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For planar parabolic coordinates and $g(u,v)=f(x(u,v),y(u,v))$, the multivariable chain rule gives $g_u=u{f}_x+v{f}_y$ and $g_v=-v{f}_x+u{f}_y$. Differentiating once more and adding cancels the first partial derivatives and the mixed partial derivatives:Consequently, on every local coordinate patch away from the origin, $f$ is a harmonic function if and only if $g$ is a harmonic function. The Laplacian changes by a positive scale factor, so solutions of the Laplace equation are preserved. The restriction away from the origin is needed to divide by the Jacobian determinant.

Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives $(\partial S/\partial P{)}_{T,N}=-\alpha V$. ^[ref]

For a simple compressible system, hold particle number $N$ and composition fixed, and allow only pressure-volume work. Write $U$ for internal energy, $H=U+PV$ for enthalpy, $F=U-TS$ for Helmholtz free energy, and $G=H-TS$ for Gibbs free energy. Their natural-variable differentials areThus the first partial derivatives, with the subscript recording the variables kept fixed, areThe symmetry of second derivatives produces all four standard Maxwell relations, including their signs:For a brief derivation of the $F$ relation, differentiate $F_T=-S$ with respect to $V$ and $F_V=-P$ with respect to $T$: $-S_V=F_{TV}=F_{VT}=-P_T$. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-$N$ constraints must remain explicit. ^[ref]

For an ideal gas with fixed $N$, the ideal gas law gives $P=N{k}_{B}T/V$. At fixed volume, heating changes the pressure at rate $N{k}_B/V$. Consequently an isothermal expansion at positive temperature hasFor one mole with $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $V_2=2V_1$, this is $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. At $300\mathrm{K}$, a reversible isothermal expansion absorbs $Q=T\Delta S\simeq 1728.85\mathrm{J}$ and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same $\Delta S$ in an insulated free expansion, although there $Q=0$ and the process is irreversible.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure, $dV=\alpha VdT$, soThus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas, $\alpha =1/T$ and ${\kappa }_T=1/P$ reduce the result to $C_P-C_V=PV/T=N{k}_B$, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also givesIntegrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently $C_V$ and $C_P$ depend only on temperature at fixed $N$, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern $G$ relation gives $\delta Q_{\mathrm{rev}}=TdS=-T\alpha VdP$. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^{[ref] [ref]}

The mixed partial derivative $f_{xy}$ measures how the slope $f_y$ changes as $x$ varies while $y$ is fixed:when this limit exists. For a small rectangle, the cross-difference $f(x+h,y+k)-f(x+h,y)-f(x,y+k)+f(x,y)$ is approximately $hk{f}_{xy}(x,y)$ if the second derivatives are continuous. It measures a change of transverse slope, rather than either first slope alone. Mere continuous differentiability does not ensure that this second derivative exists; equality with $f_{yx}$ follows under the usual hypotheses of Clairaut's theorem.

For an exact differential $Mdx+Ndy$, equality of the mixed partial derivatives requires $M_y=N_x$. Here these are $\sin x+b\cos y$ and $\sin x+3\cos y$, so $\boxed{b=3}$. Integrating $M$ with respect to $x$ gives $f=-y\cos x+3x\sin y+g(y)$. Comparison with $N$ gives $g^{\prime }=y^2$, and the specified value fixes the integration constant. ThereforeDifferentiating recovers both coefficients, and $f(0,2)=-2+8/3-2/3=0$. The explicit potential function also proves exactness globally, without relying only on a local compatibility test.

Assume $v\ne 0$ so that the change of variables is invertible. The chain rule gives ${\partial }_x={\partial }_r+{\partial }_s$ and ${\partial }_t=v({\partial }_r-{\partial }_s)$. Commuting the mixed partial derivatives for a twice continuously differentiable solution givesThus the wave equation is $F_{rs}=0$. Integrating once says that $F_r$ depends only on $r$; integrating again givesThe two arbitrary twice differentiable functions are waves moving in opposite directions at speed $|v|$. This is D'Alembert's formula, on a product domain in $(r,s)$ or globally on the usual full line.

