I have been trying to apply the method of multiple scales to a certain set of partial differential equations, but I keep having problems. I would appreciate if someone could point out my mistake, whether in my application of the method or in my interpretation of the results. I apologize about the length of the post, but I wanted to include my full derivation.
The relevant equations may be written as
\begin{gather}
\frac{\partial f}{\partial x} – i\epsilon a(x)g = 0, \\
\frac{\partial g}{\partial x} – i\epsilon b(x)f = 0,
\end{gather}
where $\epsilon\ll1$ and $a(x)$ and $b(x)$ are statistically homogeneous, positive and integrable functions (e.g., they could take the form $a(x)=1+\sin(kx)/4$). If I assume that $f(x)=f_0(x_0,x_1,x_2)+\epsilon f_1(x_0,x_1,x_2) + \epsilon^2 f_2(x_0,x_1,x_2)$ and similarly for $g(x)$, where $x_n=\epsilon^n x$, and if I further assume that $f_n$ and $g_n$ are $O(1)$, I get an series of sets of differential equations. The $O(1)$ equations may be written as
\begin{gather}
\frac{\partial f_0}{\partial x_0} = \frac{\partial g_0}{\partial x_0} = 0,
\end{gather}
and so we find that the largest-scale approximation of $f$ and $g$ does not depend on the smallest length scale, $x_0$. The $O(\epsilon)$ equations are
\begin{gather}
\frac{\partial f_1}{\partial x_0} + \frac{\partial f_0}{\partial x_1} – i a(x_0)g_0 = 0, \\
\frac{\partial g_1}{\partial x_0} + \frac{\partial g_0}{\partial x_1} – i b(x_0)f_0 = 0.
\end{gather}
Since $f_0$ and $g_0$ do not depend on $x_0$, we may integrate, yielding
\begin{gather}
f_1 + x_0\frac{\partial f_0}{\partial x_1} – i g_0\int a(x')dx' + c_1 = 0, \\
g_1 + x_0\frac{\partial g_0}{\partial x_1} – i f_0\int b(x')dx' + c_2 = 0,
\end{gather}
where $c_1$ and $c_2$ are constants of integration. These constants may be set equal to zero without loss of generality, as they may be incorporated into $f_0$ and $g_0$. We need to remove secular terms. Clearly, the $x_0$ terms are secular, but the integrals over $a$ and $b$ are also secular. If we decompose $a(x_0)$ and $b(x_0)$ such that $a(x_0)=\langle a\rangle + \Delta a(x_0)$, where $\langle a\rangle$ is the average of $a(x_0)$, and $b(x_0)=\langle b\rangle + \Delta b(x_0)$, we may then write
\begin{gather}
f_1 + x_0\left[\frac{\partial f_0}{\partial x_1} – ig_0\langle a\rangle\right] – i g_0\int_0^{x_0} \Delta a(x_0)dx_0 = 0, \\
g_1 + x_0\left[\frac{\partial g_0}{\partial x_1} – if_0\langle b\rangle\right] – i f_0\int_0^{x_0} \Delta b(x_0')dx_0' = 0.
\end{gather}
Now the integrals are not secular, and we may set the terms in the square brackets equal to zero. Thus, we may write
\begin{gather}
f_1 = i g_0\int_0^{x_0} \Delta a(x_0)dx_0, \\
g_1 = i f_0\int_0^{x_0} \Delta b(x_0)dx_0, \\
\frac{\partial f_0}{\partial x_1} – i\langle a\rangle g_0 = 0, \\
\frac{\partial g_0}{\partial x_1} – i\langle b\rangle f_0 = 0.
\end{gather}
Thus, I have equations for $f_0$ and $g_0$ that may be explicitly solved as a function of $x_1$. I can then write the solutions as
\begin{gather}
f_0(x_1,x_2) = A(x_2)e^{i\sqrt{\langle a\rangle\langle b\rangle}x_1} + B(x_2)e^{-i\sqrt{\langle a\rangle\langle b\rangle}x_1}, \\
g_0(x_1,x_2) = \sqrt{\frac{\langle b\rangle}{\langle a\rangle}}A(x_2)e^{i\sqrt{\langle a\rangle\langle b\rangle}x_1} – \sqrt{\frac{\langle b\rangle}{\langle a\rangle}}B(x_2)e^{-i\sqrt{\langle a\rangle\langle b\rangle}x_1}.
