How can I get an asymptotic solution for 2nd order differential equation.

For the 2nd order differenttiaal equation:

where α and are constants.

How can I get an asymptotic solution, where for z=0. Thanks for the help.

[Torsten]

M(z) = 0 for all values of z.

[Dereje]

only for z =0. It is an initial condition.

[Torsten]

The solution of your differential equation is M(z)=0 for all values of z. Knowing this, what do you mean by "asymptotic solution" ?

[Dereje]

Sorry, I havent mentioned in my question that M(z) = 0 only at the source or at z=0. not for all values of Z.

[Torsten]

Do you have reason to believe that there are solutions of your ODE different from M(z)=0 for all z that satisfy M(0)=0 ? Maybe there are. You could prescribe M(1)=c for a value c>0, integrate backwards and see where you come out at z=0. Maybe you can adjust "c" such that M(0)=0.

[Dereje]

A good point to begin with. Thanks a lot.

[Torsten]

function main
  % Define differential equation d^2u/dz^2 = const*u^0.25 (letting M^2 = u)
  const = 3;
  fun=@(z,u)[u(2);const*u(1)^0.25];

  % Determine u(1) and u'(1) such that u(0)=0
  u0 = [2;2];
  sol = fminsearch(@(u)optim(u,fun),u0)

  % Plot solution for M
  tspan = [1 0];
  [Z,U] = ode15s(fun,tspan,sol);
  M = U(:,1).^0.5;
  plot(Z,M(:,1))
end

function u_at_0 = optim(u0,fun)
  tspan = [1 0];
  [Z,U] = ode15s(fun,tspan,u0);
  u_at_0 = real(U(end,1))^2 + imag(U(end,1))^2 + imag(U(end,2))^2;  
end

[Dereje]

Thanks again Torsten!!

[David Goodmanson]

Hi Dereje,

the function M^2 = c*z^(8/3) is an exact solution for all z >= 0, where c is a function whose value you can work out. I think it's likely that this is also the asymptotic solution although I certainly do not know that for a fact.

[Dereje]

@David I see also in some papers similar solution like you stated above (). Can you please show me the steps how to come up with this solution.

[David Goodmanson]

HI Dereje,

There seems to be inconsistency about M^2 vs. M, so it's preferable to quit M and use Torsten's notation for a starting point:

u'' = c u^(1/4)

LIke a lot of differential equations you assume a solution and verify that it works. (It was kind of exasperating sometimes in the diffyq course, when the instructor would put up some differential equation, pull a hypothetical solution out of the air and say, let's see if this works. Which of course it always did). Anyway, assume u = a z^n, plug that in, get

n(n-1) z^(n-2) = c (a z^n)^(1/4) = c a^(1/4) z^(n/4)

Equate powers of z on each side, find out n = 8/3. Equate coefficients in front on each side and you can solve for a. You get one solution for positive z that way, and you can see that in trying to find others, expansion in powers of z^(8/3) makes some sense