User:Julius

Current focus

 * Build the bulk knowledge on calculus of variations based on Gelfand's Calculus of Variations, then recheck with a couple other books and slowly improve proofs.


 * So I just noticed that vector notation is being used in Gelfand's for higher dimensional functionals. This implies rewriting all multivariable functionals. Implement this gradually.

Lemmas and theorems for Bernstein's Theorem on Unique Extrema (1978)
Raw material

Example 1
Suppose that:


 * $J \sqbrk y = \int_1^2 \frac {\sqrt {1+y'^2} } {x} \rd x$

with the following boundary conditions:


 * $\map y 1 = 0$


 * $\map y 2 = 1$

Then the smooth minimizer of $J$ is a circle of the following form:


 * $\paren {y - 2}^2 + x^2 = 5$

Proof
$J$ is of the form


 * $J \sqbrk y = \int_a^b \map F {x, y'} \rd x$

Then we can use the "no y theorem":


 * $F_y = C$

i.e.


 * $\frac {y'} {x \sqrt {1 + y'^2} } = C$

or


 * $y' = \frac {C x} {\sqrt {1 - C^2 x^2} }$

The integral is equal to


 * $y = \frac {\sqrt {1 - C^2 x^2} } C + C_1$

or


 * $\paren {y - C_1}^2 + x^2 = C^{-2}$

From the conditions $\map y 1 = 0$, $\map y 2 = 1$ we find that


 * $C = \frac 1 {\sqrt 5}$


 * $C_1 = 2$

Example 3

 * $J \sqbrk = \int_a^b \paren {x - y}^2$

is minimized by


 * $\map y x = x$

Proof
Euler' equation:


 * $F_y = 0$

i.e.


 * $2 \paren {x - y} = 0$.

Example p31
Suppose:


 * $J \sqbrk r = \int_{\phi_0}^{\phi_1} \sqrt{r^2 + r'^2} \rd \phi$

Euler's Equation:


 * $\displaystyle \frac r {\sqrt{r^2 + r'^2} } - \dfrac \d {\d \phi} \frac {r'} {\sqrt{r^2 + r'^2} }$

Apply change of variables:


 * $x = r \cos \phi, y = r \sin \phi$

The integral becomes:


 * $\displaystyle \int_{x_0}^{x_1} \sqrt{1 + y'^2} \rd x$

Euler's equation:


 * $y'' = 0$

Its solution:


 * $y = \alpha x + \beta$

or


 * $r \sin \phi = \alpha r \cos \phi + \beta$

Example

 * $J \sqbrk = \int_{x_0}^{x_1} \map f {x,y} \sqrt {1+y'^2}\rd x$


 * $F_{y'} = \map f {x,y} \frac {y'} {\sqrt{1 + y'^2} }=\frac {y' F} {1 + y'^2}$


 * $F + \paren {\phi' - y'}F_{y'} = \frac {\paren{1+y'\phi'}F} {1+y'^2} = 0$


 * $F + \paren {\psi' - y'}F_{y'} = \frac {\paren{1+y'\psi'}F} {1+y'^2} = 0$

i.e.


 * $y' = -\frac 1 {\phi'}$


 * $y' = - \frac 1 {\psi'}$

Transversality reduces to orthogonality

Example: points on surfaces

 * $J \sqbrk {y,z} = \int_{x_0}^{x_1} \map F {x,y,z,y',z'} \rd x$

Transversality conditions:


 * $\sqbrk {F_{y'} + \dfrac {\partial \phi} {\partial y} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x0} = 0$


 * $\sqbrk {F_{z'} + \dfrac {\partial \phi} {\partial z} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x0} = 0$


 * $\sqbrk {F_{y'} + \dfrac {\partial \phi} {\partial y} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x1} = 0$


 * $\sqbrk {F_{z'} + \dfrac {\partial \phi} {\partial z} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x1} = 0$

Example: Legendre transformation

 * $\map f \xi = \frac {\xi^a} a, a>1$


 * $\map {f'} \xi = p = \xi^{a-1}$

i.e.


