Internal
problem
ID
[6394] Book
:
Ordinary
differential
equations
and
calculus
of
variations.
Makarets
and
Reshetnyak.
Wold
Scientific.
Singapore.
1995 Section
:
Chapter
1.
First
order
differential
equations.
Section
1.1
Separable
equations
problems.
page
7 Problem
number
:
31 Date
solved
:
Thursday, October 17, 2024 at 10:07:04 AM CAS
classification
:
[[_homogeneous, `class C`], _dAlembert]
\begin{equation}
\tag{6E} b_{2}-\frac {3 a_{3}}{8}+\frac {a_{2}}{2}-\frac {b_{3}}{2}+\frac {a_{3} \cos \left (2 x +2 y \right )}{2}-\frac {a_{3} \cos \left (4 x +4 y \right )}{8}+x a_{2} \sin \left (2 x +2 y \right )+x b_{2} \sin \left (2 x +2 y \right )+y a_{3} \sin \left (2 x +2 y \right )+y b_{3} \sin \left (2 x +2 y \right )-\frac {a_{2} \cos \left (2 x +2 y \right )}{2}+\frac {b_{3} \cos \left (2 x +2 y \right )}{2}+a_{1} \sin \left (2 x +2 y \right )+b_{1} \sin \left (2 x +2 y \right ) = 0
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\{x, y, \cos \left (2 x +2 y \right ), \cos \left (4 x +4 y \right ), \sin \left (2 x +2 y \right )\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\{x = v_{1}, y = v_{2}, \cos \left (2 x +2 y \right ) = v_{3}, \cos \left (4 x +4 y \right ) = v_{4}, \sin \left (2 x +2 y \right ) = v_{5}\}
\]
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\)
where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and
hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1)
gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since
\(\xi =0\) then in this special case
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of
\(R,S\) from the result obtained earlier and simplifying. This gives
\begin{align*} \frac {dS}{dR} &= 1 \end{align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts
an ode, no matter how complicated it is, to one that can be solved by integration when the
ode is in the canonical coordiates \(R,S\).
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).
\begin{align*} \int {dS} &= \int {1\, dR}\\ S \left (R \right ) &= R + c_3 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} \tan \left (x +y\right ) = x +c_3 \end{align*}
for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} p \left (x \right ) = -1\\ p \left (x \right ) = 0 \end{align*}
Solving for \(p \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} p \left (x \right )&=-1\\ p \left (x \right )&=0\\ p \left (x \right )&=-\frac {\left (x +c_1 \right )^{2}}{c_1^{2}+2 c_1 x +x^{2}+1} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*} y = -x +\frac {\pi }{2}\\ y = -x\\ y = -x +\arcsin \left (\sqrt {\frac {\left (x +c_1 \right )^{2}}{c_1^{2}+2 c_1 x +x^{2}+1}}\right )\\ \end{align*}
for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} p \left (x \right ) = -1\\ p \left (x \right ) = 0 \end{align*}
Solving for \(p \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} p \left (x \right )&=-1\\ p \left (x \right )&=0\\ p \left (x \right )&=-\frac {\left (x +c_2 \right )^{2}}{c_2^{2}+2 c_2 x +x^{2}+1} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*} y = -x -\frac {\pi }{2}\\ y = -x\\ y = -x -\arcsin \left (\sqrt {\frac {\left (x +c_2 \right )^{2}}{c_2^{2}+2 c_2 x +x^{2}+1}}\right )\\ \end{align*}
The solution
\[
y = -x
\]
was found not to satisfy the ode or the IC. Hence it is removed.
The solution
\[
y = -x
\]
was found not to satisfy the ode or the IC. Hence it is removed.
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryinghomogeneous C1storder, trying the canonical coordinates of the invariance group<-1st order, canonical coordinates successful<-homogeneous successful`