nyuuzyou/wikis huggingface dataset:
Solution of Linear Congruence/Examples/325 n = 11 mod 3

== Example of Solution of Linear Congruence ==
Let $325 n = 11 \pmod 3$.
Then:
:$x = 2 + 3 t$
where $t \in \Z$.
== Proof ==
We have that:
:$325 \equiv 1 \pmod 3$
and:
:$11 \equiv 2 \pmod 3$
Thus the congruence can be expressed as:
:$n \equiv 2 \pmod 3$
and the solution is found by inspection.
== Sources ==
* : $\text {4-2}$ Residue Systems: Example $\text {4-7}$
Category:Examples of Modulo Arithmetic
Solution of Linear Congruence/Examples/5 x = 4 mod 3

== Example of Solution of Linear Congruence ==
Let $5 x = 4 \pmod 3$.
Then:
:$x = 2 + 3 u$
where $u \in \Z$.
== Proof ==
From Solution of Linear Diophantine Equation, the general solution to $(1)$ is:
:$(2): \quad \forall t \in \Z: x = x_0 + 3 t, k = k_0 + 5 t$
where $x_0, k_0$ can be found as follows.
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {5, -3} = 1$
which is (trivially) a divisor of $4$.
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $5 x - 3 k = 4$.
Again with the Euclidean Algorithm:
and so:
is a solution.
Thus:
Setting $u = t + 2$ gives the result.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $1 \ \text{(a)}$
Category:Examples of Modulo Arithmetic
Solution of Linear Congruence/Examples/6 x = 5 mod 4

== Example of Solution of Linear Congruence ==
Let $6 x = 5 \pmod 4$.
Then $x$ has no solution in $\Z$.
== Proof ==
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {6, -4} = 2$
which is not a divisor of $5$.
So, from Solution of Linear Diophantine Equation, no solution exists.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $2 \ \text{(a)}$
Category:Examples of Modulo Arithmetic
Solution of Linear Congruence/Examples/7 x = 6 mod 5

== Example of Solution of Linear Congruence ==
Let $7 x = 6 \pmod 5$.
Then:
:$x = 3 + 5 u$
where $u \in \Z$.
== Proof ==
From Solution of Linear Diophantine Equation, the general solution to $(1)$ is:
:$(2): \quad \forall t \in \Z: x = x_0 + 5 t, k = k_0 + 7 t$
where $x_0, k_0$ can be found as follows.
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {7, -5} = 1$
which is (trivially) a divisor of $6$.
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $7 x - 5 k = 6$.
Again with the Euclidean Algorithm:
and so:
is a solution.
Thus:
Setting $u = t + 3$ gives the result.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $1 \ \text{(b)}$
Category:Examples of Modulo Arithmetic
Solution of Linear Congruence/Examples/9 x = 8 mod 7

== Example of Solution of Linear Congruence ==
Let $9 x = 8 \pmod 7$.
Then:
:$x = 4 + 7 t$
where $t \in \Z$.
== Proof ==
We have that:
:$8 \equiv 1 \pmod 7$
and so:
That should simplify the arithmetic.
Then:
From Solution of Linear Diophantine Equation, the general solution to $(1)$ is:
:$(2): \quad \forall t \in \Z: x = x_0 + 7 t, k = k_0 + 9 t$
where $x_0, k_0$ can be found as follows.
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {9, -7} = 1$
which is (trivially) a divisor of $1$.
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $9 x - 7 k = 1$.
Again with the Euclidean Algorithm:
and so:
is a solution.
Thus from $(2)$:
:$x = 4 + 7 t$
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $1 \ \text{(c)}$
Category:Examples of Modulo Arithmetic
Solution of Linear Congruence/Existence

== Theorem ==
Let $a x \equiv b \pmod n$ be a linear congruence.
$a x \equiv b \pmod n$ has at least one solution :
:$\gcd \set {a, n} \divides b$
that is, $\gcd \set {a, n}$ is a divisor of $b$.
== Proof ==
Consider the linear congruence $a x \equiv b \pmod n$.
Suppose $\exists x_0 \in \Z: a x_0 \equiv b \pmod n$.
Then $\exists y_0 \in Z: a x_0 - b = n y_0$ by definition of congruence.
Thus $x = x_0, y = y_0$ is a solution to the linear Diophantine equation $a x - n y = b$.
On the other hand, if $x = x_0, y = y_0$ is a solution to the linear Diophantine equation $a x - n y = b$, then it follows that $a x \equiv b \pmod n$.
Hence:
:the problem of finding all integers satisfying the linear congruence $a x \equiv b \pmod n$ 
is the same problem as:
:the problem of finding all the $x$ values in the linear Diophantine equation $a x - n y = b$.
From Solution of Linear Diophantine Equation:
The linear Diophantine equation $a x - n y = b$ has at least one solution :
:$\gcd \set {a, n} \divides b$
Hence the result.
== Sources ==
* 
* 
Category:Solution of Linear Congruence
Solution of Linear Congruence/Number of Solutions

