Solving a system of differential equations

[JD_PM]

Summary:: We want to find explicit functions $g(y,t)$ and $f(y,t)$ satisfying the following system of differential equations.

I attached a very similar solved example.

Given the following system of differential equations (assuming $y \neq 0$)

\begin{equation*}
-y\partial_t \left( g(y,t)\right) + f(y,t) = 0 \tag{1}
\end{equation*}

\begin{equation*}
y\partial_y \left( f(y,t)\right) -f(y,t) = 0 \tag{2}
\end{equation*}

\begin{equation*}
\frac{1}{y^2} \partial_t \left( f(y,t) \right) - \partial_y \left( \frac{1}{y^2} g(y,t) \right) - \frac{2}{y^3} g(y,t) = 0 \tag{3}
\end{equation*}

We want to find explicit functions $g(y,t)$ and $f(y,t)$ satisfying such system

Adding up $(1)$ and $(2)$

\begin{equation*}
\partial_y \left( f(y,t)\right) - \partial_t \left( g(y,t)\right) = 0 \tag{4}
\end{equation*}

Equation $(4)$ is satisfied when

\begin{equation*}
f(y,t) = f(t), \ \ \ \ g(y,t) = g(y) \tag{5}
\end{equation*}

Plugging $(5)$ into $(3)$ yields

\begin{equation*}
\partial_t f(t) - \partial_y g(y) =0 \tag{6}
\end{equation*}

But I do not see how $(6)$ is going to lead us to get explicit functions for $g(y,t)$ and $f(y,t)$.

I actually expect to get a second order differential equation out of working out the system, whose solution should provide us with explicit functions for $g(y,t)$ and $f(y,t)$.

Could you please help me out to move forward?

Thank you!

PS: I attach an analogous example. I am following the same procedure to solve our original problem

 

[BvU]

Is this homework ?

Equation (4) is satisfied when

But then (2) yields $f = 0$ !

By the way, (2) is easily solved for $y$ !

 

[JD_PM]

Is this homework ?

This is a system of differential equations I encountered while solving a specific problem I am solving to study for my exam. I guess you can call it homework

But then (2) yields $f = 0$ !

By the way, (2) is easily solved for $y$ !

So if I am not mistaken you are suggesting not to add up $(1)$ and $(2)$ (?) How to proceed then? (sorry I am a bit rusty with differential equations).

 

[berkeman]

I guess you can call it homework

Thread moved the schoolwork forums.

 

[BvU]

sorry I am a bit rusty with differential equations

solving a specific problem I am solving to study for my exam

If (2) is already a stumbling block, and the 'analogous example' is typical for the exam, I fear with great fear ...

Compare
$$
\begin{equation*}

y\partial_y \left( f(y,t)\right) -f(y,t) = 0 \tag{2}

\end{equation*}
$$
with $\ x \, f'(x) = f(x) \$

 

[JD_PM]

If (2) is already a stumbling block, and the 'analogous example' is typical for the exam, I fear with great fear ...

... let's not panic though!

Compare
$$
\begin{equation*}

y\partial_y \left( f(y,t)\right) -f(y,t) = 0 \tag{2}

\end{equation*}
$$
with $\ x \, f'(x) = f(x) \$

Thanks for the guidance! Let's start over

Focusing on $(2)$, we deduce that it takes the form

\begin{equation*}
\partial_y \left( y \ f(x,y) \right) = 0
\end{equation*}

Where we need to impose

\begin{equation*}
\partial_y f(x,y) = - f(x,y)
\end{equation*}

The simplest function I can think of that satisfies this equation (setting the constant of integration equal to zero) is $f(x,y) = e^{-\ln(y)}$ i.e.

\begin{equation*}
\partial_y \left( y \ e^{-\ln(y)} \right) = e^{-\ln(y)} - y\left( \frac{1}{y} \right) \ e^{-\ln(y)} = \frac{1}{y} - \frac{1}{y} = 0
\end{equation*}

Plugging $f(x,y) = e^{-\ln(y)}$ into $(1)$ yields

\begin{equation*}
-y\partial_t \left( g(y,t) \right) + e^{-\ln(y)} = 0 \Rightarrow \partial_t \left( g(y,t) \right) = \frac{1}{y^2}
\end{equation*}

Thus (setting the constant of integration equal to zero) we get that $g(x,y) = t/y^2$

But I have missed something as $(3)$ is not satisfied due to a factor of $2$ i.e.

\begin{equation*}
-\partial_y \left( \frac{t}{y^4} \right) - \frac{2t}{y^5} = \frac{2t}{y^5} \neq 0
\end{equation*}

What am I missing? I guess that I should impose a specific initial condition to get it done but I do not see it...

 

[BvU]

$\qquad$Funny, from $$
\begin{equation*}

\partial_y \left( y \; f(x,y) \right) = 0

\end{equation*}$$$\qquad$ I get $$
\begin{equation*}
y\,\partial_y \left( f(y,t)\right) +f(y,t) = 0
\end{equation*}$$ where $(2)$ has a minus sign !

By the way, we physicists usually write $e^{-\ln y} \$ as $\ \displaystyle{1\over y}\$

... let's not panic though!

That's the spirit !

 

[JD_PM]

OK let's make it simpler

Compare
$$
\begin{equation*}

y\partial_y \left( f(y,t)\right) -f(y,t) = 0 \tag{2}

\end{equation*}
$$
with $\ x \, f'(x) = f(x) \$

So we see that $f(y,t) = f(y) = e^{\ln y} = \frac{1}{y}$ (so everyone is happy! haha) is a solution for $(2)$

Then all we need to do is plugging it into $(1)$ and find $g(y,t)$ so that the following equation is satisfied

\begin{equation*}
\partial_t \left( g(y,t)\right) = \frac{1}{y^2}
\end{equation*}

Besides, $g(y,t)$ and $f(y,t)$ must be a solution for $(3)$ of course.

Thank you, I think I can get it done now (time to go back to study! ).

 

[BvU]

so everyone is happy! haha

Hoho! Not yet! I'll let you know when I'm happy

As long as you write

So we see that $f(y,t) = f(y) = e^{\ln y} = \frac{1}{y}$

I'm not happy by a long shot ! And the last bit $e^{\ln y} = \frac{1}{y}$ is (probably ?) a typo, so that does not count.

$f(y,t)=f(y)$ is the one that is really irking me. We solve (we try to ...) for $y$ but that doesn't mean $t$ goes away.

We are still working on $(2)$ and are nearly there ...

 

[JD_PM]

Hoho! Not yet! I'll let you know when I'm happy

As long as you write
I'm not happy by a long shot ! And the last bit $e^{\ln y} = \frac{1}{y}$ is (probably ?) a typo, so that does not count.

Oops a typo indeed. Let's go slowly

I meant that a solution for $(2)$ is given by

\begin{equation*}
f(y,t) = e^{\ln y} = y
\end{equation*}

Next step would be going to $(1)$ and solve for $g(y,t)$, wouldn't it?

 

[BvU]

No: any function of $t$ can be factored in in both terms of $(2)$ without affecting the validity of the equation.

 

[Mark44]

By the way, we physicists usually write $e^{-\ln y} \$ as $\ \displaystyle{1\over y}\$

As do mathematicians...

 

[JD_PM]

No: any function of $t$ can be factored in in both terms of $(2)$ without affecting the validity of the equation.

I better take more time to study and then reply. I do not want to exhaust your patience