Integrating the mixed partial derivative successively gives $z=x^3{y}^2/6+A(x)+B(y)$. The first boundary value gives $A(x)=x^2-1-B(0)$. Substitution at $x=1$ then gives $B(y)-B(0)=\sin y-y^2/6$. ThereforeIts mixed derivative is $x^{2}y$, and direct evaluation checks both boundary values. Their common corner is compatible: both prescribe $z(1,0)=0$.

Set $r=\sqrt{x^2+y^2}$. The bound $|xy|\le r^2/2$ and $|x^2-y^2|\le r^2$ give $|f(x,y)|\le r^2/2$. Hence $f\to 0$ at the origin, proving continuity, and $|f(x,y)|/r\to 0$, proving Fréchet differentiability there with derivative equal to the zero linear map.

Along the coordinate axes, direct difference quotients giveThese formulas include the origin. Therefore the two mixed partial derivatives differ:The explicit operator order removes any ambiguity about the notation for mixed derivatives. Also $f_{xx}(0,0)=f_{yy}(0,0)=0$, since the first derivatives restricted to their own axes vanish identically.

All four second-order partial derivatives are discontinuous at the origin. Away from it,Along $x=y\ne 0$ these are one and minus one, rather than their zero values at the origin. Meanwhile ${\partial }_y{\partial }_{x}f(x,0)=1$ for $x\ne 0$, whereas its value at the origin is minus one; and ${\partial }_x{\partial }_{y}f(0,y)=-1$ for $y\ne 0$, whereas its origin value is one. This gives a concrete unequal mixed partial derivatives example despite differentiability of the original function.

The canonical anticommutation relations realize the internal states as a fermionic Fock space ${\Lambda }^{\bullet }{\mathbb{C}}^3$. Choosing a Clifford vacuum $|0⟩$ annihilated by all ${\psi }_a$, the states are functions multiplying $|0⟩$, ${\psi }_a^{†}|0⟩$, ${\psi }_a^{†}{\psi }_b^{†}|0⟩$ with $a<b$, or ${\psi }_1^{†}{\psi }_2^{†}{\psi }_3^{†}|0⟩$. Thus the four fermion-number sectors in supersymmetric quantum mechanics arewith internal dimensions $1,3,3,1$. The fermion number operator $N={\sum }_a{\psi }_a^{†}{\psi }_a$ has value $n$ on ${\mathcal{H}}_n$. The sectors $n=0,2$ have even fermion parity, and $n=1,3$ have odd fermion parity.

Write $q={\sum }_a{\psi }_a^{†}{A}_a$, so that the supercharge is $Q=q+q^{†}$. The commutator $[A_a,A_b]$ is zero because the two mixed partial derivatives of $\chi$ agree. The canonical anticommutation relations then imply $q^2=(q^{†}{)}^2=0$ andThe Hamiltonian operator preserves each sector, while $q$ raises and $q^{†}$ lowers its degree by one. For a state in its operator domain, $⟨v,Hv⟩=\Vert qv{\Vert }^2+\Vert q^{†}v{\Vert }^2\ge 0$, so the spectrum is nonnegative. If $Hv=Ev$ with $E>0$, then $Qv/\sqrt{E}$ is a state of the same energy and opposite fermion parity, and $Q(Qv/\sqrt{E})=\sqrt{E}v$. This is an odd involution pairing bosonic and fermionic states and proves equal even and odd multiplicities at every positive eigenvalue.