== Theorem ==
Let $a x \equiv b \pmod n$ be a linear congruence.
Let $\gcd \set {a, n} = d$.
Then $a x \equiv b \pmod n$ has $d$ solutions which are given by the unique solution modulo $\dfrac n d$ of the congruence:
:$\dfrac a d x \equiv \dfrac b d \paren {\bmod \dfrac n d}$
== Proof ==
From Solution of Linear Congruence: Existence:
:the problem of finding all integers satisfying the linear congruence $a x \equiv b \pmod n$ 
is the same problem as:
:the problem of finding all the $x$ values in the linear Diophantine equation $a x - n y = b$.
From Integers Divided by GCD are Coprime:
:$\gcd \set {\dfrac a d, \dfrac n d} = 1$
So the has a unique solution modulo $\dfrac n d$, say:
:$x \equiv x_1 \paren {\bmod \dfrac n d}$
So the integers $x$ which satisfy $a x \equiv b \pmod n$ are exactly those of the form $x = x_1 + k \dfrac n d$ for some $k \in \Z$.
Consider the set of integers:
: $\set {x_1, x_1 + \dfrac n d, x_1 + 2 \dfrac n d, \ldots, x_1 + \paren {d - 1} \dfrac n d}$
None of these are congruent modulo $n$ and none differ by as much as $n$.
Further, for any $k \in Z$, we have that $x_1 + k \dfrac n d$ is congruent modulo $n$ to one of them.
To see this, write $k = d q + r$ where $0 \le r < d$ from the Division Theorem.
Then:
So these are the $d$ solutions of $a x \equiv b \pmod n$.
== Sources ==
* 
* 
Category:Solution of Linear Congruence
Solution of Linear Congruence/Unique iff Coprime to Modulus

== Theorem ==
Let $a x \equiv b \pmod n$ be a linear congruence.
$a x \equiv b \pmod n$ has a unique solution $\gcd \set {a, n} = 1$.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
From Solution of Linear Congruence: Existence:
:the problem of finding all integers satisfying the linear congruence $a x \equiv b \pmod n$ 
is the same problem as:
:the problem of finding all the $x$ values in the linear Diophantine equation $a x - n y = b$.
Let:
:$\gcd \set {a, n} = 1$
Let $x = x_0, y = y_0$ be one solution to the linear Diophantine equation:
:$a x - n y = b$
From Solution of Linear Diophantine Equation, the general solution is:
:$\forall k \in \Z: x = x_0 + n k, y = y_0 + a k$
But:
:$\forall k \in \Z: x_0 + n k \equiv x_0 \pmod n$
Hence $x \equiv x_0 \pmod n$ is the only solution of $a x \equiv b \pmod n$.
== Sources ==
* : $\S 2.3$: Congruences: Theorem $4$
Category:Solution of Linear Congruence
Solution of Linear Diophantine Equation

