Problem
Write and solve the Euler equation to make the following integral stationary
\(\int _{x_{1}}^{x_{2}}\sqrt {x}\sqrt {1+y^{\prime 2}}dx\)
Solution
Let \(F=\left (x,y,y^{\prime }\right ) =\) \(\sqrt {x}\sqrt {1+y^{\prime 2}}\)
The Euler equation is
\begin {align*} \frac {d}{dx}\left (\frac {\partial F}{\partial y^{\prime }}\right ) -\frac {\partial F}{\partial y} & =0\\ \frac {\partial F}{\partial y} & =\frac {\partial \ }{\partial y}\left ( \sqrt {x}\sqrt {1+y^{\prime 2}}\right ) =0 \end {align*}
Hence the Euler equation becomes\[ \frac {d}{dx}\left (\frac {\partial F}{\partial y^{\prime }}\right ) =0 \] This means that \(\frac {\partial F}{\partial y^{\prime }}=C\) for some constant \(C.\)\[ \frac {\partial F}{\partial y^{\prime }}=\sqrt {x}\frac {y^{\prime }}{\sqrt {1+y^{\prime 2}}}\] Hence\begin {align*} \sqrt {x}\frac {y^{\prime }}{\sqrt {1+y^{\prime 2}}} & =C\\ y^{\prime 2} & =\frac {C^{2}\left (1+y^{\prime 2}\right ) }{x}\\ x & =\frac {C^{2}+C^{2}y^{\prime 2}}{y^{\prime 2}}\\ x & =\frac {C^{2}}{y^{\prime 2}}+k\\ y^{\prime 2} & =\frac {C^{2}}{x-C^{2}}\\ y^{\prime } & =\frac {C}{\sqrt {x-C^{2}}}\\ y\relax (x) & =\frac {2C}{\sqrt {x-C^{2}}}+C_{1}\\ \frac {y\relax (x) }{2C}-\frac {C_{1}}{C} & =\frac {1}{\sqrt {x-C^{2}}} \end {align*}
Let \(\frac {C_{1}}{C}=-b\) (some constant), and Let \(\frac {1}{2C}=a\) (constant), Hence above becomes\begin {align*} a\ y+b & =\frac {1}{\sqrt {x-\frac {1}{4a^{2}}}}\\ a\ y+b & =\frac {2a}{\sqrt {4a^{2}\ x-1}} \end {align*}
This is equation of a parabola.
Problem
Write and solve the Euler equation to make the following integral stationary \(\int _{x_{1}}^{x_{2}}x\sqrt {1-y^{\prime 2}}dx\)
Solution
Let \(F=\left (x,y,y^{\prime }\right ) =\) \(x\sqrt {1-y^{\prime 2}}\). The Euler equation is\begin {align*} \frac {d}{dx}\left (\frac {\partial F}{\partial y^{\prime }}\right ) -\frac {\partial F}{\partial y} & =0\\ \frac {\partial F}{\partial y} & =\frac {\partial \ }{\partial y}\left ( x\sqrt {1-y^{\prime 2}}\right ) =0 \end {align*}
Hence Euler equation becomes\[ \frac {d}{dx}\left (\frac {\partial F}{\partial y^{\prime }}\right ) =0 \] This means that \(\frac {\partial F}{\partial y^{\prime }}=C\) for some constant \(C.\)\[ \frac {\partial F}{\partial y^{\prime }}=\frac {-x\ y^{\prime }}{\sqrt {1-y^{\prime 2}}}\] Hence\begin {align*} \frac {-x\ y^{\prime }}{\sqrt {1-y^{\prime 2}}} & =C\\ y^{\prime 2} & =\frac {C^{2}\left (1-y^{\prime 2}\right ) }{x^{2}}\\ x^{2} & =\frac {C^{2}-C^{2}y^{\prime 2}}{y^{\prime 2}}\\ x^{2} & =\frac {C^{2}}{y^{\prime 2}}-C^{2}\\ y^{\prime 2} & =\frac {C^{2}}{x^{2}+C^{2}}\\ y^{\prime } & =\frac {C}{\sqrt {x^{2}+C^{2}}}\\ y\relax (x) & =C\operatorname {arcsinh}\left (\frac {x}{C}\right ) +C_{1}\\ \frac {y-C_{1}}{C} & =\operatorname {arcsinh}\left (\frac {x}{C}\right ) \\ \frac {x}{C} & =\sinh \left (\frac {y}{C}-\frac {C_{1}}{C}\right ) \end {align*}
Let \(\frac {C_{1}}{C}=-b\) (some constant). Let \(\frac {1}{C}=a\) (some constant). Hence the above becomes\[ a\ x=\sinh \left (a\ y+b\right ) \]
Problem
Write and solve the Euler equation to make the following integral stationary \(\int _{x_{1}}^{x_{2}}\left (y^{\prime 2}+\sqrt {y}\right ) dx\)
Solution
Let \(F\left (x,y,y^{\prime }\right ) =\) \(y^{\prime 2}+\sqrt {y}\). Since \(F\) does not depend on \(x\), we change the integration variable to \(y\). Let \(y^{\prime }=\frac {1}{x^{\prime }}\), then \(dx=\frac {dx}{dy}dy\). Hence the integral becomes \[ \int _{y_{1}}^{y_{2}}\left (\frac {1}{x^{\prime 2}}+\sqrt {y}\right ) x^{\prime }\ dy=\int _{y_{1}}^{y_{2}}\left (\frac {1}{x^{\prime }}+x^{\prime }\sqrt {y}\right ) \ dy \] Now \(F\left (y,x^{\prime }\right ) =\left (\frac {1}{x^{\prime }}+x^{\prime }\sqrt {y}\right ) \). The Euler equation changes from \(\frac {d}{dx}\left ( \frac {\partial F}{\partial y^{\prime }}\right ) -\frac {\partial F}{\partial y}=0\) to \(\frac {d}{dy}\left (\frac {\partial F}{\partial x^{\prime }}\right ) -\frac {\partial F}{\partial x}=0\). Now, \(\frac {\partial F}{\partial x}=0\) since \(F\) does not depend on \(x,\)Hence the Euler equation reduces to
\begin {align*} \frac {d}{dy}\left (\frac {\partial F}{\partial x^{\prime }}\right ) & =0\\ \frac {d}{dy}\left (-\frac {1}{x^{\prime 2}}+\sqrt {y}\right ) & =0 \end {align*}
Hence \(-\frac {1}{x^{\prime 2}}+\sqrt {y}=C\) where \(C\) is some constant\begin {align*} -\frac {1}{x^{\prime 2}}\ & =C-\sqrt {y}\\ -\frac {1}{C-\sqrt {y}} & =x^{\prime 2}\\ \frac {1}{b+\sqrt {y}} & =x^{\prime 2}\qquad \text {where }b\text { is a new constant}=-C\\ \frac {1}{\sqrt {k+\sqrt {y}}} & =\frac {dx}{dy}\\ \int \frac {dy}{\sqrt {b+\sqrt {y}}} & =\int dx\\ \frac {4}{3}\left (-2b+\sqrt {y}\right ) \left (\sqrt {b+\sqrt {y}}\right ) & =x+a\qquad \text {Where }a\text { is constant of integration} \end {align*}
Hence the solution is\[ \frac {4}{3}\left (\sqrt {y}-2b\right ) \left (\sqrt {b+\sqrt {y}}\right ) =x+a \]
Problem
Write and solve the Euler equation to make the following integral stationary \(\int _{x_{1}}^{x_{2}}\frac {\sqrt {1+y^{\prime 2}}}{y^{2}}dx\)
Solution
Let \(F\left (x,y,y^{\prime }\right ) =\) \(\frac {\sqrt {1+y^{\prime 2}}}{y^{2}}\). Since \(F\) does not depend on \(x\), we change the integration variable to \(y\). Let \(y^{\prime }=\frac {1}{x^{\prime }}\), hence \(dx=\frac {dx}{dy}dy\). The integral becomes \[ \int _{y_{1}}^{y_{2}}\left (\frac {\sqrt {1+\frac {1}{x^{\prime 2}}\ }}{y^{2}}\right ) x^{\prime }\ dy=\int _{y_{1}}^{y_{2}}\frac {\sqrt {x^{\prime 2}+1\ }}{y^{2}}\ dy \] Now \(F\left (y,x^{\prime }\right ) =\frac {\sqrt {x^{\prime 2}+1\ }}{y^{2}}\). The Euler equation changes from \(\frac {d}{dx}\left (\frac {\partial F}{\partial y^{\prime }}\right ) -\frac {\partial F}{\partial y}=0\) to \(\frac {d}{dy}\left ( \frac {\partial F}{\partial x^{\prime }}\right ) -\frac {\partial F}{\partial x}=0\). But\(\frac {\partial F}{\partial x}=0\) since \(F\) does not depend on \(x,\)Hence the Euler equation reduces to\begin {align*} \frac {d}{dy}\left (\frac {\partial F}{\partial x^{\prime }}\right ) & =0\\ \frac {\partial F}{\partial x^{\prime }} & =\frac {\partial }{\partial x^{\prime }}\left (\frac {\sqrt {x^{\prime 2}+1\ }}{y^{2}}\right ) \\ & =\frac {x^{\prime }}{y^{2}\sqrt {x^{\prime 2}+1\ }} \end {align*}
