I have an inegral equation like this

$\qquad n(\phi)=\int_0^\sqrt{\phi} f(w)\sqrt{2w+\phi}dw$.

I need to find $f(w)$ analytically. Here $n(\phi)$ is known. Here $n(\phi) = -\frac{2}{\delta ^2 \phi }-\frac{2 e^{-\phi } \sqrt{\phi }}{\sqrt{\pi }}+\frac{e^{-\phi }}{\sqrt{\pi } \sqrt{\phi }}$ Is there any way for this using Mathematica.


closed as off-topic by MarcoB, m_goldberg, corey979, Edmund, Henrik Schumacher Feb 21 at 21:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – MarcoB, m_goldberg, corey979, Edmund, Henrik Schumacher
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ There are uncountably many answers. Which one do you want? The simplest one would be f(w)=const; just find the appropriate constant (depends on n and phi). Probably though you need an f(w) that is zero for w<phi/2 to make the square root real. $\endgroup$ – Roman Feb 20 at 6:41
  • $\begingroup$ You should give some information about n, \[Phi] $\endgroup$ – Ulrich Neumann Feb 20 at 7:33
  • $\begingroup$ n is a function of $\phi$ which I know... Now I have edited the question $\endgroup$ – Hari Krishnan Feb 20 at 9:40
  • $\begingroup$ I think this question is not about Mathematica. $\endgroup$ – Roman Feb 20 at 12:27
  • $\begingroup$ Yes the problem is about mathematica.. I want to know, whether such a problem can be solved using mathematica $\endgroup$ – Hari Krishnan Feb 20 at 12:39

I'll give it a try, but still think that this kind of question would be better off asked at the math stackexchange.

First, make an educated Ansatz for the shape of the function $f(w)$. As I have no idea what your $n(\phi)$ is, I'll use a Taylor sum with unknown coefficients c[i] to be determined:

f[w_] = Sum[c[i]*w^i, {i, 0, 5}]
(* c[0] + w c[1] + w^2 c[2] + w^3 c[3] + w^4 c[4] + w^5 c[5] *)

(use more terms for actual calculations).

Then, do the integration on the right-hand side to find

R[φ_] = Assuming[φ > 0, Integrate[f[w] Sqrt[2 w + φ], {w, 0, Sqrt[φ]}]];

The series-expansion of this integrated right-hand side $R(\phi)$ around $\phi=0$ is

Assuming[φ > 0, Series[R[φ], {φ, 0, 5}]]
(* lots of output *)

Now you can compare this to the series expansion of your $n(\phi)$:

Assuming[φ > 0, Series[n[φ], {φ, 0, 5}]]
(* some output *)

These series expansions can be matched term-by-term, which allows you to find the coefficients c[i] and thus the function $f(w)$.

If you give a formula for $n(\phi)$ I could maybe give more help; but as the question stands, this abstract discussion is all I can give.

  • $\begingroup$ Yeah I can give the value of n(\phi). Please see the question, I have editted it $\endgroup$ – Hari Krishnan Feb 21 at 11:28
  • $\begingroup$ In that case, I don't know how to continue. Neither any polynomial w^m with m>-1 nor any Dirac delta function (or its derivatives) gives a component proportional to 1/phi as you need it in n(phi). Again, please ask on the math stackexchange. $\endgroup$ – Roman Feb 21 at 13:01

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