How to proceed in this Differential Equation

Question

I have solved O.D.E using variable separable but I have no idea how to solve this D.E. using operator method.

$$\dfrac{dT_h}{dx}+ λ_h(T_h-T_C)=0$$ $$\dfrac{dT_c}{dx}- λ_c(T_h-T_C)=0$$

The above equations can be written as: $$(D+λ_h)T_h-λ_hT_c=0$$ $$(D+λ_c)T_c-λ_cT_h=0$$ How to proceed further?

Answer 1

This can be written as

$$T' = \begin{bmatrix} -\lambda_h & \lambda_h \\ \lambda_c & -\lambda_c \end{bmatrix} T$$

This is a first order linear system of two scalar ODEs with constant coefficients. It can be solved by the matrix exponential technique: $T(t)=e^{At} T(0)$ where $A$ is the coefficient matrix in the system. The matrix exponential can be computed by the eigendecomposition: if $A=P D P^{-1}$ for a diagonal $D$ then $e^{At}=P e^{Dt} P^{-1}$. Here $(e^{Dt})_{ii}=e^{D_{ii}t}$ and $(e^{Dt})_{ij}=0$ for $i \neq j$.

Doing this eigendecomposition symbolically is a bit ugly/tedious but is possible in the $2 \times 2$ case (especially here, where one of the eigenvalues is always zero).

Following the operator formalism, if you apply $D+\lambda_h$ to the second equation and substitute in the first equation, you get

$$(D+\lambda_h)(D+\lambda_c) T_c - \lambda_c (D+\lambda_h) T_h = (D+\lambda_h)(D+\lambda_c) T_c - \lambda_c \lambda_h T_c = 0.$$

This is now a single scalar ODE for $T_c$, which you can try to solve and then back-substitute. However, you'll find that the roots of the characteristic polynomial of this equation are just the eigenvalues of $A$, so this is really just carrying out the algebra that the eigendecomposition does "for you" by hand.

Answer 2

The two differential equations are first order ODEs and coupled. I presume $\lambda_h$ and $\lambda_c$ are constants.

Hence I would do the following:

1. Differentiate first DE with respect to $x$, and call it (3)

2. Substitute in 2nd DE into (3), and call it (4)

3. (4) is now a second order ODE in $T_h$, and can be rearranged with the RHS to be in terms of $T_c$, and LHS in terms of $T_h$. Call it (5)

4. Obtain the complementary function $T_{h_{Comp}}$ of (5) by setting RHS =0 and solving.

5. The form of the Particular Solution $T_{h_{PS}}$ of (5) to try depends on the function $T_c$ in the RHS, and may need to be adjusted if a term is the same order as in $T_{h_{Comp}}$

6. The General Solution is $T_h=T_{h_{Comp}}+T_{h_{PS}}$