== Theorem ==
The linear Diophantine equation:
:$a x + b y = c$
has solutions :
:$\gcd \set {a, b} \divides c$
where $\divides$ denotes divisibility.
If this condition holds with $\gcd \set {a, b} > 1$ then division by $\gcd \set {a, b}$ reduces the equation to:
:$a' x + b' y = c'$
where $\gcd \set {a', b'} = 1$.
If $x_0, y_0$ is one solution of the latter equation, then the general solution is:
:$\forall k \in \Z: x = x_0 + b' k, y = y_0 - a' k$
or:
:$\forall k \in \Z: x = x_0 + \dfrac b d k, y = y_0 - \dfrac a d k$
where $d = \gcd \set {a, b}$.
== Proof ==
We assume that both $a$ and $b$ are non-zero, otherwise the solution is trivial.
The first part of the problem is a direct restatement of Set of Integer Combinations equals Set of Multiples of GCD:
The set of all integer combinations of $a$ and $b$ is precisely the set of integer multiples of the GCD of $a$ and $b$:
:$\gcd \set {a, b} \divides c \iff \exists x, y \in \Z: c = x a + y b$
Now, suppose that $x', y'$ is any solution of the equation.
Then we have:
:$a' x_0 + b' y_0 = c'$ and $a' x' + b' y' = c'$
Substituting for $c'$ and rearranging:
:$a' \paren {x' - x_0} = b' \paren {y_0 - y'}$
So:
:$a' \divides b' \paren {y_0 - y'}$
Since $\gcd \set {a', b'} = 1$, from Euclid's Lemma we have:
:$a' \divides \paren {y_0 - y'}$.
So $y_0 - y' = a' k$ for some $k \in \Z$.
Substituting into the above gives $x' - x_0 = b' k$ and so:
:$x' = x_0 + b' k, y' = y_0 - a'k$ for some $k \in \Z$
which is what we claimed.
Substitution again gives that the integers:
:$x_0 + b' k, y_0 - a' k$
constitute a solution of $a' x + b' y = c'$ for any $k \in \Z$.
== Examples ==
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Theorem $\text {2-4}$
Category:Linear Diophantine Equations
Category:Greatest Common Divisor
Solution of Ljunggren Equation

== Theorem ==
The only solutions of the Ljunggren equation:
:$x^2 + 1 = 2 y^4$
are:
:$x = 1, y = 1$
:$x = 239, y = 13$
== Proof ==
Setting $x = 1$:
and so $y = 1$.
Setting $x = 239$:
and so $y = 13$.
== Also see ==
* Largest Prime Factor of $n^2 + 1$
== Sources ==
* 
* 
* 
* : $239$
Category:Diophantine Equations
Category:239
Category:13
Solution of Ljunggren Equation/Mistake

== Source Work ==
:
:The Dictionary
::$220$
== Mistake ==
:The only solution of the equation $x^2 + 1 = 2 y^2$ is $x = 239, y = 13$.
There also exists the solution $x = 1, y = 1$.
It perhaps also needs to be pointed out that the equation is Diophantine.
== Sources ==
* : $239$
Category:Mistakes/Curious and Interesting Numbers
Solution of Pell's Equation is a Convergent

== Theorem ==
Let $x = a, y = b$ be a positive solution to Pell's equation $x^2 - n y^2 = 1$.
Then $\dfrac a b$ is a convergent of $\sqrt n$.
== Proof ==
Let $a^2 - n b^2 = 1$.
Then we have:
:$\paren {a - b \sqrt n} \paren {a + b \sqrt n} = 1$.
So:
:$a - b \sqrt n = \dfrac 1 {a + b \sqrt n} > 0$
and so $a > b \sqrt n$.
Therefore:
The result follows from Condition for Rational to be Convergent.
Category:Pell's Equation
Category:Continued Fractions
Solution of Qunitic Equation using Elliptic Functions/Historical Note

== Historical Note on Solution of Qunitic Equation using Elliptic Functions ==
The Abel-Ruffini Theorem, on the general insolubility of the quintic by radicals, was proved by in $1823$.
However, demonstrated a technique for solving a general quintic using elliptic functions in $1858$.
== Sources ==
* : Chapter $\text {A}.30$: Hermite ()
Category:Historical Notes
Solution of Second Order Differential Equation with Missing Dependent Variable

== Theorem ==
Let $\map f {x, y', y} = 0$ be a second order ordinary differential equation in which the dependent variable $y$ is not explicitly present.
Then $f$ can be reduced to a first order ordinary differential equation, whose solution can be determined.
== Proof ==
Consider the second order ordinary differential equation:
:$(1): \quad \map f {x, y', y} = 0$
Let a new dependent variable $p$ be introduced:
:$y' = p$
:$y = \dfrac {\d p} {\d x}$
Then $(1)$ can be transformed into:
:$(2): \quad \map f {x, p, \dfrac {\d p} {\d x} } = 0$
which is a first order ODE.
If $(2)$ has a solution which can readily be found, it will be expressible in the form:
:$(3): \quad \map g {x, p}$
which can then be expressed in the form:
:$\map g {x, \dfrac {\d y} {\d x} } = 0$
which is likewise subject to the techniques of solution of a first order ODE.
Hence such a second order ODE is reduced to the problem of solving two first order ODEs in succession.
== Sources ==
* : $\S 2.11$: Reduction of Order
Category:Second Order ODEs
Solution of Second Order Differential Equation with Missing Independent Variable