Hence\[ \frac {d}{dy}\left (\frac {x^{\prime }}{y^{2}\sqrt {x^{\prime 2}+1\ }}\right ) =0 \] Hence \(\frac {x^{\prime }}{y^{2}\sqrt {x^{\prime 2}+1\ }}=C\) where \(C\) is some constant\begin {align*} \frac {x^{\prime 2}}{x^{\prime 2}+1\ } & =C\ y^{4}\\ \frac {x^{\prime 2}+1}{x^{\prime 2}} & =\frac {1}{C\ y^{4}}\\ 1+\frac {1}{x^{\prime 2}} & =\frac {1}{C\ y^{4}}\\ \frac {1}{x^{\prime 2}} & =\frac {1-Cy^{4}}{C\ y^{4}}\\ \frac {\sqrt {C}\ y^{2}}{\sqrt {1-Cy^{4}}} & =x^{\prime }\\ \frac {\sqrt {C}\ y^{2}}{\sqrt {1-Cy^{4}}} & =\frac {dx}{dy}\\ \int \frac {\sqrt {C}\ y^{2}}{\sqrt {1-Cy^{4}}} & =\int dx \end {align*}
The solution is\[ \frac {\sqrt {C\ }y^{3}}{3\sqrt {1-Cy^{4}}}=x+C_{1}\] Where \(C_{1}\) is constant of integration. Let \(C_{1}=a\), \(C=b\) hence solution can be written as\[ \frac {\sqrt {b\ }y^{3}}{3\sqrt {1-by^{4}}}=x+a \]
Problem
Write and solve the Euler equation to make the following integral stationary \(\int _{x_{1}}^{x_{2}}\ y\sqrt {y^{\prime 2}+y^{2}}\ dx\)
Solution
Let \(F\left (x,y,y^{\prime }\right ) =\) \(y\sqrt {y^{\prime 2}+y^{2}}\). Since \(F\) does not depend on \(x\), we change the integration variable to \(y\). Let \(y^{\prime }=\frac {1}{x^{\prime }}\) and \(dx=\frac {dx}{dy}dy\). Hence the integral becomes \[ \int _{y_{1}}^{y_{2}}\left (y\sqrt {\frac {1}{x^{\prime 2}}+y^{2}}\right ) x^{\prime }\ dy=\int _{y_{1}}^{y_{2}}\ y\sqrt {1+x^{\prime 2\ }y^{2}}\ dy \] Now \(F\left (y,x^{\prime }\right ) =y\sqrt {1+x^{\prime 2\ }y^{2}}\). The Euler equation changes from \(\frac {d}{dx}\left (\frac {\partial F}{\partial y^{\prime }}\right ) -\frac {\partial F}{\partial y}=0\) to \(\frac {d}{dy}\left ( \frac {\partial F}{\partial x^{\prime }}\right ) -\frac {\partial F}{\partial x}=0\). But\(\frac {\partial F}{\partial x}=0\) since \(F\) does not depend on \(x,\)Hence the Euler equation reduces to\[ \frac {d}{dy}\left (\frac {\partial F}{\partial x^{\prime }}\right ) =0 \]\begin {align*} \frac {\partial F}{\partial x^{\prime }} & =\frac {\partial }{\partial x^{\prime }}\left (y\sqrt {1+x^{\prime 2\ }y^{2}}\right ) \\ & =y\left (\frac {x^{\prime }\ y^{2}}{\sqrt {1+x^{\prime 2\ }y^{2}}}\right ) \ \\ & =\frac {x^{\prime }\ y^{3}}{\sqrt {1+x^{\prime 2\ }y^{2}}} \end {align*}
Hence\[ \frac {d}{dy}\left (\frac {x^{\prime }\ y^{3}}{\sqrt {1+x^{\prime 2\ }y^{2}}}\right ) =0 \] Hence \(\frac {x^{\prime }\ y^{3}}{\sqrt {1+x^{\prime 2\ }y^{2}}}=C\) where \(C\) is some constant\begin {align*} x^{\prime }\ y^{3} & =C\ \sqrt {1+x^{\prime 2\ }y^{2}}\\ x^{\prime 2}\ y^{6} & =C^{2}\ \left (1+x^{\prime 2\ }y^{2}\right ) \\ x^{\prime 2}\ y^{6} & =C^{2}+C^{2}\ x^{\prime 2\ }\ y^{2}\\ \ x^{\prime 2}\ \left (y^{6}-C^{2}\ y^{2}\right ) & =C^{2}\\ x^{\prime 2} & =\frac {C^{2}}{\left (y^{6}-C^{2}\ y^{2}\right ) }\\ x^{\prime } & =\frac {C}{y\sqrt {y^{4}-C^{2}}}\\ \int dx & =C\int \frac {1}{y\sqrt {y^{4}-C^{2}}}dy \end {align*}
The solution is (using Mathematica)\[ a\ x=-\frac {1}{2}i\log \left (\frac {-2iC+2\sqrt {-C^{2}+y^{4}}}{y^{2}}\right ) \]
Problem
Write and solve the Euler equation to make the following integral stationary \(\int _{x_{1}}^{x_{2}}\ \frac {y\ y^{\prime 2}}{1+y\ y^{\prime }}\ dx\)
Solution
Let \(F\left (x,y,y^{\prime }\right ) =\) \(\frac {y\ y^{\prime 2}}{1+y\ y^{\prime }}\). Since \(F\) does not depend on \(x\), we change the integration variable to \(y\). Let \(y^{\prime }=\frac {1}{x^{\prime }},dx=\frac {dx}{dy}dy\). Hence the integral becomes\begin {align*} \int _{y_{1}}^{y_{2}}\left (\frac {y\ \frac {1}{x^{\prime 2}}}{1+y\ \frac {1}{x^{\prime }}}\right ) x^{\prime }\ dy & =\int _{y_{1}}^{y_{2}}\left ( \frac {y\ \frac {1}{x^{\prime 2}}}{\frac {x^{\prime }+\ y}{x^{\prime }}}\right ) x^{\prime }\ dy\\ & =\int _{y_{1}}^{y_{2}}\left (\frac {y\ \frac {1}{x^{\prime }}}{x^{\prime }+\ y}\right ) x^{\prime }\ dy\\ & =\int _{y_{1}}^{y_{2}}\left (\frac {y}{x^{\prime }+y}\right ) \ dy \end {align*}
Now \(F\left (y,x^{\prime }\right ) =\left (\frac {y}{x^{\prime }+y}\right ) \). The Euler equation changes from \(\frac {d}{dx}\left (\frac {\partial F}{\partial y^{\prime }}\right ) -\frac {\partial F}{\partial y}=0\) to \(\frac {d}{dy}\left (\frac {\partial F}{\partial x^{\prime }}\right ) -\frac {\partial F}{\partial x}=0\). \(\frac {\partial F}{\partial x}=0\) since \(F\) does not depend on \(x,\)Hence the Euler equation reduces to\[ \frac {d}{dy}\left (\frac {\partial F}{\partial x^{\prime }}\right ) =0 \]\begin {align*} \frac {\partial F}{\partial x^{\prime }} & =\frac {\partial }{\partial x^{\prime }}\left (\frac {y}{x^{\prime }+y}\right ) \\ & =y\left (-\frac {1}{\left (x^{\prime }+y\right ) ^{2}}\right ) \ \\ & =\frac {-y}{\left (x^{\prime }+y\right ) ^{2}} \end {align*}
\(\ \)Hence\[ \frac {d}{dy}\left (\frac {-y}{\left (x^{\prime }+y\right ) ^{2}}\right ) =0 \] Hence\(\frac {-y}{\left (x^{\prime }+y\right ) ^{2}}=C\) where \(C\) is some constant\[ -y=C\left (x^{\prime }+y\right ) ^{2}\] Let \(C=-k\)\begin {align*} y & =k\ \ \left (x^{\prime }+y\right ) ^{2}\\ \sqrt {\frac {y}{k}} & =x^{\prime }+y\\ \sqrt {\frac {y}{k}}-y & =x^{\prime }\\ \sqrt {\frac {y}{k}}-y & =\frac {dx}{dy}\\ \int \sqrt {\frac {y}{k}}-y\ dy & =\int dx\\ -\frac {y^{2}}{2}+\frac {2}{3}y\sqrt {\frac {y}{k}} & =x+\beta \end {align*}
Where \(\beta \) is the integration constant. Let \(\frac {1}{\sqrt {k}}=\alpha \) a new constant\[ x=-\frac {1}{2}y^{2}+\frac {2}{3}\alpha \ y^{\frac {3}{2}}-\beta \ \] Let \(\frac {2}{3}\alpha =a\) a new integration constant, let \(-\beta =b\) a new constant, we get\[ x=a\ y^{\frac {3}{2}}-\frac {1}{2}y^{2}+b \]
Problem
Write and solve the Euler equation to make the following integral stationary \(\int _{\phi _{1}}^{\phi _{2}}\ \sqrt {\ \theta ^{\prime 2}\ +\sin ^{2}\theta }\ d\phi \) , \(\theta ^{\prime }=\frac {d\theta }{d\phi }\)
Solution
