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This question has been put on MSE for a while and I move it here for a better luck.

Given the ODE $$ -v''(x)+\frac{v(x)}{(1+v^2(x))^2}+v(x)=1 $$ satisfies the condition $x\in(0,1)$, $v(0)=v(1)=1$, and $v(\cdot)$ is symmetric with respect to $1/2$.


I am wondering that can we determine the solution $v$ uniquely such that the above condition satisfies? It looks to me that the solution of this ODE has a "wall" shape, but this ODE is not linear nor any standard form so I am not quiet sure about it...although the two boundary condition is given. Moreover, I know $v$ is non-negative and quasi-convex.

Typically, by quasi-convexity and symmetricity, we have $v$ is monotone decreasing in $(0,1/2)$ and monotone increasing in $(1/2,1)$.

However, I am still not so sure about the uniqueness...

Thank you!

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  • $\begingroup$ Why don't you ask for existence ? Your problem is second-order but has 4 ($>2$) boundary conditions, thus is overdetermined. It is unlikely to be solvable. If you add some freedom, like an unknown length of the interval, you could expect a solution. $\endgroup$ Nov 5 '16 at 17:56
  • $\begingroup$ @DenisSerre My bad. I edit the question. It just from the numerical solution that the derivative at the boundary is $0$. $\endgroup$
    – JumpJump
    Nov 5 '16 at 17:58
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By conservation of energy for your equation, the quantity $\dot v^2-P(v)$ is constant, with

$$ P(y):= {(1-y+y^2)^2\over 1+y^2} = -{1\over 1+y^2}-2y+y^2+2$$
(just derive to check). Moreover, for solutions $v(t)$ to your equation, symmetry w.r.to $t=1/2$ is equivalent to the condition $\dot v(1/2)=0$ (one implication is obvious, for the other observe that $v(1-t)$ is a solution too, and use the uniqueness of the Cauchy problem). Therefore, putting $v(1/2)=\lambda$, one has $\dot v(t)^2=P(v(t))-P(\lambda)\le 0$ for all $0\le t\le 1/2$, hence $P(\lambda)$ is the minimum of $P(y)$ in the interval $[\lambda, 1]$, which means that $\lambda$ lies between the minimum point of $P(y)$ on $[0,1]$, $\lambda_*=0.6823..$, and $1$.

Since you want the solution to be decreasing on $[0,1/2]$, the first order equation $$\dot v(t)= -\sqrt{P(v(t))-P(\lambda) },\qquad 0<t<1/2 $$ characterizes all symmetric solution which are decreasing on $[0,1/2]$, with $v(1/2)=\lambda$. Solving the first order equation, the condition $v(0)=1$ thus translates into an equation for $\lambda$:
$$\int_{\lambda}^{1} {dy \over\sqrt{P(y)-P(\lambda) } }={1\over 2}.$$

The integral on the LHS is strictly decreasing from $+\infty$ to $0$ w.r.to $\lambda$ in the interval $(\lambda_*, 1]$, whence the value for the solution $\lambda=0.971..$, that identifies the unique solution of the initial problem.

$$*$$

Rmk. Just existence and uniqueness are immediate from the variational nature of the equation, which is the Euler-Lagrange equation for the functional $J(u):=\int_0^1\big( \dot u^2+P(u)\big)dt$. Since $f(x):=x+{x\over 1+x^2}-1$ is an increasing function with linear growth, its antiderivative ${1\over 2}P(x)$ is a convex function with quadratic growth. So $J$ is a strictly convex, coercive functional on the affine space $H^1_0(0,1)+1$, and has a unique critical point $v $ which is symmetric by uniqueness, as $v(1-t)$ is also a minimizer.

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  • $\begingroup$ Looks good. You may want to add that the BVP are the Euler-Lagrange equations associated to the Lagrangian $L(v,\dot{v}) = \dot{v}^2 /2 + P(v)/2$ with fixed endpoint conditions $v(0)=v(1)=1$. Wouldn't the domain of the action functional be $C^1([0,1],\mathbb{R})$? $\endgroup$ Nov 5 '16 at 23:10
  • $\begingroup$ Yes, the affine space $\{v\in C^1[0,1] : v(0)=v(1)=1\}$ is a suitable domain; otherwise one can use the larger domain $\{v\in H^1(0,1) : v(0)=v(1)=1\}=H_0^1(0,1)+1$, where existence is for free, and regularity is a standard fact in one variable. $\endgroup$ Nov 5 '16 at 23:47
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Write the quasilinear, second-order problem as: $$ -v''(x) + f(v(x)) = 0 \;, \quad v(0) = v(1) = 1 $$ Since $f$ is smooth and $ f'(s) > 1/2 $ for all real inputs $s$, you can use Nagumo's theory of upper and lower solutions to get the existence of a solution which is bounded and unique. To read more about this, see

  • R. E. O’Malley. Singular perturbation methods for ordinary differential equations. Springer-Verlag, Berlin, 1991.

  • K. W. Chang and F. A. Howes. Nonlinear singular perturbation phenomena: theory and application. Springer-Verlag, Berlin, 1984.

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    $\begingroup$ In fact $v$ is the minimum of the strictly convex functional $\int_0^1 {1\over 2}\dot v(t)^2 + F(v(t)) dt$, with $F'=f$, e.g. on the affine space $H^1_0(0,1)+1$, whence existence and unicity of solutions. $\endgroup$ Nov 5 '16 at 22:13

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