Internal
problem
ID
[8378] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
4.0 Problem
number
:
60 Date
solved
:
Sunday, November 10, 2024 at 08:20:56 PM CAS
classification
:
[_quadrature]
for \(y\). 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*} y = -1\\ y = 0\\ y = 1 \end{align*}
The following diagram is the phase line diagram. It classifies each of the above
equilibrium points as stable or not stable or semi-stable.
Solving for \(y\) gives
\begin{align*}
y &= \frac {\sqrt {\left ({\mathrm e}^{2 x +2 c_1}-1\right ) {\mathrm e}^{2 x +2 c_1}}}{{\mathrm e}^{2 x +2 c_1}-1} \\
y &= -\frac {\sqrt {\left ({\mathrm e}^{2 x +2 c_1}-1\right ) {\mathrm e}^{2 x +2 c_1}}}{{\mathrm e}^{2 x +2 c_1}-1} \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= \frac {\sqrt {\left ({\mathrm e}^{2 x +2 c_1}-1\right ) {\mathrm e}^{2 x +2 c_1}}}{{\mathrm e}^{2 x +2 c_1}-1} \\
y &= -\frac {\sqrt {\left ({\mathrm e}^{2 x +2 c_1}-1\right ) {\mathrm e}^{2 x +2 c_1}}}{{\mathrm e}^{2 x +2 c_1}-1} \\
y &= -1 \\
y &= 0 \\
y &= 1 \\
\end{align*}
The next step is use the
substitution \(v = y^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be
easily solved using an integrating factor. Backsubstitution then gives the solution \(y(x)\) which is
what we want.
This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows
that
\begin{align*} f_0(x)&=1\\ f_1(x)&=-1\\ n &=3 \end{align*}
Dividing both sides of ODE (1) by \(y^n=y^{3}\) gives
for \(v \left (x \right )\). 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*} v \left (x \right ) = 1 \end{align*}
The substitution \(v = y^{1-n}\) is now used to convert the above solution back to \(y\) which results in
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that
satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
If the above condition is satisfied,
then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know
now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not
satisfied then this method will not work and we have to now look for an integrating
factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an
integrating factor to make it exact. Let
\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\)
and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function
of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives
But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant
and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as
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
\begin{align*} R = x \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{y^{3}-y}} dy \end{align*}
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 \(f(R)\).