Here \(F\left (x,y\relax (x) ,y^{\prime }\relax (x) \right ) \) becomes \(F\left (\phi ,\theta \left (\phi \right ) ,\theta ^{\prime }\left ( \phi \right ) \right ) .\) So now \(x\rightarrow \phi ,y\rightarrow \theta ,y^{\prime }\rightarrow \theta ^{\prime }\). Since \(F\left (\theta ^{\prime },\theta \right ) \) does not depend on \(\phi \), we change the integration variable to \(\theta \), so we want to change from \(\theta ^{\prime }=\frac {d\theta }{d\phi }\) to \(\phi ^{\prime }=\frac {d\phi }{d\theta }\). Let \(\theta ^{\prime }=\frac {1}{\phi ^{\prime }},d\phi =\frac {d\phi }{d\theta }d\theta \). Hence the integral becomes
\[ \int _{\theta _{1}}^{\theta _{2}}\left (\sqrt {\ \frac {1}{\phi ^{\prime 2}}\ +\sin ^{2}\theta }\right ) \phi ^{\prime }\ d\theta =\int _{\theta _{1}}^{\theta _{2}}\sqrt {\ 1\ +\phi ^{\prime 2}\sin ^{2}\theta }\ d\theta \] \(\ \)So now \[ F\left (\phi ^{\prime },\theta \right ) =\sqrt {\ 1\ +\phi ^{\prime 2}\sin ^{2}\theta }\] The Euler equation changes from \(\frac {d}{dx}\left (\frac {\partial F}{\partial y^{\prime }}\right ) -\frac {\partial F}{\partial y}=0\) to \(\frac {d}{d\theta }\left (\frac {\partial F}{\partial \phi ^{\prime }}\right ) -\frac {\partial F}{\partial \phi }=0\). \(\frac {\partial F}{\partial \phi }=0\) since \(F\) does not depend on \(\phi ,\)Hence the Euler equation reduces to\[ \frac {d}{d\theta }\left (\frac {\partial F}{\partial \phi ^{\prime }}\right ) =0 \]\begin {align*} \frac {\partial F}{\partial \phi ^{\prime }} & =\frac {\partial }{\partial \phi ^{\prime }}\left (\sqrt {1+\phi ^{\prime 2}\sin ^{2}\theta }\right ) \\ & =\frac {\phi ^{\prime }\sin ^{2}\theta }{\sqrt {1+\phi ^{\prime 2}\sin ^{2}\theta }} \end {align*}
Hence\[ \frac {d}{d\theta }\left (\frac {\phi ^{\prime }\sin ^{2}\theta }{\sqrt {1+\phi ^{\prime 2}\sin ^{2}\theta }}\right ) =0 \] Hence\(\frac {\phi ^{\prime }\sin ^{2}\theta }{\sqrt {\ 1\ +\phi ^{\prime 2}\sin ^{2}\theta }}=C\) where \(C\) is some constant\begin {align*} \phi ^{\prime }\sin ^{2}\theta & =C\sqrt {\ 1\ +\phi ^{\prime 2}\sin ^{2}\theta }\\ \phi ^{\prime 2}\sin ^{4}\theta & =C^{2}\ \left (1\ +\phi ^{\prime 2}\sin ^{2}\theta \right ) \\ \phi ^{\prime 2}\sin ^{4}\theta & =C^{2}+C^{2}\ \phi ^{\prime 2}\sin ^{2}\theta \\ \phi ^{\prime 2} & =\frac {C^{2}}{\sin ^{4}\theta -C^{2}\sin ^{2}\theta }\\ \phi ^{\prime } & =\frac {C}{\sin \theta \sqrt {\sin ^{2}\theta -C^{2}}}\\ \int d\phi & =\int \frac {C}{\sin \theta \sqrt {\sin ^{2}\theta -C^{2}}}d\theta \\ \phi +\alpha & =-\frac {C\ \tanh ^{-1}\left (\frac {\sqrt {2}\sqrt {C^{2}}\cos \left (\theta \right ) }{\sqrt {1-2C^{2}-\cos \left (2\theta \right ) }}\right ) }{\sqrt {-C^{2}}} \end {align*}
The last integral value was found using mathematica. Hence\[ \frac {\sqrt {-C^{2}}\left (\phi +\alpha \right ) }{-C}=\operatorname {arctanh}\left (\frac {\sqrt {2}\sqrt {C^{2}}\cos \left (\theta \right ) }{\sqrt {1-2C^{2}-\cos \left (2\theta \right ) }}\right ) \] Let \(\frac {\sqrt {-C^{2}}}{-C}=A,\) let \(\sqrt {2}\sqrt {C^{2}}=B,\) \(1-2C^{2}=D\), then\begin {align*} A\ \left (\phi +\alpha \right ) & =\operatorname {arctanh}\left (\frac {B\cos \left (\theta \right ) }{\sqrt {D-\cos \left (2\theta \right ) }}\right ) \\ \tanh \left (A\ \left (\phi +\alpha \right ) \right ) & =\frac {B\cos \left ( \theta \right ) }{\sqrt {D-\cos \left (2\theta \right ) }} \end {align*}
Problem
Set up Lagrange equations in cylindrical coordinates for a particle of mass m in a potential field \(V\left (r,\theta ,z\right ) \)
Solution
\(L=T-V\) where \(T\) is the K.E. and \(V\) the potential energy. \(T=\frac {1}{2}mv^{2}\), But
\[ ds^{2}=dr^{2}+r^{2}d\theta ^{2}+dz^{2}\] As shown on page 219 equation 4.4 , now differentiate both sides w.r.t. time\begin {align*} 2\ ds\ \frac {ds}{dt} & =2dr\ \dot {r}+\left (r^{2}\ 2\ d\theta \ \dot {\theta }+2r\ \dot {r}\ d\theta ^{2}\right ) +2dz\ \dot {z}\\ \frac {ds}{dt} & =\frac {dr\ \dot {r}+r^{2}\ \ d\theta \ \dot {\theta }+r\ \dot {r}\ d\theta ^{2}+dz\ \dot {z}}{\sqrt {dr^{2}+r^{2}d\theta ^{2}+dz^{2}}} \end {align*}
Hence\[ v^{2}=\frac {\left (dr\ \dot {r}+r^{2}d\theta \ \dot {\theta }+r\ \dot {r}\ d\theta ^{2}+dz\ \dot {z}\right ) ^{2}}{dr^{2}+r^{2}d\theta ^{2}+dz^{2}}\] I used Mathematica to simplify this getting\[ v^{2}=\ \dot {r}^{2}+r^{2}\ \dot {\theta }^{2}+\ \dot {z}^{2}\] Hence,\[ L=\overset {\text {K.E.}}{\overbrace {\frac {1}{2}m\ \left (\dot {r}^{2}+r^{2}\ \dot {\theta }^{2}+\ \dot {z}^{2}\right ) }}-\overset {\text {P.E.}}{\overbrace {V\left (r,\theta ,z\right ) }}\] The Lagrange equations are\begin {align*} \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {r}}\right ) -\frac {\partial L}{\partial r} & =0\\ \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {\theta }}\right ) -\frac {\partial L}{\partial \theta } & =0\\ \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {z}}\right ) -\frac {\partial L}{\partial z} & =0 \end {align*}
Hence, we get\begin {align*} \frac {d}{dt}\left (m\ \dot {r}\right ) -\left (mr\dot {\theta }^{2}-\frac {\partial V}{\partial r}\right ) & =0\\ \frac {d}{dt}\left (mr^{2}\dot {\theta }\right ) +\frac {\partial V}{\partial \theta } & =0\\ \frac {d}{dt}\left (m\dot {z}\right ) +\frac {\partial V}{\partial z} & =0 \end {align*}
Now differentiating w.r.t. time, and remembering that \(r\relax (t) \) also changes with time.\begin {align*} m\ \ddot {r}-mr\dot {\theta }^{2}+\frac {\partial V}{\partial r} & =0\\ m\left (2r\dot {r}\dot {\theta }+r^{2}\ddot {\theta }\right ) +\frac {\partial V}{\partial \theta } & =0\\ m\ddot {z}+\frac {\partial V}{\partial z} & =0 \end {align*}
Hence finally we get\begin {align*} m\left (\ddot {r}-r\dot {\theta }^{2}\right ) & =-\frac {\partial V}{\partial r}\\ m\left (2\dot {r}\dot {\theta }+r\ddot {\theta }\right ) & =-\frac {1}{r}\frac {\partial V}{\partial \theta }\\ m\ddot {z} & =-\frac {\partial V}{\partial z} \end {align*}
Problem
A particle moves on the surface of a sphere of radius \(a\) under the action of the earth gravitational field. Find the \(\theta ,\phi \) equations of motion. (this is called the spherical pendulum).
Solution
\(L=T-V\) where \(T\) is the K.E. and \(V\) the potential energy. Using spherical coordinates.\[ x=a\sin \theta \cos \phi ,\ y=a\sin \theta \sin \phi ,\ z=a\cos \theta \] Hence a position vector\[ \mathbf {r}=\mathbf {i\ }a\sin \theta \cos \phi +\mathbf {j\ }a\sin \theta \sin \phi +\mathbf {k\ }a\cos \theta \] So velocity is\begin {align*} \mathbf {\dot {r}} & =\mathbf {i\ }\frac {d}{dt}\left (a\sin \theta \cos \phi \right ) +\mathbf {j\ }\frac {d}{dt}\left (a\sin \theta \sin \phi \right ) +\mathbf {k\ }\frac {d}{dt}\left (a\cos \theta \right ) \\ & =\mathbf {i\ }\ \left (-a\sin \theta \sin \phi \ \dot {\phi }+a\cos \theta \ \dot {\theta }\ \cos \phi \right ) +\mathbf {j\ }\left (a\sin \theta \cos \phi \dot {\phi }+a\cos \theta \ \dot {\theta }\ \sin \phi \right ) +\mathbf {k\ }\left ( -a\sin \theta \ \dot {\theta }\right ) \end {align*}
Hence\[ \dot {r}=\left \Vert \mathbf {\dot {r}}\right \Vert =\ \sqrt {\left (-a\sin \theta \sin \phi \ \dot {\phi }+a\cos \theta \ \dot {\theta }\cos \phi \right ) ^{2}+\left (a\sin \theta \cos \phi \ \dot {\phi }+a\cos \theta \ \dot {\theta }\ \sin \phi \right ) ^{2}+\mathbf {\ }\left (-a\sin \theta \ \dot {\theta }\right ) ^{2}}\] Then\begin {align*} v^{2} & =\dot {r}^{2}=\left (-a\sin \theta \sin \phi \ \dot {\phi }+a\cos \theta \ \dot {\theta }\cos \phi \right ) ^{2}+\left (a\sin \theta \cos \phi \dot {\phi }+a\cos \theta \ \dot {\theta }\ \sin \phi \right ) ^{2}+\mathbf {\ }\left ( -a\sin \theta \ \dot {\theta }\right ) ^{2}\\ & \\ & =\left (a^{2}\sin ^{2}\theta \sin ^{2}\phi \ \dot {\phi }^{2}+a^{2}\cos ^{2}\theta \ \dot {\theta }^{2}\cos ^{2}\phi -\overbrace {2a^{2}\sin \theta \sin \phi \ \dot {\phi }\cos \theta \ \dot {\theta }\cos \phi }\right ) \\ & +\left (a^{2}\sin ^{2}\theta \cos ^{2}\phi \ \dot {\phi }^{2}+a^{2}\cos ^{2}\theta \ \dot {\theta }^{2}\ \sin ^{2}\phi \overbrace {+2a^{2}\sin \theta \cos \phi \dot {\phi }\ \cos \theta \ \dot {\theta }\ \sin \phi }\right ) +\mathbf {\ }\left (a^{2}\sin ^{2}\theta \ \dot {\theta }^{2}\right ) \\ & \\ & =\overbrace {a^{2}\sin ^{2}\theta \sin ^{2}\phi \ \dot {\phi }^{2}}+\widehat {a^{2}\cos ^{2}\theta \ \dot {\theta }^{2}\cos ^{2}\phi }+\overbrace {a^{2}\sin ^{2}\theta \cos ^{2}\phi \ \dot {\phi }^{2}}+\widehat {a^{2}\cos ^{2}\theta \ \dot {\theta }^{2}\ \sin ^{2}\phi }+\mathbf {\ }a^{2}\sin ^{2}\theta \ \dot {\theta }^{2}\\ & \\ & =a^{2}\dot {\phi }^{2}\sin ^{2}\theta \overbrace {\left (\sin ^{2}\phi +\cos ^{2}\phi \right ) }^{=1}+a^{2}\dot {\theta }^{2}\cos ^{2}\theta \overbrace {\left ( \cos ^{2}\phi +\sin ^{2}\phi \right ) }^{=1}+a^{2}\sin ^{2}\theta \ \dot {\theta }^{2}\\ & \\ & =a^{2}\dot {\phi }^{2}\sin ^{2}\theta \ +a^{2}\dot {\theta }\overbrace {^{2}\left (\cos ^{2}\theta +\sin ^{2}\theta \right ) }^{=1}\\ & \\ & =a^{2}\left (\dot {\phi }^{2}\sin ^{2}\theta \ +\dot {\theta }^{2}\right ) \end {align*}
Hence \(T=\frac {1}{2}mv^{2}\). For a particle, taking mass as one unit. Hence\[ T=\frac {1}{2}a^{2}\left (\dot {\phi }^{2}\sin ^{2}\theta \ +\dot {\theta }^{2}\right ) \] The P.E. is \(mga\ \cos \theta \). Hence the Lagrangian is\begin {align*} L & =T-V\\ L & =\frac {1}{2}a^{2}\left (\dot {\phi }^{2}\sin ^{2}\theta \ +\dot {\theta }^{2}\right ) -ga\ \cos \theta \end {align*}
We have 2 independent variables, hence we need 2 Lagrangian equations\begin {align*} \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {\theta }}\right ) -\frac {\partial L}{\partial \theta } & =0\\ \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {\phi }}\right ) -\frac {\partial L}{\partial \phi } & =0 \end {align*}
\begin {align*} \frac {\partial L}{\partial \dot {\theta }} & =a^{2}\dot {\theta }\\ \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {\theta }}\right ) & =a^{2}\ddot {\theta }\\ \frac {\partial L}{\partial \theta } & =a^{2}\left (\dot {\phi }^{2}\sin \theta \ \cos \theta \right ) +ga\ \sin \theta \end {align*}
Hence the first equation becomes\begin {align*} \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {\theta }}\right ) -\frac {\partial L}{\partial \theta } & =0\\ a^{2}\ddot {\theta }-a^{2}\left (\dot {\phi }^{2}\sin \theta \ \cos \theta \right ) -ga\ \sin \theta & =0\\ a\ \ddot {\theta }-a\ \left (\dot {\phi }^{2}\sin \theta \ \cos \theta \right ) -g\ \sin \theta & =0 \end {align*}
To find the second equation\begin {align*} \frac {\partial L}{\partial \dot {\phi }} & =a^{2}\left (2\dot {\phi }\sin ^{2}\theta \right ) \\ \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {\phi }}\right ) & =\frac {d}{dt}\left (a^{2}\left (2\dot {\phi }\sin ^{2}\theta \right ) \right ) \\ \frac {\partial L}{\partial \phi } & =0 \end {align*}
Hence the second equation is\begin {align*} \frac {d}{dt}\left (\frac {\partial L}{\partial \dot {\phi }}\right ) -\frac {\partial L}{\partial \phi } & =0\\ \frac {d}{dt}\left (a^{2}\left (2\dot {\phi }\sin ^{2}\theta \right ) \right ) & =0\\ \frac {d}{dt}\left (\ 2\dot {\phi }\sin ^{2}\theta \right ) & =0\\ \frac {d}{dt}\left (\ 2\dot {\phi }\sin ^{2}\theta \right ) & =0 \end {align*}
Problem
Find surface of revolution formed by rotating the curve around the x-axis that has a minimum area subject to a curve of give length \(l\) joining 2 points.
