The Null Space, also known as the kernel, is the subset of vectors which get mapped to \(0\) by some Linear Map.

constituents

Some linear map \(T \in \mathcal{L}(V,W)\)

requirements

The subset of \(V\) which \(T\) maps to \(0\) is called the “Null Space”:

\begin{equation} null\ T = \{v \in V: Tv = 0\} \end{equation}

additional information

the null space is a subspace of the domain

It should probably not be a surprise, given a Null Space is called a Null Space, that the Null Space is a subspace of the domain.

A number can be any of…

• \(\mathbb{N}\): natural number
• \(\mathbb{Z}\): integer
• \(\mathbb{Q}\): rational number
• \(\mathbb{R}\): real number
• \(\mathbb{P}\): irrational number
• \(\mathbb{C}\): complex number

Consider a general non-linear First Order ODEs:

\begin{equation} x’ = F(x) \end{equation}

Suppose we have some time interval, we have some solutions to the expression given. Is it possible for us to, given \(x(t_0) = x_0\), what \(x(t_0+T)\) would be? Can we approximate for explicit numbers?

The solutions have to exist for all time: blow-up cannot be present during numerical estimations.

Explicit Euler Method

\begin{equation} x(t+h) \approx x_{t+1} = x_{t} + h f(x_t) \end{equation}

Here’s the characteristic equation again:

\begin{equation} \pdv[2] x \qty(EI \pdv[2]{w}{x}) = -\mu \pdv{w}{t}+q(x) \end{equation}

After Fourier decomposition, we have that:

\begin{equation} EI \dv[4]{\hat{w}}{x} - \mu f^{2}\hat{w} = 0 \end{equation}

Let’s solve this!

E,I,u,f = var("E I u f")
x, L = var("x L")
w = function('w')(x)
_c0, _c1, _c2, _c3 = var("_C0 _C1 _C2 _C3")

fourier_cantileaver = (E*I*diff(w, x, 2) - u*f^2*w == 0)
fourier_cantileaver

-f^2*u*w(x) + E*I*diff(w(x), x, x) == 0

And now, we can go about solving this result.

solution = desolve(fourier_cantileaver, w, ivar=x, algorithm="fricas").expand()
w = solution

\begin{equation} _{C_{1}} e^{\left(\sqrt{f} x \left(\frac{u}{E I}\right)^{\frac{1}{4}}\right)} + _{C_{0}} e^{\left(i \, \sqrt{f} x \left(\frac{u}{E I}\right)^{\frac{1}{4}}\right)} + _{C_{2}} e^{\left(-i \, \sqrt{f} x \left(\frac{u}{E I}\right)^{\frac{1}{4}}\right)} + _{C_{3}} e^{\left(-\sqrt{f} x \left(\frac{u}{E I}\right)^{\frac{1}{4}}\right)} \end{equation}

Here’s the characteristic equation again:

\begin{equation} \pdv[2] x \qty(EI \pdv[2]{w}{x}) = -\mu \pdv{w}{t}+q(x) \end{equation}

After Fourier decomposition, we have that:

\begin{equation} EI \dv[4]{\hat{w}}{x} - \mu f^{2}\hat{w} = 0 \end{equation}

Let’s solve this!

E,I,u,f = var("E I u f")
x, L = var("x L")
w = function('w')(x)
_c0, _c1, _c2, _c3 = var("_C0 _C1 _C2 _C3")

fourier_cantileaver = (E*I*diff(w, x, 2) - u*f^2*w == 0)
fourier_cantileaver

-f^2*u*w(x) + E*I*diff(w(x), x, x) == 0

And now, we can go about solving this result.

solution = desolve(fourier_cantileaver, w, ivar=x, algorithm="fricas").expand()
w = solution

\begin{equation} _{C_{1}} e^{\left(\sqrt{f} x \left(\frac{u}{E I}\right)^{\frac{1}{4}}\right)} + _{C_{0}} e^{\left(i \, \sqrt{f} x \left(\frac{u}{E I}\right)^{\frac{1}{4}}\right)} + _{C_{2}} e^{\left(-i \, \sqrt{f} x \left(\frac{u}{E I}\right)^{\frac{1}{4}}\right)} + _{C_{3}} e^{\left(-\sqrt{f} x \left(\frac{u}{E I}\right)^{\frac{1}{4}}\right)} \end{equation}