== Theorem ==
Let $\map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$ be a second order ordinary differential equation in which the independent variable $x$ is not explicitly present.
Then $g$ can be reduced to a first order ordinary differential equation, whose solution can be determined.
== Proof ==
Consider the second order ordinary differential equation:
:$(1): \quad \map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$
Let a new dependent variable $p$ be introduced:
:$y' = p$
Hence:
:$y = \dfrac {\d p} {\d x} = \dfrac {\d p} {\d y} \dfrac {\d y} {\d x} = p \dfrac {\d p} {\d y}$
Then $(1)$ can be transformed into:
:$(2): \quad \map g {y, p, p \dfrac {\d p} {\d y} = 0}$
which is a first order ODE.
If $(2)$ has a solution which can readily be found, it will be expressible in the form:
:$(3): \quad \map g {x, p}$
which can then be expressed in the form:
:$\map g {x, \dfrac {\d y} {\d x} }$
which is likewise subject to the techniques of solution of a first order ODE.
Hence such a second order ODE is reduced to the problem of solving two first order ODEs in succession.
== Sources ==
* : $\S 2.11$: Reduction of Order
Category:Second Order ODEs
Solution to Bach or Stravinsky?

== Solution to Bach or Stravinsky? ==
== Proof ==
From the payoff table:
There are two Nash equilibria:
:$\tuple {\text {Bach}, \text {Bach} }$
:$\tuple {\text {Stravinsky}, \text {Stravinsky} }$
Thus there are two steady states:
:one in which both players always choose Bach
:one in which both players always choose Stravinsky.
Either, but not both, experience their preferred music.
== Sources ==
* : $\text I$ Strategic Games: Chapter $2$ Nash Equilibrium: $2.3$: Examples: Example $15.3$
Category:Examples of Two-Person Games
Solution to Bernoulli's Equation

== Theorem ==
Bernoulli's equation:
:$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
where:
:$n \ne 0, n \ne 1$
has the general solution:
:$\ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
where:
:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
== Proof ==
Make the substitution:
:$z = y^{1 - n}$
in $(1)$.
Then we have:
This is now a linear first order ordinary differential equation in $z$.
It has an integrating factor:
and this can be used to obtain:
:$\ds \map \mu x z = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
Substituting $z = y^{1 - n} = \dfrac 1 {y^{n - 1} }$ finishes the proof.
== Also see ==
When $n = 0$ or $n = 1$ the equation is linear, and Solution to Linear First Order Ordinary Differential Equation can be used.
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.3$: Bernoulli's equation
* : $\S 2.10$: Problem $3$
Category:Bernoulli's Equation
Solution to Card Game with Bluffing

== Solution to Card Game with Bluffing ==
== Proof ==
From the payoff table:
The solution is:
: $A$ takes strategy $A_1$ for $2/3$ of the time, and $A_2$ for $1/3$ of the time.
: $B$ takes strategy $B_1$ for $2/3$ of the time, and $B_2$ for $1/3$ of the time.
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Examples of Two-Person Zero-Sum Games
Solution to Coordination Game

== Solution to Coordination Game ==
== Proof ==
From the payoff table:
There are two Nash equilibria:
:$\tuple {\text {Mozart}, \text {Mozart} }$
:$\tuple {\text {Mahler}, \text {Mahler} }$
Thus there are two steady states:
:one in which both players always choose Mozart
:one in which both players always choose Mahler.
Just because both players have a mutual interest in reaching the preferred Nash equilibrium $\tuple {\text {Mozart}, \text {Mozart} }$, this does not rule out the steady state outcome $\tuple {\text {Mahler}, \text {Mahler} }$.
== Sources ==
* : $\text I$ Strategic Games: Chapter $2$ Nash Equilibrium: $2.3$: Examples: Example $16.1$
Category:Examples of Two-Person Games
Solution to Differential Equation/Examples

== Examples of Solutions to Differential Equations ==
=== Arbitrary Order $1$ ODE: $1$ ===
=== Arbitrary Order $1$ ODE: $2$ ===
=== Arbitrary Order $1$ ODE: $3$ ===
=== Arbitrary Order $2$ ODE ===
=== Arbitrary Order $2$, Degree $3$ ODE ===
=== Equation which is Not a Solution ===
=== Absolute Value Function ===
Category:Examples of Differential Equations
Solution to Differential Equation/Examples/Absolute Value Function

