Solution of a first order non linear differential equation

Question

I am trying to find the extrema of the integral below $$ I= \int_0^1 y^2 \mathrm dx $$ under the conditions $$ \int_0^1 \left(\frac{dy}{dx}\right)^2 \mathrm dx =1 $$ and y(0)=y(1)=0

Using Lagrange multiplier λ, equivalently, I can find the extrema of the below quantity $$ I*= \int_0^1 y^2 \mathrm dx+λ\int_0^1 \left(\frac{dy}{dx}\right)^2 \mathrm dx $$ Since there is not explicit dependence on x, by using beltrami identity $$ F-y'\frac{\partial F}{\partial y'} = C $$ the problem above reduced to the solution of the 2nd order differential equation $$ 2λ(y’)^2y’’-λ(y')^{2}- y^{2}-c=0. $$ And after the substitution u=dy/dx The second order differential equation above become the first order one below $$ 2λu^{3}u’-λ(u)^{2}- y^{2}-c=0. $$ where the differentiation refers to y. And here is where I got stuck. Can someone give me some hint on how I can procced with that equation?

Many thanks in advance!

Answer 1

The Beltrami equation can not contain a second derivative, where should it come from? It should evaluate to $$ y^2-λy'^2=C $$ which easily leads to solutions in terms of trigonometric or hyperpolic functions.

Answer 2

Taking

$$ I = \int_0^1\left(y^2+\lambda\left(y'^2-1\right)\right)dx $$

we have the Euler-Lagrange equations

$$ \cases{ y-\lambda y'' = 0\\ \int_0^1 y'^2 = 1 } $$

Solving we have

$$ y = c_1 e^{\frac{x}{\sqrt{\lambda}}}+c_2 e^{-\frac{x}{\sqrt{\lambda}}}\\ \frac{c_1^2 \left(e^{\frac{2}{\sqrt{\lambda }}}-1\right) \sqrt{\lambda }+c_2^2 \left(1-e^{-\frac{2}{\sqrt{\lambda }}}\right) \sqrt{\lambda }-4 c_2 c_1}{2 \lambda }=1 $$

now depending on your initial/boundary conditions, after determining $c_1,c_2$ we proceed with the second equation to determine the feasible value(s) for $\lambda$ and then it is done.

Example

Supposing $y(0) = 1, y(1) = 1$ we have

$$ c_1 = 0.0914543\\ c_2 = 0.908546\\ \lambda = 0.189694 $$

Follows a MATHEMATICA script making the necessary calculations

L = y[x]^2 + lambda (y'[x]^2 - 1)
EL = D[L, y[x]] - D[D[L, y'[x]], x]
yx = y[x] /. DSolve[EL == 0, y, x][[1]]
cond = Integrate[D[yx, x]^2, {x, 0, 1}] - 1
equ1 = (yx /. x -> 0) - 1;
equ2 = (yx /. x -> 1) - 1;
equs0 = {equ1, equ2, cond} /. {C[1] -> c1, C[2] -> c2};
obj = equs0.equs0
sol = NMinimize[{obj, lambda >= 0}, {c1, c2, lambda}, Reals]

NOTE

The boundary conditions $y(0) = y(1) = 0$ require a little different setup

L = y[x]^2 - lambda^2 (y'[x]^2 - 1)
EL = D[L, y[x]] - D[D[L, y'[x]], x]
yx = y[x] /. DSolve[EL == 0, y, x][[1]]
cond = Integrate[D[yx, x]^2, {x, 0, 1}] - 1
equ1 = (yx /. x -> 0);
equ2 = (yx /. x -> 1);
equs0 = {equ1, equ2, cond} /. {C[1] -> c1, C[2] -> c2};
obj = equs0.equs0
sol = NMinimize[{obj}, {c1, c2, lambda}, Reals]

this is necessary to avoid complex numbers in the calculation procedure. This last case gives

$$ \cases{ c_1 = 0\\ c_2 = -0.450158\\ \lambda = -0.31831 } $$