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enter image description hereIm trying to find the exact solution for the singular BVP $y^{\prime\prime}(t)=-\frac{2}{t}y^{\prime}(t)+y^{5}(t)$ with the Boundary Conditions $y(1)=\sqrt{\frac{3}{4}}$ and $y^{\prime}(0)=0$ and $0<t<1$. I know that the exact solution of this equation is $y(t)=\sqrt{\frac{3}{3+t^2}}$. Now I want to find the exact solution using mathemtica software. I tried the following first: (1)

DSolve[{y''[t] == -(2/t) (y'[t]) + (y[t])^5, y[1] == Sqrt[3/4], y'[0] == 0}, y[t], t]

but I cant find the exact solution like above. After this, I tried

SOL = DSolve[{y''[t] == -(2/t) (y'[t]) + (y[t])^5}, y[t],  t] (* General solution *)
F = FindRoot[{(y[t] /. SOL[[1]] /. t -> 1 /. C[1] -> c1 /.C[2] -> c2) ==Sqrt[3/4], (y'[t] /. SOL[[1]] /. t -> 0 /. C[1] -> c1 /.C[2] -> c2) == 0}, {c1, 1/2}, {c2, 1/2}] // Chop // Rationalize
(* Finding constans c1 and c2 *)
SOL /. C[1] -> c1 /.C[2] -> c2 /. F(* Paste constants c1 and c2 to general solution *)
(*{{y[t]\->Sqrt(3/(3+t)^2)}*)

In this case, I also failed to find the exact solution. Please help me any one. Advance thanks.

Junaid
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  • My first comment: in the DSolve you write f''[t] == -(2/t) (f'[t]) + (y[t])^5. It should be either all f or all y. –  Feb 23 '22 at 20:41
  • dear I have the same problem still – Junaid Feb 23 '22 at 20:43
  • Yes, I know. I just pointed that out so the OP is well written and people who copy and paste the code attempting to help to have the right expressions :-) –  Feb 23 '22 at 20:46
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    Second comment: you said that you know the analytic solution and you gave a function. Even when Mathematica cannot get the answer from DSolve it is able to verify it when we impose it. In this case, it does not. If you run y''[t] == -(2/t) (y'[t]) + (y[t])^5 /. y -> (Sqrt[3/(3 + #^2)] &) // FullSimplify you will see that it does not yield True. Could you please check the solution you provided for the y[t]? –  Feb 23 '22 at 20:47
  • yes the solution is ok and can be found in the article https://link.springer.com/article/10.1007/s40819-018-0569-8 – Junaid Feb 23 '22 at 20:56
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    The article is behind a paywall. – Bob Hanlon Feb 23 '22 at 21:33
  • Dears I'm still facing difficulty. Please any one help will be highly appreciated. – Junaid Feb 24 '22 at 05:03
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    The main problem is that "your exact solution" does not satisfy the original equation. I can say that you haven't checked if this is a solution – Artes Feb 24 '22 at 05:07
  • Actually my question is about how I find the exact solution by Mathematica? If there is any other way instead of mine approach, so that I can solve my problem. – Junaid Feb 24 '22 at 06:45
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    I would first solve it numerically to at least get a solution by figuring out what Neumann conditions satisfy your mixed boundary conditions. Maybe try the shooting method with various trial values of y'(1). Then research how to solve analytically a boundary-value problem of your type. Maybe also post your numeric work to incentivize others. – josh Feb 24 '22 at 11:26
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    Given that exact solution, your equation is wrong. You need a negative sign before y[t]^5. Here is why: If you differentiate \sqrt{3} \sqrt{\frac{1}{t^2+3}} once, you get -\sqrt{3} t \left(\frac{1}{t^2+3}\right)^{3/2}. If you differentiate twice, you get \sqrt{3} \left(\frac{1}{t^2+3}\right)^{5/2} \left(2 t^2-3\right). Now the closest form you can get it in the main equation: after multiplying the first derivative to (2/t) and dividing by y[t]^5 you will get -1. And yes, Mathematica can't solve this by normal DSolve without other advanced techniques that I am not aware of. – MathX Feb 24 '22 at 16:04
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    Many ODEs do not have symbolic solutions in terms of commonly defined functions. However, it should be noted that DSolve sometimes is unable to solve ODEs that do have symbolic solutions in terms of commonly defined functions. My code from: https://mathematica.stackexchange.com/questions/264051/solution-of-a-bvp-by-means-of-dsolve dosen't work for your nonlinear ODE. General solution exist ,but is transcentental equation can't by solved algebraically for boundary conditions. – Mariusz Iwaniuk Feb 24 '22 at 16:15
  • Dear @Mariusz Iwaniuk, I have watched many videos but im unable to solve this question. I would be very thankful, if you provide me a help as before. Thanks in advance , – Junaid Feb 24 '22 at 17:00
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    First No way to find exact solution by Mathematica. Second give us exact solution of this equation, a sceenshot from article(I do not have 42.99 euros), because as has been mentioned many times, your exact solution does not satisfy the original equation. – Mariusz Iwaniuk Feb 24 '22 at 17:51
  • dear, I have been attached the screenshot of the example of the paper. I would be very thankful if you can send me email at my mail ahmadjunaid436@gmail.com @Mariusz Iwaniuk – Junaid Feb 24 '22 at 18:05
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    If you execute this: FullSimplify[ u''[x] == -(2/x)*u'[x] + u[x]^5 /. u -> Function[{x}, Sqrt[3/(3 + x^2)]], Assumptions -> {0 < x < 1}] It should be True,but it is False. There must have been a mistake in article(a typo) or something else.But if you try a negative sign before y[t]^5 then: FullSimplify[ u''[x] == -(2/x)*u'[x] - u[x]^5 /. u -> Function[{x}, Sqrt[3/(3 + x^2)]], Assumptions -> {0 < x < 1}] is True. – Mariusz Iwaniuk Feb 24 '22 at 18:55
  • Dear @Mariusz Iwaniuk , if we assume - before u[x]^5 then the exact solution is Sqrt[3/(3 + x^2)]. Now can you offer me code of mathemtica so I can run in this case and find the exact solution. Thanks in advance – Junaid Feb 24 '22 at 19:10
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    The Singh and Kumar reference confirms Mariusz's claim above that the equation is $u''+2/x u'=u^5$: See reference (no euros needed): https://www.sciencedirect.com/science/article/abs/pii/S0010465514000034?via%3Dihub – josh Feb 24 '22 at 19:51
  • I tried also $u^{\prime\prime}=-\frac{2}{x}u^{\prime}$-$u^{5}$, but mathemtica still fails to solve it Dear@Josh and @Mariusz – Junaid Feb 26 '22 at 06:48

