Internal
problem
ID
[8184] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
46 Date
solved
:
Sunday, November 10, 2024 at 08:20:00 PM CAS
classification
:
[_quadrature]
\begin{align*} \int f d f &= dx\\ \frac {f^{2}}{2}&= x +c_1 \end{align*}
Solving for \(f\) gives
\begin{align*}
f &= \sqrt {2 c_1 +2 x} \\
f &= -\sqrt {2 c_1 +2 x} \\
\end{align*}
Summary of solutions found
\begin{align*}
f &= \sqrt {2 c_1 +2 x} \\
f &= -\sqrt {2 c_1 +2 x} \\
\end{align*}
Solved as first order Bernoulli ode
Time used: 0.089 (sec)
In canonical form, the ODE is
\begin{align*} f' &= F(x,f)\\ &= \frac {1}{f} \end{align*}
This is a Bernoulli ODE.
\[ f' = \left (1\right ) \frac {1}{f} \tag {1} \]
The standard Bernoulli ODE has the form
\[ f' = f_0(x)f+f_1(x)f^n \tag {2} \]
Comparing this to (1)
shows that
\begin{align*} f_0 &=0\\ f_1 &=1 \end{align*}
The first step is to divide the above equation by \(f^n \) which gives
\[ \frac {f'}{f^n} = f_1(x) \tag {3} \]
The next step is use the
substitution \(v = f^{1-n}\) 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 \(f(x)\) 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)&=0\\ f_1(x)&=1\\ n &=-1 \end{align*}
Dividing both sides of ODE (1) by \(f^n=\frac {1}{f}\) gives
\begin{align*} f'f &= 0 +1 \tag {4} \end{align*}
Let
\begin{align*} v &= f^{1-n} \\ &= f^{2} \tag {5} \end{align*}
Taking derivative of equation (5) w.r.t \(x\) gives
\begin{align*} v' &= 2 ff' \tag {6} \end{align*}
Substituting equations (5) and (6) into equation (4) gives
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 \(\frac {\partial M}{\partial f}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the
function \(\phi \left (x,f\right )\)
Where \(f(f)\) is used for the constant of integration since \(\phi \) is a function
of both \(x\) and \(f\). Taking derivative of equation (3) w.r.t \(f\) gives
Where \(c_1\) is constant of integration. Substituting result found above for \(f(f)\) into equation
(3) gives \(\phi \)
\[
\phi = -x +\frac {f^{2}}{2}+ c_1
\]
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
\[
c_1 = -x +\frac {f^{2}}{2}
\]
Solving for \(f\) gives
\begin{align*}
f &= \sqrt {2 c_1 +2 x} \\
f &= -\sqrt {2 c_1 +2 x} \\
\end{align*}
Summary of solutions found
\begin{align*}
f &= \sqrt {2 c_1 +2 x} \\
f &= -\sqrt {2 c_1 +2 x} \\
\end{align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,f\right ) \to \left ( R,S \right )\)
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 \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial f}\right ) S(x,f) = 1\). 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}{-\frac {1}{f}}} dy \end{align*}
Which results in
\begin{align*} S&= -\frac {f^{2}}{2} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,f\) 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 \(f(R)\).
This ODE is now solved for \(x \left (p \right )\). Integrating gives
\begin{align*} x \left (p \right )&=\int -\frac {1}{p^{3}}d p +c_1\\ x \left (p \right )&=\frac {1}{2 p^{2}}+c_1\tag {5} \end{align*}
Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve
for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\)
from Eq. (5) and substitute the result into (1A).
Eliminating \(p\) from the following two equations
\begin{align*}
x &= \frac {1}{2 p^{2}}+c_1 \\
f &= \frac {1}{p} \\
\end{align*}
results in
\begin{align*}
p &= \operatorname {RootOf}\left (1+\left (2 c_1 -2 x \right ) \textit {\_Z}^{2}\right ) \\
\end{align*}
Substituting the above into Eq
(1A) and simplifying gives
\begin{align*}
f &= \frac {1}{\operatorname {RootOf}\left (1+\left (2 c_1 -2 x \right ) \textit {\_Z}^{2}\right )} \\
\end{align*}
Summary of solutions found
\begin{align*}
f &= \frac {1}{\operatorname {RootOf}\left (1+\left (2 c_1 -2 x \right ) \textit {\_Z}^{2}\right )} \\
\end{align*}