There is also a more precise adjacent-sector pairing. On the positive-energy eigenspace,The two summands are orthogonal since $q^2=(q^{†}{)}^2=0$. A state $u$ in the image of $q^{†}$ obeys $q^{†}u=0$, $q^{†}qu=Eu$, so $q/\sqrt{E}$ maps it to a normalized partner in the next sector, with inverse $q^{†}/\sqrt{E}$. Thus positive-energy doublets lie in ${\mathcal{H}}_0\oplus {\mathcal{H}}_1$, ${\mathcal{H}}_1\oplus {\mathcal{H}}_2$, or ${\mathcal{H}}_2\oplus {\mathcal{H}}_3$. In particular, every positive-energy state of ${\mathcal{H}}_0$ has a partner in ${\mathcal{H}}_1$, and every one of ${\mathcal{H}}_3$ has a partner in ${\mathcal{H}}_2$; the remaining middle-sector states pair with each other. There is no requirement that all four individual sector spectra coincide. Zero-energy states satisfy $qv=q^{†}v=Qv=0$ and can be unpaired. No confinement assumption was given: a continuous spectrum is possible, with the same even/odd pairing on its positive spectral subspaces.

For real $\chi$, the adjoint operator is $A_a^{†}=-{\partial }_a-{\partial }_a\chi$. In ${\mathcal{H}}_0$, every ${\psi }_a^{†}{\psi }_b$ vanishes; in the one-dimensional internal space of ${\mathcal{H}}_3$, it equals ${\delta }_{ab}$. The two scalar Hamiltonian operators are consequentlyThese are the extreme sectors of the gradient superpotential Hamiltonian in supersymmetric quantum mechanics.

For radial $\chi$ and a normalized state $v_0=f(r)|0⟩$ with $E>0$, $q^{†}{v}_0=0$ and $A_{a}f=(x_a/r)(f^{\prime }-{\chi }^{\prime }f)$. The radial superpartner in three-dimensional supersymmetric quantum mechanics therefore has the simple formIt belongs to ${\mathcal{H}}_1$; $H{v}_1=E{v}_1$ follows from $[H,Q]=0$, and its wavefunction normalization follows from $\Vert Q{v}_0{\Vert }^2=⟨v_0,H{v}_0⟩=E$. Also $Q{v}_1=\sqrt{E}{v}_0$. At the origin the formula is interpreted by regular extension; its degree-one components carry the radial unit vector rather than being three independent radial scalars.

For a zero mode in the fermion-vacuum sector, the factorization of $H_0$ gives $0={\sum }_a\Vert A_{a}f{\Vert }^2$. Hence ${\partial }_{a}f=({\partial }_a\chi )f$, or ${\partial }_a(e^{-\chi }f)=0$. The requested wavefunction isA nonzero normalizable wavefunction exists only if this integral is finite. The positive sign in the exponential is fixed by the given convention $A_a={\partial }_a-{\partial }_a\chi$.

For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, the necessary and sufficient condition for a conservative vector field is $\nabla \times \mathbf{F}=0$. Necessity follows from equality of mixed partial derivatives of a potential; sufficiency uses the ${\mathbb{R}}^3$ being a simply connected space. On a general domain the topology cannot be omitted.

Here the relevant mixed partial derivatives areThus the curl vanishes. Integrating the first component in $x$ gives $\Phi =x^2\cos y-2x{z}^3+A(y,z)$. Matching the second component gives $A_y=3+2y{e}^z$, hence $A=3y+y^2{e}^z+B(z)$. Matching the last component forces $B^{\prime }=0$. A potential of a conservative vector field with the convention $\mathbf{F}=\nabla \Phi$ is consequentlyIf a physical potential is defined instead through $\mathbf{F}=-\nabla V$, it is $V=-\Phi$.

Fix $(x,y)\in U$ and a sufficiently small rectangle inside $U$. Its rectangular incrementcan be evaluated twice using the fundamental theorem of calculus, givingDivide by $hk$ and let $h,k\to 0$. Continuity of the mixed partial derivatives makes the two limits $D_2{D}_{1}f(x,y)$ and $D_1{D}_{2}f(x,y)$. They are equal everywhere on $U$.

