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diff --git a/mindmap/differential equation.org b/mindmap/differential equation.org index 8229587..c509da0 100644 --- a/mindmap/differential equation.org +++ b/mindmap/differential equation.org @@ -17,10 +17,301 @@ initial value problems. Differential equations are often used to model real world systems, and are the main tool in numerical simulations of said systems. ** ODE +:PROPERTIES: +:ID: 5ef63bef-2d8f-4e00-b292-8206cf69469a +:END: An ODE is a differential equation involving single variable function solution classes and their derivatives. ** PDE +:PROPERTIES: +:ID: 365190d8-0f3a-4728-9b09-83a216292256 +:END: A PDE is a differential equation involving multivariable function solution classes and generally involve partial derivatives of the unknown function. ** initial value problem +:PROPERTIES: +:ID: bc7e9e01-9721-4b3e-a886-74a2fd27daf3 +:END: An initial value problem is a problem where one is given a differential equation and particular values of the unknown function and particular values of its derivatives, and the result is a particular solution. +** separable differential equation +:PROPERTIES: +:ID: 8e9c975a-cd75-447e-b094-16258147d83c +:END: +For [[id:5ef63bef-2d8f-4e00-b292-8206cf69469a][ODEs]], separable differential equations are differential equations of the form: +\begin{align} +\label{} +\frac{dy}{dx} = f(y)g(x) +\end{align} +which can be solved by taking an integral on both sides: +\begin{align} +\label{} +\frac{dy}{f(y)} = g(x)dx \\ +\int\frac{dy}{f(y)} = \int g(x)dx +\end{align} +evaluating the integrals and solving for $y$, you obtain solutions for $y$ in terms of $x$. +** [[id:ab024db7-6903-48ee-98fc-b2a228709c04][linear]] differential equation +:PROPERTIES: +:ID: 32a116d9-b813-4b5a-a2e8-6dd7b767ec16 +:END: +Linear differential equations are differential equations of the form: +\begin{align} +\label{} +[\sum_{i}f_{i}(x)D^{i}]y(x) = g(x) +\end{align} +where $D$ is the [[id:31d3944a-cddc-496c-89a3-67a56e821de3][derivative]] operator. They are linear because the unknown function $y(x)$ is being operated on by some +linear operator, and common methods in linear algebra can be used in order to analyze equations of this kind. For +example, a first-order linear differential equation would look like this: +\begin{align} +\label{} +[f(x)D + g(x)]y(x) = h(x) +\end{align} +which can be easily solved in the following way where $G(x) = \frac{g(x)}{f(x)}$ and $H(x) = \frac{h(x)}{f(x)}$: +\begin{align} +\label{} +[D + G(x)]y(x) = H(x) \\ +\mu(x)[D + G(x)]y(x) = \mu(x)H(x) \\ +\mu'(x) := G(x)\mu(x) \\ +D(\mu(x)y(x)) = \mu(x)H(x) \\ +y(x) = \frac{\int\mu(x)H(x)dx}{\mu(x)} +\end{align} +Now to solve for $\mu(x)$, using [[id:8e9c975a-cd75-447e-b094-16258147d83c][separable differential equation]] methods: +\begin{align} +\label{} +\frac{d\mu}{dx} = G(x)\mu(x) \\ +\frac{1}{\mu}d\mu = G(x)dx \\ +\int\frac{1}{\mu}d\mu = \int G(x)dx \\ +ln(\mu) = \int G(x)dx \\ +e^{\int G(x)dx} = \mu +\end{align} +Therefore: +\begin{align} +\label{} +y(x) = \frac{\int e^{\int G(x)dx}H(x)dx}{e^{\int G(x)dx}} +\end{align} +Then, to model any particular first order system, plug in functions for $G(x)$ and $H(x)$. +*** superposition principle +:PROPERTIES: +:ID: 422653e2-daa4-422a-9cb7-3983a5a72554 +:END: +The principle of superposition states that any solutions $f_i(x)$ add to a new solution: +\begin{align} +\label{} +\sum_{i=0}^{N}f_{i}(x) = f_{new}(x) +\end{align} +that also satisfies the linear differential equation. This works because the operator is [[id:ab024db7-6903-48ee-98fc-b2a228709c04][linear]], so additive properties +work over this space. +*** Higher Order Linear Differential Equations +Solving higher order linear differential equations requires a couple of tricks. For example, transforms such +as the [[id:e73baa24-1a29-4f35-9d3d-0fad4a3a8e59][Laplace Transform]] or the [[id:262ca511-432f-404f-8320-09a2afe1dfb7][Fourier Transform]] may