Using the Fredholm alternative to find a solvabilty condition for a PDE

Question

I'm trying to understand how to use the Fredholm alternative to solve a system of PDEs. The specific problem is as follows

Answer 1

I'll separate this post into sections. Hopefully, by following them in the order written, you can understand how to solve systems of equations asymptotically, what solvability conditions are and how they come about. I've written this in the hope that others may also find it useful.

---

The Fredholm alternative says that, given an equation $Lu = f$, a solution to this problem exists iff $f$ is orthogonal to the homogeneous adjoint solution $v$ i.e $v$ satisfies $L^{*} v = 0$ with $L^{*}$ the adjoint operator of $L$. This is because

\begin{align} Lu = f \implies \langle Lu, v \rangle &= \langle u, L^{*} v \rangle \\ &= \langle u, 0 \rangle \\ &= 0 \\ &= \langle f, v \rangle \tag 1 \end{align}

for some suitable inner product $\langle \cdot, \cdot \rangle$.

---

Note that at leading order, you would have had the following problem

\begin{align} (1+\Gamma)^{1/2} \partial_{t} \zeta_{0} - \partial_{y} \phi_{0} &= 0 \tag 2 \\ (1+\Gamma)^{1/2} \partial_{t} \phi_{0} + (1-\Gamma \partial_{x}^{2}) \zeta_{0} &= 0 \tag 3 \end{align}

We will call this equation set $M \vec{u}_{0} = \vec{0}$ where $M$ is a matrix of derivative operators and $\vec{u}_{0} = (\zeta_{0}, \phi_{0})$. We can eliminate $\phi_{0}$ and make a single PDE by taking

$$(1+\Gamma)^{1/2} \partial_{t}(2) + \partial_{y} (3) \tag 4$$

assuming the solution is regular enough to do so and to interchange derivatives, to get

$$(1+\Gamma) \partial_{t}^{2} \zeta_{0} + (1-\Gamma \partial_{x}^{2}) \partial_{y} \zeta_{0} = 0 \implies L \zeta_{0} = 0 \tag 5$$

assuming $\Gamma$ is a constant. Solving this PDE yields $\zeta_{0}$ and backsolving through $(2)$ and $(3)$ then yields $\phi_{0}$.

---

We now want to find the adjoint of $(5)$ i.e we want to find $L^{*}$ such that $\langle L \zeta_{0}, v \rangle = \langle \zeta_{0}, L^{*} v \rangle$. As we are working with differential equations, we choose the inner product

$$\langle A, B \rangle = \int_{\Omega} A \bar{B} dx$$

where $\bar{B}$ denotes the complex conjugate of $B$. Computing the adjoint, using integration by parts, we find

$$L^{*} v = (1+\Gamma) \partial_{t}^{2} v - (1-\Gamma \partial_{x}^{2}) \partial_{y} v \tag 6$$

where I have assumed the evaluated terms vanish as you didn't provide the boundary conditions or even the domain. Solving $L^{*} v = 0$ fixes the homogeneous adjoint solution $v$, which we need to compute the solvability condition.

---

Finally we have all the necessary ingredients to compute the solvability condition. In your problem, at second order, you have the system

\begin{align} (1+\Gamma)^{1/2} \partial_{t} \zeta_{2} - \partial_{y} \phi_{2} &= f_{1}(\zeta_{1}, \phi_{1}, \zeta_{0}, \phi_{0}) \\ (1+\Gamma)^{1/2} \partial_{t} \phi_{2} + (1-\Gamma \partial_{x}^{2}) \zeta_{2} &= f_{2}(\zeta_{1}, \phi_{1}, \zeta_{0}, \phi_{0}) \end{align}

which we can write as

$$M \vec{u}_{2} = \vec{f}(\vec{u}_{1}, \vec{u}_{0}) \tag 7$$

with $M$ as before and $\vec{f} = (f_{1}, f_{2})$. As before, we apply the same operations in $(4)$ to $(7)$ to get

$$L \zeta_{2} = (1+\Gamma)^{1/2} \partial_{t} f_{1} + \partial_{y} f_{2} := f \tag 8$$

In this form, hopefully you recognise why the Fredholm alternative is needed; a solution to $(8)$ exists iff $f$ is orthogonal to the homogeneous adjoint solution $v$. Therefore, provided

$$\langle (1+\Gamma)^{1/2} \partial_{t} f_{1} + \partial_{y} f_{2}, v \rangle = 0 \tag 9$$

then you can continue to compute the asymptotic solutions down further scales. Equation $(9)$ is what we call the solvability condition because, clearly, the system is solvable only if that condition holds. Also notice that, as the function forms of $\zeta_{1}, \phi_{1}, \zeta_{0}, \phi_{0}$ are known by this point except for the amplitudes $A$ and $\bar{A}$, the solvability condition $(9)$ yields an evolution equation for the amplitudes themselves.