kb autocommit

content/posts/KBhpset_6.md

@@ -224,6 +224,10 @@ This allows us to write out:
 
 ### Problem 16.1, part b {#problem-16-dot-1-part-b}
 
+Explicit Euler's method relates to the slope field works by starting at the initial point, and taking the slope at that point, and "tracing" out to a distance of \\(h\\) on that slope + reevaluating.
+
+In essence, explicit Euler traces out piecewise linear segments of the function on the slope field.
+
 In Explicit Euler's method, we leverage the fact that:
 
 \begin{equation}
@@ -232,20 +236,24 @@ x\_{0}+ h f(x\_0) \approx x(t\_0 + h)
 
 to obtain our solution at some point beyond the initial point. A graphical way of doing this would involve drawing piecewise linear line segments which span \\(h\\) in \\(t\\) and has slope \\(f(x\_0)\\), starting at point \\(x\_0\\). In problem \\(b\\), specifically, we have:
 
-{{< figure src="/ox-hugo/2024-02-14_17-33-18_screenshot.png" >}}
+{{< figure src="/ox-hugo/2024-02-20_14-43-08_screenshot.png" >}}
 
 
 ### Problem 16.2, part a {#problem-16-dot-2-part-a}
 
-In Explicit Euler's method, we seek a point such that:
+In Implicit Euler's method, we seek a point such that:
 
 \begin{equation}
 x\_{i+1} - h f(x\_{i+1}) = x\_i
 \end{equation}
 
-to obtain our solution at some point beyond the initial point. Essentially, we go backwards from each \\(x\_{i+1}\\) to "connect" the best line such the slope at that line can reach backwards into hitting our previous point \\(x\_{i}\\). We keep going ahead by steps \\(h\\), and "connecting" backwards to where we last finished computation. For \\(y' = -3y\\), we have:
+to obtain our solution at some point beyond the initial point. Essentially, we go backwards from each \\(x\_{i+1}\\) in distance \\(h\\) from our previous point to "connect" the best line such the slope at that line can reach backwards into hitting our previous point \\(x\_{i}\\). The slopes chosen at each step should be the slope at the destination step, and not the source step.
+
+We keep going ahead by steps \\(h\\), and "connecting" backwards to where we last finished computation. For \\(y' = -3y\\), we have:
+
+This method may get stuck if, after a time period ahead of your current point \\(f(x\_{t+1})\\) results in a value for which there's no valid solution for \\(x\_{t+1}\\) which is connected to your previous point.
 
-{{< figure src="/ox-hugo/2024-02-14_17-41-29_screenshot.png" >}}
+{{< figure src="/ox-hugo/2024-02-20_14-45-48_screenshot.png" >}}
 
 
 ## Chapter 17 {#chapter-17}

content/posts/KBhsu_math53_feb212024.md

@@ -4,7 +4,7 @@ author = ["Houjun Liu"]
 draft = false
 +++
 
-A [Partial Differential Equation]({{< relref "KBhpartial_differential_equations.md" >}}) is a [Differential Equation]({{< relref "KBhdiffeq_intro.md" >}}) which has more than one **independent variable**.
+A [Partial Differential Equation]({{< relref "KBhpartial_differential_equations.md" >}}) is a [Differential Equation]({{< relref "KBhdiffeq_intro.md" >}}) which has more than one **independent variable**: $u(x,y), u(t,x,y), ...$
 
 For instance:
 
