# Anyone ever work with a DE like this?

• Mosis
In summary, the conversation discusses a difficult differential equation with a cubic root term on the right-hand side. The participants consider different methods for solving the equation and determining the nature of its solutions. They mention using a perturbation method and plotting the equation to gain insights into its behavior. Ultimately, they conclude that most solutions of the equation are unbounded, with the exception of some special cases.
Mosis
This has come up in my research, and my supervisor and I don't really know how to proceed. It reads

$$r^2\frac{d^2Y}{dr^2} + r\frac{dY}{dr} - \left(\frac{3}{2}\right)^2Y = Cr^3Y^{1/3}$$

I know the RHS is an equidimensional DE which has the nice solution $$Y = r^{\pm 3/2}$$, but I have no idea how this could help with that nasty $$Y^{1/3}$$ term sitting on the RHS.

Really, we just want to learn SOMETHING about this DE. What do the solutions look like? Does a bounded solution exist? What can we do if C is small? etc.

Any ideas?

wow, 46 views, no reply - is this really that uncommon?

The problem is that that is a seriously non-linear equation. Since solutions of non-linear equations do not add to another solution, Knowing a solution to LHS= 0 does not help. For C small, you might try a perturbation method, writing the solution as a series in powers of C: $y_0+ Cy_1+ C^2y_2+ \cdot\cdot\cdot$. Substituting that into the equation and equating like powers of C will give, for the "C0" term, the homgeneous equation as the equation for $y_0$. Then the "C1 term will gvie a non-homogeneous equation for $y_1$ with the (now known) $y_0$ on the right side, etc.

Last edited by a moderator:
I was thinking about that kind of perturbation scheme, but the fact that we have Y^1/3 makes that intractable i.e. what to do with $$\left(Y_0 + CY_1\right)^{1/3}$$?

You could express it as $$Y_0^{1/3} (1+C(Y_1/Y_0)+C^2(Y_2/Y_0)+...)^{1/3} = Y_0^{1/3}(1 + \frac{1}{3}(C(Y_1/Y_0)+C^2(Y_2/Y_0)...) - \frac{1}{9}(C(Y_1/Y_0)+C^2(Y_2/Y_0)...)^2 ...$$

because $$(1+x)^a = 1 + xa + x^2 a(a-1)/2! + x^3 a(a-1)(a-2)/3! + ..., x \ll 1$$

It will be incredibly messy beyond the first two terms, but at least you can get some results.

By the way, it would simplify your calculations if you did a substitution $$z = \ln{r}$$.

Last edited:
Further thoughts.

There are two distinct cases, C>0 and C<0.

Once you rewrite your expression as r = ln z, $$Y'' = 9/4 Y + C e^{3z} Y^{1/3}$$,

it's easy to see that, as long as C>0, all nontrivial solutions are unbounded (because the second derivative is always positive, therefore the graph of the function is convex).

Every individual solution will look like a bunch of exponentials a*exp(b*z) stitched together, depending on which of the two terms on the right hand side dominates (most of the time, one of the two terms would be negligible). Draw a plot of the solution of the equation $$9/4 Y = C e^{3z} Y^{1/3}$$. Use log scale on the vertical axis. Think about the behavior of a solution of your DE that passes through a point to the left of the curve. See if you can prove (or at least convince yourself) that almost all solutions either blow up because Y crosses below zero, or diverge as exp(-1.5 z) as $$z\rightarrow -\infty$$ and as exp(4.5 z) as $$z\rightarrow +\infty$$.

If C<0, things are more complicated, and many solutions would blow up at some point because there's nothing to keep Y from going negative.

Last edited:
hamster143 said:
Further thoughts.

There are two distinct cases, C>0 and C<0.

Once you rewrite your expression as r = ln z, $$Y'' = 9/4 Y + C e^{3z} Y^{1/3}$$,

it's easy to see that, as long as C>0, all nontrivial solutions are unbounded (because the second derivative is always positive, therefore the graph of the function is convex).

why are you assuming Y is positive? also, exp(-r) is convex but decays as r goes to infinity. Of course, it blows up in the other direction, but in my case, Y = Y(r), and r > 0, so I'd really need some information about the first derivative to conclude all nontrivial solutions are unbounded.

hamster143 said:
Further thoughts.
Draw a plot of the solution of the equation $$9/4 Y = C e^{3z} Y^{1/3}$$. Use log scale on the vertical axis. Think about the behavior of a solution of your DE that passes through a point to the left of the curve. See if you can prove (or at least convince yourself) that almost all solutions either blow up because Y crosses below zero, or diverge as exp(-1.5 z) as $$z\rightarrow -\infty$$ and as exp(4.5 z) as $$z\rightarrow +\infty$$.

also, I'm not sure why you suggest plotting the solution to that equation - how can that help me?

