PDE
Newtons method
When given the equation f(x) = 0, and f is a noninvertable function. We can apply newtons method. Take x_k to be close to our solution. Then we use the following x_{k+1} = x_k + \delta
Where \delta is the error. By taylor expanding f(x), we can get the following formula
x_{k+1} = x_k - \frac{f(x_k)}{f^\prime(x_k)}
Second Order Linear Constant Coefficients
The general equation is
\begin{cases} (p(x)y^{\prime})^\prime + q(x)y + \lambda r(x)y = 0 \\ \alpha_1 y(a) + \beta_1 y^\prime(a) = 0 \\ \alpha_2 y(b) + \beta_2 y^\prime(b) = 0 \end{cases}Take the equation
\begin{cases} y^{\prime\prime} + \lambda y = 0 \\ y(0) = 0 \\ y(1) + y^\prime(1) = 0 \end{cases}Solving the equation gets us
y_i(x) = \sin\left(x\sqrt{\lambda}\right)
Let s_i = \sqrt{\lambda_i}. We can then plug this into our integral and get
\int_0^1 \sin(s_i x) \sin(s_j x)\, dx.
Using wolframalpha,
\frac{b\cos(b)\sin(a) - a\cos(a)\sin(b)}{a^2 - b^2}
Using the equation
s_i\cos(s_i) + \sin s_i = 0
We can simplify and get
\frac{b\cos(b)\sin(a) - a\cos(a)\sin(b)}{a^2 - b^2} = 0
Laplaces Equation
Given the PDE
u_{xx} + u_{yy} = 0
We rewrite using the \overline{\nabla} operator, defined
\overline{\nabla} = \frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k}
Recall
\overline{\nabla}^2u = 0
We have from the divergence theorem
-k\iint\limits_S \overline{\nabla}u \cdot \hat{u}\,dS = -k\iiint\limits_V \overline{\nabla}u \, dV
Example of an ill posed problem
If given a problem like
\begin{cases} u_{xx} + u_{yy} = 0 \\ u(0,y) = u(\pi,y) = 0 \\ u(x,0) = 1\\ \frac{\partial x}{\partial y}(x,0) = 0 \end{cases}We see that the fourier series diverges, and there is no solution for y > 0.
Ellipitcal equations need to have boundary conditions upon their boundary, unlike parabloic or hyperbolic equations.
Occasionally when we have a problem with a shperical boundary like a circle, we will need to set up a boundary problem in polar coordinates.
Wave function
The general solution for a wave equation is in the form
u(x,t) = \frac{f(x+ct)+f(x-ct)}{2} + \frac{1}{2c}\int_{x-ct}^{x+ct}g(s)ds
When given a wave function in the form
\begin{cases} u_{tt} = c^2u_{xx}\\ u(x,0) = f(x) \\ u_t(x,0) = g(x)\\ u(0,t) = 0 \end{cases}We can invent a function f(x) such that f(x) is odd, forcing the solution to be zero. For example
\begin{cases} u_{tt} = c^2u_{xx}\\ u(x,0) = x^2\\ u_t(x,0) = 0\\ u(0,t) = 0 \end{cases}Clearly in this case the function x^2 is even. Because we are restricted to the positive realm, we can write f(x) = x|x|, which perfectly models x^2 when x>0, but is an odd function. Thus we can easily find the solution.