Euler Lagrange Equations

Question

Let f(x,y,y') = y*[1 + ( (y')^2 ) ] , where y = y(x)

Solve for y

Answer 1

According to Ashley's notes in the 9 comments made in the question, what Ashley gave for f(x,y,y') contains a typo.

It should read f(x,y,y') = y√[1+(y')²] = y[1+(y')²]^1/2, where y=y(x).

This makes it possible to explain the notes that Ashley gives in the 9 comments. I will do that now.

We are setting up the Euler-Lagrange equation ∂f/∂y - d(∂f/∂y')/dx = 0.

Because f(x,y,y') = y√[1+(y')²]

∂f/∂y = [1+(y')²]^1/2

∂f/∂y' = (2yy')/(2√[1+(y')²]) = yy'/√[1+(y')²]

Then

d( ∂f/∂y' )/dx = d( yy'/√[1+(y')²] )/dx

We can use the quotient rule of calculus to get the derivative of the expression on the right-hand side of the line above. Because this expression is messy, I will write the numerator and the denominator of that expression separately.

The denominator is [1+(y')²]

The numerator is (y'y' + yy'')√[1+(y')²] - yy'(2y'y'')/(2√[1+(y')²]) = [(y')² + yy'']√[1+(y')²] - yy''(y')²/√[1+(y')²]

We can factor out 1/√[1+(y')²] from the numerator.

This makes the denominator [1+(y')²]^3/2

The numerator becomes

[(y')² + yy''][1+(y')²] - yy''(y')² = (y')² + yy'' + (y')⁴ + yy''(y')² - yy''(y')² = (y')² + yy'' + (y')⁴

Combining the numerator and denominator,

d(∂f/∂y')/dx = [(y')² + yy'' + (y')⁴]×[1+(y')²]^-3/2

∂f/∂y - d(∂f/∂y')/dx = [1+(y')²]^1/2 - [(y')² + yy'' + (y')⁴]×[1+(y')²]^-3/2 = 0

Multiplying through by [1+(y')²]^3/2,

[1+(y')²]² - [(y')² + yy'' + (y')⁴] = 1 + 2(y')² + (y')⁴ - (y')² - yy'' - (y')⁴ = 1 + (y')² - yy'' = 0

The equation 1 + (y')² - yy'' = 0 is the equation that Ashley wanted to see the derivative of.

This is a non-linear second-order differential equation, which I don't see how to solve. Anyone have any ideas?

Fortunately, there's another way to tackle this problem, using an alternate formulation of the Euler-Lagrange equation that can be used when f(x,y,y') has no x-dependence, it depends only on y and y':

y∂f/∂y' - f = c, where c is a constant. This gives us

y'yy'[1+(y')²]^-1/2 - y[1+(y')²]^1/2 = c

Multiplying through by [1+(y')²]^1/2 gives us y(y')² - y[1+(y')²] = c[1+(y')²]^1/2

The left-hand side y(y')² - y[1+(y')²] = y(y')² - y - y(y')² = y

Thus, y = c[1+(y')²]^1/2

Squaring both sides gives us y² = c²[1+(y')²].

Then

y²/c² = 1+(y')²

This is the last entry in Ashley's notes.

This can be solved. •Take the derivative of both sides of this equation:

2yy' = 2c²y'y''

Cancelling out the twos and gathering everything on one side of the equation,

yy' - c²y'y'' = 0

Factoring out y',

y'(y - c²y'') = 0

Either y' = 0 or y - c²y'' = 0.

y' = 0 implies y = a constant, which is a trivial solution.

y - c²y'' = 0 implies y'' = y/c², so y = a•sinh(x/c) + b•cosh(x/c), where a and b are constants.