be used. Such transforms reduce differential equation +problems to algebraic problems, thus simplifying their solution methods. Other methods include guessing (I'm not +pulling your leg here, this is real), formulation as an eigenvalue problem, and taylor polynomial solutions. We will +take a look at all of these in this section. +*** Homogeneous Case +Take the case $Ay'' + By' + Cy = 0$, the substitute the form $y = De^{kt}$. Then: +\begin{align} +\label{} +De^{kt}(Ak^{2} + Bk + C) = 0 \\ +Ak^{2} + Bk + C = 0 +\end{align} +Then, use the quadratic formula to solve for $k$ in terms of the other constants. Such a polynomial is called the +characteristic polynomial of this differential equation. +*** Eigenvalue Problems +Eigenvalue problems can be solved just like in the familiar linear algebra case. For instance, take some differential +equation in this form: +\begin{align} +\label{} +A(f) = \lambda f +\end{align} +where $A$ is a linear operator in [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] space, and $\lambda$ is any constant. Traditionally, one would solve such an eigenvalue +problem like so: +\begin{align} +\label{} +\det{(A - \lambda I)} = 0 +\end{align} +In the simple example of a polynomial basis, this function $f$ can be represented as some linear combination of linearly +independent polynomials. A simple basis to choose could be the Taylor series basis i.e. $\vec{e_{n}} = x^{n}$ where $e_{n}$ is the nth +basis vector. Note that there are many polynomial bases that are an orthogonal basis and span this subset of function +space, but this is a simple example. In this case, the matrix $A$ would represent an operation on an infinite polynomial, +and the $\lambda I$ tells you to subtract $\lambda$ from all its diagonals. You can interpret this literally, using the following example. +**** Example +\begin{align} +\label{} +D(r^{2}D(f(r)) = \lambda f(r) +\end{align} +is such an example of an eigenvalue problem. Now, using the Taylor basis, we need to know two things: what $D$ is as an +infinite dimensional matrix in this basis, and what $t^{2}$ is as an infinite dimensional matrix. $f$ is some unknown vector +we are trying to solve for in this system. Note this observation: +\begin{align} +\label{} +\begin{pmatrix} +0 & 1 & 0 & 0 & 0 \\ +0 & 0 & 2 & 0 & 0 \\ +0 & 0 & 0 & 3 & 0 \\ +0 & 0 & 0 & 0 & 4 \\ +\end{pmatrix} +\begin{pmatrix} +a \\ +b \\ +c \\ +d \\ +e +\end{pmatrix} += +\begin{pmatrix} +b \\ +2c \\ +3d \\ +4e +\end{pmatrix} +\end{align} + +That this matrix encodes the power rule for the Taylor eigenbasis for 4 dimensions; each entry in the vectors encodes the nth +power monomial term, which means, for example: +\begin{align} +\label{} +\begin{pmatrix} +a \\ +b \\ +c \\ +d \\ +e +\end{pmatrix} := a + bx + cx^{2} + dx^{3} + ex^{4} +\end{align} +then the derivative of this vector would be: +\begin{align} +\label{derivative} +b + 2cx + 3dx^{2} + 4ex^{3} +\end{align} +which is exactly the coefficients in the resultant vector! Now, if we generalize this to an infinite amount of dimensions +(where the vector has an infinite length and the matrix has infinite entries), this corresponds to the same effect. + +Thus, the infinite matrix with the off-diagonal increasing entries is the $D$ matrix, or $D$ operator. But what is the +$r^{2}$ operator? We know it must be a matrix operation that shifts the entire vector up by two, and pads the first two +entries of the vector with two zeros. If we find this matrix, the matrix multiplication $DRD$ should yield a new +infinite matrix $M$, which we can use in order to solve the eigenvalue problem $\det{(M - \lambda I)} = 0$. Now this matrix is: +\begin{align} +\label{R matrix} +\begin{pmatrix} +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +1 & 0 & 0 & 0 & 0 \\ +0 & 1 & 0 & 0 & 0 \\ +0 & 0 & 1 & 0 & 0 \\ +\end{pmatrix} +\end{align} +and so on. This matrix does the exact same thing to a polynomial vector as what multiplying $r^{2}$ does to a polynomial. +We then multiply the two matrices to get this new matrix: +\begin{align} +\label{new} +\begin{pmatrix} +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +1 & 0 & 0 & 0 & 0 \\ +0 & 1 & 0 & 0 & 0 \\ +0 & 0 & 1 & 0 & 0 \\ +\end{pmatrix} +\begin{pmatrix} +0 & 1 & 0 & 0 & 0 \\ +0 & 0 & 2 & 0 & 0 \\ +0 & 0 & 0 & 3 & 0 \\ +0 & 0 & 0 & 0 & 4 \\ +0 & 0 & 0 & 0 & 0 +\end{pmatrix} = +\begin{pmatrix} +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +0 & 1 & 0 & 0 & 0 \\ +0 & 0 & 2 & 0 & 0 \\ +0 & 0 & 0 & 3 & 0 \\ +\end{pmatrix} +\end{align} +and so on, as you can see the pattern. Now we multiply in another $D$: +\begin{align} +\label{DS} +\begin{pmatrix} +0 & 1 & 0 & 0 & 0 \\ +0 & 0 & 2 & 0 & 0 \\ +0 & 0 & 0 & 3 & 0 \\ +0 & 0 & 0 & 0 & 4 \\ +0 & 0 & 0 & 0 & 0 +\end{pmatrix} +\begin{pmatrix} +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +0 & 1 & 0 & 0 & 0 \\ +0 & 0 & 2 & 0 & 0 \\ +0 & 0 & 0 & 3 & 0 \\ +\end{pmatrix} = +\begin{pmatrix} +0 & 0 & 0 & 0 & 0 \\ +0 & 2 \cdot 1 & 0 & 0 & 0 \\ +0 & 0 & 3 \cdot 2 & 0 & 0 \\ +0 & 0 & 0 & 4 \cdot 3 & 0 \\ +0 & 0 & 0 & 0 & 5 \cdot 4 +\end{pmatrix} +\end{align} +of course the $5 \cdot 4$ isn't actually resultant from the image above, but it is from the infinite version +of this process. Now, we can finally subtract $\lambda$ from this infinite matrix: +\begin{align} +\label{lambda} +\begin{pmatrix} +-\lambda & 0 & 0 & 0 & 0 \\ +0 & 2 \cdot 1 - \lambda & 0 & 0 & 0 \\ +0 & 0 & 3 \cdot 2 - \lambda & 0 & 0 \\ +0 & 0 & 0 & 4 \cdot 3 - \lambda & 0 \\ +0 & 0 & 0 & 0 & 5 \cdot 4 - \lambda +\end{pmatrix} +\end{align} +Now, you might be wondering how we're going to take the determinant of this infinite matrix. We can take +a [[id:122fd244-ffeb-47d0-89ce-bf9bc6f01b70][limit]] of finite matrices to find out what the generalization might be. For instance, the 3d case might look like this: +\begin{align} +\label{} +\lambda^{2}(2\cdot 1 - \lambda) +\end{align} +(as the very last diagonal entry in the finite case does not extend infinitely, there is no $3 \cdot 2$ term). Now taking some +higher dimensions: +\begin{align} +\label{higher dimensions} +\lambda^{2}(2 \cdot 1 - \lambda)(3 \cdot 2 - \lambda) \\ +\lambda^{2}(2 \cdot 1 - \lambda)(3 \cdot 2 - \lambda)(4 \cdot 3 - \lambda) \\ +\lambda^{2}(2 \cdot 1 - \lambda)(3 \cdot 2 - \lambda)(4 \cdot 3 - \lambda)(5 \cdot 4 - \lambda) +\end{align} +using inductive reasoning we should expect the infinite form to be: +\begin{align} +\label{} +det(M) = -\lambda(2 \cdot 1 - \lambda)(3 \cdot 2 - \lambda)(4 \cdot 3 - \lambda)(5 \cdot 4 - \lambda)(6 \cdot 5 - \lambda)\dots +\end{align} +(note that it isn't $\lambda^{2}$ because the very last $-\lambda$ never gets multiplied, and it's negative for that reason too). Note +that if we want to set $det(M) = 0$, $\lambda = n(n - 1)$ where $n$ is a natural number (including zero). Then we substitute +back in the $\lambda$ for some $n$, let's use $n = 2$ as an example: +\begin{align} +\label{n=2} +\begin{pmatrix} +-2 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 4 & 0 & 0 \\ +0 & 0 & 0 & 10 & 0 \\ +0 & 0 & 0 & 0 & 18 +\end{pmatrix} +\begin{pmatrix} +a_{1} \\ +a_{2} \\ +a_{3} \\ +a_{4} \\ +a_{5} \\ +\end{pmatrix} = +\begin{pmatrix} +0 \\ +0 \\ +0 \\ +0 \\ +0 +\end{pmatrix} +\end{align} +clearly, when we choose $n = 2$, the second value $a_{2}$ is free, and the rest for the given eigenfunction +must be zero, meaning for a given $n$, the resulting eigenvector is $ax^{n}$ for any value $a$. This is one of the solutions +to this differential equation. + +It turns out there's another solution in a space that the Taylor space does not span, but I'll leave it as an exercise +to find the other solution using this method, by extending it to include other kinds of functions. Note that for your +eigenbasis one can use the [[id:262ca511-432f-404f-8320-09a2afe1dfb7][Fourier Transform]] to make a Fourier basis, but that's also easy to generalize. |