@@ -13,6 +13,200 @@ For instance:
 \end{equation}
 
 
+## Key Intuition {#key-intuition}
+
+-   [PDE]({{< relref "KBhpartial_differential_equations.md" >}})s may have no solutions (unlike [Uniqueness and Existance]({{< relref "KBhuniqueness_and_existance.md" >}}) for [ODE]({{< relref "KBhordinary_differential_equations.md" >}})s)
+-   yet, usually, there are too many solutions---so... how do you describe all solutions?
+-   usually, there are no explicit formulas
+
+
+## Laplacian of \\(u(x,y)\\) {#laplacian-of-u--x-y}
+
+\begin{equation}
+\pdv[2]{u}{x} + \pdv[2]{u}{y}
+\end{equation}
+
+Think about a Gaussian distribution, a bell shape curve. One important intuition is that the entire thing is curving down.
+
+
+## Examples {#examples}
+
+
+### Heat Equation {#heat-equation}
+
+heat distributes by "diffusing"; this is heat \\(u\\) diffusing across a plate
+
+\begin{equation}
+\pdv{u}{t} = \alpha \qty( \pdv[2]{u}{x} + \pdv[2]{u}{y})
+\end{equation}
+
+
+#### Removing a constant {#removing-a-constant}
+
+Consider a function:
+
+\begin{equation}
+t = c \tau
+\end{equation}
+
+you can remove the constant by finanglisng because the constant drops out when scaled (i.e. you can just scale your results back TODO check this).
+
+
+#### Solution {#solution}
+
+We can get a solution:
+
+\begin{equation}
+u(t,x) = \frac{1}{\sqrt{t}}e^{-\frac{x^{2}}{4t}}
+\end{equation}
+
+we can check that it satisfy the equation through route algebra.
+
+
+#### Bell Curves {#bell-curves}
+
+\begin{equation}
+y = a e^{-\frac{x^{2}}{b}}
+\end{equation}
+
+this function takes maximum at \\(a\\), centered at \\(0\\), and as \\(b\\) increases the bell becomes more
+
+<!--list-separator-->
+
+-  Integrating a Bell Curve
+
+    \begin{equation}
+    \int\_{-\infty}^{\infty} \frac{1}{\sqrt{t}} e^{-\frac{x^{2}}{4t}} \dd{x}
+    \end{equation}
+
+    let us declare: \\(u = \frac{x}{2 \sqrt{t}}\\)
+
+    meaning:
+
+    \begin{equation}
+    2 \int\_{-\infty}^{\infty} e^{-u^{2}}\dd{u}
+    \end{equation}
+
+    which is a constant. So, no matter the \\(t\\), we have a constant value of energy.
+
+
+### Wave Equation {#wave-equation}
+
+\begin{equation}
+u(t,x), u(t,x,y)
+\end{equation}
+
+we describe it:
+
+\begin{equation}
+\pdv[2]{u}{t} = c^{2} \qty(\pdv[2]{u}{x} + \pdv[2]{u}{y})
+\end{equation}
+
+If we write it in a single set of variables:
+
+\begin{equation}
+\pdv[2]{u}{t} = \pdv[2]{u}{x}
+\end{equation}
+
+One particular solution:
+
+\begin{equation}
+u(t,x) = \cos t \sin x
+\end{equation}
+
+If you consider traveling across \\(t\\), you will note that we begin at \\(\cos (1) = 1\\), then slowly travel to \\(\cos (\frac{\pi}{2}) = 0\\), and back and forth.
+
+
+#### General Standing Wave Solution {#general-standing-wave-solution}
+
+Because the [PDE]({{< relref "KBhpartial_differential_equations.md" >}}) given is linear, solutions compose, and we note that any scale of \\(\cos kt \sin kx\\) will compose.
+
+\begin{equation}
+u(t,x) = \sum\_{k=0}^{\infty} a\_{k} \cos kt \sin kx
+\end{equation}
+
+
+#### Fourier Series {#fourier-series}
+
+\begin{equation}
+u(o,x) \sum\_{k} a\_{k}\sin kx
+\end{equation}
+
+BIG **stunning conclusion**: **every single function, including wack ones, can be decomposed** :
+
+
+#### General Traveling Wave Solution {#general-traveling-wave-solution}
+
+\begin{equation}
+u(t,x) = \sin (x-t) w(x-t)
+\end{equation}
+
+as long as \\(w\\) is a valid twice-differentiable solution, plugging its derivative in will resolve as well.
+
+<!--list-separator-->
+
+-  Composition
+
+    \begin{equation}
+    \sin (x-t) + \sin (x+t) = \sin x \cos t - \cos x \sin t + \sin x \cos t + \cos x \sin t = 2 \sin x \cos t
+    \end{equation}
+
+
+### Transport Equation {#transport-equation}
+
+\begin{equation}
+\pdv{u}{t} = \pdv{u}{x}
+\end{equation}
+
+generally any \\(u = w(x+t)\\) should solve this
+
+
+### Schrodinger Equation {#schrodinger-equation}
+
+We have some:
+
+\begin{equation}
+u(x,t)
+\end{equation}
+
+and its a complex-valued function:
+
+\begin{equation}
+i \pdv{u}{t} = \pdv[2]{u}{x}
+\end{equation}
+
+which results in a superposition in linear equations
+
+
+### Nonlinear Example {#nonlinear-example}
+
+\begin{equation}
+\pdv{u}{t} = \pdv[2]{u}{x} + u(1-u)
+\end{equation}
+
+this is a [PDE]({{< relref "KBhpartial_differential_equations.md" >}}) variant of the [logistic equation]({{< relref "KBhlogistic_equations.md" >}}): this is **non-linear**
+
+
+### Monge-Ampere Equations {#monge-ampere-equations}
+
+\begin{equation}
+u(x,y)
+\end{equation}
+
+
+#### Hessian {#hessian}
+
+\begin{equation}
+Hess(u) = \mqty(\pdv[2]{u}{x} & \frac{\partial^{2} u}{\partial x \partial y}  \\\ \frac{\partial^{2} u}{\partial x \partial y} & \pdv[2]{u}{y})
+\end{equation}
+
+If we take its determinant, we obtain:
+
+\begin{equation}
+\pdv[2]{u}{x} \pdv[2]{u}{y} - \qty(\frac{\partial^{2} u}{\partial x \partial y})^{2}
+\end{equation}
+
+
 ## Linear Partial Differential Equation {#linear-partial-differential-equation}
 
 A PDE is a [Linear PDE](#linear-partial-differential-equation) when it takes on the form of: