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How to simplify/solve this differential equation

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JulieK
#1
Dec29-12, 12:10 PM
P: 40
I have the following equation

[itex]\frac{\partial}{\partial y}\left(y\frac{dm}{dx}+m\frac{dy}{dx}\right)-\frac{dm}{dx}=0[/itex]

where [itex]m[/itex] is a function of [itex]y[/itex] (say [itex]m=f\left(y\right)[/itex]) and [itex]y[/itex] is a function of [itex]x[/itex] (say [itex]y=g\left(x\right)[/itex]). Are there any conditions under which [itex]\frac{dm}{dx}[/itex] becomes identically zero and hence this equation can be reduced to the follwoing form which is easier to solve:

[itex]\frac{\partial}{\partial y}\left(m\frac{dy}{dx}\right)=0[/itex]

If such condtions do not exist, what is the best and easiest method to solve the original equation?

Note: I know [itex]f(y)[/itex] and I want to find [itex]g(x)[/itex] which is the function of interest to me.

I also wish to know if this equation can be solved numerically for g(x) if analytical solution is not possible.
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mfb
#2
Dec29-12, 04:23 PM
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P: 11,911
[itex]\frac{dm}{dx}=0[/itex] gives [itex]\frac{df(g(x))}{dx}=0[/itex] - this is trivial for constant f or constant g, and can happen for specific non-constant f and g.

What about the product rule?
[tex]\frac{\partial}{\partial y}\left(y\frac{dm}{dx}+m\frac{dy}{dx}\right)-\frac{dm}{dx}=y\frac{\partial}{\partial y}\frac{dm}{dx}+\frac{\partial}{\partial y}\left(m\frac{dy}{dx}\right)[/tex]

I would expect that a numerical solution is possible. Depending on m, an analytic solution might be possible, too.
JulieK
#3
Dec30-12, 04:51 AM
P: 40
Many thank!
What sort of numeric solution would you suggest? Can you be more specific?

I like Serena
#4
Dec30-12, 05:15 AM
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How to simplify/solve this differential equation

I find your usage of the partial derivative with respect to y confusing.

Which occurrences of y is it supposed to differentiate?
This is subject to more than 1 interpretation.
Can you clarify?


If I substitute g(x) for the first ##y## that suggests it should not be derived.
I suspect that was not your intention...
Actually I would suggest to write it as y(x) instead of g(x) if it should be derived.


I would have to rewrite ##{dm \over dx}## as ##{d \over dy}f(y) \cdot {d \over dx}g(x)## before I can derive it with respect to y.
Is that the intention?
Or should I assume there is no explicit dependency on y at all?
Or only in part of the expression?


What should happen to ##dy \over dx##?
Should I assume it has no explicit dependency on y, so the partial derivative is zero?
JulieK
#5
Dec30-12, 06:45 AM
P: 40
"Which occurrences of y is it supposed to differentiate?
This is subject to more than 1 interpretation.
Can you clarify?"

Any occurrence where it is acted on by a differential operator. There is no ambguity.

"If I substitute g(x) for the first y that suggests it should not be derived.
I suspect that was not your intention...
Actually I would suggest to write it as y(x) instead of g(x) if it should be derived."

It should be derived where it is acted on by a differential operator.
There is no harm in writing y(x) instead of g(x), but I don't see how this will help.

"I would have to rewrite.......... only in part of the expression?"

You could write it in this form.
Yes it is my intention.
m is explicitly dependent on y.

"What should happen to............ partial derivative is zero?"
dy/dx has no explicit dependency on y. It depends on x only.
mfb
#6
Dec30-12, 07:50 AM
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P: 11,911
Quote Quote by I like Serena View Post
Which occurrences of y is it supposed to differentiate?
This is subject to more than 1 interpretation.
Can you clarify?
That confused me as well, and it did not become clearer with the last post.

Quote Quote by JulieK
dy/dx has no explicit dependency on y. It depends on x only.
I think you can simplify the derivative a lot with that rule - but I see an y there, clearly d/dx (y) depends on y.
I like Serena
#7
Dec30-12, 09:28 AM
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Hi JulieK!

I suggest you use the Quote button to retain the quotes and the formulas.

Quote Quote by JulieK View Post
"If I substitute g(x) for the first y that suggests it should not be derived.
I suspect that was not your intention...
Actually I would suggest to write it as y(x) instead of g(x) if it should be derived."

It should be derived where it is acted on by a differential operator.
There is no harm in writing y(x) instead of g(x), but I don't see how this will help.
Let me clarify the ambiguity.
We have:
$${\partial \over \partial y}y=1$$
versus
$${\partial \over \partial y}g(x) = 0$$
Which of the two is intended?


"I would have to rewrite.......... only in part of the expression?"

You could write it in this form.
Yes it is my intention.
m is explicitly dependent on y.
A partial derivative ##{\partial \over \partial y}## is defined on a function of the form ##f(x,y)##, where ##y## is an explicit dependency.
The expression ##{\partial \over \partial y}({d \over dx}m(y(x)))## is not well defined, since the dependency on y is not explicit as it should be to apply the definition of a partial derivative.



