Solving coupled differential equations

  • Thread starter Avarus
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  • #1
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Ok guys, I've got an issue with a coupled differential equation and I just can't get to solve it:

[itex]\frac{\partial r}{\partial t} = Q\frac{\partial c}{\partial t}[/itex]

Obviously, r depends on c and visa versa, but they both depend on time. Is there a way to uncouple these variables and solve the equation, so that I get r(t) and c(t)?

To me, this really seemed trivial at first, but I've spent hours solving this using decoupling recipes I found all over the internet, but I just can't get it to work (which is frustrating). Could you please help me out?
 

Answers and Replies

  • #2
fluidistic
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Ok guys, I've got an issue with a coupled differential equation and I just can't get to solve it:

[itex]\frac{\partial r}{\partial t} = Q\frac{\partial c}{\partial t}[/itex]

Obviously, r depends on c and visa versa, but they both depend on time. Is there a way to uncouple these variables and solve the equation, so that I get r(t) and c(t)?

To me, this really seemed trivial at first, but I've spent hours solving this using decoupling recipes I found all over the internet, but I just can't get it to work (which is frustrating). Could you please help me out?
With such an equation I think the best you can reach is that [itex]r(t)=Qc(t)+K[/itex] where K is a constant. There's not enough information to get r(t) or c(t). They could in principle be any differentiable function with respect to time.
I'd wait a more experienced opinion on this though, but this is what I think (as a student).
 
  • #3
lanedance
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i think you can just integrate this w.r.t. t
[tex]
\frac{\partial r}{\partial t} = Q\frac{\partial c}{\partial t}
[/tex]
[tex]
\int_0^t \frac{\partial r}{\partial t} dt= \int_0^t Q\frac{\partial c}{\partial t}dt
[/tex]
[tex]
r(t)|_0^t dt= Qc(t)_0^t
[/tex]
[tex]
r(t)-r(0)= Qc(t)-Qc(0)
[/tex]

Then we can roll up all the constants into a single constant b=-Qc(0)+r(0)
[tex]
r(t)= Qc(t)-Qc(0)+r(0)
[/tex]

Though i note you use partial differntiation sign rather than total - is that intended?
 
Last edited:
  • #4
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Well, problem still is that I don't know either r(t) or c(t). But I do know r(0) and c(0), and that dc/dt = k @ t = 0, perhaps that helps?
 
  • #5
lanedance
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yeah that helps

For example, the following is a solution to your problem
c(t)=c(0)e^(kt/c(0))
r(t)=c(0)Q(e^(kt/c(0))-1)+r(0)

not unique though, as here is another
c(t)=kt+c(0)
r(t)=Qkt+r(0)
 
Last edited:
  • #6
HallsofIvy
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This is not, by the way, "coupled differential equations". "Coupled differential equations" would be two ordinary differential equations of two dependent variables with derivatives with respect to one independent variable. This is a "partial differential equation" with a single dependent variable depending on two independent variables.

In general, where the general solution to a single first order ordinary differential equation involves a single undetermined constant, the solution to a single first order partrial differential equation involves a single undetermined function.
 
  • #7
lanedance
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actually I wouldn't call this a partial differential equation...

Not too sure what it is, but wouldn't partial imply differention by at least 2 different & independent variables?

Not really sure what this is besides related rates
 
  • #8
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This is not, by the way, "coupled differential equations".

Oh I'm sorry, I thought coupled would be appropiate terminology, since one is related to the other and visa verse. But thanks for putting that straight :)

yeah that helps

For example, the following is a solution to your problem
c(t)=c(0)e^(kt/c(0))
r(t)=c(0)Q(e^(kt/c(0))-1)+r(0)

not unique though, as here is another
c(t)=kt+c(0)
r(t)=Qkt+r(0)

How did you find those solutions? (Btw, c(o) = 0 in this particular case)

I have the teacher's numerical solution attached as a graph, so I guess the analytical solutions should look like that aswell.
 

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  • #9
HallsofIvy
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Oh, I see. your use of "[itex]\partial[/itex]" rather than "d" fooled me. I see now that both derivatives are with respect to t. Here, the difficulty is that you have only one equation. Just as you cannot solve one algebraic equation for two unknown numbers, so you cannot solve one differentrial equation for two unknown functions. The best you can do is arrive at a relation between the two functions- the relation that fluidistic and lanedance got.
 
  • #10
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Oh, I see.

So it is a coupled differential equation then?

Just as you cannot solve one algebraic equation for two unknown numbers, so you cannot solve one differentrial equation for two unknown functions.

Such a shame :(. I should have known...

Thanks everyone!
 
  • #11
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Ok guys, this problem keeps bugging me and I've revised the whole thing (again, sigh...) and came to the conclusion that I've made some errors along the way (again, sigh...). Now I have these sets of equations:

[itex]\frac{dr}{dt} = -\frac{M}{ρ}k(1-\frac{c}{c_{eq}})[/itex]

and

[itex]\frac{dc}{dt} = \frac{4π}{V}r^{2}k(1-\frac{c}{c_{eq}})[/itex]

in which r is a function of c and t, and c a function of r and t; M, ρ, k, [itex]c_{eq}[/itex], π and V are constants.

I appear to have two independent equations now, so could this be solved?

(btw, I'm sorry to have bugged you with the wrong equation before, I hope these ones are correct)
 
  • #12
lanedance
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the second equation is separable so you can integrate directly to find c(t) and r will follow
 
  • #13
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the second equation is separable so you can integrate directly to find c(t) and r will follow

When I separate the 2nd equation, I get the following (if I'm correct):

[itex]\frac{dc}{1-\frac{c}{c_{eq}}} = \frac{4πk}{V}r^{2}dt[/itex]

Which solves to:

[itex]-c_{eq}ln(1-\frac{c}{c_{eq}}) = \frac{4πk}{V}\int r^{2}dt + constant[/itex]

But I do not yet know what [itex]\int r^{2}dt[/itex] is, only [itex]\frac{dr}{dt}[/itex]. Did I do something wrong?
 
  • #14
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Could anyone still give me a hand on my post above?
 

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