Calculus of Variation (3) - 2nd Form of the Euler Equation

4. 2nd Form of the Euler Equation

 

to get better form of equation for $ \frac{\partial f}{\partial x}=0 $

 

$ df=\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial y'}dy'+\frac{\partial f}{\partial x}dx $

$ \frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}f(y,y';x)=\frac{\partial f}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{\partial f}{\partial y'}\frac{\mathrm{d} y'}{\mathrm{d} x}+\frac{\partial f}{\partial x}=y'\frac{\partial f}{\partial y}+y''\frac{\partial f}{\partial y'}+\frac{\partial f}{\partial x} $

$ \rightarrow y''\frac{\partial f}{\partial y'}=\frac{\mathrm{d} f}{\mathrm{d} x}-\frac{\partial f}{\partial x}-y'\frac{\partial f}{\partial y} $

 

$ \frac{\mathrm{d} }{\mathrm{d} x}(y'\frac{\partial f}{\partial y'})=y''\frac{\partial f}{\partial y'}+y'\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\partial f}{\partial y'}) $

   $ =\frac{\mathrm{d} f}{\mathrm{d} x}-\frac{\partial f}{\partial x}-y'\frac{\partial f}{\partial y}+y'\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\partial f}{\partial y'}) $

   $ =\frac{\mathrm{d} f}{\mathrm{d} x}-\frac{\partial f}{\partial x}+y'(\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y}) $

   $ =\frac{\mathrm{d} f}{\mathrm{d} x}-\frac{\partial f}{\partial x} $

$ \therefore \frac{\partial f}{\partial x}-\frac{\mathrm{d} }{\mathrm{d} x}(f-y'\frac{\partial f}{\partial y'})=0 $: 2nd Form of Euler's Equation

 

f does not depend explicitly on x, and $ \frac{\partial f}{\partial x}=0 $

$ \Rightarrow f-y'\frac{\partial f}{\partial y'}=0 $