Is it possible to solve such a differential equation?

In summary, the equation is saying that the change in depth (z) is equal to the change in distance (x) multiplied by 1 minus the change in depth (dz/z).
  • #1
rhcp89
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4
Hello,

I would like to is it possible to solve such a differential equation (I would like to know the z(x) function):

[tex] \displaystyle{ \frac{z}{z+dz}= \frac{(x+dx)d(x+dx)}{xdx}} [/tex]

I separated variables z,x to integrate it some way. Then I would get this z(x) function.

My idea is to find such a function f(z) (for the left hand side of above equation) that:

[tex] f(z)dz=\frac{z}{z+dz}[/tex]

And for the right side of this first equation I think I should also find similar function g(x) that:

[tex]g(x)dx=\frac{(x+dx)d(x+dx)}{xdx}[/tex]

If I would get these f(z) and g(x) I could get the wanted function z(x) this way:

[tex]\int f(z)dz=\int g(x)dx[/tex]

I think f(z) ang g(x) can be found as numerical approximations. But how to do it in this specific case?

..............
The right hand side of first equation seems for me as pretty hard to find the g(x) so I simplified it this way (however I am not sure if such methods of transformations are ok):

[tex] \frac{(x+dx)d(x+dx)}{xdx}=\frac{(x+dx)(dx+d^2 x)}{xdx}=\frac{(x+dx)dx(1+d)}{xdx}=\frac{x+2dx+d^2 x}{x} [/tex]Thanks for your help.
 
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  • #2
rhcp89 said:
I would like to is it possible to solve such a differential equation (I would like to know the z(x) function):

[tex] \displaystyle{ \frac{z}{z+dz}= \frac{(x+dx)d(x+dx)}{xdx}} [/tex]
I taught college-level courses in differential equations for many years, but I've never seen a problem that looked like this, particularly where you have z + dz or x + dx.

Is this a problem in some textbook? If not, is there some context for the question, such as how the question arose?
 
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  • #3
Mark44 said:
but I've never seen a problem that looked like this, particularly where you have z + dz or x + dx.
I totally agree with you at this, usually differentials are added (or subtracted) together like ##zdx+dx## or ##zdx+xdz## for example. Those ##z+dz## and ##x+dx## look pretty weird

However the equation seems to have some consistency: If we take the limit dx->0 (and hence dz->0) we end up with 1=1.

@rhcp89 if this differential equation can be solved, I don't think it will be with the method of separation of variables. I just don't think we can find the functions f,g with the property you suggest.
 
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  • #4
@Mark44

Here is the context for this first equation in starter thread:

We have open pit mine (to excavate sand). Excavation area is circular with radius of [tex]r_{0}[/tex]Groundwater flows through the sand to excavation area from the height of [tex]H[/tex] Maximum radius of depression cone is [tex]R[/tex]Water level in excavation area is [tex]h_{0}[/tex]
rys na forum fizykow.jpg


Assumption:

Q= const (Q- volumetric flow rate it equals area times velocity in x direction[tex]Q= Av_{x} [/tex]
so:
[tex]Q_{1}=Q_{2}[/tex]
[tex]A_{1}v_1=A_{2}v_{2}[/tex]
[tex] 2 \pi x_{1}z_{1} \frac{dx_{1}}{dt}=2 \pi x_{2}z_{2} \frac{dx_{2}}{dt} [/tex]
[tex]x_{1}z_{1} \frac{dx_{1}}{dt}= (x_{1}+dx)(z_{1}+dz) \frac{d(x_{1}+dx)}{dt}[/tex]
I accept that point [tex](x_{1};z_{1})[/tex] is any point that belongs to cone of depression curve. it means that:
[tex]x_{1}=x[/tex]
[tex]z_{1}=z[/tex]
Then I canceled [tex]dt[/tex] from the equation, however I would prefer to get z(t) and x(t) function but it seemed easier to get the z(x) functionwhat's why I did this cancellation.
Next I separated variables z,x to prepare the equation for integration and that's how I got this first equation in starter thread:
[tex]\frac{z}{z+dz}= \frac{(x+dx)d(x+dx)}{xdx}[/tex]
 
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  • #5
Well I have to say that I am surprised that this equation is from the study of a physical system, I thought you came up with this equation in an artificial way, by combining differential quantities and see how it goes...
 
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  • #6
@Delta2 No. It's the nature itself :)
 
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  • #7
Here's what I would do: From the drawing, find (by approximation) a formula for the part of the curve identified as "depression curve." This should be relatively easy to do - a cubic spline would do nicely. After a formula is found, you'll have a functional relationship between x and z; i.e. ##z = f(x) = a_3x^3 + a_2x^2 + a_1x + a_0##.

Then you would be able to write an equation in terms of ##x, z, dx,## and ##dz## to come up with a differential equation that is actually solvable.
 