 * $\xi = p^{\frac {1} {a-1} }$


 * $H = - \frac {\xi^a} {a} + p\xi = - \frac {p^{\frac {a} {a-1} } } a + p p^{\frac {a} {a-1} } = p^{\frac {a} {a-1} } \paren{1 - \frac 1 a}$

Hence:


 * $\map H p = \frac {p^b} b$

where:


 * $\frac 1 a + \frac 1 b = 1$

Example

 * $J \sqbrk y = \int_a^b \paren {Py'^2 + Q y^2} \rd x$


 * $p = 2 P y', H = P y'^2 - Q y^2$

Hence:


 * $H = \frac {p^2} {4 P} - Q y^2$

Canonical equations:


 * $\dfrac {\d p} {\d x} = 2 Q y$


 * $\dfrac {\d y} {\d x} = \frac p {2 P}$

Euler's Equation:


 * $2 y Q - \dfrac \d {\d x} \paren {2 P y'} = 0$

Example: Noether's theorem 1

 * $J \sqbrk y = \int_{x0}^{x1} y'^2 \rd x$

is invariant under the transformation:


 * $x^* = x + \epsilon, y^* = y$


 * $y^* = \map y {x^* - \epsilon} = \map {y^*} {x^*}$

Then:


 * $J \sqbrk {\gamma^*} = \int_{x0^*}^{x1^*} \sqbrk { \dfrac {\d \map {y^*} {x^*} } {\d x^*} } \rd x^* = \int_{x0+\epsilon}^{x_1 + \epsilon} \sqbrk { \dfrac {\d \map y {x^* - \epsilon} } {\d x^*} }^2 \rd x^* = \int_{x0}^{x1} \sqbrk { \dfrac {\d \map y x} {\d x} }^2 \rd x = J \sqbrk \gamma$

Example: Neother's theorem 2

 * $J \sqbrk y = \int_{x_0}^{x_1} x y'^2 \rd x$

Example: Noether's theorem 3

 * $J \sqbrk y = \int_{x_0}^{x_1} \map F {y, y'} \rd x$

Invariant under $x^* = x + \epsilon, y_i^* = y_i$

I.e. $\phi = 1, \psi_i = 0$

reduces to $H = \const$

Momentum of the system:

 * $P_x = \sum_{y = 1}^n p_{ix}, P_y = \sum_{y = 1}^n p_{iy}, P_z = \sum_{z = 1}^n p_{iz}$

(Examples: attraction to a fixed point, attraction to a homogenous distribution on an axis)

Geodetic distance:Examples
If $J$ is arclength, $S$ is distance.

If $J$ is a moment of time to pass a segment of optical medium, then $S$ is the time needed to pass the whole optical body.

If $J$ is action, then $S$ is the minimal action.

Examples of quadratic functionals
1) $B \sqbrk {x, y} = \int_{t_0}^{t_1} \map x t \map y t \rd t$

Corresponding quadratic functional

$A \sqbrk x = \int_{t_0}^{t_1} \map {x^2} t$

2) $B \sqbrk {x, y} = \int_{t_0}^{t_1} \map \alpha t \map x t \map y t \rd t$

Corresponding quadratic functional

$A \sqbrk x = \int_{t_0}^{t_1} \map \alpha t \map {x^2} t \rd t$

3)

$A \sqbrk x = \int_{t_0}^{t_1} \paren {\map \alpha t \map {x^2} t + \map \beta t \map x t \map {x'} t+ \map \gamma t \map {x'^2} t} \rd t$

4)

$B \sqbrk {x, y} = \int_a^b \int_a^b \map K {s, t} \map x s \map y t \rd s \rd t$

Functional Analysis
$\paren{C \closedint a b,\norm{\cdot}_\infty }$ is a Banach space.

Let $\sequence{x_n}_{n \in \N}$ be a Cauchy sequence.


 * $\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n, m > N : \norm{x_n − x_m}_\infty < \epsilon$

Suppose, all the elements of $\sequence{x_n}_{n \in \N}$ are additionally indexed with $t$:


 * $\sequence{x_n}_{n \in \N} = \sequence{\map {x_n} t }_{n \in \N}$

Let $t \in \closedint a b$.