== Examples of Solutions to Differential Equations ==
Consider the real function defined as:
:$\map f x = \size x$
where $\size x$ is the absolute value function.
Then $\map f x$ cannot be the solution to a differential equation.
However, by suitably restricting $\map f x$ to a domain which does not include $x = 0$, there may well exist differential equations for which the resulting real function is a solution.
== Proof ==
By definition, for $\map f x$ to be a solution to a differential equation $Q$, it must satisfy $Q$ at all elements of its domain.
But by Derivative of Absolute Value Function, the derivative of $\size x$ $x$ does not exist at $x = 0$.
So at $x = 0$, $\map {f'} x$ is not defined.
However,let $\map f x$ be restricted to $x \ne 0$:
:$\forall x \in \R_{\ne 0}: \map g x := \size x$
Then it is seen that $\map g x$ is a solution to the differential equation:
:$y' = \begin {cases} 1 & : x > 0 \\ -1 & : x < 0 \end {cases}$
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: Comment $3.53$
Category:Examples of Differential Equations
Solution to Differential Equation/Examples/Arbitrary Order 1 ODE: 1

== Examples of Solutions to Differential Equations ==
Consider the real function defined as:
:$y = \ln x + C$
defined on the domain $x \in \R_{>0}$.
Then $\map f x$ is a solution to the first order ODE:
:$(1): y' = \dfrac 1 x$
defined on the domain $x \in \R_{>0}$.
== Proof ==
It is noted that $\map f x$ is not defined in $\R$ when $x \le 0$ because for those values of $x$ the logarithm is not defined.
It is also noted that $(1)$ is indeed defined for all $x \in \R_{>0}$.
Having established that, we continue:
and it is seen immediately that $(2)$ is the first order ODE $(1)$.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: Example $3.51$
Category:Examples of Differential Equations
Solution to Differential Equation/Examples/Arbitrary Order 1 ODE: 2

== Examples of Solutions to Differential Equations ==
Consider the real function defined as:
:$y = \tan x - x$
defined on the domain $S := \set {x \in \R: x \ne \dfrac {\paren {2 n + 1} \pi} 2, n \in \Z}$.
Then $\map f x$ is a solution to the first order ODE:
:$(1): y' = \paren {x + y}^2$
when $x$ is restricted to $S$.
== Proof ==
It is noted that $\map f x$ is not defined in $\R$ when $x = \dfrac {\paren {2 n + 1} \pi} 2$ because for those values of $x$ the tangent is not defined.
Hence the restriction of $x$ to $S$
It is also noted that $(1)$ is indeed defined for all $x \in S$.
Having established that, we continue:
For ease of manipulation we rewrite $(1)$ as:
:$(2): y' - \paren {x + y}^2 = 0$
Then:
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: Example $3.52$
Category:Examples of Differential Equations
Solution to Differential Equation/Examples/Arbitrary Order 1 ODE: 3

== Examples of Solutions to Differential Equations ==
Consider the equation:
:$(1): \quad y = \dfrac {x^2} 2$
for all $x \in \R$.
Then $(1)$ is a solution to the first order ODE:
:$y' = x$
== Proof ==
We have that:
== Sources ==
* 
* 
Category:Examples of Differential Equations
Solution to Differential Equation/Examples/Arbitrary Order 2 Degree 3 ODE

== Examples of Solutions to Differential Equations ==
Consider the equation:
:$(1): \quad y = x^2$
where $x \in \R$.
Then $(1)$ is a solution to the second order ODE:
:$(2): \quad \paren {y' '}^3 + \paren {y'}^2 - y - 3 x^2 - 8 = 0$
defined on the domain $x \in \R$.
== Proof ==
Then:
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: Example $3.5$
Category:Examples of Differential Equations
Solution to Differential Equation/Examples/Arbitrary Order 2 ODE

== Examples of Solutions to Differential Equations ==
Consider the real function defined as:
:$y = \map f x = \ln x + x$
defined on the domain $x \in \R_{>0}$.
Then $\map f x$ is a solution to the second order ODE:
:$(1): \quad x^2 y' ' + 2 x y' + y = \ln x + 3 x + 1$
defined on the domain $x \in \R_{>0}$.
== Proof ==
It is noted that $\map f x$ is not defined in $\R$ when $x \le 0$ because that is outside the domain of the natural logarithm $\ln$.
The same constraint applies to $(1)$.
Having established that, we continue:
Then:
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation
Category:Examples of Differential Equations
Solution to Differential Equation/Examples/Equation which is Not a Solution

