Question

Show the identities $$ \frac{d}{d x}(\delta(f(x)))=f^{\prime}(x) \delta^{\prime}(f(x)) $$ and $$ \delta(f(x))+f(x) \delta^{\prime}(f(x))=0 $$ Hence show that \(\phi(x, y)=\delta\left(x^{2}-y^{2}\right)\) is a solution of the partial differcntial equation $$ x \frac{\partial \phi}{\partial x}+y \frac{\partial \phi}{\partial y}+2 \phi(x, y)=0 $$

Short answer

The identities are correct, and substituting \( \phi(x, y) = \delta(x^2 - y^2) \) into the PDE shows that it indeed holds true, hence \( \phi(x, y) \) is a solution to the PDE.

Step-by-step solution

Identity verification

Let's verify the identities. We'll need the definition of the Dirac delta function: it's zero everywhere except at zero, and its integral over any range containing zero is one.\n\nNow take the derivative of \( \delta(f(x)) \) with respect to \( x \). By the chain rule of calculus, this is given by \( f'^{\prime}(x) \delta^{\prime}(f(x)) \) .\n\nFor the second identity, consider the definition of the derivative of the delta function: \( \delta^{\prime}(x) = -\delta(x)/x \). Substituting this in, we get \( \delta(f(x)) + f(x) \delta^{\prime}(f(x)) = 0 \). Our identities are verified.

Substituting into the PDE

Now substitute \( \phi(x, y) = \delta (x^2 - y^2) \) into the partial differential equation \( x\partial\phi/\partial{x} + y\partial\phi/\partial{y} + 2\phi(x, y) = 0 \).\n\nNote that \( \partial\phi/\partial{x} = 2x\delta^{\prime}(x^2 - y^2) \) and \( \partial\phi/\partial{y} = -2y\delta^{\prime}(x^2 - y^2) \).\n\nSubstituting these in, we get \( 2x^2\delta^{\prime}(x^2 - y^2) - 2y^2\delta^{\prime}(x^2 - y^2) + 2\delta(x^2 - y^2) = 0 \), which simplifies to \( \delta(x^2 - y^2) + (x^2 - y^2)\delta^{\prime}(x^2 - y^2) = 0 \).\n\nThis matches our second identity, so \( \phi(x, y) \) is indeed a solution to the PDE.

Key concepts

Dirac Delta Function

The Dirac Delta Function is a fascinating mathematical construct. It's not like typical functions, as it's defined to be zero everywhere except at one specific point. At that point, its value becomes infinite. However, the true magic lies in its integral: if you integrate the Dirac Delta Function over a range that includes the specific point, the result is one. This property makes it extremely useful for modeling scenarios in physics and engineering where a point impact or charge is being considered.

The Dirac Delta Function is often used in integral calculations to "pick out" values of functions. So, when \( \int_{-\infty}^{\infty} f(x) \delta(x-a) \, dx = f(a) \), it effectively samples the function at \( x = a \). This is particularly valuable in equations involving impulse forces or electrical signals in circuits, where changes occur instantaneously.

Chain Rule of Calculus

The Chain Rule is a cornerstone of calculus that allows us to differentiate composite functions. When you have a function nested within another function, the Chain Rule is your ticket to finding the derivative. In simple terms, it states that you take the derivative of the outer function, multiply it by the derivative of the inner function, and then put it all together.

For more clarity, if you have a function \( g(x) = f(u(x)) \), then the derivative, \( g'(x) \), is calculated as \( f'(u(x)) \cdot u'(x) \). This rule isn't just limited to simple functions; it extends beautifully to more complex scenarios like the differentiation of the Dirac Delta Function. When we differentiate \( \delta(f(x)) \) with respect to \( x \), the Chain Rule tells us to take the derivative of \( \delta \) and multiply it by the derivative of \( f(x) \). This results in the expression \( f'(x) \delta'(f(x)) \).

Derivative

Derivatives are fundamental in calculus, representing how a function changes as its input changes. In simpler terms, a derivative describes the rate of change or the slope of a function's graph at any given point. If a function \( f(x) \) is changing quickly at a certain point, its derivative will have a larger magnitude at that point.

Calculating derivatives involves applying rules like the power rule, product rule, and chain rule, depending on the function's form. The derivative of simple functions might be straightforward, but complexities arise when dealing with intricate functions or when derivatives themselves are differentiated.

For example, the derivative of the Dirac Delta Function, denoted as \( \delta^{\prime}(x) \), is unique. It's not just about computing rates and slopes but captures how the function behaves under transformation, providing deeper insights, especially when applied to identities or to verify solutions of differential equations.

Solution Verification

Solution verification is a key aspect of problem solving in mathematics, particularly with differential equations. It's about confirming that a proposed solution actually satisfies the original equation. This often involves substituting the solution back into the equation and simplifying to check if the equation holds true.

For complex mathematical problems, like partial differential equations involving the Dirac Delta Function, verifying the solution ensures your approach is correct and identifies any potential errors. In our exercise, we start with the function \( \phi(x, y) = \delta(x^2 - y^2) \).

By substituting this into the given partial differential equation and performing differentiations, we check our work against the identities that were verified earlier. This meticulous process confirms that the function isn't just a guess but indeed the solution, providing confidence and clarity in mathematical results.