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Let $U_m^n=u(mh,nk)$ be the sampled exact solution and let $A_h$ denote the update matrix of the Forward Euler diffusion scheme. For $0\le \mu \le 1/2$,is a convex combination, with the homogeneous boundary values included. ThereforeThis is the parabolic discrete maximum principle and proves max-norm stability directly.

The Taylor theorem and the second-order central difference give the exact-grid residualfor a sufficiently smooth solution on a fixed interval $0\le t\le T$. If $e^n=U^n-u^n$, thenIteration for $nk\le T$ yieldsThus $\mu \le 1/2$ implies convergence when the initial grid values converge to the initial data.

If $p>1/2$, complementing every received bit converts the channel into a binary symmetric channel with crossover probability $1-p<1/2$. This deterministic and invertible output relabelling does not change the channel capacity. Since $h_2(p)=h_2(1-p)$,Thus the capacity rises again from zero as $p$ increases beyond $1/2$, reaching one at $p=1$, where every bit is flipped deterministically and can be recovered by complementation.

At $p=1/2$, the output of the binary symmetric channel is independent of its input. It conveys no mutual information, and therefore

A reliable transmission rate $R$ means that length-$n$ codes can carry $M_n$ messages with $n^{-1}{\log }_2{M}_n$ tending to at least $R$ while the probability measure of the decoding-error event tends to zero. By the Shannon second coding theorem, the supremum of reliable rates is the binary symmetric channel capacityFor small positive $p$, the binary entropy function has the asymptotic expansionso $C(p)$ starts at one bit per channel use and decreases rapidly from one as errors are introduced.

Let $Z_j={\rho }_j{c}_j$, $k_j=\omega /c_j$, and $\varphi =k_{1}L$. Use the harmonic convention $e^{-i\omega t}$ and write the pressure and velocity amplitudes asContinuity of pressure and normal velocity at both interfaces is equivalently the layer transfer relationSubstituting $p(0)=1+R$, $u(0)=(1-R)/Z_0$, $p(L)=T$, and $u(L)=T/Z_0$, then adding the two resulting equations, givesHence the acoustic transmission through a uniform layer isFor either $\lambda \ll 1$ or $\lambda \gg 1$, the strong acoustic impedance mismatch makes transmission small except near $k_{1}L=n\pi$, where $|T|=1$. Away from those resonances,At the antiresonant points $k_{1}L=(n+\frac{1}{2})\pi$, the minimum is $2/(\lambda +{\lambda }^{-1})$.

For a progressive plane wave, writeThe momentum equation gives ${\rho }_0\omega U=kP$, while the dispersion relation is $\omega =c_{0}k$. Therefore the pressure and longitudinal velocity amplitudes obeyThe constant $Z$ is the acoustic impedance.

The linear homentropic acoustic equations areDifferentiate the first equation in time, take the divergence of the second, and use the third. This eliminates ${\rho }^{\prime }$ and $\mathbf{u}$ and gives the acoustic pressure wave equation

Use the axisymmetric streamfunction conventionFor the similarity solution $\psi =\nu zg(\eta )$, $\eta =r/z$, regularity on the axis requires $g(0)=g^{\prime }(0)=0$, while matching to still ambient fluid requires $g^{\prime }(\eta )/\eta \to 0$ as $\eta \to \infty$. The given equation has the first integralThe axis conditions give $C=0$, and separation yieldsfor some $a>0$.

The axial velocity isNormalizing by the conserved momentum flux givesso $a=3M/(32{\nu }^2)$. The similarity solution for a laminar round jet is therefore

Use the continuity equation to write the axial momentum equation in conservative form:Integrating from the axis to infinity, regularity at $r=0$ and decay of $u,w,w_r$ as $r\to \infty$ eliminate the boundary terms. Therefore the conserved momentum flux of a laminar round jet satisfies

Let $W(z)$ be the centre-line velocity scale and $\delta (z)$ the jet width. Conservation gives $M\sim W^2{\delta }^2$, hence $W\delta \sim M^{1/2}$. The axial inertial and viscous terms scale asso $W{\delta }^2\sim \nu z$. Combining the two relations givesThus the width grows linearly and the centre-line speed decays as $z^{-1}$. The volume flux scales asso it increases downstream. By fluid entrainment, the extra flux must be ambient fluid drawn radially into the jet.

The states $|2⟩$ and $|3⟩$ are degenerate and couple to $|1⟩$ with amplitudes $\epsilon V_0$ and ${\epsilon }^2{V}_0$. The bright and dark state reduction of a star-coupled Hamiltonian gives one exact dark eigenstate proportional to $-\epsilon |2⟩+|3⟩$, with energy $E$. The orthogonal bright state couples to $|1⟩$ with magnitudeDiagonalizing the remaining two-dimensional Hermitian matrix gives the exact spectrumWriting $\Delta =E_1-E\ne 0$ and expanding the branches according to the eigenvalue to which they converge,These are exactly the three perturbative energies from part b through order ${\epsilon }^2$.

Now put $E_2=E_3=E$ and $E_1\ne E$. All diagonal matrix elements of $V$ vanish. The leading coupling is $⟨1|V|2⟩=\epsilon V_0$, whereas $⟨1|V|3⟩={\epsilon }^2{V}_0$. Applying the second-order formula givesThe degeneracy causes no singular denominator through this order because $⟨2|V|3⟩=0$. Equivalently, the effective perturbation on the degenerate subspace begins asIts off-diagonal mixing is only order ${\epsilon }^3$, so degenerate perturbation theory is unnecessary for energies through order ${\epsilon }^2$.

Put $E_1=E_2=E$. Within the degenerate eigenspace spanned by $|1⟩,|2⟩$, the perturbation isDegenerate perturbation theory diagonalizes this matrix. Its normalized eigenstates $(|1⟩\pm |2⟩)/\sqrt{2}$ have first-order shifts $\pm \epsilon V_0$. The coupling to $|3⟩$ has order ${\epsilon }^2$, so it cannot affect an energy through order $\epsilon$. Therefore