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how to solve set of differential equation Warning: Unable to find explicit solution. > In dsolve (line 190)

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syms A1(z) A2(z)
n1=1;
n2=1;
L=100;
deltak=2*pi/L;
ode1 = diff(A1) == i*n1*A1*A2*exp(-i*deltak*z);
ode2 = diff(A2) == i*n2*A1^2*exp(i*deltak*z);
cond1=A(0)==1
cond2=A(1)==0
dsolve([ode1,ode2],[cond1,cond2])
  2 Comments
Walter Roberson
Walter Roberson on 10 Mar 2020
A is not defined in your boundary conditions. Probably A(0) should be A1(0) and A(1) should be A2(0) but I cannot be sure.
Waqar Ahmed
Waqar Ahmed on 10 Mar 2020
the answer is same for the the below question
syms A1(z) A2(z)
n1=1;
n2=1;
L=100;
deltak=2*pi/L;
ode1 = diff(A1) == i*n1*A1*A2*exp(-i*deltak*z);
ode2 = diff(A2) == i*n2*A1^2*exp(i*deltak*z);
cond1=A1(0)==1;
cond2=A2(1)==0;
dsolve([ode1,ode2],[cond1,cond2])

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Accepted Answer

Waqar Ahmed
Waqar Ahmed on 10 Mar 2020
This question is solved numerically and the result is gotten as expected using ode45 function

More Answers (1)

Walter Roberson
Walter Roberson on 10 Mar 2020
Ignoring the questionable boundary conditions for the moment:
That pair of equations has one closed form solution:
A1(z) = 0
A2(z) = constant that depends on boundary condition
There is also a more general solution that cannot be expressed explicitly -- not really a solution as such, but rather a set of conditions that A1 and A2 would have to satisfy in order for them to be solutions.
  2 Comments
Walter Roberson
Walter Roberson on 10 Mar 2020
Note that your boundary condition of A1(0)=1 rather than A1(0) = 0, rules out the degenerate form.
if the conditon are not been used it does not have even the solution
It has no closed form solution, just a description of what a solution would have to look like. The description looks like
MATLAB is not programmed to be able to find these general descriptions of solutions.
TL;DR: You cannot solve those symbolically.

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