But

$\displaystyle \forall n, m > N : \norm {\map {x_n} t - \map {x_m} t}_\infty < \max_{\tau \in \closedint a b}\norm {\map {x_n} \tau - \map {x_m} \tau}_\infty = \norm {x_n - x_m}_\infty < \epsilon$

Hence, $\sequence{\map {x_n} t}_{n \in \N}$ is a Cauchy sequence in $\R$.

$\R$ is complete.

Therefore, $\sequence{\map {x_n} t}_{n \in \N}$ is convergent with limit $L = \map L t$.

Choose $N$ such that $\forall n,m > N : \norm{x_n - x_m} \le \frac \epsilon 3$

Let $\tau \in \closedint a b$.

Then $\forall n > N : \norm {\map {x_n} \tau - \map {x_{N + 1}} \tau } \le \norm {x_n - x_{N + 1} }_\infty \le \frac \epsilon 3$

Take the limit $n \to \infty$:


 * $\lim_{n \to \infty} \norm {\map {x_n} \tau - \map {x_{N + 1}} \tau } = \norm {\map x \tau - \map {x_{N + 1}} \tau } \le \frac \epsilon 3$

which holds for all $\tau \in \closedint a b$.

Now $\map {x_{N+1} } \tau \in C \closedint a b$

$\exists \delta > 0: \norm {\tau - t} < \delta \implies \norm {\map {x_{N+1} } t - \map {x_{N+1} } \tau} \le \frac \epsilon 3$

Thus:


 * $\norm {\map x \tau - \map x t} = \norm {\map x \tau - \map {x_{N+1}} \tau + \map {x_{N+1}} \tau - \map {x_{N+1}} t + \map {x_{N+1}} t - \map x t} \le$


 * $\norm {\map x \tau - \map {x_{N+1}} \tau} + \norm {\map {x_{N+1}} \tau - \map {x_{N+1}} t} + \norm {\map {x_{N+1}} t - \map x t} \le \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3 = \epsilon$

Hence $x$ is continuous at $t$.

Since $t \in C \closedint a b$, $t$ is continuous in whole interval.

Finally, show that $\sequence {x_n}_{n \in \N}$ converges to $x$.

Let $\epsilon > 0$.

Choose $N$ such that $\forall n,m > N : \norm{x_n - x_m}_\infty < \epsilon$

Fix $n > N$.

Let $t \in \closedint a, b$.

Then $\forall m > N: \norm {\map {x_n} t - \map {x_m} t} \le \norm {x_n - x_m}_\infty < \epsilon$

Thus $\norm{\map {x_n} t - \map x t} = \lim_{n \to \infty} \norm {\map {x_n} t - \map {x_m} t} \le \epsilon$

Since $t$ was arbitrary: $\norm {x_n - x}_\infty = \max_{t \in \closedint a b } \norm{\map {x_n} t - \map x t} \le \epsilon$

This could also have been achieved by fixing $n > N$.

So, $\forall n > N \norm {x_n - x}_\infty \le \epsilon$.

Therefore $\lim_{x \to \infty} x_n = x$ in $C \closedint a b$

Theorem
Let $f: \R \times \R \rightarrow \R$ be a real function.

Suppose, there exists $r > 0$ and $L > 0$ such that:


 * $\forall t \ge 0, \forall x, y \in \R : \size {\map f {x, t} - \map f {y, t}} \le L \size {x - y}$

Then:


 * $\forall x_0 \in \R : \exists T > 0: \exists x \in C^1 \closedint 0 T$

such that for all $t \in \closedint 0 T$ the following is satisfied.


 * $\map {\dfrac {dx}{dt}} t = \map f {\map x t, t}$


 * $\map x 0 = x_0$

Furthermore, the solution is unique.

Uniqueness

Let $x_1, x_2$ be solutions to IVP for $t \in \closedint 0 T$ for some $T > 0$.