\end{gather}
To this point I do not think there is an issue, but I would like to extend the results to higher orders. The $O(\epsilon^2)$ equations are given by
\begin{gather}
\frac{\partial f_2}{\partial x_0} + \frac{\partial f_1}{\partial x_1} + \frac{\partial f_0}{\partial x_2} – i a(x_0)g_1 = 0, \\
\frac{\partial g_2}{\partial x_0} + \frac{\partial g_1}{\partial x_1} + \frac{\partial g_0}{\partial x_2} – i b(x_0)f_1 = 0.
\end{gather}
Substituting in the solutions to $f_1$ and $g_1$ and their $x_1$ gradients then yields
\begin{gather}
\frac{\partial f_2}{\partial x_0} + \frac{\partial f_0}{\partial x_2} + \left[ a(x_0)\int_0^{x_0} \Delta b(x_0')dx_0' – \langle b\rangle \int_0^{x_0} \Delta a(x_0')dx_0' \right] f_0 = 0, \\
\frac{\partial g_2}{\partial x_0} + \frac{\partial g_0}{\partial x_2} + \left[ b(x_0)\int_0^{x_0} \Delta a(x_0')dx_0' – \langle a\rangle \int_0^{x_0} \Delta b(x_0')dx_0' \right] g_0 = 0.
\end{gather}
We may integrate with respect to $x_0$ again, and eliminating the secular terms leads to the conclusion
\begin{gather}
\frac{\partial f_0}{\partial x_2} + \left[ \left\langle a(x_0)\int_0^{x_0} \Delta b(x_0')dx_0'\right\rangle – \langle b\rangle \left\langle \int_0^{x_0} \Delta a(x_0')dx_0' \right\rangle \right] f_0 = 0, \\
\frac{\partial g_0}{\partial x_2} + \left[ \left\langle b(x_0)\int_0^{x_0} \Delta a(x_0')dx_0'\right\rangle – \langle a\rangle \left\langle \int_0^{x_0} \Delta b(x_0')dx_0' \right\rangle \right] g_0 = 0.
\end{gather}
These equations may also be solved explicitly. If we let
\begin{gather}
\alpha \equiv \left\langle a(x_0)\int_0^{x_0} \Delta b(x_0')dx_0'\right\rangle – \langle b\rangle \left\langle \int_0^{x_0} \Delta a(x_0')dx_0' \right\rangle, \\
\beta \equiv \left\langle b(x_0)\int_0^{x_0} \Delta a(x_0')dx_0'\right\rangle – \langle a\rangle \left\langle \int_0^{x_0} \Delta b(x_0')dx_0' \right\rangle,
\end{gather}
then the solutions are given by
\begin{gather}
f_0(x_1,x_2) = f_{01}(x_1)e^{-\alpha x_2}, \\
g_0(x_1,x_2) = g_{01}(x_1)e^{-\beta x_2}.
\end{gather}
Combining this result with the $O(\epsilon)$ result leads to the conclusions
\begin{gather}
A(x_2) = B(x_2) = C_0e^{-\alpha x_2}, \\
A(x_2) = B(x_2) = D_0e^{-\beta x_2}.
\end{gather}
However, both of these statements can only be true if $\alpha=\beta$, which is not always the case, or if $C_0=D_0=0$, which is the trivial solution. I know that a non-trivial solution can exist for the original equations, even when $\alpha\ne\beta$.
Did I make a mistake somewhere? Is there a restriction on the method of multiple scales that makes may analysis invalid? Any assistance would be greatly appreciated.
Best Answer
Bottom Line Up Front
I think I have figured out the issues; there are a couple of them. First, I discarded ``constants'' of integration when I should have retained them, and then I messed up the identification of secular terms.