1 Answers1

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Using perturbation method for the equation:

$$y''(x)+\frac{2 y'(x)}{x}+y(x)^5=0$$

If solution is y[x]=Sqrt[3/(3 + x^2)] for x=0 is: 1.

Then we have initial conditions:y[0]=1,and y'[0]=0

ClearAll["`*"]; Remove["`*"]

eq = y''[x] + 2/xy'[x] + [Delta]y[x]^5;


(Perturbation method)


ord = 10;
eqs = CoefficientList[
Normal[Series[
 eq /. y -> 
   Function[x, 
    Evaluate[Sum[y[i][x] [Delta]^i, {i, 0, ord}]]], {[Delta], 0,
   ord}]], [Delta]];yiList = {};
 Do[AppendTo[yiList, 
MapAt[Factor, 
DSolve[{(eqs[[k + 1]] /. yiList /. Log[Exp[x]] -> x) == 0, 
   y[k][0] == If[k == 0, 1, 0], y[k]'[0] == If[k == 0, 0, 0]}, 
  y[k][x], x][[1, 1]], 2]], {k, 0, ord - 2}];


(Series solution:)


sol2 = Sum[y[i][x] [Delta]^i, {i, 0, ord - 2}] /. yiList /. [Delta] -> 1


(Converting series solution to function:)


FullSimplify[FindGeneratingFunction[
Table[SeriesCoefficient[sol2 + O[x]^18, n], {n, 0, 10}], x], 
Assumptions -> x > 0]


(Sqrt[3]/Sqrt[3 + x^2])

With this metod we can solve:$$y''(x)+\frac{2 y'(x)}{x}-y(x)^5=0$$ with: y[0]=1,and y'[0]=0.

Exact solution is:$$y(x)=\frac{1}{\sqrt{1-\frac{x^2}{3}}}$$

Mariusz Iwaniuk
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