hadaasyj

2022-04-23

How to solve ${X}_{\times}=(1+\delta \left\{\left(x\right)\right\})X$

I am trying to find the vibration modes of a string that has a uniform mass density, plus some point mass somewhere attached to it, modelled by an additional Dirac delta function in the mass density. The wave equation is of the form

${u}_{xx}=(1+\delta \left(x\right)){u}_{tt}$

where u is the deformation, and $(1+\delta \left(x\right))$ the mass density. After separation of variables we find

${X}_{\times}=(1+\delta \left\{\left(x\right)\right\})X$

where X is the spatial part of the solution. Is there any analytical solution for X?

smachttenbem

Beginner2022-04-24Added 18 answers

Step 1

We want to solve the differential equation$y{}^{\u2033}\left(x\right)=(1+\delta \left(x\right))y\left(x\right)$ , where $\delta$ is the Dirac delta "function".

Since$f\left(x\right)\delta \left(x\right)=f\left(0\right)\delta \left(x\right)$ for every f continuous at $x=0$ , the equation reduces to $y{}^{\u2033}\left(x\right)=y\left(x\right)+y\left(0\right)\delta \left(x\right)$ , i.e.

$y{}^{\u2033}\left(x\right)-y\left(x\right)=y\left(0\right)\delta \left(x\right)$ . A solution to this is continuous, satisfies $y{}^{\u2033}-y=0$ on $(-\mathrm{\infty},0)$ and on $(0,\mathrm{\infty})$ , and its derivative makes a jump at $x=0$ which is equal to y(0).

Step 2

Let$y\left(x\right)=A{e}^{x}+B{e}^{-x}$ be the solution on $(-\mathrm{\infty},0)$ and $y\left(x\right)=C{e}^{x}+D{e}^{-x}$ be the solution on $(0,\mathrm{\infty})$ . Then, for continuity we shall have $A+B=C+D=:E$ , and to get $y{}^{\u2033}-y=y\left(0\right)\delta$ we shall have $(C-D)-(A-B)=y\left(0\right)=E$ .

Now you only have to find all$A,B,C,D\in \mathbb{R}$ such that $A+B=C+D=(C-D)-(A-B)$ .

We want to solve the differential equation

Since

Step 2

Let

Now you only have to find all

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