== Examples of Solutions to Differential Equations ==
Consider the equation:
:$(1): \quad y = \sqrt {-\paren {1 + x^2} }$
where $x \in \R$.
Consider the first order ODE:
:$(2): \quad x + y y' = 0$
Then despite the fact that the formal substition for $y$ and $y'$ from $(1)$ into $(2)$ yields an identity, $(1)$ is not a solution to $(2)$.
== Proof ==
First we note that:
Then:
However, by Domain of Real Square Root Function, $\sqrt {-\paren {1 + x^2} }$ is defined for $-\paren {1 + x^2} \ge 0$.
But by Square of Real Number is Non-Negative:
:$1 + x^2 > 0$
and so:
:$-\paren {1 + x^2} < 0$
So there exists no $x \in \R$ for which $y = \sqrt {-\paren {1 + x^2} }$ is defined.
So $(1)$ does not define a real function and so $(1)$ is not a solution to $(2)$.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: Comment $3.43$
Category:Examples of Differential Equations
Solution to Distributional Ordinary Differential Equation with Constant Coefficients

== Theorem ==
Let $D$ be an ordinary differential operator with constant complex coefficients:
:$\ds D = \sum_{k \mathop = 0}^n a_k \paren {\dfrac \d {\d x}}^k$
Let $f \in \map {C^\infty} \R$ be a smooth real function.
Let $T \in \map {\DD'} \R$ be a distribution.
Let $T_f$ be a distribution associated with $f$.
Suppose $T$ is a distributional solution to $D T = T_f$.
Then $T = T_F$ where $F \in \map {C^\infty} \R$ is a classical solution to $D F = f$.
== Proof ==
Let $\map P \xi$ be a polynomial over complex numbers such that:
:$\ds \map P \xi = \sum_{k \mathop = 0}^n a_k \xi^k = a_n \prod_{k \mathop = 0}^n \paren {\xi - \lambda_k}$
where $a_n \ne 0$.
Then there exists a polynomial $\map Q \xi$ such that:
:$\map P \xi = \paren {\xi - \lambda_n} \map Q \lambda$
Let:
:$\ds D = \sum_{k \mathop = 0}^n a_k \paren {\dfrac \d {\d x}}^k$
Then:
:$D = \map P {\dfrac \d {\d x} }$
Furthermore:
:$D = \paren {\dfrac \d {\d x} - \lambda_n} D_1$
where:
:$D_1 := \map Q {\dfrac \d {\d x} }$
Now we will use the principle of mathematical induction to show that:
:$\paren {DT = T_f, f \in \map {C^\infty} \R} \implies \paren {T = T_F, F \in \map {C^\infty} \R}$
=== Basis for the Induction ===
$n = 1$ case is covered by Dx - k is Hypoelliptic.
=== Induction Hypothesis ===
Suppose for $D$ of the order $n \in \N$ it holds that:
:$\paren {DT = T_f, f \in \map {C^\infty} \R} \implies \paren {T = T_F, F \in \map {C^\infty} \R}$
We will show that the same property holds for $D$ of the order $n + 1$.
=== Induction Step ===
Let $D$ be of the order $n + 1 \in \N$.
Let:
:$D = \paren {\dfrac \d {\d x} - \lambda_{n + 1}} D_1$
where $D_1$ is of the order $n \in \N$.
Then:
:$ \paren {\dfrac \d {\d x} - \lambda_{n + 1}} D_1 T = T_f$
By Dx - k is Hypoelliptic:
:$\paren {\paren {\dfrac \d {\d x} - \lambda_{n + 1} } D_1 T = T_f, f \in \map {C^\infty} \R} \implies \paren {D_1 T = T_g, g \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - \lambda_{n + 1} } g = f}$
By the induction hypothesis:
:$\paren {D_1 T = T_g, g \in \map {C^\infty} \R} \implies \paren {T = T_F, F \in \map {C^\infty} \R : D_1 F = g}$
Just to make sure:
Hence, in the distributional sense we have that $D T = T_f$ is solved by $T = T_F$ where $D F = f$ and $F \in \map {C^\infty} \R$.
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Hypoelliptic Operators
Category:Examples of Distributional Solutions
Category:Distributional Derivatives