# Problem in solving a set of 5 odes and getting Nan values

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Ghs Jahanmir on 15 Sep 2021
Commented: Ghs Jahanmir on 15 Sep 2021
Dear All,
I am trying to solve a set of 5 ode equaions using ode45 solver by using a separate independent function (named "Multi_func" contains my equations. However, they are solved with NaN values in final "eq" matrix.
any help is appreciated.
%%%%% Parameters
Q=[1100 350 1240 930 3620];
V_i=[3180 1696 1065 27000 2670];
V_t=[2290 1610 300 26649 0 ];
k_el=[0 0 90 0 0];
g=0;
tpan=[0 1];
Vm=[0 1.19e-4 2e-5 0 0];
K_M=[0 27 32 0 0];
W=6;
initial_conditions=zeros(1,W-1);
initial_conditions(5)=10; % ug/ml
[t,eqs]=ode45(@(t,eqs) Multi_func(t,eqs,Q,V_i,k_el,V_t,Vm,K_M,g,W),tpan,initial_conditions);
Time2=t./60; %convert to hr
cc=jet(W);
for i=1:W-1
plot(Time2,eqs(:,i),'color', cc(i,:));
hold on
end
my Multi_func function is:
function eq = Multi_func(t,eqs,Q,V_i,k_el,V_t,Vm,K_M,g,W)
eq=zeros(W-1,1);
eq(1)=((Q(1)*eqs(5)-Q(1)*eqs(1))/V_i(1))-(k_el(1)*eqs(5)/V_i(1))-V_t(1)*(((Vm(1)*eqs(1)))/(K_M(1)+eqs(1)))/V_i(1);
eq(2)=((Q(2)*eqs(5)-Q(2)*eqs(2))/V_i(2))+(Q(1)*eqs(1)/V_i(2))-(k_el(2)*eqs(5)/V_i(2))-V_t(2)*(((Vm(2)*eqs(2)))/(K_M(2)+eqs(2)))/V_i(2);
eq(3)=((Q(3)*eqs(5)-Q(3)*eqs(3))/V_i(3))-(k_el(3)*eqs(5)/V_i(3))-V_t(3)*(((Vm(3)*eqs(3)))/(K_M(3)+eqs(3)))/V_i(3);
eq(4)=((Q(4)*eqs(5)-Q(4)*eqs(4))/V_i(4))-(k_el(4)*eqs(5)/V_i(4))-V_t(4)*(((Vm(4)*eqs(4)))/(K_M(4)+eqs(4)))/V_i(4);
eq(5)=(Q(1)*eqs(1)+Q(2)*eqs(2)+Q(3)*eqs(3)+Q(4)*eqs(4)-Q(5)*eqs(5))/V_i(5)+V_t(5)*(((Vm(5)*eqs(5)))/(K_M(5)+eqs(5)))/V_i(5)+g/V_i(5);
end

Walter Roberson on 15 Sep 2021
K_M=[0 27 32 0 0];
First K_M is 0.
W=6;
initial_conditions=zeros(1,W-1);
initial_conditions(5)=10; % ug/ml
First boundary condition is 0
eq(1)=((Q(1)*eqs(5)-Q(1)*eqs(1))/V_i(1))-(k_el(1)*eqs(5)/V_i(1))-V_t(1)*(((Vm(1)*eqs(1)))/(K_M(1)+eqs(1)))/V_i(1); ^^^^^^^^^^^^^^^^
Notice the /(K_M(1)+eqs(1)) . As pointed out above, K_M(1) is 0 an eqs(1) starts out 0, and sum of those two is 0, so you have a division by 0 there.
Ghs Jahanmir on 15 Sep 2021
Yes, I got the point.
Thanks a lot for your time.

R2020b

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