Let $t_* := \max \set{t \in \closedint 0 T : \map {x_1} {\tau} = \map {x_2} {\tau}, \forall \tau \le t }$

Then:


 * $ \map {x_1} t - \map {x_1} {t_*} = \int_{t_*}^t \map {x_1'} \tau \rd \tau = \int_{t_*}^t \map {f_1} {\map {x_1} \tau, \tau} \rd \tau$


 * $ \map {x_2} t - \map {x_2} {t_*} = \int_{t_*}^t \map {x_2'} \tau \rd \tau = \int_{t_*}^t \map {f_2} {\map {x_2} \tau, \tau} \rd \tau$

So:


 * $\map {x_1} t - \map {x_2} t = \int_{t_*}^t \paren{\map {f_1} {\map {x_1} \tau, \tau} - \map {f_2} {\map {x_2} \tau, \tau}} \rd \tau$

Let

$N > \max \set {1, \frac 1 L, \frac 1 {L \paren{t - t_*}}}$

and

$\displaystyle M = \max_{t \in \closedint {t_*} {t_* + \frac 1 {LN}}} \size {\map {x_2} t - \map {x_1} t}$

Note that: $t_* + \frac 1 {LN} < T$

Then $\forall t \in \closedint {t_*} {t_* + \frac 1 {LN}}$

Thus $\forall t \in \closedint {t_*} {t_* + \frac 1 {LN}} : \size {\map {x_1} \tau - \map {x_2} \tau} \le \frac M N$

so $M \le \frac M N$ or $N \le 1$. Contradiction.

Sequence of recursively defined functions being increasingly better approximations.

Note that:


 * $x_{n + 1} = \Sigma_{k = 0}^n \paren {x_{n + 1} - x_n}$

For $0 \le t \le \frac {1}{2L}$

Therefore:


 * $\norm {x_{n+1} - x_n} \le \frac 1 2 \norm {x_n - x_{n-1}}$

and


 * $\norm {x_{n+1} - x_n} \le \frac 1 {2^n} \norm {x_1 - x_0}$

Therefore:


 * $\sum_{n = 0}^\infty \norm {x_{n+1} - x_n} \le \norm {x_1 - x_0} \sum_{n = 0}^\infty \frac 1 {2^n} < \infty$

and $x_0 + \sum_{n = 0}^\infty \paren{x_{n+1} -x_n}$ converges in $\struct{C \paren{0, T}, \norm{\cdot}_{\infty} }$ to $x \in C \paren{0, T}$.

We know that :$x_{n + 1} = x_0 + \int_0^t \map f {\map x \tau, \tau} \rd \tau$ for $t \in \sqbrk{0, T}$.

Define $\map {g_n} t = \map f {\map {x_n} t, t}$

Then the sequence $g_0, g_1, g_2$ is a sequence of partial sums $g_0 + \sum_{k = 0}^n \paren{g_{k + 1} - g_k}$.

We have $\norm{\map {g_{k+1}} t - \map {g_k} t} = \norm {\map f {\map {x_{k+1}} t, t} - \map f {\map {x_k} t, t}} \le L \norm{\map {x_{k+1}} t - \map {x_k} t} \le L \norm{x_{k+1} - x_k}_\infty \le \frac 1 {2^k} \norm{x_1 - x_0}_\infty$

So the partial sum converges to some $g$ in $\struct{C\paren{0, T}, \norm{\cdot}_\infty}$

We have :$\map g t = \lim_{n \to \infty} \map {g_n} t = \lim_{n \to \infty} \map f {\map {x_n} t, t}$

Since $\sum_{k = 1}^n \map {f_k} t$ converges in $\struct {C \paren{a,b}, \norm{\cdot}_\infty}$ then


 * $\sum_{k=1}^\infty \int_a^b \map {f_k} t \rd t = \int_a^b \map f t \rd t$

Using this, :$\lim_{n \to \infty} \int_0^T \map f {\map {x_n} t, t} \rd t = \lim_{n \to \infty} \int_0^T \paren {\map {g_0} t + \sum_{k = 0}^{n - 1} \paren {\map {g_{k+1}} t - \map {g_k} t}} \rd t = \int_0^T \map f {\map x t, t} \rd t$

Thus $\map x 0 = x_0 + 0 = x_0$ and by fundamental theorem of calculus $\map {x'} t = 0 + \map f {\map x t, t}$ for all $t \in \sqbrk{0, T}$.

-

Proof
Let us define the following series of functions:

What we are going to do is prove that $\displaystyle \map y x = \lim_{n \to \infty} \map {y_n} x$ is the required solution.

The curve lies in the rectangle
We will show that for $a - h \le x \le a + h$, the curve $y = \map {y_n}x$ lies in the rectangle $R$.