Corrected Derivation
I think it may be easier to state writing everything in matrix form. Let $\vec F=[f,g]^T$ and $\mathsf A=[0,a(x);b(x),0]$, such that the initial problem may be written as $$ \frac{d\vec F}{dx} = i\epsilon\mathsf A\cdot\vec F. $$ By the multiple scales approximation we write $$\vec F = \vec F_0(x_0,x_1,\cdots) + \epsilon\vec F_1(x_0,x_1,\cdots) + \cdots, $$ leading to the first three equations being written as \begin{gather} \frac{d\vec F_0}{dx_0} = 0, \tag{1} \\ \frac{d\vec F_0}{dx_1} + \frac{d\vec F_1}{dx_0} = i\mathsf A\cdot\vec F_0, \tag{2} \\ \frac{d\vec F_0}{dx_2} + \frac{d\vec F_1}{dx_1} + \frac{d\vec F_2}{dx_0} = i\mathsf A\cdot\vec F_1. \tag{3} \end{gather} Equation (1) simply states that $\vec F_0$ is independent of $x_0$. Using the notation of the original post we write $\mathsf A = \left\langle\mathsf A\right\rangle + \Delta\mathsf A$, and so we may solve Eq. (2) as \begin{equation} \vec F_1 = -x_0\left[ \frac{d\vec F_0}{dx_1} - i\left\langle\mathsf A\right\rangle\cdot\vec F_0 \right] + i\int_0^{x_0}\Delta\mathsf A(x_0')dx_0'\cdot\vec F_0 + \vec F_{10}, \end{equation} where $\vec F_{10}$ is the constant of integration. In the original post I assumed that $\vec F_{10}$ may be set equal to zero, but this assumption was my first mistake. In reality, $\vec F_{10}$ is a function of $x_1$, $x_2$, etc., and including it will be critical to continue.
As in the original post, the secular terms are set equal to zero, leading to the equation $$ \frac{d\vec F_0}{dx_1} - i \left\langle\mathsf A\right\rangle\cdot\vec F_0 = 0. $$ To solve this we diagonalize $\left\langle\mathsf A\right\rangle$, such that we may write $\left\langle\mathsf A\right\rangle = \mathsf V\cdot\mathsf\Lambda\cdot\mathsf V^{-1},$ where $\mathsf \Lambda$ is the diagonal matrix of eigenvalues and $\mathsf V$ is the matrix of eigenvectors, which may be written as \begin{gather} \mathsf\Lambda = \left[\begin{matrix} \sqrt{\langle a\rangle\langle b\rangle} & 0 \\ 0 & -\sqrt{\langle a\rangle\langle b\rangle} \end{matrix}\right], \\ \mathsf V = \left[\begin{matrix} 1 & 1 \\ \sqrt{\langle b\rangle/\langle a\rangle} & -\sqrt{\langle b\rangle/\langle a\rangle} \end{matrix}\right], \\ \mathsf V^{-1} = \frac{1}{2}\sqrt{\frac{\langle a\rangle}{\langle b\rangle}}\left[\begin{matrix} \sqrt{\langle b\rangle/\langle a\rangle} & 1 \\ \sqrt{\langle b\rangle/\langle a\rangle} & -1 \end{matrix}\right]. \end{gather} Defining $\vec G_0=\mathsf V^{-1}\cdot\vec F_0$, the equation then becomes $$ \frac{d\vec G_0}{dx_1} - i\mathsf\Lambda\cdot\vec G_0 = 0, $$ which is a set of two independent equations, and be may solve this by $$ \vec G_0 = \exp\left[i\mathsf\Lambda x_1\right]\cdot\vec G_{00}, $$ where $\vec G_{00}$ is independent of $x_0$ and $x_1$. This conclusion is identical to that given in the original post.
