How to solve this equation with NDSolve?

Question

Here is the code:

a=4;b=10;
s = NDSolve[{y^2 v[y]^3 v''[y] + v[y]^3 y v'[y] + 2 y^2 v[y]^2 v'[y]^2 - a v[y]^4 +b== 0, 
v[1.0] == 5.0, v'[50.0] == 0.0}, v, {y, 1.0, 50.0}]

v1[y_] = Evaluate[v[y] /. s];

LogLinearPlot[v1[y], {y, 1, 50}]

For a=4; b=10; mathematica can't solve this equation. However, if I change the values of a and b (say a=1; b=1;), I can obtain the results. For b=0, mathematic can solve this equation with any given positive value of a. What's the problem here? Thanks!

Answer 1

NDSolve It has problems with equation.Why? I don't No.Solution with $MAPLE.$

Let's try analytically.

$$-a v(y)^4+b+y^2 v(y)^3 v''(y)+2 y^2 v(y)^2 v'(y)^2+y v(y)^3 v'(y)=0 \tag{1}$$

By making the substitution $y=exp(t)$. When subbing for the independent variable, we need to make careful use of the chain rule to express $v(y),v′(y),v″(y)$ in terms of $v(t),v′(t),v″(t)$.

 ode = y^2*v[y]^3*v''[y] + v[y]^3 *y*v'[y] + 2 y^2*v[y]^2*v'[y]^2 - 
 a*v[y]^4 + b;
 subs = Exp[t]; newode = 
 ode /. {v[y] -> v[t], v'[y] -> v'[t]/D[subs, t], 
 v''[y] -> (v''[t] - D[subs, {t, 2}] v'[t]/D[subs, t])/D[subs, t]^2,
 y -> subs};
 newode2 = Expand[Collect[FullSimplify[newode], v''[t]]]

$-a v(t)^4+b+v(t)^3 v''(t)+2 v(t)^2 v'(t)^2=0 \tag{2}$

Using the substitution $v'(t)=k(v)$,where $v$ plays the role of the independent variable, and taking into account the relations $v''(t)=k'(t)=k'(v)*v'(t)=k'(v)*k(v)$, can reduce to a first-order equation:

newode3 = newode2 /. {v'[t] -> k[v], v''[t] -> k[v]*k'[v], v[t] -> v}

$-a v^4+b+v^3 k(v) k'(v)+2 v^2 k(v)^2=0 \tag{3}$

ODE = DSolve[newode3 == 0, k[v], v]

$\left\{k(v)\to -\frac{\sqrt{a v^6-3 b v^2+3 c_1}}{\sqrt{3} v^2},k(v)\to \frac{\sqrt{a v^6-3 b v^2+3 c_1}}{\sqrt{3} v^2}\right\} \tag{4}$

then $v'(t)=k(v)$ and $t=Log(y)$ :

{Integrate[1/k[v] /. ODE[[1]], v] == Log[y] + C[2], 
Integrate[1/k[v] /. ODE[[2]], v] == Log[y] + C[2]}

We have analyticall solution:

$\left\{-\sqrt{3} \int \frac{v^2}{\sqrt{a v^6-3 b v^2+3 c_1}} \, dv=c_2+\log (y),\sqrt{3} \int \frac{v^2}{\sqrt{a v^6-3 b v^2+3 c_1}} \, dv=c_2+\log (y)\right\}$

$c_1$ and $c_2$ integration constants. Mathematica can't find this Integrated's $:($

Try's with Series.

We need a equation $(2)$.By making the substitution $y=exp(t)$ and solve with parametr $t$ $t=Log(y)$ we have a new boundary conditions:

$$\{v(0)=5,v(\log (50))=0\}$$

a = 4;
b = 10;
n = 13;
v = Sum[c[i]*t^i, {i, 0, n}] + O[t]^(n + 1);
ODE = v^3*D[v, {t, 2}] + 2*D[v, t]^2*v^2 - a*v^4 + b == 0;
V = FullSimplify@
Normal[v /. Solve[LogicalExpand[ODE], Table[c[i], {i, 1, n}]]] // 
Quiet;
Sol = First@
Solve[{V == 5 /. t -> 0, D[V, t] == 0 /. t -> Log[50]}, {c[0], 
c[1]}];
VV = V /. Sol /. t -> Log[y] // N

LogLinearPlot[VV, {y, 1, 5}, PlotRange -> All, 
PlotLegends -> {"Series"}, PlotStyle -> {Thick, Dashed}]

Solution is not accurates ,you must increase $n=20$,mayby $50$.

Answer 2

This equation also can be solved using the substitution v[y] -> u[y]^(1/4).

Clear[a, b];
eq = y^2 v[y]^3 v''[y] + v[y]^3 y v'[y] + 2 y^2 v[y]^2 v'[y]^2 - a v[y]^4 + b;
First@Solve[{u''[y] == D[v[y]^4, {y, 2}], u'[y] == D[v[y]^4, {y, 1}]}, {v''[y], v'[y]}];
equ = eq /. % /. v[y] -> u[y]^(1/4)
(* 10 - 4*u[y] + (y*Derivative[1][u][y])/4 + (y^2*Derivative[1][u][y]^2)/(8*u[y]) + 
   (y^2*(-3*Derivative[1][u][y]^2 + 4*u[y]*Derivative[2][u][y]))/(16*u[y]) *)

a = 4; b = 10;
su = NDSolveValue[{equ == 0, u[1] == 5^4, u'[50] == 0}, u, {y, 1, 50}, 
    WorkingPrecision -> 30];
Plot[su[y]^(1/4), {y, 1, 50}, PlotRange -> All, AxesLabel -> {v, y}]

Addendum: Method -> "Shooting"

Method -> "Shooting" also works, but here only with an excellent initial guess.

sv = NDSolveValue[{eq == 0, v[1] == 5, v'[50] == 0}, v, {y, 1, 50}, 
  Method -> {"Shooting", "StartingInitialConditions" -> {v[50] == 1.25745, v'[50] == 0}}]