0
$\begingroup$

I have two coupled ODEs for $T(x)$ and $t(x)$:

$$\frac{d^2 T}{d x^2}-\beta (T-t)+K=0 \tag 1$$

$$\frac{d t}{dx}-\alpha(T-t)=0 \tag 2$$

$\alpha, \beta$ and $K$ are constants $>0$. Also, it is known that $t(x=0)=t_i$. Additionally, for $(1)$ we know:

$$\frac{d T(x=0)}{d x}=\frac{d T(x=L)}{d x} = 0$$

The mathematica code I use is:

ode1 = D[T[x], x, x] - β (T[x] - t[x]) + K == 0

ode2 = D[t[x], x] - α (T[x] - t[x]) == 0

sol1 = DSolve[{ode1, D[T[0], x] = 0, D[T[L], x] = 0}, T[x], x]

On executing the third command i.e. DSolve I get an error like:

DSolve::deqn: Equation or list of equations expected instead of True in the first argument {K-β (-t[x]+T[x])+(T^′′)[x]==0,True,True}.

How can I get a symbolic solution to this coupled system ?

$\endgroup$
2
$\begingroup$

Add the second ode2 and you need == instead of = to define the boundary conditions T'[0] == 0, T'[L] == 0 .

Try

sol1 = DSolve[{ode1, ode2, T'[0] == 0, T'[L] == 0}, {T, t}, x]
(* symbolic solution.. *)
| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks ! I did not forget the second ODE but was planning to use the solution of the first to solve the second. I wasn't aware that two ODEs can be simultaneously given as an input to DSolve. I should have gone through the documentation more diligently. Coming to the solution, it has an undetermined constant c1, which I think can be resolved using the b.c. on $t$ i.e. $t(x=0)=0$ (I am considering $t_i=0$). So I used an additional term t[0]=0 and found an exact solution with no constants $\endgroup$ – Indrasis Mitra Aug 19 at 6:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.