Second order homogenous coupled differential equations

Question

I am trying to solve the following coupled differential equations, \begin{align} f''(x)+c_1(f(x)+g(x))&=0\\ g''(x)+c_2(f(x)+g(x))&=0 \end{align}

where $c_1$ and $c_2$ are positive constants. The problem is I don't know any boundary conditions. The only condition that I can think of is $f(\infty)=g(\infty)=0$. How can I solve this?

Answer 1

Equation 1 $\times c_2$ - Equation 2 $\times c_1$: $$ f''(x) = \frac{c_1}{c_2} g''(x)$$ Hence, $$f(x) = \frac{c_1}{c_2} g(x) + Ax + B$$

Apply some boundary conditions to solve for $A,B$.

This decouples both the differential equations. Now second equation becomes: $$g''(x) = -c_2(\frac{c_1}{c_2} g(x) + Ax + B + g(x))$$ $$g''(x) = -(c_1+c_2) g(x) - A'x - B'$$

One possible solution is $$g(x) = e^{i \sqrt{c_1+c_2} x} + C_1 x^3 + C_2 x^2.$$ Substituting above $g(x)$ in differential equation, we get: $$6C_1 = -(c_1+c_2) C_1 -A'$$ $$2C_2 = -(c_1+c_2) C_2 -B'$$

Use $A',B'$ from boundary conditions and solve above equation to get the solution.

Answer 2

Summing the two equations, one obtains $$ (f+g)''+(c_1+c_2)(f+g)=0, \tag{1} $$ the general solution of which is $$ f(x)+g(x)=A\cos(kx)+B\sin(kx),\qquad k=\sqrt{c_1+c_2}. \tag{2} $$ Also, notice that $$ c_2f''-c_1g''=0, \tag{3} $$ which, integrated twice, yields $$ c_2f(x)-c_1g(x)=Cx+D. \tag{4} $$ Solving the system of linear equations $(2)$ and $(4)$ one finally obtains \begin{align} f(x)&=\frac{1}{c_1+c_2}[c_1(A\cos(kx)+B\sin(kx))+Cx+D], \\ g(x)&=\frac{1}{c_1+c_2}[c_2(A\cos(kx)+B\sin(kx))-Cx-D]. \tag{5} \end{align}