  • #8
rhcp89 said:
My idea is to find such a function f(z) (for the left hand side of above equation) that:

[tex] f(z)dz=\frac{z}{z+dz}[/tex]

I don't think this can make sense since when ##dz\to 0##, the LHS goes to ##0## but the RHS goes to ##1##.

The way I'd informally interpret it is: to first order ##\frac{z}{z+dz}=\frac{1}{1+dz/z}=1-\frac{dz}{z}.## With the same "reasoning" on the RHS, $$\frac{(x+dx)(dx+d^2x)}{xdx}=\frac{xdx+(x+1)d^2x+d^3x}{xdx}=1+\left(\frac{x+1}{x}\right)dx,$$

This gives ##\frac{dz}{z}=-\left(\frac{x+1}{x}\right)dx## which is a meaningful differential equation.

Edit: I don't know if this is sensible because I've identified ##d^2x## and ##(dx)^2##, which I guess should be distinct objects, with ##d^2x=0##. If we keep track of that better, then

##(x+dx)(dx+d^2x)=xdx+xd^2x+(dx)^2+dxd^2x.## Dividing by ##xdx## gives ##1+\frac{1}{x}dx,## and the equation would be ##dz/z=-dx/x.## This might just be nonsense though. I didn't look at your physical setup to see if it fits.
 
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  • #9
Mark44 said:
Here's what I would do: From the drawing, find (by approximation) a formula for the part of the curve identified as "depression curve." This should be relatively easy to do - a cubic spline would do nicely. After a formula is found, you'll have a functional relationship between x and z; i.e. ##z = f(x) = a_3x^3 + a_2x^2 + a_1x + a_0##.

Then you would be able to write an equation in terms of ##x, z, dx,## and ##dz## to come up with a differential equation that is actually solvable.
@Mark44 I don't understand what you mean:
I can suppose that you tell me to find the exact formula which must be an approximation of the real depression curve. You suppose it could be cubic spline. But how can I find these ##a_{0},a_{1},a_{2},a_{3}## coefficients If I don't know the real depression curve? That curve is what I'm looking for. I can tell you also that this curve is not tangent to any horizontal or vertical line at any point.

Then I thought that you just say to substitute the z, dz, dx with this 3rd degree polynomial (whose coeffiecients are unknown) to my equation which I struggle to solve. So I would have then 1 equation with 6 unknown variables: ##x,x_{2},a_{0},a_{1},a_{2},a_{3}## (such that ##x_{2}-x=dx## and ##z_{2}=f(x_{2})##). It still doesn't gives me the values for these coefficients.

I also doubt that one 3rd degree polynomial would be good approximation of real depression curve, because the picture is not about the true scale - the real curve is highly stretched in x- direction in comparison to that picture ( "R" value is like 200- 1000 meters when "H" is only about 6 meters).
 
  • #10
Depression cone calculations are well known to hydrologists. I searched groundwater level depression cone calculation, and found this MS thesis: http://www.math.clemson.edu/~warner/ground.pdf. It has a number of references that appear to directly answer your question.
 
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  • #11
rhcp89 said:
You suppose it could be cubic spline. But how can I find these a0,a1,a2,a3 coefficients If I don't know the real depression curve? That curve is what I'm looking for.
You showed a graph of the depression cone on graph paper -- you can estimate x and z coordinates directly from the graph paper, assuming that the graph is drawn to scale. @jrmichler also gave a link to a paper that talks about groundwater depression cones.
 
  • #12
You showed a graph of the depression cone on graph paper -- you can estimate x and z coordinates directly from the graph paper, assuming that the graph is drawn to scale. @jrmichler also gave a link to a paper that talks about groundwater depression cones.

That drawing was just a draft to give you the idea of the problem and as I said it's completely not in scale.
 
  • #13
rhcp89 said:
@Mark44

Assumption:

Q= const (Q- volumetric flow rate it equals area times velocity in x direction[tex]Q= Av_{x} [/tex]
What is it that has a velocity in the ##x## direction?

If ##x(t)## represents a flux then, at a given location, the assumption is that the flux at that location is constant. Writing the flux as a function ##f## of two variables ##f(x,t)##, we have ##\frac{\partial f}{\partial t} = 0##

so:
[tex]Q_{1}=Q_{2}[/tex]
[tex]A_{1}v_1=A_{2}v_{2}[/tex]
[tex] 2 \pi x_{1}z_{1} \frac{dx_{1}}{dt}=2 \pi x_{2}z_{2} \frac{dx_{2}}{dt} [/tex]

The equation says that we have functions ##x(t), z(t)## such that:
##x(t)x'(t)z(t) = C ## for some constant ##C##. If you pick a function ##x(t)## such that ##x(t)## and ##x'(t)## are nonzero in the time interval in question, then you can set ##z(t) = \frac{C}{x(t)x'(t)} ## So how can the equation determine ##z## as a specific function of ##x## since we have some freedom of choice in picking ##x(t)##?
 
