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Yes. The differential operator $\mathcal{L}=2y{\partial }_x-x{\partial }_y$ is linear, even though its coefficients depend on position. For constants $A,B$,If each homogeneous right-hand side is zero, so is their combination. Hence the superposition principle holds.

Yes. The wave equation is linear and homogeneous. For $\mathcal{W}={\partial }_x^2-c^{-2}{\partial }_t^2$ and constants $A,B$,If both terms on the right vanish, their combination is also a solution. Thus the superposition principle applies, and it also preserves homogeneous fixed-end boundary conditions.

The superposition principle gives the general solution as $y_c+2y_{p,2}+y_{p,1}$:The first boundary condition gives $A=2/(3\sqrt{3})-1/3$. At $x=\pi$, the cosine vanishes and the sine equals $-1$. Comparing with the second boundary condition then givesThe unique solution satisfying both boundary conditions is thereforeUniqueness follows because a homogeneous solution zero at both endpoints would have $A=0$ from the first and $B=0$ from the second.

For a separated nonzero solution $T=f(x)g(y)$, substitution and division by $fg$ givesThe homogeneous Dirichlet boundary conditions impose $g(0)=g(1)=0$. For a nonzero spatial factor, integration by parts gives ${\lambda }^2{\int }_0^1{g}^{2}dy={\int }_0^1(g^{\prime }{)}^{2}dy$, so the separation value is nonnegative. The zero value gives only a trivial linear $g$ under these boundary conditions. The nontrivial separated factors, for $a\ne 0$, are thereforeFor the intended forward diffusion equation, assume $a>0$. The superposition principle gives a Fourier sine series, whose coefficients must expand the initial ramp in the interior:Consequently the standard heat-flow solution isIt satisfies both lateral boundary conditions for $x>0$, and tends to $y$ as $x\downarrow 0$ for $0<y<1$, also converging to the ramp in the L2 norm. The exponential coefficient decay justifies differentiating the series for positive $x$.

There is a genuine incompatibility in the printed data: at $(0,1)$ the lateral condition demands zero while the initial condition demands one. Thus no function satisfies every stated value at that corner. The displayed series solves the usual consistent interpretation with initial data on $0<y<1$ and lateral data for $x>0$; it cannot be continuous at the upper initial corner. This discrepancy is present in the original PDF.

The sign of $a$ is also unstated. If $a=0$, the spatial equation with two zero endpoints forces $T=0$, incompatible with the ramp. If $a<0$, the separated Fourier coefficients grow like $e^{n^2{\pi }^{2}x/|a|}/n$ and fail to be square summable for every $x>0$; these data therefore have no forward solution in the standard L2 space class. The positive-$a$ assumption is essential to the stated heat-flow interpretation.

For $t>0$, differentiate the printed ansatz:Matching the independent coefficients in $h_t=(h{h}_x{)}_x-1$ givesThere are two parameter pairs, including the degenerate spatially constant case:Their corresponding solutions areThe request for “the general form” cannot mean the complete solution set of this nonlinear partial differential equation: substitution of one ansatz cannot provide arbitrary initial data, and the superposition principle does not apply.

A useful wider family follows by allowing an independent time-dependent constant term and quadratic coefficient, $h=A(t)+B(t)x^2$. Coefficient comparison gives $B^{\prime }=6B^2$ and $A^{\prime }=2AB-1$. For $B\ne 0$, write $\tau =t-t_0>0$ to obtain $B=-1/(6\tau )$ and $A=C{\tau }^{-1/3}-3\tau /4$. Spatial translation, also a symmetry of the equation, therefore givesHere $C,x_0,t_0$ are arbitrary constants; the displayed nontrivial ansatz corresponds to $C=1/6$ and $x_0=t_0=0$. There are also affine solutions $h=px+(p^2-1)t+q$, including the spatially constant branch. These are valid solution families, rather than a claim of a full general solution.

The wave equation is a homogeneous linear partial differential equation. Consequently, if ${\psi }_1$ and ${\psi }_2$ solve it, every constant linear combination $A{\psi }_1+B{\psi }_2$ also solves it. This is the superposition principle, because partial differentiation distributes over the linear combination.

The same statement holds for the homogeneous Neumann boundary conditions. Initial conditions combine in the same way; an arbitrary combination need not retain a fixed nonzero initial displacement.