The expression is equal to:
$${\partial \over \partial y}({d \over dx}m(y(x))) = {\partial \over \partial y}({d \over dy}m(y) \cdot {dy \over dx})$$
I assume you intend this to be equal to:
$$={\partial \over \partial y}({d \over dy}(m(y))) \cdot {dy \over dx} + {d \over dy}(m(y)) \cdot {\partial \over \partial y}({dy \over dx}) $$
$$= {d^2 \over dy^2}(m(y)) \cdot {dy \over dx} + {d \over dy}(m(y)) \cdot {\partial \over \partial y}({dy \over dx})$$
$$= m''(y) \cdot y' + m'(y) \cdot {\partial \over \partial y}({dy \over dx})$$
Right?


"What should happen to............ partial derivative is zero?"
dy/dx has no explicit dependency on y. It depends on x only.
Okay.
So:
$${\partial \over \partial y}({dy \over dx}) = 0$$
Agreed?


Quote Quote by mfb View Post
I think you can simplify the derivative a lot with that rule - but I see an y there, clearly d/dx (y) depends on y.
Yes, there is an implicit dependency on y.
My confusion is that it's unclear whether that is explicit.

For a total derivative ##{d \over dy}({dy \over dx})## we certainly need to take the dependency on y into account.

For a partial derivative ##{\partial \over \partial y}({dy \over dx})## one can argue either way.
If we consider x to be constant then obviously y will be constant as well, so the partial derivative would be zero.
However, "y" is in there, so what should we do with it?
Should we assume that y(x) is invertible and make the dependency on y explicit?
That would mean we say:
$${\partial \over \partial y}({dy \over dx}) = {\partial \over \partial y}({d \over dx}(g(x(y))))$$
But this is starting to look a bit convoluted to me.
JulieK
#8
Dec30-12, 10:18 AM
P: 40
Hi All

Many thanks to everyone who replied. The problem is becoming messy and I find it difficult to address these posts point by point. I therefore put my problem in the clearest possible way as follow:

I have the following differential equation:

[itex]\frac{\partial}{\partial y}\left(y\frac{dm}{dx}+m\frac{dy}{dx}\right)-\frac{dm}{dx}=0[/itex]

where

[itex]m\left(y\right)=ay^{b}[/itex]

with [itex]a[/itex] and [itex]b[/itex] being constants, and where [itex]y[/itex] is a function of [itex]x[/itex] which I want to find.

I have two boundary conditions as well:

[itex]y\left(x=0\right)=0[/itex]

and

[itex]y\left(x=l\right)=h[/itex]

where [itex]l[/itex] and [itex]h[/itex] are known constants.

Can you suggest an analytic or numeric solution to this problem?

Many thanks!
I like Serena
#9
Dec30-12, 10:27 AM
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Hi JulieK

Do you have a context for the problem?
Is it related to some physics or medical experiment?

In particular, how does the partial derivative tie in?
Where did it come from?
Often it only comes in after other calculations in which a multi variable chain rule was applied.
Or should it really be a total derivative (##{d \over dy}##)?
JulieK
#10
Dec30-12, 10:38 AM
P: 40
Hi I like Serena

It is partial not total derivative. My formulation is accurate and represents a physical model that arises in several contexts such as biological systems.
mfb
#11
Dec30-12, 11:02 AM
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P: 11,911
Quote Quote by JulieK View Post
I have the following differential equation:

[itex]\frac{\partial}{\partial y}\left(y\frac{dm}{dx}+m\frac{dy}{dx}\right)-\frac{dm}{dx}=0[/itex]

where

[itex]m\left(y\right)=ay^{b}[/itex]
That is a nice m, as it has a simple derivative.
$$\frac{dm}{dx}=\frac{dm}{dy}\frac{dy}{dx}=aby^{b-1}\frac{dy}{dx} = sy^t \frac{dy}{dx}$$
with new parameters s and t.

Those two boundary conditions look problematic, as they do not apply to the same point.

Could you clarify the problem of ##\frac{\partial}{\partial y} y = 1## <-> ##\frac{\partial}{\partial y} g(x) = 0##?
I like Serena
#12
Dec30-12, 11:03 AM
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Okay.
Based on my best guess assumptions, you have:

$$\frac{\partial}{\partial y}\left(y\frac{dm}{dx}+m\frac{dy}{dx}\right)-\frac{dm}{dx} = 0$$
$$\Rightarrow 1 \cdot m'(y)y' + y \cdot m''(y)y' + m'(y)y' - m'(y)y' = 0$$
$$\Rightarrow (m'(y) + y m''(y))y' = 0$$

So ##y'=0## or ##m'(y) + y m''(y) = 0##.

In the first case the solution is ##y = C##, where ##C## is an arbitrary constant.

In the second case it follows that:
$$m'(y) + y m''(y) = 0$$
$$\Rightarrow {d \over dy}(y m'(y)) = 0$$
$$\Rightarrow y m'(y) = C$$

If we substitute your expression for m(y), we get:
$$y {d \over dy}(ay^b) = C$$
$$y ab y^{b-1} = C$$
$$y = ({C \over ab})^{1 \over b}$$
or in other words, y is just a constant function.

In combination with the boundary conditions, it follows that there is no solution if ##h \ne 0##.
JulieK
#13
Jan1-13, 04:11 PM
P: 40
Thank you!
JulieK
#14
Jan1-13, 04:13 PM
P: 40
Thank you mfb and I like Serena!
I like Serena
#15
Jan1-13, 05:18 PM
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P: 6,188
You're welcome! ;)


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