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  • #14
rhcp89 said:
That drawing was just a draft to give you the idea of the problem and as I said it's completely not in scale.
Then you could either make some measurements to get a more accurate drawing (I assume this would be possible), or you could take a look at the link that @jrmichler provided.
 
  • #15
Mark44 said:
Then you could either make some measurements to get a more accurate drawing

Yes, if ##z(x)## is a given function.

We need to establish whether the problem expects us to treat ##z(x)## as a unknown function.
 
  • #16
Stephen Tashi said:
Yes, if ##z(x)## is a given function.

We need to establish whether the problem expects us to treat ##z(x)## as a unknown function.
##z(x)## is not given, and this fact is well established in this thread. My point has been that with a reasonably accurate graph of the boundary of the depression cone, based on a small number of measurements, z(x) can be approximated by, say, a cubic spline.

Also, and as already mentioned, jrmichler provided a link to a PDF that could help with determining the groundwater depression cone.
 
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  • #17
rhcp89 said:
@Mark44

Assumption:

Q= const (Q- volumetric flow rate it equals area times velocity in x directionQ=Avx
What is it that has a velocity in the x direction?
This is a velocity of some specific water particles which flow through the specific A area. Speed of this water is constant for that given area. This area has a shape of side surface of the cylinder. The given side surface has a radius of x.
If x(t) represents a flux then, at a given location, the assumption is that the flux at that location is constant. Writing the flux as a function f of two variables f(x,t), we have ∂f∂t=0
EDIT: "x"quantity is neither volumetric flow rate nor volumetric flux (wiki says the "flux" could refer to those 2 terms). ##x_{i}## stands for momentary position (radius) of specific water particles. At that moment they occupy surface ##A_{i}## with the height of ##z_{i}##. I made a mistake when I said that I wanted to find some general x(t) function, because each ##x_{i}## also depends on initial position. I assumed that such an initial position for given ##i## water particles can be such a location when these particles start to take place under that curve of depression:
$$x_{i}=R(ti)-f(t-ti) $$
$$where R(ti)=R(t=ti)$$
- R(ti) is momentary maximum radius of depression cone at time ##t=ti##
- that ##x_{i}## should be something like that however there may be some difficulties with the formula for ##R(ti)##.
Then you assume that flux (I suppose you mean volumetric flux) is constant at given location - it could only be true if the shape of depression would not change in time. But it is not.
so:
Q1=Q2
A1v1=A2v2
2πx1z1dx1dt=2πx2z2dx2dt
For me these equation are only right when shape of depression cone doesn't changes. They are for specific moment "ti". Q changes with time. As you know I canceled out these ##dt## differentials later, so I think here we don't take care that depression curve changes.

The equation says that we have functions x(t),z(t) such that:
x(t)x′(t)z(t)=C for some constant C. If you pick a function x(t) such that x(t) and x′(t) are nonzero in the time interval in question, then you can set z(t)=Cx(t)x′(t) So how can the equation determine z as a specific function of x since we have some freedom of choice in picking x(t)?
EDIT: Here you are right - that was my mistake - ##x ## (I mean ##x_{i}##) is not only the function of ##t## but also of initial position ##R(ti)##. So for every ##z_{i}## I think we would have formula: ##z_{i}=f(x_{i})## and I hope it means that we can have have function z(x) for given moment ##ti##.
 
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  • #18
rhcp89 said:
I think if someone was able to solve my equation from the starter thread by integration he/she would get something like ##z=f(x)+C## (C- integration constant). C depends on time because boundary condition is that the function goes through (R,H) point and I know in someway the H(t) function. So I would get ##z=f(x)+C(t)##. It means that I would have that z(x,t).

I am unfamiliar with the physical science involved in the problem , but the document
https://geology.humboldt.edu/courses/geology556/556_handouts/well_flow_equations.pdf (page 2, eq. 1) begins with the equations

##Q = 2\pi b K \frac{dh}{dr}##

##dh = \frac{Q}{2 \pi b K dr}##

and derives the integral equation

##\int_{h_w}^{h_0} dh = \int_{r_w}^{r_0} Q 2 \pi k b \frac{dr}{r}##

which is solved to obtain

## Q = 2 \pi k b \frac{h_0 - h}{\ln (\frac{r_0}{r})} ##

It says:
This is the equilibrium or Theim equation.
 
  • #19
@Stephen Tashi

I edited my previous post so it seems more correct. Could you check it?