A causal Green function vanishes before the impulse. For $t>\tau$ it solves the homogeneous linear ordinary differential equation, whose characteristic roots are $-q\pm 2i$. Thus it has the formContinuity at $t=\tau$ is necessary to avoid a delta derivative in $G_{tt}$, so $A=0$. Integrating its forced equation across the impulse gives the jump condition for an impulse, $G_t({\tau }^{+},\tau )-G_t({\tau }^{-},\tau )=1$. The derivative from below is zero and the derivative from above is $2B$, givingHere $\Theta$ is the Heaviside step function. The zero initial data exclude any additional homogeneous response. By the superposition principle, the output for a general locally integrable input isThe upper endpoint causes no unwanted term in the first derivative because $G(t,t)=0$; in the second derivative its unit derivative jump reproduces the input.

Apply separation of variables with $\rho (x,t)=X(x)T(t)$ and temporal factor $T=e^{-\lambda t}$. The spatial eigenvalue equation isThe nontrivial solutions of these Dirichlet boundary conditions are sine eigenfunctions. Normalizing in the usual inner product on $(0,L)$ giveswith ${\int }_0^L{\psi }_n{\psi }_{m}dx={\delta }_{nm}$. The superposition principle and the Fourier sine series then giveThe constants are the initial-data coefficients $a_n={\int }_0^L\rho (x,0){\psi }_n(x)dx$. Here physical diffusion means $D>0$.

For separation of variables, take $n=T(t)X(x)Y(y)Z(z)$ and divide the reaction-diffusion equation by this product wherever it is nonzero:Each spatial term depends on a different variable and must be constant. Write $X^{\prime \prime }=-k_x^{2}X$ and similarly for the other two factors. The Dirichlet boundary conditions give $X(0)=X(L)=0$. A zero separation constant permits only a linear function and hence the zero solution; a positive value of $X^{\prime \prime }/X$ likewise permits no nonzero function vanishing at both endpoints. The nontrivial choices are thereforeThe temporal ordinary differential equation isso, absorbing its constant multiplier into $a_{p,q,r}$,The superposition principle and completeness of the Fourier sine series give the stated threefold expansion with these factors. For an initial density $n_0$, orthogonality also determines the constants:

Set $t=y-1$. Then $y^2-1=t(t+2)$, and a Frobenius solution has the formIn the differential equation, the lowest power is $t^{\sigma -1}$, whose coefficient isThe indicial equation and its indicial roots are consequentlyTheir difference is not an integer, so the Frobenius method produces two independent series without a logarithmic term:Choose a branch of $(y-1{)}^{1/2}$, or take $y>1$ for real solutions. A general local solution is $C_1{x}_1+C_2{x}_2$ by linear independence and the superposition principle. The analytic factors converge at least for $|y-1|<2$, up to the next singular point at $y=-1$.

Take the electric potential to vanish at infinity. The distance from the centre to each point charge is half the diagonal of a polygon of the square, namely $a/\sqrt{2}$. The superposition principle for the electric potential therefore gives, for four charges $+q$,Strictly, specifying only the magnitude of each electric charge leaves its sign undetermined. For signed charges $q_j$, the general answer is $V=\sqrt{2}{\sum }_{j=1}^4{q}_j/(4\pi {ϵ}_{0}a)$. For example, two positive and two negative charges of equal magnitude give zero electric potential at the centre. The boxed result uses the usual interpretation that all four charges have the same positive sign.

Use the geometric cross-section shown in the source: the outer cylinder has radius $2a$ and centre $O$, and the circular cavity has radius $a$ and centre $C=a\hat{\mathbf{x}}$. Let the electric current be in the $+z$ direction. The conducting cross-sectional area is $3\pi a^2$, so the uniform electric current density is $\mathbf{J}=I\hat{\mathbf{z}}/(3\pi a^2)$.

To obtain the exterior magnetic field of a wire with an eccentric cylindrical cavity, apply the superposition principle to a full cylinder of electric current density $\mathbf{J}$ centred at $O$ and a second cylinder of electric current density $-\mathbf{J}$ centred at $C$. Their electric currents are $4I/3$ and $-I/3$. This reproduces the actual electric current density everywhere, including zero in the cavity. Outside a long circular cylinder, Ampère's law gives $\mathbf{B}={\mu }_0{I}_{\mathrm{cyl}}\hat{\mathbf{z}}\times \mathbf{r}/(2\pi r^2)$. Every point outside the outer boundary is outside both component cylinders. Consequently, writing $\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}$,In Cartesian coordinates, the same magnetic flux density isThe cavity breaks rotational symmetry. Thus the magnetic flux density is generally neither constant in magnitude nor tangential on a circle centred at $O$; replacing its line integral by $2\pi rB$ would be unjustified. Its line integral is nevertheless ${\mu }_{0}I$, as Ampère's law requires. At distances $r\gg a$, the leading term becomes ${\mu }_{0}I\hat{\mathbf{\varphi }}/(2\pi r)$, with corrections of order ${\mu }_{0}Ia/r^2$.

