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First Integrals of the Euler Lagrange Equations

April 9th, 2009 |

Author: Merrin

Calculus of variations problems which cannot be solved analytically are not as satisfying. Technical problems arise when the differential equations generated by the Euler Lagrange equations are too complicated to solve. We should look out for any tricks to solve the differential equations because in general these problems are very hard. The form of the function in the integrand determines the differential equation that extremizes the integral. It turns out if x or y is missing from the functional there are some simplifications to the differential equations we need to solve. For each cases, there will be what is called a first integral of the Euler Lagrange equations. They are a tremendous simplification.

$J=\displaystyle \int _{x_{1} }^{x_{2} }f(x,y,\dot{y})dx$

$\cfrac{d}{dx} \cfrac{\partial f}{\partial \dot{y}} -\cfrac{\partial f}{\partial y} =0$

If the function f doesn’t explicitly depend on y then we have a very nice simplification of the Euler Lagrange equations. The constant c_1 is called the first integral.

$\cfrac{d}{dx} \cfrac{\partial f}{\partial \dot{y}} =0$

$\cfrac{\partial f}{\partial \dot{y}} =c_{1}$

On the other hand the functional integral might depend on y but not on x. Then we can convert between x and y in the integration and still get a first integral.

$J=\displaystyle \int _{y_{1} }^{y_{2} }g(y,x,\dot{x})dy$

$\cfrac{d}{dy} \cfrac{\partial g}{\partial \dot{x}} -\cfrac{\partial g}{\partial x} =0$

When

$\cfrac{\partial g}{\partial x} =0$

we have a first integral of

$\cfrac{\partial g}{\partial \dot{x}} =c_{1}$

Often the integral will be in a form that needs the variables to be reversed to find a first integral. This transformation reverses the integration variables.

$J=\displaystyle \int _{x_{1} }^{x_{2} }f(x,y,\dot{y})dx=\displaystyle \int _{y_{1} }^{y_{2} }f(y,x,\left(\cfrac{dx}{dy} \right)^{-1} )\cfrac{dx}{dy} dy$

$J=\displaystyle \int _{y_{1} }^{y_{2} }f(y,x,\dot{x}^{-1} )\dot{x}dy$

The transformation in the reverse direction is easy to derive also.

Example 1. Convert the following integral into an integral with respect to y then find the first integral.

$J=\displaystyle \int _{x_{1} }^{x_{2} }\sqrt{1+y^{2} \dot{y}^{2} } dx$

Solution 1. Since this integrand depends on y there is not a first integral right off because the right hand side of Euler’s equation is not zero.

$\cfrac{d}{dx} \cfrac{\partial f}{\partial \dot{y}} -\cfrac{\partial f}{\partial y} =0$

However since the integral doesn’t depend on x we can reverse the variables and then get a first integral.

$J=\displaystyle \int _{x_{1} }^{x_{2} }\sqrt{1+y^{2} \dot{y}^{2} } dx=\displaystyle \int _{y_{1} }^{y_{2} }\sqrt{1+y^{2} \dot{x}^{-2} } \dot{x}dy=\displaystyle \int _{y_{1} }^{y_{2} }\sqrt{\dot{x}^{2} +y^{2} } dy$

Now we apply

$\cfrac{d}{dy} \cfrac{\partial g}{\partial \dot{x}} -\cfrac{\partial g}{\partial x} =0$

This gives us a simpler differential equation to solve via a first integral.

${\cfrac{d}{dy} \cfrac{\dot{x}}{\sqrt{\dot{x}^{2} +y^{2} } } =0} \\ {\cfrac{\dot{x}}{\sqrt{\dot{x}^{2} +y^{2} } } =c_{1} } \\ {\dot{x}^{2} =c_{1}^{2} (\dot{x}^{2} +y^{2} )} \\ {\dot{x}=2Ay} \\ {x=Ay^{2} +B}$

Example 2. If a functional integrand doesn’t depend on x

$J=\displaystyle \int _{x_{1} }^{x_{2} }L(y,y')dx$

Show that

$y'\cfrac{\partial L}{\partial y'} -L={\rm const}$

is an alternative method to calculate the first integral known as the Beltrami identity.

Solution 2. We can prove this by differentiating both sides of the equation.

${\cfrac{dL}{dx} -\cfrac{dy'}{dx} \cfrac{\partial L}{\partial y'} -y'\cfrac{d}{dx} \cfrac{\partial L}{\partial y'} =0} \\ {\left(\cfrac{\partial L}{\partial x} +\cfrac{\partial L}{\partial y} y'+\cfrac{\partial L}{\partial y} \cfrac{\partial y'}{\partial x} \right)-\cfrac{dy'}{dx} \cfrac{\partial L}{\partial y'} -y'\cfrac{d}{dx} \cfrac{\partial L}{\partial y'} =0} \\ {\cfrac{\partial L}{\partial y} y'+\cfrac{\partial L}{\partial y} \cfrac{dy'}{dx} -\cfrac{dy'}{dx} \cfrac{\partial L}{\partial y'} -y'\cfrac{d}{dx} \cfrac{\partial L}{\partial y'} =0} \\ {y'(\cfrac{\partial L}{\partial y} -\cfrac{d}{dx} \cfrac{\partial L}{\partial y'} )=0}$

Since L obeys the Euler Lagrange equations we have proved the identity.

Example 3. Verify that the Beltrami identity gives the same differential equation as in the previous example where

$J=\displaystyle \int _{x_{1} }^{x_{2} }\sqrt{1+y^{2} \dot{y}^{2} } dx$

Solution 3. If we plug the integrand into the Beltrami identity

$y'\cfrac{\partial L}{\partial y'} -L= \text{const}$

With a bit of algebra we can show the Beltrami identity gives an equivalent differential equation.

$\dot{y}\cfrac{y^2\dot{y}}{\sqrt{1+y^2\dot{y}^2}}-\cfrac{{1+y^{2} \dot{y}^{2} }}{\sqrt{1+y^{2} \dot{y}^{2} }}=-B=\cfrac{-1}{\sqrt{1+y^{2} \dot{y}^{2}}}=\cfrac{-\dot{x}}{\sqrt{\dot{x}^2+y^2}}$