MATLAB: Problem with dsolve: Unable to find explicit solution

differential equationsdsolveMATLAB

Hi everyone,

I am trying to solve the following ODE:

consider that [3] = a, [4] = b and [5] = c

I am using the following code to solve this:

answer = dsolve('Da=-k1*a + km1*b - k3*a*b','Db=k1*a - km1*b - k3*a*b','Dc=k3*a*b','a(0)=0.26','b(0)=0','c(0)=0')

However when I run this code I get the following:

Warning: Unable to find explicit solution. 
> In dsolve (line 201)
  In Paper2_Figure4 (line 13) 
 
answer =
 
[ empty sym ]

I don't understand what I'm doing wrong, any help would be greatly appreciated! Thanks 🙂

Best Answer

I found a way to do this, but I will need to wait until I get to my desktop to write through all of the steps as simplify() gave some unexpected results that are not understandable as-is.

The approach is to notice that the last sub-expression of the first two equations is the same as the expression for Dc. The first two are like Da=P+Dc , Db=-P+Dc in form. Select down to those two with the sub-expression replaced by Dc and dsolve. You will get solutions for a and c.

Take the expression for c and substitute t=0 to get c(0). You will get an expression in C1, C2, and some other terms and you know that c(0)=0 so solve() the expression for C2. The result will involve b(0) but we know that b(0)=0 so subs() that to get C2.

Now subs that C2 into the solution for a. subs t=0, solve() that equal 0.26 for C1, subs b(0)=0 and you get out C1

Now go back to the solutions for a and c and substitute in C1 and C2 and b(0)=0 to get the final a and c. Do not simplify() c!!! -- you will get weird results if you do!

Those in hand, go back to the second equation giving the expression for Db and substitute a and c and derivative of c to get Db.

At this point I simplify() and the dB term disappeared, and it was then that I noticed unexpected functions in the output, which I traced back to my having simplify() after I subs C1 into c. So at the moment I do not have the final production for b, but it should be down to a single equation and b(0)=0 will clearly be important. This resolution will have to wait until I get to my desktop.

Related Solutions

MATLAB: Unable to find explicit solution to differential eq

Hi Ditlev,

I believe this does have an analytic solution that is not overly complicated, of the form

x = (a+be^(-ct)) / (f+ge^(-ct))

First of all the scaling of this problem is not too good so I went to the variables (I used y in place of your x2)

u = km*t         x = (kp/km) x_old        y = (kp/km) y_old

in which case the equations are

x' = 2y - 2x^2
y' = -y + x^2

where differentiation is d/du. The starting value for the new x is

(kp/km)*1e-9 = 740.7

The most complicated the algebra gets is the roots of a quadratic eqn. The code below plots both a numerical and analytic solution.

Eventually the solution gets to the fixed point

x = r1 = 18.9966      y = -r1/2 + C = 360.8721

at which point x^2 = y and both derivatives are zero.

kp = 2*10^7;
km = 0.27*10^-4;
x0 = (kp/km)*1e-9   % starting point for x
y0 = 0;
% numerical version
[unum xynum] = ode45(@(u,xy) fun(u,xy), [0 .01], [x0 y0]);         % xy = [x; y]
%analytic version
C = x0/2;
r = roots([1,1/2,-C]);
r1 = max(r);
r2 = min(r);
alpha = 2*(r1-r2);
D = (x0-r1)/(x0-r2);
u = 0:1e-4:.01;
x = (r1-D*r2*exp(-alpha*u))./(1-D*exp(-alpha*u));   % solution
y = -x/2 + C;
figure(1)
plot(unum,xynum,u,x,'o',u,y,'o')
grid on
function dxy = fun(u,xy)
  dxy = zeros(2,1);
  dxy(1) = 2*xy(2)-2*xy(1)^2;
  dxy(2) = -xy(2) + xy(1)^2;
end

MATLAB: Assign a vector

variant with loop with example

Ac = randi(100,[1 5 2 2]);
D = zeros(3,3,2,2,2,2)
for k=1:K 
    for m=1:M
        C = num2cell(Ac(1,:,k,m));
        [D(1,2,2,2,k,m) D(2,1,2,2,k,m) D(1,3,2,2,k,m) D(3,1,2,2,k,m) D(2,2,2,2,k,m)] = C{:};
    end
end

Best Answer

Related Solutions

MATLAB: Unable to find explicit solution to differential eq

MATLAB: Assign a vector

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