Solution
Area is\begin {equation} I=\int _{x_{1}}^{x_{2}}2\pi y\sqrt {1+y^{\prime 2}}dx \tag {1} \end {equation} Since integrand does not depend on \(x\) we change the independent variable to \(y\). \(dx=\frac {dx}{dy}dy,y^{\prime }=\frac {1}{x^{\prime }}\). Hence \((1)\) becomes\begin {align} I & =\int _{y_{1}}^{y_{2}}2\pi y\sqrt {1+\ \frac {1}{x^{\prime 2}}}\ x^{\prime }dy\tag {1}\\ & =\int _{y_{1}}^{y_{2}}2\pi y\sqrt {x^{\prime 2}+\ 1}\ dy\nonumber \end {align}
Hence \(F\left (y,x^{\prime },x\right ) =2\pi y\sqrt {x^{\prime 2}+\ 1}\). Now finding the constraint\begin {align*} g & =\int ds=l\\ & =\int _{x_{1}}^{x_{2}}\sqrt {1+y^{\prime 2}}dx \end {align*}
Since integrand does not depend on \(x\) we change the independent variable to \(y\). \(dx=\frac {dx}{dy}dy,y^{\prime }=\frac {1}{x^{\prime }}\). Hence\begin {align*} g & =\ \int _{y_{1}}^{y_{2}}\sqrt {1+\frac {1}{x^{\prime 2}}}\ x^{\prime }dy\\ & =\int _{y_{1}}^{y_{2}}\sqrt {x^{\prime 2}+1}\ \ dy \end {align*}
So \(G=\sqrt {x^{\prime 2}+1}\). Hence we get\[ F+\lambda G=\left (2\pi y\sqrt {x^{\prime 2}+\ 1}\right ) +\lambda \sqrt {x^{\prime 2}+1}\] As the new Euler equation (with constrains). Solving\begin {align*} \frac {d}{dy}\left (\frac {\partial }{\partial x^{\prime }}\left (F+\lambda G\right ) \right ) -\overset {0\text { since does not depend on }x}{\overbrace {\frac {\partial }{\partial x}\left (F+\lambda G\right ) }} & =0\\ \frac {d}{dy}\left (\frac {\partial }{\partial x^{\prime }}\left (2\pi y\sqrt {x^{\prime 2}+\ 1}+\lambda \sqrt {x^{\prime 2}+1}\right ) \right ) & =0\\ \frac {d}{dy}\left (\frac {2\pi yx^{\prime }}{\sqrt {x^{\prime 2}+\ 1}}+\frac {\lambda x^{\prime }}{\sqrt {x^{\prime 2}+1}}\ \right ) & =0 \end {align*}
Hence\begin {align*} \frac {2\pi yx^{\prime }}{\sqrt {x^{\prime 2}+1}}+\frac {\lambda x^{\prime }}{\sqrt {x^{\prime 2}+1}} & =c\\ \frac {2\pi yx^{\prime }+\lambda x^{\prime }}{\sqrt {x^{\prime 2}+1}} & =c\\ x^{\prime }\left (2\pi y+\lambda \right ) \ & =c\sqrt {x^{\prime 2}+1}\\ x^{\prime 2}\left (2\pi y+\lambda \right ) ^{2}\ & =c^{2}\left (x^{\prime 2}+\ 1\right ) \\ \frac {x^{\prime 2}}{\left (x^{\prime 2}+1\right ) } & =\frac {c^{2}}{\left ( 2\pi y+\lambda \right ) ^{2}}\\ \frac {\left (x^{\prime 2}+1\right ) }{x^{\prime 2}} & =\frac {\left (2\pi y+\lambda \right ) ^{2}}{c^{2}}\\ 1+\frac {1}{x^{\prime 2}} & =\frac {\left (2\pi y+\lambda \right ) ^{2}}{c^{2}}\\ \frac {1}{x^{\prime 2}} & =\frac {\left (2\pi y+\lambda \right ) ^{2}-c^{2}}{c^{2}}\\ \frac {c^{2}}{\left (2\pi y+\lambda \right ) ^{2}-c^{2}} & =x^{\prime 2}\\ \frac {c}{\sqrt {\left (2\pi y+\lambda \right ) ^{2}-c^{2}}} & =x^{\prime }\\ \frac {dx}{dy} & =\frac {c}{\sqrt {\left (2\pi y+\lambda \right ) ^{2}-c^{2}}}\\ \int dx & =\int \frac {c}{\sqrt {\left (2\pi y+\lambda \right ) ^{2}-c^{2}}}dy\\ \int dx & ={\displaystyle \int } \frac {1}{\sqrt {\left (\frac {2\pi y+\lambda }{c}\right ) ^{2}-1}}dy\\ \ x & =\frac {c}{2\pi }\operatorname {arccosh}\left (\frac {2\pi y+\lambda }{c}\right ) +c_{1} \end {align*}
To express this as \(y\) a function of \(x\) we get\begin {align*} \frac {2\pi }{c}\left (x-c_{1}\right ) & =\operatorname {arccosh}\left ( \frac {2\pi y+\lambda }{c}\right ) \\ \cosh \left (\frac {2\pi }{c}\left (x-c_{1}\right ) \right ) & =\frac {2\pi y+\lambda }{c}\\ \frac {c\cosh \left (\frac {2\pi }{c}\left (x-c_{1}\right ) \right ) -\lambda }{2\pi } & =y \end {align*}
We have 3 unknowns\(,c,c_{1},\lambda \) that we can use boundary conditions, and length \(l\) to determine.
Problem
Find the equation of the curve subject to a curve of give length \(l\) joining 2 points so that the plane area between the curve and straight line joining the points is a maximum.
Solution
Area is \(\int y\ dx\). Hence area is \(I=\int _{x_{1}}^{x_{2}}y\ dx\) subject to constraint that \(\int \ ds=l\) or \(\ g=\int _{x_{1}}^{x_{2}}\sqrt {1+y^{\prime 2}}dx=l.\) Hence the Euler equation with constrains now becomes\[ F+\lambda G=y\ +\lambda \sqrt {y^{\prime 2}+1}\] Therefore\begin {align*} \frac {d}{dx}\left (\frac {\partial }{\partial y^{\prime }}\left (F+\lambda G\right ) \right ) -\frac {d}{dy}\left (F+\lambda G\right ) & =0\\ \frac {d}{dy}\left (\frac {\lambda \ y^{\prime }}{\sqrt {y^{\prime 2}+1}}\right ) -1 & =0\\ \frac {\lambda \ y^{\prime }}{\sqrt {y^{\prime 2}+1}} & =x+c \end {align*}
This simplifies to\begin {align*} \int dy & =\int \frac {\left (x+c\right ) }{\sqrt {\lambda ^{2}-\left ( x+c\right ) ^{2}}}dx\\ y+c_{1} & =-\sqrt {\lambda ^{2}-\left (x+c\right ) ^{2}}\\ \left (y+c_{1}\right ) ^{2} & =\lambda ^{2}-\left (x+c\right ) ^{2}\\ \left (y+c_{1}\right ) ^{2}+\left (x+c\right ) ^{2} & =\lambda ^{2} \end {align*}
This is the equation of a circle.
