Can This Differential Equation Be Simplified or Solved Numerically?

In summary, the original equation can be reduced to a simpler form if specific conditions are met, but if not, a numerical solution may be possible. However, there may be ambiguity in the interpretation of the partial derivative when considering the dependency on y.
  • #1
JulieK
50
0
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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  • #2
[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.
 
  • #3
Many thank!
What sort of numeric solution would you suggest? Can you be more specific?
 
  • #4
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?
 
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  • #5
"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.
 
  • #6
I like Serena said:
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.

JulieK said:
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.
 
  • #7
Hi JulieK!

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

JulieK said:
"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?


mfb said:
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.
 
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  • #8
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!
 
  • #9
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}##)?
 
  • #10
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.
 
  • #11
JulieK said:
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##?
 
  • #12
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##.
 
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  • #13
Thank you!
 
  • #14
Thank you mfb and I like Serena!
 
  • #15
You're welcome! ;)
 

Related to Can This Differential Equation Be Simplified or Solved Numerically?

1. What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It describes the relationship between the rate of change of a quantity and the quantity itself.

2. How do I know if a differential equation can be simplified?

You can determine if a differential equation can be simplified by looking at its form. If it can be written in a simpler form, such as a first-order or second-order equation, then it can be simplified.

3. What are the steps to solving a differential equation?

The steps to solving a differential equation depend on its type and complexity. In general, the steps involve separating variables, integrating both sides, and solving for the constant of integration. Additional techniques, such as substitution or using a differential equation solver, may also be necessary.

4. What are initial conditions and why are they important in solving a differential equation?

Initial conditions are values given for the dependent variable and its derivatives at a specific point. They are important because they help determine the specific solution to the differential equation. Without initial conditions, the solution would have an arbitrary constant, making it a general solution rather than a specific one.

5. Can all differential equations be solved analytically?

No, not all differential equations can be solved analytically. Some equations are too complex or do not have a known solution. In these cases, numerical methods or approximation techniques may be used to find an approximate solution.

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