How to plot a function which is defined on different subintervals

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Cris19 2021-3-7

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评论： Walter Roberson 2021-3-10

采纳的回答： Walter Roberson

I am trying to plot the function

But I don't know how to write the code for the definition of the function f which is given on different subintervals.

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KALYAN ACHARJYA 2021-3-7

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Cris19 2021-3-7

@KALYAN ACHARJYA Thank you, but didn't help.

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Walter Roberson 2021-3-7

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There are several methods available. The most straight forward is to write a function that loops over the inputs, testing each one to decide what the result should be.

function y = f(X)
   y = zeros(size(X));
   for K = 1 : numel(X)
       x = X(K);
       if x < 2
          y(K) = 0;
       elseif x <= 5
           y(K) = x.^2;
       elseif x <= 8
           y(K) = x-x.^3;
       else
           y(K) = 0;
       end
end

With regards to those intervals you need, think about floor(x+1/2)

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Walter Roberson 2021-3-8

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for i= 2:500

i is a positive integer

Bounds_1 = [i - (1/i^2),i];

With i being real valued, 1/i^2 is real valued and positive (or infinite and positive), and subtracting a positive amount from a real value always gives you a result that is less than the original amount.

Bounds_2 = [i,i + (1/i^2)];

With i being real-valued, 1/i^2 is real valued and positive (or infinite and positive), and adding a positive amount to a real value always gives you a result that is greater than the original amount

Bounds = [Bounds_1 Bounds_2];

That would be a list [i - (1/i^2),i,i,i + (1/i^2)]

Min = min(Bounds);

As demonstrated above, i-1/i^2 is always less than i, and i+1/i^2 is always greater than i, so we can quickly see that the minimum value must be i-1/i^2

Max = max(Bounds);

and the maximum value must be 1+1/i^2 . So there is no point in creating the Bounds list and taking the min and max: we already know which will be the min and which will be the max

end

You do not use Min and Max inside the loop, so you overwrite the value each time. Therefore at the end of the loop, the value will be whatever was calculated on the last iteration, the one for i=500

fplot(f);
xlim([Min-2 Max+2]);

and you create your xlim according to only that iteration.

f1 = @(x) (i^2.*x - i^3 + 1).*(Bounds_1(1) < x & x <= Bounds_1(2));
f2 = @(x) (-i^2.*x + i^3 + 1).*(Bounds_2(1) < x & x <= Bounds_2(2));
f = @(x) f1(x) + f2(x); 
end

You overwrite all of f1, f2, and f each iteration, so in that code there is no point doing any iteration except the last one.

Cris19 2021-3-9

编辑：Cris19 2021-3-9

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One more question, if possible... I try to solve to second order ODE y''+f(x)y'+y=0, whre f(x) is the function above. I am interested to plot on [0,10] the solution of this ODE. for the initial data y(0)=y'(0)=0.1. I used the function (where f(x) is defined with the code you wrote)

function dwdt=susp(t,w)
f = @(x) zeros(size(x));
for i= 2:10
Bounds_1 = [i - (1/i^2),i];
Bounds_2 = [i,i + (1/i^2)];
f1 = @(x) (i^2.*x - i^3 + 1).*(Bounds_1(1) < x & x <= Bounds_1(2));
f2 = @(x) (-i^2.*x + i^3 + 1).*(Bounds_2(1) < x & x <= Bounds_2(2));
f = @(x) f(x) + f1(x) + f2(x);
end
dwdt=zeros(2,1);
dwdt(1)=w(2);
dwdt(2)=-f*w(2)-w(1);
end 

and the commands

tspan = [0 10];
z0=[0.1 0.1];
[t,z] = ode45(@(t,z) susp(t,z), tspan, z0);
figure
plot(t,z(:,1),'r');
legend('x(t)');

But I get several errors...

Unary operator '-' is not supported for operand of type 'function_handle'.
Error in susp (line 12)
dwdt(2)=-f*w(2)-w(1);
Error in @(t,z)susp(t,z)
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:});   % ODE15I sets args{1} to yp0.
Error in ode45 (line 115)
  odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);

How could I fix them?

Cris19 2021-3-10

Thank you. I wonder if there is an alternative solution. I am interested to find the asymptotic behavior at +infinity of the solution of that ODE. So is this why I think it is interesting to see the plotting of the solution on intervals larger and larger, such as [0,10], [0,500] etc.

I really don't know how to code this...

Walter Roberson 2021-3-10

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At asymptopic behaviour is

syms n f(x)
D = symsum(dirac(x-n), n, 1, 200)
df = diff(f)
d2f = diff(df)
eqn = d2f + D*df + f == 0
dsolve([eqn, f(0)==0])

but in practice you will not get any solution. Even if you reduce it down to dirac at one particular integer you are not going to get a solution.

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