The Root of All Calculus

Calculus or mathematical analysis is built up from 2 basic ingredients: integration and differentiation. Differentiation is concerned with things like speeds and accelerations, slopes and curves ect. These are Rates of Change, they are things that are defined locally. The Fundamental Theorem of Calculus is that Integration and Differentiation are the inverse of each other.

(1)   \begin{equation*} \frac{d}{dx}x^r = rx^{r-1} \end{equation*}

(2)   \begin{equation*} \frac{d}{dx}(uv) = v\frac{du}{dx}+u\frac{dv}{dx} \end{equation*}

(3)   \begin{equation*} \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2} \end{equation*}

(4)   \begin{equation*} \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} \end{equation*}

(5)   \begin{equation*} \frac{d}{dx}e^x = e^x \end{equation*}

Separation of Variables

Wave equations can be difficult to solve as they can have more than one solution and can look hideously difficult. One way of solving partial differential wave equations is to use separation of variables.

To separate the variables you have to assume that the wave equation you are trying to solve can be split up into two different functions, one of x and one of t. So you assume your equation can be written in this form,

(6)   \begin{equation*} y(x,t) = X(x)T(t) \end{equation*}

Where X is a function of distance (x) only, and T is a function of time (t) only. So lets try this method on a wave equation,

(7)   \begin{equation*} \frac{d^2y}{dx^2} = \frac{1}{c^2}\frac{d^2y}{dt^2}T \end{equation*}

So if we take the left hand side first we need to differentiate our assumed equation with respect to x twice, which is

(8)   \begin{equation*} \frac{d^2y}{dx^2} = \frac{d^2X}{dx^2}T \end{equation*}

The T sticks around as it is a constant with respect to x. For the right hand we get

(9)   \begin{equation*} \frac{d^2y}{dt^2} = \frac{d^2T}{dt^2}X \end{equation*}

Once again, the X doesn’t disappear as it is just classed as a constant. Now we can equate these two things like they are in the wave equation

(10)   \begin{equation*} T\frac{d^2X}{dx^2} = \frac{X}{c^2T}\frac{d^2T}{dt^2} \end{equation*}

Rearranging them gives

(11)   \begin{equation*} \frac{1}{X}\frac{d^2X}{dx^2} = \frac{1}{c^2T}\frac{d^2T}{dt^2} \end{equation*}

The reason we rearranged them was to get all the X’s on one side and all the T’s on the other. This shows that a function of x on the left equals something that has nothing to do with x on the right. This can only happen if both sides of the equation equal a constant. You can choose whatever symbol you want for this constant but unless you pick it as negative you’ll have imaginary numbers and complex numbers to deal with. So you get

(12)   \begin{equation*} \frac{d^2X}{dx^2} = -\alpha^2X \end{equation*}


(13)   \begin{equation*} \frac{d^2T}{dt^2} = -\alpha^2c^2T \end{equation*}

I have taken the X and T to the other side with the constant so now we have two relatively simple differential equations.

For X we have something that when differentiated twice with respectt to x gives us the original thing plus a factor of -α2, and for T we have something that when differentiated twice with respect to t gives us the original plus a factor of -α2c2. The only way we can get these is with sin or cos functions, so the options are

(14)   \begin{equation*} X = \cos(\alpha x) \text{ or } \sin(\alpha x) \end{equation*}

(15)   \begin{equation*} T = \cos(\alpha ct) \text{ or } \sin(\alpha ct) \end{equation*}

Without any boundary conditions there is no way to get rid of any of the answers, so the solutions are

    \begin{equation} \[ y(x,t)=\left\{    \begin{array}{l l}     \cos(\alpha x)\cos(\alpha ct)\\     \cos(\alpha x)\sin(\alpha ct}\\ \sin(\alpha x)\cos(\alpha ct)\\ \sin(\alph ax)\sin(\alpha ct)   \end{array} \right.\] \end{equation}

Div, Grad & Curl

When you need to find the Divergence, Gradient or Curl of a vector field or scalar field you basically need to know one main operator. This operator is called Del, and looks like this

(16)   \begin{equation*} \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right) = \text{An Operator} \end{equation*}

You get the div, grad or curl depending on how you use del. If for example you just use it to operate on a scalar field T the you get

(17)   \begin{equation*} \text{Gradient} = \nabla T = \left(\frac{\partial T}{\partial x},\frac{\partial T}{\partial y},\frac{\partial T}{\partial z}\right) = \text{A Vector} \end{equation*}

Del operating on a scalar give a vector answer corresponding to the divergence of the field. If however you decide to take the dot product of del with a vector field you would get a scalar answer, like so

(18)   \begin{equation*} \text{Divergence} = \nabla \cdot h = \left[\frac{\partial h_x}{\partial x}+\frac{\partial h_x}{\partial y}+\frac{\partial h_x}{\partial z}\right] = \text{A Scalar} \end{equation*}

Taking the cross product of del with a vector field gives a vector answer that corresponds to the curl of the field

(19)   \begin{equation*} \text{Curl} = \nabla \times h = \left[\frac{\partial h_y}{\partial z}-\frac{\partial h_z}{\partial y},\frac{\partial h_z}{\partial x}-\frac{\partial h_x}{\partial z},\frac{\partial h_x}{\partial y}-\frac{ \partial h_y}{\partial x}\right] = \text{A Vector} \end{equation*}

You can also use del multiple times. In the below example the gradient of the scalar field T was calculated, then dotted with del to calculate the divergence of the gradient

(20)   \begin{equation*} \nabla \cdot (\nabla T) \equiv \nabla^2T \equiv \nabla T \end{equation*}

This is usually referred to as del-squared or the Laplace operator and is equal to

(21)   \begin{equation*} \nabla^2 = \left[ \frac{d^2}{dx^2}+\frac{d^2}{dy^2}+\frac{d^2}{dz^2} \right] \end{equation*}

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