I have this equation f(g)=1+cos(g)*cosh(g)-a*g*(cos(g)*sinh(g)-sin(g)*cosh(g))=0. I am trying to plot it for (a,g).
I have the initial condition that when a=0 then g=1.8751.
I am not sure how to code it as an equation with two unknowns and two initial values ?
My suggestion would be to plot the graph by using the solve function by using a=a+da (where da is small increment of a) and using the previous value of g as an initial value to find g for the new value of a.
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I'm writing a function that calculates the Taylor series of any Function.
syms x
y=cos(x);
y0=0;
a=0;
for i=0:25
diff(y,i); %%Gives the derivative formula
y0=y0+diff(y,i)*((x-a)^i)/factorial(i); %%sums every new element of the series
end
x=0:0.1:2*pi;
res = subs(y0,x);
plot(x,res,x,cos(x))
This is the Matlab code.
My problem is that it graphs cos(2x) instead of cos(x), similarly it graphs ln(2x) instead of ln(x) and so on.
I have checked the factorials and they appear to be correct.
What could be the problem, have I messed up the series or have I made a Matlab mistake?
You are constructing the Taylor polynomial around the point x with incrementx-a, that is, you are computing an approximation for
f(x+(x-a))=f(2*x-a)
Now as a=0, this means that as observed, you get f(2*x).
You would need to evaluate the derivatives at a to get the correct coefficients.
y0=y0+subs(diff(y,i),a)*((x-a)^i)/factorial(i); %%sums every new element of the series
I have to solve a second-degree differential equation and I specifically need the value of the first derivative of y at the final time point. My code is the following:
[T Y]=ode45(#(t y)vdp4(t,y,0.3),[0 1],[0.3/4,((3*0.3)^0.5)/2]);
I know the output will contain the values at which ode45 evaluated the function. To get the y values at specific time value at have it has been advised to give more than two time points in the MATLAB documentation. I did that too.
tspan=[0:0.01:1]
[T Y]=ode45(#(t y)vdp4(t,y,0.3),tspan,[0.3/4,((3*0.3)^0.5)/2]);
The T vector still does not have all the values from 0 to 1 (The last value is 0.39). This happens especially after multiple executions of ode45 function. I found something else in the MATLAB documentation: using "sol" structure to deval the values for specific t values. Is that the right way to go?
For reference, my differential equation is in the following function.
function dy = vdp4(t,y,k)
dy = zeros(2,1); % a column vector
dy(1)=y(2);
dy(2)=(y(2)^2-2*t*y(2)+2*y(1))/k+2;
end
Edit: I provided the parameter value. It should now be executable.
Try to plot your solution, you will find the answer
Hi i have data in the form of two column vectors (z,y) which forms an exponential curve.
And I want to find 1 unknown for it by fitting a curve of the form i.e. solve for b in the equation
y= (1-0.05*x^(b))/(1-0.05*x^b)^2
I want to solve for b which gives a good fit for data (z,y) however the only known parameter is z and y. Is there possible way I could solve for b? any tips on how to proceed with this equation? The x values are unknown and x seem to be the main problem!
Thanks!
I solved a PDE using Matlab solver, pdepe. The initial condition is the solution of an ODE, which I solved in a different m.file. Now, I have the ODE solution in a matrix form of size NxM. How I can use that to be my IC in pdepe? Is that even possible? When I use for loop, pdepe takes only the last iteration to be the initial condition. Any help is appreciated.
Per the pdepe documentation, the initial condition function for the solver has the syntax:
u = icFun(x);
where the initial value of the PDE at a specified value of x is returned in the column vector u.
So the only time an initial condition will be a N x M matrix is when the PDE is a system of N unknowns with M spatial mesh points.
Therefore, an N x M matrix could be used to populate the initial condition, but there would need to be some mapping that associates a given column with a specific value of x. For instance, in the main function that calls pdepe, there could be
% icData is the NxM matrix of data
% xMesh is an 1xM row vector that has the spatial value for each column of icData
icFun = #(x) icData(:,x==xMesh);
The only shortcoming of this approach is that the mesh of the initial condition, and therefore the pdepe solution, is constrained by the initial data. This can be overcome by using an interpolation scheme like:
% icData is the NxM matrix of data
% xMesh is an 1xM row vector that has the spatial value for each column of icData
icFun = #(x) interp1(xMesh,icData',x,'pchip')';
where the transposes are present to conform to the interpretation of the data by interp1.
it is easier for u to use 'method of line' style to define different conditions on each mesh rather than using pdepe
MOL is also more flexible to use in different situation like 3D problem
just saying :))
My experience is that the function defining the initial conditions must return a column vector, i.e. Nx1 matrix if you have N equations. Even if your xmesh is an array of M numbers, the matrix corresponding to the initial condition is still Nx1. You can still return a spatially varying initial condition, and my solution was the following.
I defined an anonymous function, pdeic, which was passed as an argument to pdepe:
pdeic=#(x) pdeic2(x,p1,p2,p3);
And I also defined pdeic2, which always returns a 3x1 column vector, but depending on x, the value is different:
function u0=pdeic2(x,extrap1,extrap2,extrap3)
if x==extrap3
u0=[extrap1;0;extrap2];
else
u0=[extrap1;0;0];
end
So going back to your original question, my guess would be that you have to pass the solution of your ODE to what is named 'pdeic2' in my example, and depending on X, return a column vector.
I am trying to model a system of three differential equations. This is a droplet model, parametrized vs the arc length, s.
The equations are:
dx/ds=cos(theta)
dz/ds=sin(theta)
(theta)/ds=2*b+c*z-sin(theta)/x
The initial conditions are that x,z, and theta are all 0 at s=0. To avoid the singularity on d(theta)/ds, I also have the condition that, at s=0, d(theta)/ds=b. I have already written this code:
[s,x]=ode23(#(s,x)drpivp(s,x,p),sspan,x0);
%where p contains two parameters and x0 contains initial angle theta, x, z values.
%droplet ODE function:
function drpivp = drpivp(s,x,p);
%x(1)=theta
%x(2)=x
%x(3)=z
%b is curvature at apex
%c is capillarity constant
b=p(1);
c=p(2);
drpivp=[2/p(1)+p(2)*x(3)-sin(x(1))/x(2); cos(x(1)); sin(x(1))];
Which yields a solution that spirals out. Instead of creating one droplet profile, it creates many. Of course, here I have not initialized the equation properly, because I am not certain how to use a different equation for theta at s=0.
So the question is: How do I include the initial condition that d(theta)/ds=b instead of it's usual at s=0? Is this possible using the built-in solvers on matlab?
Thanks.
There are several ways of doing this, the easiest is to simply add an if statement into your equation:
function drpivp = drpivp(s,x,p);
%x(1)=theta
%x(2)=x
%x(3)=z
%b is curvature at apex
%c is capillarity constant
b=p(1);
c=p(2);
if (s == 0)
drpivp=[b; cos(x(1)); sin(x(1))];
else
drpivp=[2/p(1)+p(2)*x(3)-sin(x(1))/x(2); cos(x(1)); sin(x(1))];
end