Use Fraunhofer diffraction, appropriate to observation in the far field or in the focal plane of a lens, and uniform illumination across the slit. At observation angle $\theta$, points separated by transverse coordinate $u$ acquire a phase difference $ku\sin \theta$, where $k=2\pi /\lambda$ is the wavenumber. The superposition principle for the coherent wave amplitudes givesThe intensity is proportional to the squared modulus of the wave amplitude, soThe limit at $\beta =0$ is one. The noncentral zeros satisfy $w\sin \theta =m\lambda$ for nonzero integers $m$ for which the angle is physically allowed. In the Rayleigh criterion, two mutually incoherent point images are just resolved when the central maximum of one diffraction pattern lies at the first minimum of the other. For single-slit diffraction this givesThe exact statement here takes one image on the slit normal; for nearby paraxial images the angular resolution is $\lambda /w$. The first propagating minimum exists only for $w\ge \lambda$. The numerical factor $1.22$ belongs to a circular aperture and does not apply to this slit.

The electric dipole moment is $\mathbf{p}=2aq\hat{\mathbf{z}}$. By the superposition principle, the exact electric potential isFor $a/r\ll 1$, the binomial expansion gives each reciprocal distance as $r^{-1}(1\pm (a/r)\cos \theta +O(a^2/r^2))$. The leading $1/r$ terms cancel because the total electric charge is zero. The next even terms also cancel between the two signs. ConsequentlyThis is the leading electric dipole term of the multipole expansion; it describes the far field, not the electric potential arbitrarily close to either point charge.

Take $\mathbf{d}$ from the negative to the positive point charge. Their electric dipole moment is $\mathbf{p}=q\mathbf{d}$. Place them at $\pm \mathbf{d}/2$ and use the superposition principle for the electric potential:For $s\gg d$, the first-order Taylor expansion of the reciprocal distances givesto leading electric dipole order. In spherical coordinates aligned with $\mathbf{p}$, the gradient gives $E_s=-{\partial }_{s}V$, $E_{\theta }=-s^{-1}{\partial }_{\theta }V$ and $E_{\varphi }=0$. ThusEquivalently, the electric dipole field is $[3(\mathbf{p}\cdot \hat{\mathbf{s}})\hat{\mathbf{s}}-\mathbf{p}]/(4\pi {ϵ}_0{s}^3)$. The omitted terms vanish relative to these leading expressions as $d/s\to 0$.

Interpret the coordinates as a plane in ordinary three-dimensional electrostatics. Set $k_e=1/(4\pi {ϵ}_0)$, where ${ϵ}_0$ is the vacuum permittivity. The work required to bring the second point charge in quasistatically is its electric charge times the first charge's electric potential:The individual point-charge self-energies are excluded; this is the electrostatic potential energy of point charges relative to infinite separation.

By the superposition principle, for $\mathbf{r}=(x,y)$ and $\mathbf{d}=(d,0)$ the electric field isAt the first position both contributions point toward positive $x$, whereas at the second they oppose one another. ThusAt $(d/2,d)$ the two distances are $\sqrt{5}d/2$. The vertical electric field components cancel and the horizontal components add:Dimensional convention. The source's description of space as two-dimensional cannot literally mean two-dimensional electrostatics with the same inverse-square formula. In a genuinely two-dimensional Gauss's law, the circumference grows as $r$, giving an electric field proportional to $1/r$ and a logarithmic electric potential. For example, integrating $1/r$ from $d$ to infinity diverges, so the same finite reference potential energy at infinite separation does not exist. The displayed results use the conventional three-dimensional Coulomb's law restricted to the stated plane.

