## Values of intersection points of plot. Print results.

on 22 Apr 2013

### Teja Muppirala (view profile)

Suppose I have a code that numerically integrates a set of ODEs, where my dependent variable is t and the vector of variables is x. I wish to print and plot values of x(1) for whenever x(2)=1. How might I achieve that? Some sort of "If" function?

In other words, graphically speaking, I can plot x(1) against x(2) and draw a vertical line at x(2)=1 and I want to know the x(1) values at the intersections of the plot and the vertical line.

Thanks.

Cedric Wannaz

### Cedric Wannaz (view profile)

on 22 Apr 2013

I am not sure to fully understand your question.. you get a vector x and you want to print elements x(i-1) for all i such as x(i) equals 1?

Pauline

### Pauline (view profile)

on 23 Apr 2013

Hi, Cedric. Not quite, perhaps my edit to the question would help clarify my problem?

Walter Roberson

### Walter Roberson (view profile)

on 23 Apr 2013

Saying x(:,1) and x(:,2) would make the question clearer.

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### Teja Muppirala (view profile)

on 23 Apr 2013

If your x is the output of an ODE solver, then it might not hit x2 = 1 exactly and you will need to interpolate.

You could use a line intersection finder from the file exchange or the Mapping Toolbox's POLYXPOLY function, or you could kind of do it manually:

```% Just making some test data
A = [-0.1 -2; 2 -0.2];
F = @(t,x) A*x;
[T,X] = ode45(F,[0 10],[5; 2]);
plot(X(:,2),X(:,1)); xlabel('X(2)'),ylabel('X(1)');
hold on;
```
```x2 = X(:,2); % Get the second state
x2goal = 1; % Position of vertical line
```
```% Find zero crossings by changes in sign
E = x2-x2goal; % Find when E == 0
dE = diff(sign(E));
t = find(abs(dE)==2);
t = [t + E(t)./(E(t)-E(t+1)); find(E==0)]; %Linear Interpolation
t = sort(t); %Not necessary
```
```crossings = interp1(X(:,1),t)
```
```% Plot the result
plot(x2goal,crossings,'r.','markersize',16)
grid on;
```

Pauline

on 23 Apr 2013

Thank you, Teja!

### Cedric Wannaz (view profile)

on 23 Apr 2013
Edited by Cedric Wannaz

### Cedric Wannaz (view profile)

on 23 Apr 2013

EDIT: this answer is wrong, I answered too quickly without paying attention to the fact that your x is the output of an ODE solver. Please jump directly to Teja's answer.

Ok, then you can proceed as follows:

``` id = x(:,2) == 1 ;
x(id,1)                        % This will provide you with all x(:,1) that
% correspond to a x(:,2) equal to 1.```

id is a vector of logicals (outcome of the test of equality on x(:,2)), that we use for indexing x(:,1). If you want to have the positions numerically, you can use

` find(id)`

Walter Roberson

on 23 Apr 2013
Cedric Wannaz

### Cedric Wannaz (view profile)

on 23 Apr 2013

Ah yes, if x gets out of an ODE solver, Walter's comment and Teja's answer are really important; I should have thought a little further.

EDIT: and my answer is even wrong if you don't interpolate, because it is absolutely not trivial to quantify how far from 1 points that are from each side of 1 can be. Setting a large tolerance would be wrong too as it could lead to taking points that are not around any 1 crossing.

Pauline

### Pauline (view profile)

on 23 Apr 2013

Thanks, Cedric, Walter.

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