Mosis said:
why are you assuming Y is positive? also, exp(-r) is convex but decays as r goes to infinity. Of course, it blows up in the other direction, but in my case, Y = Y(r), and r > 0, so I'd really need some information about the first derivative to conclude all nontrivial solutions are unbounded.

Because Y^(1/3) is only defined for positive Y.

Yes, exp(-r) is convex. But notice my variable change: from r that goes 0..infinity, to z that goes from minus infinity to infinity.

If you have a function whose argument goes from minus infinity to infinity and its graph is convex, it must be unbounded on either side.

I'm not sure why you suggest plotting the solution to that equation

Because either of two terms on the right hand side would dominate, depending on which side of the graph you are. So, as long as you're sufficiently far to the left, solutions of DE would behave as if there's no pesky cubic root term.

I agree with hamster143, even it does not help you in terms of the details it will give you a lot of information about the vector field. My suggestion is setup your DE in MATLAB or your favorite solver while keeping it nonlinear, and let the solutions from many (I mean many) initial conditions and get a feeling for the differential equation. You can even read off if there is a bounded solution or not.

You never know! It might act like almost linear in the region of interest. (Happened to me also)

hamster143 said:
Further thoughts.

There are two distinct cases, C>0 and C<0.

Once you rewrite your expression as r = ln z, $$Y'' = 9/4 Y + C e^{3z} Y^{1/3}$$,

it's easy to see that, as long as C>0, all nontrivial solutions are unbounded (because the second derivative is always positive, therefore the graph of the function is convex).

you can further write that using the fact that $$Y=exp(9z/2)$$ is an exact solution, and write $$Y=exp(9z/2)y(z)$$, with :
$$y''+9y'+18y-Cy^{1/3}$$
which is autonomous, and proceed by local analysis.

you can check that the critical points are
$$y = y'=0$$, and $$y=(-C/18)^{3/2}, y'=0$$
the second stationary points leads to a non bounded solution Y(z). The jacobian at (0,0) is not defined, but we have the inequality
$$[y'^{2}/2+9y^{2}-Cy^{4/3}3/4]'=-9y'^{2}$$
So, for C<0, y=0=y' is stable, and you might have bounded solutions
* beware when simulating that in matlab, y^(1/3) does not give the proper solution. try sign(y)*abs(y)^1/3 instead.

what do you mean by a convex graph

## 1. What is a DE?

A DE, or differential equation, is a mathematical equation that describes how a quantity changes over time. It involves the rate of change of the quantity and one or more variables.

## 2. What types of problems can be solved using DEs?

DEs are commonly used to model physical phenomena such as heat transfer, population growth, and fluid dynamics. They can also be used in economics, biology, and other fields to describe the relationship between different variables.

## 3. How are DEs solved?

There are various methods for solving DEs, including analytical, numerical, and graphical methods. Analytical solutions involve finding an exact mathematical expression for the solution, while numerical solutions involve using algorithms to approximate the solution. Graphical methods involve plotting the solution to visually understand the behavior of the system.

## 4. What is the importance of DEs in science?

DEs are important in science because they allow us to mathematically model and understand complex systems. They are used in many fields, including physics, engineering, and biology, to make predictions and gain insights into the behavior of systems over time.

## 5. Are there any real-world applications of DEs?

Yes, there are many real-world applications of DEs. For example, they can be used to model the spread of diseases, predict the stock market, and design efficient transportation systems. DEs are also used in engineering to design aircrafts, control systems, and optimize processes in manufacturing.

• Differential Equations
Replies
1
Views
2K
• Differential Equations
Replies
3
Views
2K
• Differential Equations
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Differential Equations
Replies
4
Views
2K
• Differential Equations
Replies
7
Views
2K
• Differential Equations
Replies
1
Views
2K
• Differential Equations
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
681
• Differential Equations
Replies
3
Views
2K