How do I solve the differential equation numerically which is of the form : (z"/[1+(z'​)^2]^(3/2)​)+(0.19285​29/x)=2+(5​0.99203)*z ? I need to get the x and z values.

Initial conditions are x=0 and z=0

[Birdman]

y is a function of x or z?

[Purush otham]

Hi,

I am sorry that was an error. It is not 'y' it is 'z' . Now it has been corrected.

Answer 1

[James Tursa]

Start with this:

(z"/[1+(z')^2]^(3/2))+(0.1928529/x)=2+(50.99203)*z

Solve for the highest order derivative z"

z" = [1+(z')^2]^(3/2) * (2 + (50.99203)*z - (0.1928529/x))

And then follow the examples in the ode45 doc for solving a 2nd order ODE. E.g., the derivative function

dzdx = @(x,z) [z(2);  ((1+z(2)^2)^(3/2)) * (2 + (50.99203)*z(1) - (0.1928529/x))

The initial vector assuming the initial value z' = 0 (you didn't specify)

z0 = [0;0];

[Purush otham]

Hi I already tried this method. The initial conditions are: when x=0 then z=0 and z'=x.

On using ODE45 I must get the x and z value such that when x= 0.1283 then z = 0.0813 (approx).

How do I do that? Can you exhibit the code for the same?

Thank you

[Torsten]

You can only prescribe two initial conditions, not three. Thus you can prescribe z(0)=0 and z'(0)=0 and hope that z(0.1283)=0.0813. Or you can prescribe only one condition at x=0 and the condition at x=0.1283 and use "bvp4c" instead of "ODE45".

Best wishes

Torsten.

[Purush otham]

Hi,

What if I am using the time period as [0 1] then why don't I get the solution?

Does not time period represent the range for within which I want the x value?

I get NaN as my output.

What do I do?

[Purush otham]

Hi,

So this is the code I used:

    function myoutput=myode45function(x,y)
myoutput=[y(2);(2+50.9920*y(1)-0.21572*y(2)/(x*(1+(y(2))^2)^(1/2))*(1+(y(2))^2)^(3/2))];

And:

time_period=[0 1];
initial=[0,0];
[x,y]=ode45(@myode45function,time_period,initial)

As z here is a function of x( represented here as y(x)). And I need the value of y for x within 0 and 1.

I would also like to select step size.

Thank you

[Torsten]

You divide by x - and x=0 at the start of your integration ...

[Purush otham]

Hi can you elaborate it? How does it change the code? How do I divide?

[James Tursa]

@Torsten: +1 good catch

@Purush: Since you specify the initial condition x=0, your derivative function (which divides by x) is not defined at that point. So the problem, as posed, has no solution. If this represents some real world problem, either you have a mistake in your derivative equation or in your initial condition.

[Purush otham]

Ok.

If the initial conditions remain the same, i.e z(0)=0 and z'(0)=x.

Then: How do I solve the parent equation of the above, given in image:

[Purush otham]

This is the Laplace pressure equation and represents the shape of a drop. The existing literature gives x and z value for the above equation as:

x=0.1142,z=0.1127

x=0.1150,z=0.1121

x=0.1171,z=0.1101 and so on...

My question is besides the method given in the literature, by using MATLAB's numerical integration, how do I get the exact results?

[Torsten]

My guess is that initial conditions are not specified at x=0 to obtain the results from above.

But maybe you can include a link to the publication you are referering to.

Best wishes

Torsten.

[Purush otham]

 Hi,
I am trying to replicate the results obtained in the paper.
Its called "an attempt to test...-by adam and bashforth"
okay so the link for the same is:

Adam and bashforth drop shape

Thank you