1
$\begingroup$

I would like to solve the integro-differential equation of the form

$$\left( -n \int_0^b b db + \frac{i \Lambda l_P^2}{9V_c}\frac{d}{db}+b^2+k\right)\psi(b)=0.$$

I followed the steps in Solve an Integro-Differential Equation but it did not give any output. Is there a way to solve it either numerically or analytically? If $b \in [0,\infty]$, would the solution blow up?

Improve this question
$\endgroup$
5
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign $\endgroup$
    – Dunlop
    Commented Aug 8, 2020 at 4:12
  • $\begingroup$ Can you show us what you have tried? $\endgroup$
    – Dunlop
    Commented Aug 8, 2020 at 4:13
  • 3
    $\begingroup$ $∫^b_0bd b$ evaluates to $b^2/2$, doesn't it? $\endgroup$
    – xzczd ♦
    Commented Aug 8, 2020 at 5:27
  • 1
    $\begingroup$ @xzczd the integral is probably meant to include $\psi(b)$ as well: I'd assume that the entire parenthesis is an operator acting on the function $\psi$. nelly could you please make the function more explicit to reduce ambiguities? $\endgroup$
    – Roman
    Commented Aug 8, 2020 at 11:34
  • $\begingroup$ @Roman yes $\psi$ is a function of $b$ so it sits inside the integral. Sorry for the unclarity. $\endgroup$
    – nelly
    Commented Aug 8, 2020 at 17:22

2 Answers 2

Reset to default
4
$\begingroup$

Deriving the IDE we get

$$ -n b\psi(b) + c_0 \psi''(b) + (b^2+k)\psi'(b) + 2b\psi(b) = 0 $$

or

$$ c_0 \psi''(b) + (b^2+k)\psi'(b)+b(2-n)\psi(b) = 0 $$ Now if $n < 2,\ \ c_0 > 0,\ \ k > 0$ it looks as an stable ODE.

Follows a possible script to solve the complex case ($\psi = \psi_r+i\psi_i$) according to the comment.

parms = {c0 -> 1/10, k -> -1/10, n -> 3, psir0 -> 4, dpsir0 -> 0, psii0 -> -1, dpsii0 -> 0};
bmax = 3;
sol = NDSolve[{c0 psir''[b] + (b^2 + k) psii'[b] + b (n - 2) psii[b] == 0, -c0 psii''[b] + (b^2 + k) psir'[b] + b (n - 2) psir[b] == 0, psir[0] == psir0, psir'[0] == dpsir0, psii[0] == psii0, psii'[0] == dpsii0} /. parms, {psir, psii}, {b, 0, bmax}]
ParametricPlot[Evaluate[{psir[b], psii[b]} /. sol], {b, 0, bmax}]

enter image description here

Improve this answer
$\endgroup$
6
  • $\begingroup$ It looks like $c_0$ is complex $c_0=\frac {i \Lambda l^2_p}{9 V_c}$ $\endgroup$
    – Alex Trounev
    Commented Aug 8, 2020 at 15:47
  • $\begingroup$ If $c_0$ is imaginary then $\psi = \psi_r + i \psi_i$ and $$\cases{c_0\psi_r'' + (b^2+k)\psi_i'+b(n-2)\psi_i = 0\\ -c_0\psi_i''+(b^2+k)\psi_r'+b(n-2)\psi_r = 0}$$ $\endgroup$
    – Cesareo
    Commented Aug 8, 2020 at 15:57
  • $\begingroup$ @Cesareo Thank you. I will put this in mathematica and see what comes out. $\endgroup$
    – nelly
    Commented Aug 8, 2020 at 17:48
  • $\begingroup$ @Cesareo I tried to solve the coupled DE in mathematica but it does not give anything. I am new to the forum and trying to figure out how to post my code neatly on the comment section... p.s. I also tried for the case where $c_0$ is real just to see whether it can be solved, and mathematica found $\psi(b)$ to be a Heun function. $\endgroup$
    – nelly
    Commented Aug 8, 2020 at 18:02
  • $\begingroup$ @nelly Attached a possible case numerical solution. $\endgroup$
    – Cesareo
    Commented Aug 8, 2020 at 18:29
0
$\begingroup$

This is not an integro-differential equation. The operator factor complete. It is a wave equation with an operator that includes an integral and a differential term in a sum.

It is usual to differentiate between boundaries and integral variable. So the upper boundary of the integral may not be b?

(−𝑛∫𝑏0𝑏𝑑𝑏+𝑖Λ𝑙2𝑃9𝑉𝑐𝑑𝑑𝑏+𝑏2+𝑘)𝜓(𝑏)=0

(cd/db+d)Ff==0

where d=-n∫𝑏0𝑏𝑑𝑏+b^2+k.

This can be solved with ParametricNDSolve.

It depends on how the integral should be treated. In doubt differentiate again to get rid of the integral. Therefore the integral should be Integrate[b', {b', 0, b}]. So only the upper boundary depends on b. Then the operator can be differentiated and the wave function. Both have to be added up again to give the new wave equation.

This is something like: (((1 - n^2/2) b^2 + k) d/db + (2 g) d^2/db^2 + ((2 - n) b))f(𝑏)=0

ParametricNDSolve can deals with coefficient function dependent on b. It requires boundary values for a meaningful integration.

Example from the documentation ParametricNDSolveValue:

pfun = ParametricNDSolveValue[{-y''[x] + x^2 y[x] == \[Lambda] y[x], 
    y[0] == 0, y'[0] == 1}, y, {x, -15, 15}, {\[Lambda]}];

Plot[Evaluate@
  Table[pfun[\[Lambda]][x], {\[Lambda], 0, 13, 1}], {x, -5, 5}, 
 PlotRange -> {-2, 2}]

Plot

Improve this answer
$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.