= Solution
\b[Use the original scanned <boundary condition>.] The scan fixes the <Magnetic field>[normal-field amplitude] at the corner $x=z=0$. The supplied TeX drops the condition $x=0$ and would instead prescribe a constant normal field on the entire bottom. Those are different <Boundary value problem>[boundary-value problems]; the solution below uses the legible original scan.
Choose a positive integer $n$, put $k=n\pi/a$, and seek a <Separation of variables>[separated decaying mode]. A suitable <Cartesian magnetic flux function>[flux function] is
$$
A(x,z)=\frac{B_0}{k}\sin(kx)e^{-qz},\qquad
\mathbf B=(-\partial_z A,\lambda A,\partial_x A).
$$
This is a <decaying linear force-free magnetic arcade>. The first and third <Curl>[curl components] then satisfy the <Force-free magnetic field>[force-free equation] automatically; its second component requires $-(\partial_x^2+\partial_z^2)A=\lambda^2A$. Thus
$$
\boxed{q=\sqrt{k^2-\lambda^2},\qquad
\mathbf B=B_0e^{-qz}
\left(\frac qk\sin kx,\frac\lambda k\sin kx,\cos kx\right).}
$$
Direct <Derivative>[differentiation] gives $\nabla\cdot\mathbf B=0$ and $\nabla\times\mathbf B=\lambda\mathbf B$. Both horizontal components vanish at $x=0,a$, and $B_z(0,0)=B_0$. The <magnetic field> on the bottom is $B_0\cos kx$, not a uniform $B_0$.
A strictly decaying real <normal mode> requires $q>0$, so
$$
\boxed{|\lambda|<n\pi/a.}
$$
For the fundamental, single-arcade choice $n=1$, this becomes $|\lambda|<\pi/a$. Equality gives no vertical <exponential decay>, while a larger magnitude gives an oscillatory rather than decaying vertical dependence for that <Normal mode>[mode]. The data do not explicitly fix the <Normal mode>[horizontal mode] number. If arbitrarily fine higher-mode arcades are permitted, every finite real $\lambda$ admits some integer $n$ with $n\pi/a>|\lambda|$. The fundamental restriction must therefore not be advertised as a necessary bound for every possible field satisfying only the stated pointwise and wall data.
For a fundamental <Magnetic field line>[field line], $A$ is constant along its projection in the $(x,z)$ <plane>. If $0<s<1$ denotes its normalized flux-function value, its arch is
$$
z=\frac1q\log\frac{\sin(kx)}s,
\qquad x_-=\frac{\arcsin s}{k},\quad
x_+=\frac{\pi-\arcsin s}{k}.
$$
The maximum height is $q^{-1}\log(1/s)$ at $x=a/2$. In the horizontal projection,
$$
\boxed{\frac{dy}{dx}=\frac{B_y}{B_x}=\frac\lambda q,
\qquad y-y_-=\frac\lambda q(x-x_-).}
$$
Thus the vertical projections are arches and the horizontal projections are straight sheared segments joining their paired footpoints. Positive $\lambda$ gives positive horizontal slope, negative $\lambda$ reverses the shear, and $\lambda=0$ gives an unsheared <Magnetic scalar potential>[potential arcade]. Higher $n$ repeats such cells across the width.
\Image[/past-exam-of-the-advanced-fluid-dynamics-course-of-the-university-of-oxford/2017-force-free-arcade-projections.png]
{title=Vertical arches and horizontal shear of a decaying linear force-free arcade}
{description=An original fundamental-mode example has lambda divided by k equal to 0.6 and q divided by k equal to 0.8. Paired colours identify the same projected field lines; arrows follow the magnetic field for positive B0. The horizontal segments display the slope lambda divided by q.}
The transcription's alternative uniform-bottom condition would be inconsistent with <exponential decay> and zero side flux for $B_0\ne0$: <Solenoidal vector field>[solenoidality] gives $d\int_0^aB_z\,dx/dz=-B_x(a,z)+B_x(0,z)=0$, whereas a uniform nonzero bottom would give a <Magnetic flux>[conserved flux] $aB_0$ that cannot decay to zero. This integral check is an independent reason not to silently solve that mistranscribed problem.
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