Repeated interchange of adjacent partial derivatives reduces every order-$m$ derivative of a smooth function to $D_1^j{D}_2^{m-j}f$, $0\le j\le m$. Hence there are at most $m+1$ distinct functions. This maximum is attained by $f(x,y)=e^{x+2y}$, whose listed partial derivatives are the distinct functions $2^{m-j}{e}^{x+2y}$. The answer is $m+1$, rather than the $2^m$ possible written orders.

For the supplied $f$, $f(t,t)=1/2$ for $t\ne 0$, whereas $f(0,0)=0$. It is not even continuous at the origin, so $f$ is neither differentiable nor infinitely differentiable there.

For $g$, $|g(x,y)|\le |xy|\le (x^2+y^2)/2$. Thus $g(x,y)=o(\sqrt{x^2+y^2})$, proving Fréchet differentiability at the origin with derivative zero. However,including zero at the origin. Therefore $D_2{D}_{1}g(0,0)=-1$ and $D_1{D}_{2}g(0,0)=1$. $g$ is differentiable but not infinitely differentiable at the origin: it cannot have continuous second partial derivatives near that point.

Use the inverse change of variables $u=(x+y)/2$ and $v=(x-y)/2$. The chain rule gives the differential operatorsThe mixed partial derivatives cancel, assuming the twice continuously differentiable solution appropriate here. Consequently $f_{xy}=(g_{uu}-g_{vv})/4$.

The transformed equation isThe factors of $1/2$ belong to the inverse Jacobian matrix; using the forward transformation instead would give an incorrect factor.

The multivariable chain rule first givesTaking a second partial derivative in each variable, with the partial derivatives of $f$ evaluated at $(x(u,v),y(u,v))$, givesThe mixed partial derivatives cancel even without using their equality. Hence the Laplacians satisfyAway from the origin the factor $u^2+v^2$ is positive, so $g$ is a harmonic function if and only if $f$ is a harmonic function on the corresponding local coordinate patch. This is the Laplacian in planar parabolic coordinates.

If $Pdx+Qdy=dF$ for a twice continuously differentiable potential $F$, then $P=F_x$ and $Q=F_y$. Equality of mixed partial derivatives gives the necessary condition for an exact differential:This condition alone does not ensure global exactness on a domain with holes.

For the Helmholtz free energy $A=U-TS$, use the stated identity for internal energy and the product rule:Thus $A_T{|}_V=-S$ and $A_V{|}_T=-p$. Equality of mixed partial derivatives yields the Maxwell relationThe temperature and volume are the independent variables here, with composition and amount of gas fixed.

Next express $S=S(T,V)$ and substitute its total differential into $dU=TdS-pdV$:Matching the coefficient of $dV$ givesThis is the thermodynamic internal pressure identity: an equation of state determines how internal energy changes under an isothermal change of volume.

For the Dieterici equation, set $c=an/(VR)$ while differentiating at fixed $V$. ThenSubstitute into the thermodynamic internal pressure identity. The term proportional to $T$ cancels $p$, leavingThe physical gas domain has $T>0$ and $V>nb$.

For the ideal gas instead, $p=nRT/V$ gives $T(\partial p/\partial T{)}_V-p=0$. Therefore $U$ is independent of volume at fixed amount of gas. Integrating the remaining condition $(\partial U/\partial T{)}_V=C_V$, with constant heat capacity at constant volume, giveswhere $U_0$ is independent of both $T$ and $V$ and fixes the reference for internal energy.

Differentiate $v_t=y{u}_x-x{u}_y$ with respect to $t$, remembering that its coefficients also depend on $t$. The product rule and multivariable chain rule giveFor a twice continuously differentiable $u$, equality of mixed partial derivatives gives $u_{yx}=u_{xy}$. HenceThe first-order terms come from differentiating $x$ and $y$ themselves; treating $y{\partial }_x-x{\partial }_y$ as an operator with constant coefficients would miss them.