Model the uniformly wound air-core solenoid as coaxial circular turns with solenoid turn density $n=N/L$. Put its ends at $x^{\prime }=0,L$ and an internal axial point at $x\in [0,L]$. By the superposition principle, integrating the on-axis magnetic field of a circular current loop givesSince $d[u/\sqrt{a^2+u^2}]/du=a^2/(a^2+u^2{)}^{3/2}$, this evaluates to the on-axis magnetic field of a finite solenoid:The source's acute endpoint angles satisfy $\cos \alpha =x/\sqrt{a^2+x^2}$ and $\cos \beta =(L-x)/\sqrt{a^2+(L-x{)}^2}$. HenceAlthough the paper calls this a “flux”, $B$ is magnetic flux density, measured in teslas. Magnetic flux is an surface integral ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{A}$, measured in webers, and cannot be assigned to a point. For example, two differently sized small disks at the same point have the same limiting $B$ but different magnetic fluxes, approximately $B$ times their normal areas. The displayed plus sign uses the internal-point angle convention shown in the source. Outside the coil, the integral expression still works, but one must retain signed axial projections rather than adding two acute-angle cosines.

In vacuum, Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:Take opposite electric charges $\pm Q$ on the two electrical conductors, with surface charge densities $\pm \sigma =\pm Q/A$. A Gaussian pillbox crossing one isolated large sheet gives $2ES=\sigma S/{ϵ}_0$, so each sheet contributes $\sigma /(2{ϵ}_0)$ on either side. By the superposition principle, the electric fields add between the sheets and cancel outside. Ignoring the fringing field of a capacitor, the gap electric field is $E=\sigma /{ϵ}_0$. The electric potential difference and capacitance of this parallel-plate capacitor are consequentlyThe approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with permittivity $ϵ$ replaces ${ϵ}_0$ by $ϵ$.

In SI units, the Maxwell equations in vacuum with prescribed charge density $\rho$ and current density $\mathbf{J}$ areWhen $\rho =0$ and $\mathbf{J}=0$, take the curl of the third equation and substitute the fourth. Using the curl of the curl identity and $\nabla \cdot \mathbf{E}=0$ gives the electromagnetic wave equation:For example, $\mathbf{E}=E_0\hat{\mathbf{x}}\cos (kz-\omega t)$, $\mathbf{B}=(E_0/c)\hat{\mathbf{y}}\cos (kz-\omega t)$ is a plane electromagnetic wave provided $\omega =ck$. The electric field, magnetic field and propagation direction are mutually perpendicular.

The Maxwell equations are linear, so the superposition principle applies to fields. However, the Poynting vector is quadratic:One may add the average wave intensities only when the cross terms vanish under the actual detector average. For co-propagating plane electromagnetic waves of parallel linear polarization, averaging over optical cycles but potentially resolving the difference frequency givesIf the detector integrates for $T$, the last term acquires $\mathrm{sinc}[({\omega }_1-{\omega }_2)T/2]$, with sinc function convention $\mathrm{sinc}x=\sin x/x$, evaluated at the midpoint phase. Thus $T\gg 1/|f_1-f_2|$ suppresses beats, and the infinite-time average adds the wave intensities exactly for distinct fixed frequencies. Random mutually uncorrelated wave phases also remove the cross terms in an ensemble average; orthogonal linear polarizations remove them for co-propagating waves in polarization-insensitive detection. Merely having different frequencies does not remove a resolved beat signal: two nearly equal frequencies with fixed relative wave phase are a concrete counterexample. For different propagation directions, the relevant addition is of average Poynting vectors, or their components through the chosen detector surface.

Let the original cylinder axis be $\mathbf{r}=0$, and let the hole axis be at transverse vector $\mathbf{a}$, with $|\mathbf{a}|=a$. The charge density is reproduced by a full cylinder of radius $R_1$ and charge density $+\rho$, superposed with a cylinder of radius $R_2$ centred at $\mathbf{a}$ and charge density $-\rho$. This uses the superposition principle for fixed insulating electric charges; the hole is not a hole in an electrical conductor.

For a complete uniformly charged infinite cylinder of radius $R$, a coaxial Gaussian surface of length $\ell$ gives, by Gauss's law,Consequently defineThe complete electric field, including the exterior if desired, isIn the insulating material, $|\mathbf{r}|<R_1$ and $s=|\mathbf{r}-\mathbf{a}|>R_2$, soInside the hole both contributions use their interior formulas, and all dependence on $\mathbf{r}$ cancels:The electric field in the hole is uniform, points from the original axis towards the hole axis for $\rho >0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed surface charge density there. Its uniformity is the useful uniform electric field in an off-axis cylindrical cavity result.