Problem
Given surface area of solid of revolution, finds its shape to make its volume a maximum.
Solution
Volume is \(\int \pi y^{2}ds\) where \(ds\) is a small segment of the curve length. Hence\begin {equation} I=\int _{x_{1}}^{x_{2}}\pi y^{2}\sqrt {1+y^{\prime 2}}dx\tag {1} \end {equation} Constraint is that area is given, say \(A\). Hence\begin {equation} g=\int _{x_{1}}^{x_{2}}2\pi y\sqrt {1+y^{\prime 2}}dx=A\tag {2} \end {equation} Since both integrands do not depend on \(x\) we change the independent variable to \(y\). \(dx=\frac {dx}{dy}dy,y^{\prime }=\frac {1}{x^{\prime }}\). Hence \((1)\) becomes\begin {align*} I & =\int _{x_{1}}^{x_{2}}\pi y^{2}\sqrt {1+\frac {1}{x^{\prime 2}}}x^{\prime }dy\\ & =\int _{x_{1}}^{x_{2}}\pi y^{2}\sqrt {x^{\prime 2}+1}dy \end {align*}
And (2) becomes\begin {align*} g & =\int _{y_{1}}^{y_{2}}2\pi y\sqrt {1+\frac {1}{x^{\prime 2}}}\ x^{\prime }dy\\ & =\int _{y_{1}}^{y_{2}}2\pi y\sqrt {x^{\prime 2}+1}\ dy \end {align*}
Hence we get\[ F+\lambda G=\left (\pi y^{2}\sqrt {x^{\prime 2}+1}\right ) +2\lambda \pi y\sqrt {x^{\prime 2}+\ 1}\] as the new Euler equation (with constrains) to solve.\begin {align*} \frac {d}{dy}\left (\frac {\partial }{\partial x^{\prime }}\left (F+\lambda G\right ) \right ) -\overset {0\text { since does not depend on }x}{\overbrace {\frac {\partial }{\partial x}\left (F+\lambda G\right ) }} & =0\\ \frac {d}{dy}\left (\frac {\partial }{\partial x^{\prime }}\left (\pi y^{2}\sqrt {x^{\prime 2}+1}+2\lambda \pi y\sqrt {x^{\prime 2}+1}\right ) \right ) & =0\\ \frac {d}{dy}\left (\frac {\pi y^{2}x^{\prime }}{\sqrt {x^{\prime 2}+1}}+\frac {2\lambda \pi yx^{\prime }}{\sqrt {x^{\prime 2}+1}}\right ) & =0 \end {align*}
Hence\begin {align*} \frac {\pi y^{2}x^{\prime }}{\sqrt {x^{\prime 2}+1}}+\frac {2\lambda \pi yx^{\prime }}{\sqrt {x^{\prime 2}+1}} & =c\\ \frac {\pi y^{2}x^{\prime }+2\lambda \pi yx^{\prime }}{\sqrt {x^{\prime 2}+1}} & =c\\ \pi y^{2}x^{\prime }+2\lambda \pi yx^{\prime } & =c\sqrt {x^{\prime 2}+\ 1}\\ x^{\prime 2}\ \left (\pi y^{2}+2\lambda \pi y\right ) ^{2} & =c^{2}\left ( x^{\prime 2}+\ 1\right ) \\ \frac {x^{\prime 2}}{\left (x^{\prime 2}+1\right ) } & =\frac {c^{2}}{\left ( \pi y^{2}+2\lambda \pi y\right ) ^{2}}\\ \frac {\left (x^{\prime 2}+1\right ) }{x^{\prime 2}} & =\frac {\left (\pi y^{2}+2\lambda \pi y\right ) ^{2}}{c^{2}}\\ 1+\frac {1}{x^{\prime 2}} & =\frac {\left (\pi y^{2}+2\lambda \pi y\right ) ^{2}}{c^{2}}\\ \frac {1}{x^{\prime 2}} & =\frac {\left (\pi y^{2}+2\lambda \pi y\right ) ^{2}-c^{2}}{c^{2}}\\ \frac {c^{2}}{\left (\pi y^{2}+2\lambda \pi y\right ) ^{2}-c^{2}} & =x^{\prime 2}\\ \frac {c}{\sqrt {\left (\pi y^{2}+2\lambda \pi y\right ) ^{2}-c^{2}}} & =x^{\prime }\\ \frac {dx}{dy} & =\frac {c}{\sqrt {\left (\pi y^{2}+2\lambda \pi y\right ) ^{2}-c^{2}}}\\ \int dx & =\int \frac {c}{\sqrt {\left (\pi y^{2}+2\lambda \pi y\right ) ^{2}-c^{2}}}dy\\ x & =\int \frac {c}{\sqrt {\left (\pi y^{2}+2\lambda \pi y\right ) ^{2}-c^{2}}}dy\\ x & =\int \frac {1}{\sqrt {\left (\frac {\pi y^{2}+2\lambda \pi y}{c}\right ) ^{2}-1}}dy \end {align*}
Hence\[ x=\left (\frac {c}{2y\pi +2\lambda \pi }\right ) \cosh ^{-1}\left (\frac {\pi y^{2}+2\lambda \pi y}{c}\right ) \]
Problem
Solve \(y^{\prime \prime }+y=f\relax (x) \) with \(y\relax (0) =y\left (\frac {\pi }{2}\right ) =0\) using 8.17:\[ y\relax (x) =-\cos x\int _{0}^{x}\sin \left (x^{\prime }\right ) f\left ( x^{\prime }\right ) \ dx^{\prime }-\sin x\int _{x}^{\frac {\pi }{2}}\ \cos \left ( x^{\prime }\right ) f\left (x^{\prime }\right ) \ dx^{\prime }\] when \(f\relax (x) =\sec x\)
Solution
\[ y\relax (x) =-\cos x\int _{0}^{x}\sin \left (x^{\prime }\right ) \sec x^{\prime }\ dx^{\prime }-\sin x\int _{x}^{\frac {\pi }{2}}\cos \left (x^{\prime }\right ) \sec x^{\prime }\ dx^{\prime }\] Since \(\sec x^{\prime }=\frac {1}{\cos x^{\prime }}\) we get\[ y\relax (x) =-\cos x\int _{0}^{x}\tan x^{\prime }\ dx^{\prime }-\sin x\int _{x}^{\frac {\pi }{2}}dx^{\prime }\] But \(\int _{0}^{x}\tan x^{\prime }\ dx^{\prime }=-\log \left (\cos \left ( x\right ) \right ) \), Hence\begin {align*} y\relax (x) & =\cos \relax (x) \log \left (\cos \left ( x\right ) \right ) -\sin x\ \left (\frac {1}{2}\pi -x\right ) \\ & =\cos \relax (x) \log \left (\cos \relax (x) \right ) -\frac {1}{2}\pi \sin x+x\sin x \end {align*}
Problem
Use Green function method and the given solutions of the homogeneous equation to find a particular solution to \(y^{\prime \prime }-y=\sec h\relax (x) \), where \(y_{1}\relax (x) =\sinh \relax (x) ,\ \ y_{2}\left ( x\right ) =\cosh \relax (x) \)
Solution
\begin {equation} y_{p}=y_{2}\int \frac {y_{1}\ f}{W}dx-y_{1}\int \frac {y_{2}\ f}{W}dx \tag {1} \end {equation} Where \(f=\sec h\relax (x) \)\begin {align*} W & =\left \vert \begin {array} [c]{cc}y_{1}^{\prime } & y_{2}^{\prime }\\ y_{1} & y_{2}\end {array} \right \vert \\ & =\left \vert \begin {array} [c]{cc}\cosh x & \sinh x\\ \sinh x & \cosh x \end {array} \right \vert \\ & =\cosh ^{2}x-\sinh ^{2}x\\ & =1 \end {align*}
So from (1) we get\[ y_{p}=\cosh \relax (x) \int \sinh \relax (x) \ \sec h\left ( x\right ) \ dx-\sinh \relax (x) \int \cosh \relax (x) \sec h\left ( x\right ) \ dx \] But \(\sec h\relax (x) =\frac {1}{\cosh x}\), Hence\begin {align*} y_{p} & =\cosh \relax (x) \int \sinh \relax (x) \ \frac {1}{\cosh x}\ dx-\sinh \relax (x) \int \cosh \relax (x) \frac {1}{\cosh x}\ dx\\ & =\cosh \relax (x) \int \tan sh\relax (x) \ dx-\sinh \left ( x\right ) \int dx \end {align*}