That is, that $b - k < y < b + k$.

Suppose $y = \map {y_{n-1}} x$ lies in $R$.

Then:

Clearly $y_0$ lies in $R$, and the argument holds for $y_1$.

So by induction, $y = \map {y_n} x$ lies in $R$ for all $n \in \N$.

Existence
Note that:


 * $\displaystyle y_{n + 1} = y_0 + \sum_{k = 0}^n \paren {y_{k + 1} - y_k}$

The theorem contains more variables and parameters than inequality constraints.

Thus, more relations between them can be chosen without affecting the constraints.

Choose $\displaystyle h = \frac A 2$.

By taking supremum norm of both sides, we get:


 * $\displaystyle \norm {y_{n+1} - y_n}_{\infty} \le \frac 1 2 \norm {y_n - y_{n-1}}_{\infty}$

By induction, the inequality can be extended:


 * $\paren {\star} \displaystyle \norm {y_{n+1} - y_n} \le \frac 1 {2^n} \norm {y_1 - y_0}$

Therefore:

Hence,


 * $\displaystyle y_{n + 1} = y_0 + \sum_{k = 0}^\infty \paren{y_{k+1} - y_k}$

converges in $\struct{\map {C^1} {\size{x - a} < h}, \norm{\cdot}_{\infty}}$ to $y \in \map {C^1} {\size {x - a} < h}$ absolutely.

Therefore, the sequence is convergent:


 * $\displaystyle \map y x = \lim_{n \to \infty} \map {y_{n + 1}} x = x_0 + \lim_{n \to \infty} \int_a^x \map f {\map {y_n} x, x} \rd x$

To find the limit, consider the following sequence:


 * $\displaystyle \map {g_n} x = \map f {\map {x_n} x, x}$

The sequence $\sequence {g_n}_{n \in \N_0}$ is a sequence of partial sums $\displaystyle g_0 + \sum_{k = 0}^n \paren{g_{k + 1} - g_k}$.

It follows that:

So $\sequence {g_n}_{n \in \N_0}$ converges to some $g$ in $\struct{C\paren{\size {x - a} < h}, \norm{\cdot}_\infty}$ absolutely.

It follows that

Since $\displaystyle \sum_{k = 0}^{\infty} g_k$ converges to $g$ in $\struct {C \paren{\size {x - a} < h}, \norm{\cdot}_\infty}$ then


 * $\displaystyle \sum_{k=1}^\infty \int_a^x \map {g_k} t \rd t = \int_a^x \map g t \rd t$

Using this:


 * $\displaystyle \map y x = y_0 + \int_a^x \map f {\map y t, t} \rd t$

Thus $\map y 0 = y_0 + 0 = y_0$ and by fundamental theorem of calculus $\map {y'} x = 0 + \map f {\map y x, x}$ for all $x \in \sqbrk{\size {x - a} < h}$.

Bounded Nature of Adjacent Differences
We will show that:
 * $\displaystyle \left|{y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right| \le \frac {M A^{n-1}} {n!}\left|{x - a}\right|^n$

This is also to be proved by induction.

Suppose that this holds for $n-1$ in place of $n$.

Let this be the induction hypothesis.

We have:
 * $\displaystyle y_n \left({x}\right) - y_{n-1} \left({x}\right) = \int_a^x \left({f \left({t, y_{n-1} \left({t}\right)}\right) - f \left({t, y_{n-2} \left({t}\right)}\right)}\right) \, \mathrm d t$

We also have that:
 * $\left|{f \left({t, y_{n-1} \left({t}\right)}\right) - f \left({t, y_{n-2} \left({t}\right)}\right)}\right| \le A \left|{y_{n-1} \left({t}\right) - y_{n-2} \left({t}\right)}\right|$

by the Lipschitz condition.

By the induction hypothesis, it follows that:

$\displaystyle \left|{f \left({t, y_{n-1} \left({t}\right)}\right) - f \left({t, y_{n-2} \left({t}\right)}\right)}\right| \le \frac {M A^{n-1} \left|{t - a}\right|^{n-1}} {\left({n - 1}\right)!}$

So:

For the base case, we use $n = 1$:


 * $\displaystyle \left|{y_1 \left({x}\right) - b}\right| \le \left|{\int_a^x f \left({t, b}\right) \, \mathrm d t}\right| \le M \left|{x - a}\right|$

Thus by induction:
 * $\displaystyle \left|{y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right| \le \frac {M A^{n-1} } {n!}\left|{x - a}\right|^n$

for all $n$.