At fixed particle number, the differential of the internal energy identifies $T=(\partial U/\partial S{)}_V$ and $p=-(\partial U/\partial V{)}_S$. Equality of continuous mixed partial derivatives gives the first Maxwell relation:For the enthalpy, $dH=dU+pdV+Vdp=TdS+Vdp$. Thus $T=(\partial H/\partial S{)}_p$, $V=(\partial H/\partial p{)}_S$, and the same argument gives the enthalpy Maxwell relationThe smoothness and independent thermodynamic coordinates are understood within a single phase.

First $f_x=2x{y}^3$ and $f_y=3x^2{y}^2$. Differentiating again gives the four second-order partial derivatives:The two mixed partial derivatives agree, as expected for this smooth polynomial.

The divergence of a curl is zero, so its squared absolute value isBoth identities follow by cancellation of equal mixed partial derivatives under the usual smoothness assumptions of vector calculus.

The relevant mixed partial derivatives agree:The vector field is smooth on the entire simply connected plane, so this condition establishes that it is a conservative vector field. We can also construct its vector-field potential explicitly. From ${\varphi }_x=y^2-2x$, take an antiderivative in $x$:Then ${\varphi }_y=2xy+h^{\prime }(y)=2xy-y$ requires $h^{\prime }(y)=-y$. HenceThe fundamental theorem for line integrals checks the previous answer: $\varphi (1,1)-\varphi (0,0)=-1/2$.

Harmonic parts. Write the holomorphic function as $f=u+iv$. The Cauchy-Riemann equations say $u_x=v_y$ and $u_y=-v_x$. Equality of mixed partial derivatives therefore givesThus both the real and imaginary parts are harmonic functions and satisfy two-dimensional Laplace's equation.

Single-valued logarithm. A convenient choice isusing the principal logarithm for each term. Its branch cuts in the $z$ plane are $(-\infty ,-1]$ and $[1,\infty )$. The branch points are $-1$ and $1$: going once around the first adds $2\pi i$, while going once around the second subtracts $2\pi i$. Removing the two exterior rays makes this a consistent branch of a multivalued function and in particular leaves the whole open unit disk available.

Potential inside the cylinder. Put $z=r{e}^{i\theta }$ with $r<1$. Multiplication by $1-\overline{z}$ givesThe denominator is positive and the real part of the numerator is positive. Thus its argument lies strictly between $-\pi /2$ and $\pi /2$, andAs $r\uparrow 1$, this tends to $\pi /2$ on the upper open semicircle and $-\pi /2$ on the lower open semicircle. Scale the imaginary part to obtainIt is a harmonic function, has the required boundary values, and gives $\Phi (0)=0$. The two joining points of the boundary arcs have discontinuous boundary data, so no common continuous boundary value is demanded there. Along the real diameter, including radial approaches to either joining point, the displayed solution is zero. This is the disk potential with two semicircular boundary values.

For a second check, the Taylor series of the logarithms gives $f(z)=2{\sum }_{j\ge 0}{z}^{2j+1}/(2j+1)$. Taking the imaginary part yields the Fourier serieswhose boundary limit is exactly the required alternating pair of constant semicircle values.

Compute the complex derivative along horizontal and vertical increments. Along the real direction, $f^{\prime }=u_x+i{v}_x$; along the imaginary direction, $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equality gives the Cauchy-Riemann equationsThe real and imaginary parts of a holomorphic function are smooth. Equality of their mixed partial derivatives then gives $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$, proving that both are harmonic functions.

For the first real part, $u_x=2x$ and $u_y=-2y$. The Cauchy-Riemann equations require $v_y=2x$ and $v_x=2y$, so its harmonic conjugate is $v=2xy+C$. ThereforeFor the second real part, $u=x/(x^2+y^2)$ is the real part of $1/z$. Its harmonic conjugate is $-y/(x^2+y^2)+C$, givingThese are the most general answers on a connected open domain: the difference of two candidates has zero real part, so the Cauchy-Riemann equations make its imaginary part constant. The second function cannot extend holomorphically through zero.