Problem 21 - Applied Partial Differential Equations

Question

Solve $$ \frac{d G}{d x}=\delta\left(x-x_{0}\right) \quad \text { with } \quad G\left(0, x_{0}\right)=0 $$ Show that \(G\left(x, x_{0}\right)\) is not symmetric even though \(\delta\left(x-x_{0}\right)\) is.

Step-by-Step Solution

Answer

The solution to the differential equation \(G(x, x_{0})\) is a step function, where the function is 0 for \(xx_{0}\). This is clearly not symmetric around \(x_{0}\), despite the Dirac Delta function itself being symmetric around \(x_{0}\).

Step 1: Solving the Differential Equation

To solve the differential equation, integrate both sides with respect to x from a point slightly before \(x_{0}\) to a point slightly after \(x_{0}\). On the left, you obtain \(G(x, x_{0})\), on the right, the integral of \(\delta(x-x_{0})\) gives 1.

Step 2: Boundary Conditions

Next, employ the given boundary condition \(G(0, x_{0}) = 0\). This implies that for \(x < x_{0}\), the function \(G(x, x_{0})\) will be 0, as the Dirac delta function has not 'fired' yet. For \(x > x_{0}\), the function \(G(x, x_{0})\) will be 1, as the Dirac delta function has 'fired' at \(x_{0}\). The function is therefore a step function that steps up from 0 to 1 at \(x_{0}\).

Step 3: Analyzing Symmetry

Now, analyze the symmetry properties of \(G(x, x_{0})\). A step function which steps from 0 to 1 at some point \(x_{0}\) is clearly not symmetric around that point, as it is 0 before the point and 1 after. Hence, despite the Dirac Delta function being symmetric, its integral up to any point \(x\) is not symmetric around \(x_{0}\), so the differential equation solution \(G(x, x_{0})\) is not symmetric either.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are used to describe a variety of physical phenomena such as heat, sound, fluid dynamics, and more. In the context of the given exercise, we are dealing with a first-order differential equation that involves the derivative of a function \( G(x, x_0) \) and the Dirac Delta function \( \delta(x-x_0) \).

To solve this equation, one must integrate both sides with respect to \( x \). The Dirac Delta function plays a crucial role here; it acts as a 'sifting' function, effectively 'picking out' the value at \( x_0 \) during integration. Because of this property, solving the equation gives us a step function solution:

• Before \( x_0 \), the function \( G(x, x_0) \) is 0.
• Exactly at \( x_0 \), there's an instantaneous jump to 1.
• For values of \( x \) greater than \( x_0 \), \( G(x, x_0) \) remains 1.

This outcome showcases the linkage between differential equations and the behavior of functions that they describe.

Symmetry in Functions

Symmetry in functions is an important concept in mathematics, referring to how a function's graph can be reflected or rotated and still look the same. Often, symmetric functions can be even (symmetric about the y-axis) or odd (symmetric about the origin).

In the exercise, even though the Dirac Delta function \( \delta(x-x_0) \) is symmetric, the solution \( G(x, x_0) \) to the differential equation is not symmetric. Why is this the case?

• The Dirac Delta function is a perfect symmetric spike at \( x_0 \).
• However, when integrated, it results in a step increase from 0 to 1 at \( x_0 \).
• The step function is not mirrored around any point; it has a distinct directionality, being different before and after \( x_0 \).

Thus, despite the symmetry of the Dirac Delta function itself, its integral as seen in \( G(x, x_0) \) shows asymmetry.

Step Functions

Step functions are defined as functions that jump from one constant value to another. The solution \( G(x, x_0) \) in this exercise is an excellent example of a step function.

Let’s delve into how step functions work with the Dirac Delta function involved:

• Before the point \( x_0 \), the function \( G(x, x_0) \) is zero because it ‘anticipates’ the delta effect.
• At the exact point \( x_0 \), the Dirac Delta causes an immediate step from 0 to 1.
• After \( x_0 \), the value of the function remains 1.

This behavior is crucial in many signal processing applications, where sudden changes need to be captured. Step functions help model such instantaneous shifts accurately.