91Ó°ÊÓ

In calculus, a partial derivative represents the rate at which a function changes as one of its variables changes, keeping all other variables constant. This concept is essential in multivariable calculus. For the given function \( f(x, t) = (4 \pi D t)^{-1/2} \exp(-x^2 / 4Dt) \), we need to find how the function changes with respect to the variables \( x \) and \( t \).
The first partial derivative of \( f \) with respect to \( x \) is:

• \( \frac{\partial f}{\partial x} = -\frac{x}{2Dt} \cdot f(x, t) \)

This derivative tells us how \( f \) changes as x varies, with \( t \) remaining the same.
The second partial derivative, \( \frac{\partial^2 f}{\partial x^2} \), is:

• \( \left(\frac{x^2}{4D^2t^2} - \frac{1}{2Dt}\right) f(x, t) \)

Similarly, for \( t \):

• \( \frac{\partial f}{\partial t} = \left(-\frac{1}{2t} - \frac{x^2}{4Dt^2}\right) f(x, t) \)

These calculations are crucial in proving that the function \( f(x, t) \) satisfies a particular differential equation.

A differential equation involves functions and their derivatives. They're much like instructions telling you how a function behaves. For the function \( f(x, t) \), part of the exercise requires verifying that it satisfies the following differential equation: \( D\left(\frac{\partial^2 f}{\partial x^2}\right) = \frac{\partial f}{\partial t} \).
Substituting the partial derivatives we found earlier into this expression shows that both sides of the equation equate perfectly:

• Left side: \( D \cdot \frac{\partial^2 f}{\partial x^2} = D\left(\frac{x^2}{4D^2t^2} - \frac{1}{2Dt}\right) f(x, t) \)
• Right side: \( \left(-\frac{1}{2t} - \frac{x^2}{4Dt^2}\right) f(x, t) \)

This confirms the differential equation is true for our function, illustrating the relationship between the behavior of \( f \) over \( x \) and \( t \).

Integrating a function provides the accumulated value over an interval, which is foundational in calculus. Part of the task is to show that the integral of \( f(x, t) \) over all \( x \) results in unity—for all time \( t \). This indicates a normalization condition typical for probability functions.
We calculate:
\[ \int_{-\infty}^{+\infty} f(x, t) \, dx = \int_{-\infty}^{+\infty} (4 \pi D t)^{-1 / 2} \exp \left(-\frac{x^{2}}{4Dt}\right) dx \] By substituting \( u = \frac{x}{\sqrt{4Dt}} \), we simplify our calculations to:
\[ \int_{-\infty}^{+\infty} \exp(-u^2) \, du = \sqrt{\pi} \] The result shows the total equals 1, confirming that \( f(x, t) \) is normalized correctly.

The Dirac delta function, \( \delta(x) \), is a special function in mathematics modeled to represent an idealized point source or point impulse. It is infinitely high at \( x = 0 \) and zero elsewhere, with an integral over its domain equal to one. As \( t \rightarrow 0 \), our Gaussian function \( f(x, t) \) gets narrower and taller, resembling \( \delta(x) \).
In the limit, as \( t \) approaches zero, \( f(x, t) \) becomes sharply peaked, mathematically approaching the characteristics of a delta function. This property is handy for signals and systems in physics and engineering, where impulse responses and initial conditions are often modeled using \( \delta(x) \).
Thus, as \( t \rightarrow 0 \), \( f(x,t) \) not only maintains its integral of one but also reflects the distribution properties of a Dirac delta function.