Equation (3) now reads $$ \frac{d\vec F_0}{dx_2} - \int_0^{x_0}\Delta\mathsf A(x_0')dx_0'\cdot\left\langle\mathsf A\right\rangle\cdot\vec F_0 + \frac{d\vec F_{10}}{dx_1} + \frac{d\vec F_2}{dx_0} = -\mathsf A\cdot\int_0^{x_0}\Delta\mathsf A(x_0')dx_0'\cdot\vec F_0 + i\mathsf A\cdot\vec F_{10}. $$ Again, we desire to remove secular terms, and so we conclude $$ \frac{d\vec F_{10}}{dx_1} - i\left\langle\mathsf A\right\rangle\cdot\vec F_{10} = - \frac{d\vec F_0}{dx_2} + \mathsf B\cdot\vec F_0 , $$ where $$ \mathsf B = \left\langle\int_0^{x_0}\Delta\mathsf A(x_0')dx_0'\right\rangle\cdot\left\langle\mathsf A\right\rangle - \left\langle\mathsf A\cdot\int_0^{x_0}\Delta\mathsf A(x_0')dx_0'\right\rangle. $$ This last equation is the same as in the original post except for the inclusion of the $\vec F_{10}$ terms. We may transform this equation as we did the $\vec F_0$ equation above using $\mathsf\Lambda$ and $\mathsf V$, yielding $$ \frac{d\vec G_{10}}{dx_1} - i\mathsf \Lambda\cdot\vec G_{10} = - \frac{d\vec G_0}{dx_2} + \mathsf V^{-1}\cdot\mathsf B\cdot\mathsf V\cdot\vec G_0 , $$ where $\vec G_{10}=\mathsf V^{-1}\cdot\vec F_{10}$. Substituting in the solution to $\vec G_0$ then yields $$ \frac{d\vec G_{10}}{dx_1} - i\mathsf \Lambda\cdot\vec G_{10} = - \exp\left[i\mathsf\Lambda x_1\right]\cdot\frac{d\vec G_{00}}{dx_2} + \mathsf V^{-1}\cdot\mathsf B\cdot\mathsf V\cdot\exp\left[i\mathsf\Lambda x_1\right]\cdot\vec G_{00} . $$ The homogeneous solution for $\vec G_{10}$ is similar to that for $\vec G_0$.
At this point I am struggling to present a completely general explanation in terms of matrices and vectors. The basic idea is that if there are terms including $e^{i\lambda_i x_1}$ in the $i$th equation they will lead to secular solutions. Thus, it is not the entire right-hand side that must be equal to zer, but simply these terms. (This is the second major error I made in the original post.) Clearly, the first term on the right leads to secular terms, but only the diagonal terms of the matrix of the second term must. Note that $\mathsf B$ is diagonal, and so we may write $\mathsf B=[c_1,0;0,c_2]$. Explicitly performing the matrix multiplication then yields $$ \mathsf V^{-1}\cdot\mathsf B\cdot\mathsf V\cdot\exp\left[i\mathsf\Lambda x_1\right] = \frac{1}{2} \left[\begin{matrix} (c_1+c_2)e^{i\lambda_1 x_1} & (c_1-c_2)e^{i\lambda_2x_1} \\ (c_1-c_2)e^{i\lambda_1 x_1} & (c_1+c_2)e^{i\lambda_2x_1} \end{matrix}\right]. $$ Thus, the actual constraint to avoid secular terms is given by $$ \frac{d\vec G_{00}}{dx_2} - \frac{1}{2}\mathrm{tr}\left[\mathsf B\right]\vec G_{00} = 0, $$ where $\mathrm{tr}[\mathsf B]$ is the trace of $\mathsf B$. Solving this equation yields $$ \vec G_{00} = e^{(\mathrm{tr}[\mathsf B]/2)x_2}\vec G_{000}, $$ where $\vec G_{000}$ is not a function of $x_0$, $x_1$, or $x_2$. Substituting this into $\vec G_0$ we obtain $$ \vec G_0 = \exp\left[i\mathsf\Lambda x_1 + \mathsf I\frac{\mathrm{tr}[\mathsf B]}{2}x_2\right]\cdot\vec G_{000}, $$ where $\mathsf I$ is the identity matrix, and finally we obtain the contradiction-free solution $$ \vec F_0 = \mathsf V\cdot\exp\left[i\mathsf\Lambda x_1 + \mathsf I\frac{\mathrm{tr}[\mathsf B]}{2}x_2\right]\cdot\mathsf V^{-1}\cdot\vec F_{000}, $$ where $\vec F_{000}$ is not a function of $x_0$, $x_1$, or $x_2$. We could break this solution out into individual expressions for $f_0$ and $g_0$, but I feel that it is unnecessary and tedious, so I will not do that here.