# Transforms¶

## Fourier Transform¶

Fourier transforms are quite useful in solving differential equations. By decomposing the functions using Fourier transform, we might be able to simplify many differential equations.

Suppose we have a differential equation

$\frac{\partial}{\partial x} f(x) = a g(x).$

To solve the equation, we decompose $$f(x)$$ using its Fourier transform,

$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(k) e^{ikx} dk.$

Then we get

$\frac{1}{2\pi}\frac{\partial}{\partial x} \int_{-\infty}^{\infty} f(k) e^{ikx} dk = a \frac{1}{2\pi}\int_{-\infty}^{\infty} g(k) e^{ikx} dk.$

The equation is then simplified into

$ik f(k) = a g(k).$

Note

To summarize, we simple do replacement of the differential operators.

$\begin{split}\frac{\partial}{\partial x} e^{ikx}=ike^{ikx} &\implies \frac{\partial}{\partial x} \to ik \\ \frac{\partial^2}{\partial x^2} e^{ikx} = -k^2 e^{ikx} & \implies \frac{\partial^2}{\partial x^2} \to -k^2\end{split}$

## Laplace Transform¶

Similar to Fourier transform, Laplace transform is also useful in equation solving.

Laplace transform is a transform of a function of $$t$$, e.g. $$f(t)$$, to a function of $$s$$,

$\mathscr{L}[f(t)] = \int_0^\infty f(t) e^{ - s t} dt .$

Some useful properties:

1. $$\mathscr{L}[\frac{d}{dt}f(t)] = s \mathscr{L}[f(t)] - f(0)$$;

2. $$\mathscr{L}[\frac{d^2}{dt^2}f(t) = s^2 \mathscr{L}[f(t)] - s f(0) - \frac{d f(0)}{dt}$$;

3. $$\mathscr{L}[\int_0^t g(\tau) d\tau ] = \frac{\mathscr{L}[f(t)]}{s}$$;

4. $$\mathscr{L}[\alpha t] = \frac{1}{\alpha} \mathscr{L}[s/\alpha]$$;

5. $$\mathscr{L}[e^{at}f(t)] = \mathscr{L}[f(s-a)]$$;

6. $$\mathscr{L}[tf(t)] = - \frac{d}{ds} \mathscr{L}[f(t)]$$.

Some useful results:

1. $$\mathscr{L}[1] = \frac{1}{s}$$;

2. $$\mathscr{L}[\delta] = 1$$;

3. $$\mathscr{L}[\delta^k] = s^k$$;

4. $$\mathscr{L}[t] = \frac{1}{s^2}$$;

5. $$\mathscr{L}[t^n] = \frac{n!}{s^{n+1}}$$

6. $$\mathscr{L}[e^{at}]= \frac{1}{s-a}$$.

A very nice property of Laplace transform is

$\begin{split}\mathscr{L}_s [e^{at}f(t)] &= \int_0^\infty e^{-st} e^{-at} f(t) dt \\ & = \int_0^\infty e^{-(s+a)t}f(t) dt \\ & = L_{s+a}[f(t)]\end{split}$

which is very useful when dealing with master equations.

Two useful results are

$\mathscr{L}[I_0(2Ft)] = \frac{1}{\sqrt{ \epsilon^2 - (2F)^2 }}$

and

$\mathscr{L}[J_0[2Ft]] = \frac{1}{\sqrt{\epsilon^2 + (2F)^2}},$

where $$I_0(2Ft)$$ is the modified Bessel functions of the first kind. $$J_0(2Ft)$$ is its companion.

Using the property above, we can find out

$\mathscr{L}[I_0(2Ft)e^{-2Ft}] = \frac{1}{\sqrt{(\epsilon + 2F)^2 - (2F)^2}} .$

Example: Solving Differential Equations

For a first order differential equation

$f'(x) = x,$

we apply Laplace transform,

$s F(s) - f(0) = \frac{1}{s^2},$

from which we solve

$F(s) = \frac{1}{s^3} + \frac{f(0)}{s}.$

Then we lookup in the transform table, we find that

$f(x) = x^2/2 + f(0).$

## Legendre Transform¶

The geometrical meaning of Legendre transformation in thermodynamics can be illustrated by the following graph.

Fig. 4 Legendre transform

In the above example, we know that entropy $$S$$ is actually a function of temperature $$T$$. For simplicity, we assume that they are monotonically related like in the graph above. When we are talking about the quantity $$T \mathrm d S$$ we actually mean the area shaded with blue grid lines. Meanwhile the area shaded with orange line means $$S \mathrm d T$$.

Let’s think about the change in internal energy. For this example, we only consider the thermal part,

$\mathrm d U = T \mathrm d S .$

Internal energy change is equal to the the area shaded with blue lines. The area shaded with orange lines is the Helmholtz free energy,

$\mathrm d A = S \mathrm d T .$

The two quantities $$T \mathrm d S$$ and $$S \mathrm d T$$ sum up to $$d(TS)$$. This is also the area change of the rectangle determined by two edges $$0$$ to $$T$$ and $$0$$ to $$S$$.

This is a Legendre transform,

$\mathrm d U \to \mathrm d A,$

or

$T\mathrm dS \to S \mathrm d T.$

The point is that $$S(T)$$ is a function of $$T$$. However, if we know the blue area, we can find out the orange area. This means that the two functions $$A(T)$$ and $$U(S)$$ are somewhat like a pair. Choosing one of them for a specific calculation is a choice of freedom but we carry all the information in either one once the relation between $$T$$ and $$S$$ is know.

The above example sheds light on Legendre transform. The mathematical form is a little bit tricky so we will illustrate it using an example. For a function $$U(T, X)$$, we find its differential as

$\mathrm d U(T, X) = \frac{\partial U}{\partial T} \mathrm d T + \frac{\partial U}{\partial X} \mathrm d X.$

For convinience, we define

$\begin{split}S =& \frac{\partial U}{\partial T} \\ Y =& \frac{\partial U}{\partial X}.\end{split}$

The differential of function becomes

$\mathrm d U(T, X) = S \mathrm dT + Y \mathrm d X,$

where $$S$$ ($$Y$$) and $$T$$ ($$X$$) are a conjugate pair.

A Legendre transform says that we change the variable of the differential from $$T$$ ($$X$$) to $$S$$ ($$Y$$). For example, we know that

$S \mathrm d T = \mathrm d (ST) - T \mathrm d S.$

Plugging this into $$\mathrm d U$$, we get

$\mathrm d U(T, X) - \mathrm d(ST) = - T \mathrm dS + Y \mathrm d X.$

The left hand side is defined as a new differential

$\mathrm d A(S, X) = \mathrm d ( U(T, X) - ST ).$

In these calculations, $$U$$ is the internal energy and $$A$$ is the Helmholtz free energy. The transform that changes the variable from $$X$$ to $$Y$$ gives us enthalpy $$H$$. If we transform both variables then we get Gibbs free energy $$G$$. More about these thermodynamic potentials will be discussed in the following chapters.

## Refs & Note¶

1. Zia, Royce K. P., Edward F. Redish and Susan R. McKay. “Making sense of the Legendre transform.” (2009).

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