But \(\int \tan sh\relax (x) \ dx=\log \left (\cosh \relax (x) \right ) \), Hence\[ y_{p}=\cosh \relax (x) \log \left (\cosh \relax (x) \right ) -x\ \sinh \relax (x) \]
Problem
Use Green function method and the given solutions of the homogeneous equation to find a particular solution to \(y^{\prime \prime }-2\left (\csc ^{2}\left ( x\right ) \right ) y=\sin ^{2}\relax (x) \) , where \(y_{1}\left ( x\right ) =\cot x,\ \ y_{2}\relax (x) =1-x\ \cot \relax (x) \)
Solution
Note \(\cot \relax (x) =\frac {1}{\tan \relax (x) }=\frac {\cos \relax (x) }{\sin \relax (x) },\csc \relax (x) =\frac {1}{\sin \relax (x) }\)\begin {equation} y_{p}=y_{2}\int \frac {y_{1}\ f}{W}dx-y_{1}\int \frac {y_{2}\ f}{W}dx\tag {1} \end {equation} Where \(f=\sin ^{2}\relax (x) \).\begin {align*} y_{1}^{\prime } & =\frac {d}{dx}\left (\cot \relax (x) \right ) =-\cot ^{2}x-1\\ & =-\frac {1}{\sin ^{2}\relax (x) } \end {align*}
And\begin {align*} y_{2}^{\prime } & =\frac {d}{dx}\left (1-x\ \cot \relax (x) \right ) \\ & =-\frac {\cos \relax (x) }{\sin \relax (x) }+\frac {x}{\sin ^{2}\relax (x) } \end {align*}
Therefore\begin {align*} W & =\left \vert \begin {array} [c]{cc}y_{1}^{\prime } & y_{2}^{\prime }\\ y_{1} & y_{2}\end {array} \right \vert \\ & =\left \vert \begin {array} [c]{cc}-\frac {1}{\sin ^{2}\relax (x) } & -\frac {\cos \relax (x) }{\sin \relax (x) }+\frac {x}{\sin ^{2}\relax (x) }\\ \frac {\cos \relax (x) }{\sin \relax (x) } & 1-\frac {x\ \cos \left ( x\right ) }{\sin \relax (x) }\end {array} \right \vert \\ & =\left (-\frac {1}{\sin ^{2}\relax (x) }\right ) \left ( 1-\frac {x\ \cos \relax (x) }{\sin \relax (x) }\right ) -\left ( -\frac {\cos \relax (x) }{\sin \relax (x) }+\frac {x}{\sin ^{2}\relax (x) }\right ) \frac {\cos \relax (x) }{\sin \left ( x\right ) }\\ & =-\frac {1}{\sin ^{2}\relax (x) }+\overbrace {\frac {x\ \cos \left ( x\right ) }{\sin ^{3}\relax (x) }}+\frac {\cos ^{2}\relax (x) }{\sin ^{2}\relax (x) }-\overbrace {\frac {x\cos \relax (x) }{\sin ^{3}\relax (x) }}\\ & =-\frac {1}{\sin ^{2}\relax (x) }+\frac {\cos ^{2}\relax (x) }{\sin ^{2}\relax (x) } \end {align*}
So from (1) we get\begin {align} y_{p} & =\left (1-\frac {x\cos x}{\sin x}\right ) \int \frac {\frac {\cos x}{\sin x}\ \sin ^{2}\relax (x) }{-\frac {1}{\sin ^{2}\relax (x) }+\frac {\cos ^{2}\relax (x) }{\sin ^{2}\relax (x) }}dx-\frac {\cos x}{\sin x}\int \frac {\left (1-\frac {x\cos x}{\sin \relax (x) }\right ) \ \sin ^{2}\relax (x) }{-\frac {1}{\sin ^{2}\relax (x) }+\frac {\cos ^{2}\relax (x) }{\sin ^{2}\relax (x) }}dx\nonumber \\ & =\left (1-\frac {x\cos x}{\sin x}\right ) \int \frac {\cos x\ \sin x}{\ \frac {-1+\cos ^{2}x}{\sin ^{2}\relax (x) }}dx-\frac {\cos x}{\sin x}\int \frac {\sin ^{2}x-x\ \cos x\ \sin x\ \ }{\frac {-1+\cos ^{2}x}{\sin ^{2}\relax (x) }}dx\nonumber \\ & =\left (1-\frac {x\cos x}{\sin x}\right ) \int \frac {\cos x\ \sin ^{3}x}{\ -1+\cos ^{2}x}dx-\frac {\cos x}{\sin x}\int \frac {\sin ^{4}x-x\ \cos x\ \sin ^{3}x\ \ }{-1+\cos ^{2}x}dx\nonumber \end {align}
but \(I=\) \(\int \frac {\cos x\ \sin ^{3}x}{\ \cos ^{2}x-1}=\int \frac {\cos x\ \sin ^{3}x}{\ -\sin ^{2}x}=\int -\cos x\ \sin x=\frac {1}{2}\cos ^{2}x\) And \begin {align*} I & =\int \frac {\sin ^{4}x-x\ \cos x\ \sin ^{3}x\ \ }{-1+\cos ^{2}x}\ \\ & =\int \frac {\sin ^{4}x-x\ \cos x\ \sin ^{3}x\ \ }{-\sin ^{2}x}\\ & =\int -\sin ^{2}x+x\ \cos x\ \sin x\ \\ & =-\int \sin ^{2}\relax (x) \ dx+\int x\ \cos \relax (x) \ \sin \relax (x) \ dx \end {align*}
But \(\int \sin ^{2}\relax (x) \ dx=\frac {x}{2}-\frac {1}{4}\sin \left ( 2x\right ) \) and \(\int x\ \cos \relax (x) \ \sin \relax (x) \ dx=-\frac {1}{4}x\cos \left (2x\right ) +\frac {1}{8}\sin \left (2x\right ) \), therefore \begin {align*} -\int \sin ^{2}\relax (x) \ dx+\int x\ \cos \relax (x) \ \sin \relax (x) \ dx & =\left (-\frac {x}{2}+\frac {1}{4}\sin \left ( 2x\right ) \right ) +\left (-\frac {1}{4}x\cos \left (2x\right ) +\frac {1}{8}\sin \left (2x\right ) \right ) \\ & =-\frac {x}{2}+\frac {1}{4}\sin \left (2x\right ) -\frac {1}{4}x\cos \left ( 2x\right ) +\frac {1}{8}\sin \left (2x\right ) \\ & =\frac {3}{8}\sin 2x-\frac {1}{2}x-\frac {1}{4}x\cos 2x \end {align*}
Hence (2) becomes\begin {align*} y_{p}\relax (x) & =\left (1-\frac {x\cos x}{\sin x}\right ) \left ( \frac {1}{2}\cos ^{2}x\right ) -\frac {\cos x}{\sin x}\left (\frac {3}{8}\sin 2x-\frac {1}{2}x-\frac {1}{4}x\cos 2x\right ) \\ & =\left (\frac {1}{2}\cos ^{2}x-\frac {1}{2}\frac {x\cos ^{3}x}{\sin x}\right ) -\left (\frac {3}{8}\sin 2x\frac {\cos x}{\sin x}-\frac {1}{2}x\frac {\cos x}{\sin x}-\frac {1}{4}x\cos 2x\frac {\cos x}{\sin x}\right ) \\ & =\frac {1}{2}\cos ^{2}x-\frac {1}{2}\frac {x\cos ^{3}x}{\sin x}-\frac {3}{8}\sin 2x\frac {\cos x}{\sin x}+\frac {1}{2}x\frac {\cos x}{\sin x}+\frac {1}{4}x\cos 2x\frac {\cos x}{\sin x}\\ & =\frac {1}{4}\cot x\ \left (x-\cos x\ \sin x\right ) \end {align*}
Problem
Solve \(y^{\prime \prime }+\omega ^{2}y=f\relax (t) \) using \(y\left ( t\right ) =\int _{0}^{t}\frac {1}{\omega }\sin \omega \left (t-t^{\prime }\right ) \ f\left (t^{\prime }\right ) \ dt^{\prime }\) when \(f\relax (t) =\sin \omega t\)