Stephen Tashi said:
I am unfamiliar with the physical science involved in the problem , but the document
https://geology.humboldt.edu/courses/geology556/556_handouts/well_flow_equations.pdf (page 2, eq. 1) begins with the equations

##Q = 2\pi b K \frac{dh}{dr}##

##dh = \frac{Q}{2 \pi b K dr}##

and derives the integral equation

##\int_{h_w}^{h_0} dh = \int_{r_w}^{r_0} Q 2 \pi k b \frac{dr}{r}##

which is solved to obtain

## Q = 2 \pi k b \frac{h_0 - h}{\ln (\frac{r_0}{r})} ##

Yes I know it. It is modification of Dupuit's formula for some specific case. Such formula give you an information of volumetric flow rate ##Q## so you can apply a proper pump that can handle with amount of water. I've already used Dupuit's formula. It is one of my 4 equations. I think I need one more to know the behaviour of depression cone curve in terms of time.
 
  • #20
rhcp89 said:
I edited my previous post so it seems more correct. Could you check it?

I need to understand the physics.

If we are dealing with a situation where the shape of the depression cone changes in time, then I need to understand the physical law that explains why it would change.

The only physical law you used in the first post is that volumetric flow is constant in time through each cylindrical subsurface of the cone. Given some initial cone shape ##z(x)## and some velocity formula ##x'(t)## that implies the constant flow, what is the explanation for why ##z(x)## would change as function of x?
Q changes with time

Taking ##Q = Q(x,t)##, I assume you mean ##\frac{\partial Q}{\partial t} \ne 0 ## at some locations ##x_i## (or possibly all of them).

My intuitiive picture (which may be incorrect) is that a change in the the shape of the depression cone at location ##x_1## at time ##t_1## would be associated with a change of the volume flow ##Q(x_1,t_1)## through the surface corresponding to that location. This change would not be propagated instantaneously to other locations. The change would propagate to other locations at a finite speed, so it would be like a wave.
 
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  • #21
Stephen Tashi said:
The only physical law you used in the first post is that volumetric flow is constant in time through each cylindrical subsurface of the cone. Given some initial cone shape z(x) and some velocity formula x′(t) that implies the constant flow, what is the explanation for why z(x) would change as function of x?
I suppose that you wanted to say: "what is the explanation for why z(x) would change as function of t?"
The "z" function must change in time because the well is deepened at constant speed. This forces the depression cone curve to change in time.
Stephen Tashi said:
Taking Q=Q(x,t), I assume you mean ∂Q∂t≠0 at some locations xi (or possibly all of them).

My intuitiive picture (which may be incorrect) is that a change in the the shape of the depression cone at location x1 at time t1 would be associated with a change of the volume flow Q(x1,t1) through the surface corresponding to that location. This change would not be propagated instantaneously to other locations. The change would propagate to other locations at a finite speed, so it would be like a wave.

Because as I said depression cone curve changes in time so it means that Q must also change (this is fact - for example Dupuit's or Theim's formula says so). But I assumed that for any moment "t" Q is the same for all position ##x_{i}## so I thought it could be accepted simplification that information of a Q change propagates to every locations at infinite speed. And I was wrong and you were right: shape of depression cone changes together with Q because (based on http://www.math.clemson.edu/~warner/ground.pdf thanks to @jrmichler):
- when there is increase in mass for elementary cube at location ##x_{i}## then ##z_{i}## also rises (but q (=Q but for this elementary cube) falls)
- when there is decrease in mass for elementary cube at location ##x_{i}## then ##z_{i}## also falls (but q rises).

It is said in this link that such change happens at infinitely small period of time but this not "zero". These observations gives you equation of continuity which can be transformed to governing equation when Darcy law is applied. It means that my first equation from starter thread is based on wrong assumptions and there is no use thinking on how to solve it.

Nevertheless the governing equation from above link provided by @jrmichler does not meet the conditions of my physical setup and yet I can't find it but that will be topic for another thread.
 
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Related to Is it possible to solve such a differential equation?

1. Can all differential equations be solved?

Not all differential equations have analytical solutions, meaning that they cannot be solved using traditional algebraic methods. However, many differential equations can be solved using numerical methods or approximations.

2. How do you know if a differential equation can be solved?

There are certain characteristics of a differential equation that can indicate whether it can be solved analytically. These include linearity, separability, and exactness. If a differential equation does not have these characteristics, it may not have an analytical solution.

3. What are some common techniques for solving differential equations?

Some common techniques for solving differential equations include separation of variables, substitution, and the use of integrating factors. Other methods such as Laplace transforms and power series expansions are also commonly used.

4. Can computer programs be used to solve differential equations?

Yes, computer programs can be used to solve differential equations numerically. These methods involve using algorithms to approximate the solution to the differential equation, which can be more accurate than analytical solutions in some cases.

5. Are there real-world applications for solving differential equations?

Yes, differential equations are used in many fields of science and engineering to model and understand various natural phenomena. They are commonly used in physics, chemistry, biology, and economics, among others.

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