right\rvert} and {\left\lvert \beta_1 \right\rvert > \left\lvert \beta_2 \right\rvert}, and pick {\theta=\alpha\beta^{-1}}.

Now assume {n \ge 3}; it suffices to construct {\theta_1}. By induction, there is a {\gamma} such that

\displaystyle  \left\lvert \gamma \right\rvert_1 > 1 \quad\text{and}\quad \left\lvert \gamma \right\rvert_i < 1 \text{ for } i = 2, \dots, n-1.

Also, there is a {\psi} such that

\displaystyle  \left\lvert \delta \right\rvert_1 > 1 \quad\text{and}\quad \left\lvert \delta \right\rvert_n < 1.

Now we can pick

\displaystyle  \theta_1 = \begin{cases} \gamma & \left\lvert \gamma \right\rvert_n < 1 \\ \phi^r\gamma & \left\lvert \gamma \right\rvert_n = 1 \\ \frac{\gamma^r}{1+\gamma^r} & \left\lvert \gamma \right\rvert_n > 1 \\ \end{cases}

for sufficiently large {r}. \Box