Solution\begin {align} y\relax (t) & =\int _{0}^{t}\frac {1}{\omega }\sin \omega \left ( t-t^{\prime }\right ) f\left (t^{\prime }\right ) \ dt^{\prime }\nonumber \\ & =\int _{0}^{t}\frac {1}{\omega }\sin \omega \left (t-t^{\prime }\right ) \sin \omega t^{\prime }\ dt^{\prime }\tag {1} \end {align}
But \(\sin \alpha \sin \beta =\frac {1}{2}\cos \left (\alpha -\beta \right ) -\frac {1}{2}\cos \left (\alpha +\beta \right ) \), hence\begin {align*} \sin \omega \left (t-t^{\prime }\right ) \sin \omega t^{\prime } & =\frac {1}{2}\cos \left (\omega \left (t-t^{\prime }\right ) -\omega t^{\prime }\right ) -\frac {1}{2}\cos \left (\omega \left (t-t^{\prime }\right ) +\omega t^{\prime }\right ) \\ & =\frac {1}{2}\cos \left (t\omega -2\omega t^{\prime }\right ) -\frac {1}{2}\cos \omega t \end {align*}
Hence (1) becomes\begin {align*} y\relax (t) & =\int _{0}^{t}\frac {1}{\omega }\frac {1}{2}\cos \left ( \omega t-2\omega t^{\prime }\right ) -\frac {1}{2}\cos \omega t\ dt^{\prime }\\ & =\frac {1}{2\omega }\int _{0}^{t}\cos \left (\omega t-2\omega t^{\prime }\right ) \ dt^{\prime }-\frac {1}{2}\cos \omega t\int _{0}^{t}dt^{\prime }\\ & =\frac {1}{2\omega }\left [ \frac {\sin \left (\omega t-2\omega t^{\prime }\right ) }{-2\omega }\right ] _{0}^{t}-\frac {1}{2}t\cos t\omega \\ & =\frac {-1}{4\omega ^{2}}\left (\sin \left (\omega t-2\omega t\right ) -\sin \left (\omega t\right ) \right ) -\frac {1}{2}t\cos t\omega \\ & =\frac {1}{2\omega ^{2}}\sin t\omega -\frac {1}{2}t\cos t\omega \\ & =\frac {1}{2\omega ^{2}}\left (\sin t\omega -\omega t\cos t\omega \right ) \end {align*}
\[ \ y\relax (t) =\frac {1}{2\omega ^{2}}\left (\sin t\omega -\omega t\cos t\omega \right ) \]
Problem
Solve \(y^{\prime \prime }+\omega ^{2}y=f\relax (t) \) using \(y\left ( t\right ) =\int _{0}^{t}\frac {1}{\omega }\sin \omega \left (t-t^{\prime }\right ) \ f\left (t^{\prime }\right ) \ dt^{\prime }\) when \(f\relax (t) =e^{-t}\)
Solution\begin {align} y\relax (t) & =\int _{0}^{t}\frac {1}{\omega }\sin \omega \left ( t-t^{\prime }\right ) \ f\left (t^{\prime }\right ) \ dt^{\prime }\nonumber \\ & =\frac {1}{\omega }\int _{0}^{t}\sin \omega \left (t-t^{\prime }\right ) \ e^{-t^{\prime }}\ dt^{\prime }\tag {1} \end {align}
\[ \text {Let \ }I=\int _{0}^{t}\sin \omega \left (t-t^{\prime }\right ) \ e^{-t^{\prime }}\ dt^{\prime }\] Integrate by part, let \(u=\sin \left (\omega t-\omega t^{\prime }\right ) \,\ ,v=-e^{-t^{\prime }}\)\begin {align*} I & =[\sin \omega \left (t-t^{\prime }\right ) \ \left (-e^{-t^{\prime }}\right ) ]_{0}^{t}-\omega \int _{0}^{t}\cos \left (\omega t-\omega t^{\prime }\right ) \ e^{-t^{\prime }}\ dt^{\prime }\\ & =\sin \omega t\ -\omega \int _{0}^{t}\cos \left (\omega t-\omega t^{\prime }\right ) \ e^{-t^{\prime }}\ dt^{\prime } \end {align*}
Integrate by parts again. \(u=\cos \left (\omega t-\omega t^{\prime }\right ) \,\ ,v=-e^{-t^{\prime }}\)\begin {align*} I & =\sin \omega t-\omega \left (\lbrack \cos \left (\omega t-\omega t^{\prime }\right ) \left (-e^{-t^{\prime }}\right ) ]_{0}^{t}+\omega \int _{0}^{t}\sin \omega \left (t-t^{\prime }\right ) e^{-t^{\prime }}\ dt^{\prime }\right ) \\ I & =\sin \omega t-\omega \left (\lbrack -e^{-t}+\cos \left (\omega t\right ) ]+\omega I\right ) \\ I & =\ \sin \omega t+\omega e^{-t}-\omega \cos \left (\omega t\right ) -\omega ^{2}I\\ I+\omega ^{2}I & =\ \sin \omega t+\omega e^{-t}-\omega \cos \left (\omega t\right ) \\ I & =\frac {\sin \omega t+\omega e^{-t}-\omega \cos \left (\omega t\right ) }{1+\omega ^{2}} \end {align*}
Hence from (1)\[ y\relax (t) =\frac {1}{\omega }\frac {\omega e^{-t}-\omega \cos \left ( \omega t\right ) +\sin \left (\omega t\right ) }{1+\omega ^{2}}\]
Problem
Change the independent variable to simplify the Euler equation and then find the first integral of it. \(\int _{x_{2}}^{x_{1}}\ y^{\frac {3}{2}}ds\)
Solution
\[ ds=\sqrt {\left (dx\right ) ^{2}+\left (dy\right ) ^{2}}=dx\sqrt {1+\left ( \frac {dy}{dx}\right ) ^{2}}=dx\sqrt {1+y^{\prime 2}}\] Hence\[ I=\int _{x_{2}}^{x_{1}}\ y^{\frac {3}{2}}ds=\int _{x_{2}}^{x_{1}}\ y^{\frac {3}{2}}\sqrt {1+y^{\prime 2}}dx \] Since integrand does not depend on \(x,\) changing the independent variable to \(y\) in order to simplify solution. Using \(dx=\frac {dx}{dy}dy\rightarrow y^{\prime }=\frac {1}{x^{\prime }}\). The integral now becomes\begin {align*} I & =\int _{x_{2}}^{x_{1}}\ y^{\frac {3}{2}}\sqrt {1+\frac {1}{x^{\prime 2}}}x^{\prime }\ dy\\ & =\int _{x_{2}}^{x_{1}}\ y^{\frac {3}{2}}\sqrt {x^{\prime 2}+1}\ dy \end {align*}
\[ F\left (y,x^{\prime },x\right ) =y^{\frac {3}{2}}\sqrt {x^{\prime 2}+1}\] The Euler equation is\begin {align*} \frac {d}{dy}\left (\frac {\partial F}{\partial x^{\prime }}\right ) -\overset {0}{\overbrace {\frac {\partial F}{\partial x}}} & =0\\ \frac {d}{dy}\left (\frac {\partial F}{\partial x^{\prime }}\right ) & =0\\ \frac {\partial F}{\partial x^{\prime }} & =c\\ y^{\frac {3}{2}}\frac {x^{\prime }}{\sqrt {x^{\prime 2}+1}} & =c \end {align*}
Simplifying gives\begin {align*} x^{\prime } & =\frac {c}{\sqrt {y^{3}-c^{2}}}\\ \frac {dx}{dy} & =\frac {c}{\sqrt {y^{3}-c^{2}}}\\ x & =\int \frac {1}{\sqrt {\frac {y^{3}}{c^{2}}-1}}dy \end {align*}
We can stop here as the problem did not ask to fully solve the integral.