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Chapter 12: Problem 9

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

Expert verified
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

01

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.
02

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.

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Most popular questions from this chapter

Which of the following is a distribution? (a) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi^{(n)}(0) \quad\left(\lambda_{n} \in \mathbb{R}\right)\) (b) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi\left(x_{n}\right) \quad\left(\lambda_{n}, x_{n} \in \mathbb{R}\right)\). (c) \(T(\phi)=(\phi(0))^{2}\). (d) \(T(\phi)=\sup \phi\) (c) \(T(\phi)=\int_{-\infty}^{\infty}|\phi(x)| \mathrm{d} x\).

Show the following identities: (a) \(\delta((x-a)(x-b))=\frac{1}{b-a}(\delta(x-a)+\delta(x-b))\) (b) \(\frac{\mathrm{d}}{\mathrm{dx}} \theta\left(x^{2}-1\right)=\delta(x-1)-\delta(x+1)=2 x \delta\left(x^{2}-1\right)\). (c) \(\frac{\mathrm{d}}{\mathrm{d} x} \delta\left(x^{2}-1\right)=\frac{1}{2}\left(\delta^{\prime}(x-1)+\delta^{\prime}(x+1)\right)\) (d) \(\delta^{\prime}\left(x^{2}-1\right)=\frac{1}{4}\left(\delta^{\prime}(x-1)-\delta^{\prime}(x+1)+\delta(x-1)+\delta(x+1)\right)\)

Prove that $$ x^{m} \delta^{(n)}(x)=(-1)^{m} \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \quad \text { for } n \geq m $$ Hence show that the Fourier trarsform of the distrobution $$ \sqrt{2 \pi} \frac{k !}{(m+k) !} x^{m} \delta^{(m+k)}(-x) \quad(m, k \geq 0) $$ is \((-2 y)^{k}\)

Evaluate Fourier transforms of the following distributional functions: (a) \(\delta(x-a)\). (b) \(\delta^{\prime}(x-a)\). (c) \(\delta^{(n)}(x-a)\). (d) \(\delta\left(x^{2}-a^{2}\right)\) (c) \(\delta^{\prime}\left(x^{2}-a^{2}\right)\)

Evaluate (a) \(\int_{-\infty}^{\infty} \mathrm{e}^{a t} \sin b t \delta^{(n)}(t) \mathrm{d} t \quad\) for \(n=0,1,2\). (b) \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta^{(n)}\left(t^{3}+t^{2}+t\right) d t \quad\) for \(n=0,1\).
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