# Differential equation involving exponents

$$\begin{cases}&-z^{\prime\prime}(t)=\lambda(1+(N-2)t)^{\frac1{2-N}(2(N-1)+\alpha)}f(z(t)),\quad t\in(0,+\infty)\\&z(0)=z^\prime(+\infty)=0\end{cases}$$

I tried to solve the equation using the code below, but got a null solution. If you change the values of p for 2 and alpha to -1.99, a smooth and nonzero solution is obtained. But I need to respect the condition:

p<=(n+alpha)/(n-2)

. Can anyone help? Thanks!

n = 3;
alpha = 1;
xf = 10000;
p = 3;
s = NDSolve[{-z''[t] == (1 + (n - 2) t)^(1/(2 - n) (2 (n - 1) + alpha)) z[t]^p,
z == 0, z'[xf] == 0}, z[t], {t, 0, xf}];
xt = Plot[Evaluate[z[t] /. s], {t, 0, xf}, PlotRange -> All, PlotStyle -> Thick]

• What do you mean by "got a null solution"? With the code as it is I get a plot without problems. Dec 13, 2015 at 15:14
• A very small number in all the domain. Dec 13, 2015 at 15:15
• And that is not correct? Dec 13, 2015 at 15:17
• I don't belive it is correct. If you look to the boundary conditions you will see that both are not satisfied. And also the solution should looks like a polinomial with higher order than 3. Dec 13, 2015 at 15:28
• Umm, no. z=0 satisfies both z==0 and z'[xf]==0 Dec 13, 2015 at 15:29

The desired solution can be obtained, but the computation is temperamental. Use

n = 3;
alpha = 1;
xf = 10000;
p = 3;
s = NDSolveValue[{-z''[t] == (1 + (n - 2) t)^(1/(2 - n) (2 (n - 1) + 1)) z[t]^p,
z == 0, z'[xf] == 0}, z, {t, 0, xf}}}, WorkingPrecision -> 30,
Method -> {"Shooting", "StartingInitialConditions" -> {z == 0, z' == 5]
Plot[s[t], {t, 0, 10}, PlotRange -> All, AxesLabel -> {z, t}] Although the computation is carried to xf = 10000, only 0 < t < 10 is plotted to show key behavior near t = 0.

Apparently, there are many solutions to this nonlinear ODE for the parameters given in the question. For instance, with

"StartingInitialConditions" -> {z == 0, z' == 2} "StartingInitialConditions" -> {z == 0, z' == 189/100} "StartingInitialConditions" -> {z == 0, z' == 9/5} Finding further such solutions becomes progressively more difficult as the number of oscillations increases. Note also that, for each solution with positive z', a companion solution with negative z' exists. More solutions to this equation are available here.