I Non-constant coefficient

  • Thread starter dRic2
  • Start date

dRic2

I was attempting to solve a differential equation using separation of variables' method. After some calculations I got to this:

$$ \frac 1 {a(r)} \frac k r \frac d {dr} \left(r \frac {da(r)}{dr} \right) = f $$

and I rearranged to this:

$$\frac k r \frac {da(r)}{dr} + \frac{d^2a(r)}{dr^2} = fa(r)$$

##k## and ##f## are constants. In a more elegant way:

$$ \ddot a + \frac k r \dot a - fa = 0 $$

Then I realized I don't know how to solve it because of ## \frac 1/r ##. I never studied how to solve differential equation with non-constant coefficients. Can anybody help me? Or suggest me some reading that can help me?
Thank you
 

dRic2

Thank you. I'll be studying it in the next days... see if I can get it done.
 
19,270
3,815
I was attempting to solve a differential equation using separation of variables' method. After some calculations I got to this:

$$ \frac 1 {a(r)} \frac k r \frac d {dr} \left(r \frac {da(r)}{dr} \right) = f $$

and I rearranged to this:

$$\frac k r \frac {da(r)}{dr} + \frac{d^2a(r)}{dr^2} = fa(r)$$

##k## and ##f## are constants. In a more elegant way:

$$ \ddot a + \frac k r \dot a - fa = 0 $$

Then I realized I don't know how to solve it because of ## \frac 1/r ##. I never studied how to solve differential equation with non-constant coefficients. Can anybody help me? Or suggest me some reading that can help me?
Thank you
This looks a lot like a Bessel equation. Are you familiar with Bessel functions?
 

Orodruin

Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
15,841
5,828
This looks a lot like a Bessel equation. Are you familiar with Bessel functions?
Not only looks like. It is Bessel's differential equation for ##k = 0## (or modified Bessel's differential equation depending on the signs of ##f## and ##k##), which is the reason I linked to the Wikipedia page on Bessel functions.
 
19,270
3,815
Not only looks like. It is Bessel's differential equation for ##k = 0## (or modified Bessel's differential equation depending on the signs of ##f## and ##k##), which is the reason I linked to the Wikipedia page on Bessel functions.
Oops. Missed that. Sorry.
 

dRic2

Hi, thanks for the replies.

I'm not familiar with Bessel's equation at all (I just happened to see it somewhere some times ago, so I knew there was a solution but I didn't even remember what to search). I checked the wikipedia page and I found out it's used to solve heat/mass transfer problems and it's funny because it is what I'm actually trying to do. So i think I'm on the right spot! :)

Anyway I think it is a bit too advanced to me right know, but I looked it up on "Mathematical methods for physics and engineering" and I found, like wikipedia says, Bessel's equation is:
$$ x^2 \ddot y + x \dot y + (x^2-k^2)y = 0 $$

My function looks instead like:

$$ kr^2 \ddot a + kr \dot a + fr^2a = 0 → r^2 \ddot a + r \dot a + \frac f k r^2a = 0$$

Since I started studying it this morning (with no special background in differential equations) I'd like to know if ##\frac f k ## affects the solutions of the equation and how. Thank you

Ps: since I got here by separating variables ##ƒ## is a constant which I no nothing about, I have to use B.C. later on to find it
 
Last edited by a moderator:

Orodruin

Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
15,841
5,828
Make a variable substitution ##r = x \sqrt{k/f}##.
 

dRic2

sorry I don't get it. I still have something different from bessel equation
 

Orodruin

Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
15,841
5,828
sorry I don't get it. I still have something different from bessel equation
Please quote what you get, otherwise it is impossible to know where you go wrong.
 

dRic2

Sorry, here:

(In order to make it easier I will put ##h = f/k##)

$$ \ddot a(r) + \frac 1 r \dot a(r) + ha(r) = 0$$
$$ r = x \sqrt \frac 1 h $$
Chain's rule:
$$ \frac {da(r)} {dr} = \frac d {dx} a \left(x \sqrt {\frac 1 h} \right) \frac {dx} {dx \sqrt {\frac 1 h}} = \sqrt h \frac d {dx} a \left(x \sqrt \frac 1 h \right) $$
$$ \frac {d^2a(r)} {dr^2} = \frac {d^2} {dx^2} a \left(x \sqrt {\frac 1 h} \right) \left( \frac {dx} {dx \sqrt {\frac 1 h}} \right)^2 + \frac d {dx} a \left(x \sqrt \frac 1 h \right) \frac {d^2x} {d \left( x \sqrt \frac 1 h \right)^2 } = h \frac {d^2} {dx^2} a \left( x sqrt \frac 1 h \right)$$
finally:
$$h \frac {d^2} {dx^2} a \left(x \sqrt \frac 1 h\right) + \frac 1 {x \frac 1 h} \sqrt h \frac d {dx} a \left(x \sqrt \frac 1 h \right) + h a \left(x \sqrt \frac 1 h \right) = 0$$
$$ \frac {d^2} {dx^2} a \left(x \sqrt \frac 1 h\right) + \frac 1 x \frac d {dx} a \left(x \sqrt \frac 1 h \right) + a \left(x \sqrt \frac 1 h \right) = 0 $$

How should I proceed ?
 

Orodruin

Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
15,841
5,828
How is that not Bessel's differential equation?
 

dRic2

But there is ##\sqrt \frac 1 h ## inside the parenthesis... isn't it a problem?

I mean the solution of Bessel's equation is ##a(x) = c_1J_v(x)+c_2J_{-v}(x) ##, should I replace ##x## with ##x\sqrt \frac 1 h##?
 
Last edited by a moderator:

Orodruin

Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
15,841
5,828
No, this is not a problem. Just introduce a new function ##h(x) = a(x/\sqrt{h})##. This will lead to ##a(r) = h(r\sqrt{h})##.
 

dRic2

Oh thank you. Just to be sure, the solution is then ##a(r) = cJ_v(r\sqrt h)##?
 
263
70

dRic2

Yes, thank you for mentioning @eys_physics. I'll be studying that too in the future. Btw I didn't understand if you are implying that my solution is wrong because I'm like in the middle of the ocean right now ahahah :frown:
 
263
70
No, your solutions is not wrong. I said that there are in general more solutions. The ones I mentioned, ##Y_v## are singular at the origin (r=0). So, if you want a regular solution they should be disregarded. But, it depends on the actual problem you want to solve, which wasn't clear from your post.

By the Frobenius method you can derive the solution of your equation in terms of series expansion. One can then compare with series expansions existing for the Bessel functions. It takes some effort but can be useful to understand better how to solve this kind of differential equations.
 

Orodruin

Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
15,841
5,828
The ones I mentioned, ##Y_v## are singular at the origin (r=0). So, if you want a regular solution they should be disregarded.
This is a common but not completely kosher argument. The question becomes why you want your solution to be regular at r=0. In the end it boils down to the behaviour of the inhomogeneity in your differential equation (typically r=0 is actually just a coordinate singularity in your domain). For example, the Green’s function of the Helmholtz equation in two dimensions involves a Bessel function of the second kind in order to take care of the delta inhomogeneity at r=0. However, as long as your inhomogeneity is sufficiently nice you will be able to assume regularity (and therefore throw away the Bessel functions of the second kind).

Of course, if r=0 is not part of your domain you cannot apply these arguments and you need to include both linearly independent solutions.
 
263
70
Thanks, Orodruin for your clarification. I was a bit sloppy with the wording in my post. The word "should" in the quoted sentence wasn't appropriate. I meant that in some cases the irregular solution could be disregarded (depending on the problem statement), but in other cases not as you mention.
 

Want to reply to this thread?

"Non-constant coefficient" You must log in or register to reply here.

Related Threads for: Non-constant coefficient

  • Posted
Replies
2
Views
2K
Replies
2
Views
3K
Replies
3
Views
4K
Replies
1
Views
619
Replies
4
Views
4K
Replies
15
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top