Uniform Convergence of Sequence
Next we show that the sequence $\left \langle {y_n \left({x}\right)} \right \rangle$ converges uniformly to a limit for $a - h \le x \le a + h$.

From Bounded Nature of Adjacent Differences above, we have:

From Radius of Convergence of Power Series over Factorial, it follows that $b + M h + \cdots + \dfrac {M A^{n-1} h^n} {n!} + \cdots$ is absolutely convergent for all $h$.

Hence, by the Weierstrass M-Test:
 * $b + \left({y_1 \left({x}\right) - b}\right) + \cdots + \left({y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right) + \cdots$

converges uniformly for $a - h \le x \le a + h$.

Since its terms are continuous functions of $x$, its sum $\displaystyle \lim_{n \to \infty} y_n \left({x}\right) = y \left({x}\right)$ is also continuous from Combination Theorem for Sequences.

Solution Satisfies Differential Equation
We now show that $y = y \left({x}\right)$ satisfies the differential equationn $y' = f \left({x, y}\right)$.

Since:
 * $y_n \left({x}\right)$ converges uniformly to $y \left({x}\right)$ in the open interval $\left({a - h \,.\,.\, a + h}\right)$ from Uniform Convergence of Sequence above
 * $\left|{f \left({x, y}\right) - f \left({x, y_n}\right)}\right| \le A \left|{y - y_n}\right|$ from the Lipschitz condition in $y$

it follows that $f \left({x, y_n \left({x}\right)}\right)$ tends uniformly to $f \left({x, y \left({x}\right)}\right)$.

Letting $n \to \infty$ in:
 * $\displaystyle y_n \left({x}\right) = b + \int_a^x f \left({t, y_{n-1} \left({t}\right)}\right) \, \mathrm d t$

we get:
 * $\displaystyle y \left({x}\right) = b + \int_a^x f \left({t, y \left({t}\right)}\right) \, \mathrm d t$

The integrand $f \left({t, y \left({t}\right)}\right)$ is a continuous function of $t$.

Therefore the integral has the derivative $f \left({x, y}\right)$.

Also, we have that $y \left({a}\right) = b$.

Uniqueness of Solution
We now show that the solution $y = y \left({x}\right)$ that we have found is the only solution where $y \left({a}\right) = b$.

Suppose there is another such solution, $y = Y \left({x}\right)$, say.

Let $\left|{Y \left({x}\right) - y \left({x}\right)}\right| \le B$ when $\left|{x - a}\right| \le h$. (Certainly we could take $B = 2k$.)

Then:
 * $\displaystyle Y \left({x}\right) - y \left({x}\right) = \int_a^x \left({f \left({t, Y \left({t}\right)}\right) - f \left({t, y \left({t}\right)}\right)}\right) \, \mathrm d t$

But:
 * $\left|{f \left({t, Y \left({t}\right)}\right) - f \left({t, y \left({t}\right)}\right)}\right| \le A \left|{Y \left({t}\right) - y \left({t}\right)}\right| \le AB$

So:
 * $\left|{Y \left({t}\right) - y \left({t}\right)}\right| \le AB \left|{x - a}\right|$

Repeating the argument, we can get successive estimates for the upper bound of $\left|{Y \left({x}\right) - y \left({x}\right)}\right|$ in $\left({a - h \,.\,.\, a + h}\right)$.

This gives:
 * $\displaystyle \frac {A^2 B}{2!} \left|{x - a}\right|^2, \ldots, \frac {A^n B}{n!} \left|{x - a}\right|^n, \ldots$

But this sequence tends to $0$.

So $Y \left({x}\right) = y \left({x}\right)$ in $\left({a - h \,.\,.\, a + h}\right)$.

It is also known as the Picard-Lindelöf Theorem or the Cauchy-Lipschitz Theorem, after, and.

Some sources give this as Picard's Theorem but there are other theorems with this appellation so it is better to disambiguate.