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A constant added to a function changes its reference level without changing its derivative:Geometrically, it is a vertical translation of the graph. For $F(x)=x^3/3$, changing $C$ moves every plotted value by the same amount while preserving every tangent slope. This is the geometric reason that antiderivatives form a family.

When $F$ represents a physical quantity, the physical dimension of $C$ must match that of $F$. When position is obtained from velocity, $C$ is a position offset. An unknown starting position remains unknown even when the rate of change is known exactly.

When describing the family of antiderivatives on an interval, the arbitrary additive constant is the constant of integration. On a disconnected open domain, the corresponding freedom is a locally constant function, as in antiderivatives on disconnected domains, rather than necessarily one constant on the whole domain.

An antiderivative recovers a function from its local rate of change. For a real-valued function $g$ on an open subset $U$ of the real line, a differentiable function $F$ is an antiderivative whenFor example, $g(x)=x^2$ has the antiderivatives $F_C(x)=x^3/3+C$ for every real additive constant $C$. Their derivatives agree, while their values can differ. On an interval, the constant of integration describes the entire freedom in the answer. ^[ref]

The indefinite-integral notation $\int g(x)dx=F(x)+C$ records this family on an interval. For the illustrated family, the extra condition $F(1)=2$ selects $C=5/3$. Thus reversing differentiation requires both a rate and a reference value if a unique function is wanted.

The fundamental theorem of calculus supplies a construction when $g$ is continuous on an interval $I$. Choose $x_0\in I$ and set $F(x)={\int }_{x_0}^{x}g(t)dt$; then $F^{\prime }=g$. Integrability alone does not guarantee a classical antiderivative. The Heaviside step function is Riemann integrable on bounded intervals, but cannot be a derivative throughout an interval containing its jump: Darboux theorem for derivatives requires every intermediate value. ^[ref]

On a connected open interval, if $F^{\prime }=g$, every antiderivative of $g$ has the form $F+C$, where $C$ is a constant function. Indeed, two solutions satisfy $(G-F{)}^{\prime }=0$: the zero derivative on a connected open set principle makes their difference constant. In one dimension, the mean value theorem proves this on any closed subinterval. ^[ref]

An initial condition or boundary condition can determine $C$. If a particular antiderivative is $F_0$ and $F(x_0)=y_0$, thenFor $g(x)=x^2$, $F_0(x)=x^3/3$ and $F(1)=2$, this gives $C=5/3$. Without that reference value, choosing $C=0$ simply selects one member of the family.

The freedom has a physical meaning in a simple kinematic model. A velocity $v(t)=at$, with constant acceleration $a$, determines position only as $x(t)=a{t}^2/2+C$. The initial condition $x(0)=x_0$ selects $C=x_0$; changing the starting position does not change the prescribed velocity.

For the differential equation $T^{\prime }=-B/r^2$ on $r>0$, the antiderivative is $T=B/r+C$. The boundary condition $T(R_s)=T_s$ gives $C=T_s-B/R_s$. Discarding the constant of integration therefore needs a specified condition or an approximation that quantifies its effect. On a domain with several connected components, the freedom is larger: antiderivatives on disconnected domains require a separate constant on each component.

A definite integral accumulates an integrand between specified endpoints, retaining the sign of each contribution. For a real-valued function on a finite real interval, a finite definite integral is a scalar; an indefinite integral describes a family of antiderivatives. The lower and upper limits of integration $a,b$ specify where accumulation starts and ends. The integration variable is a placeholder: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(u)du$.

This is useful whenever a varying rate must be converted into a total change. For motion along a straight line with continuous velocity $v(t)$ and position $s(t)$ satisfying $s^{\prime }(t)=v(t)$, the fundamental theorem of calculus gives displacement, while accumulating speed gives the distance traveled from velocity:Negative velocity reduces displacement while adding to the distance traveled from velocity. The physical dimension of a definite integral is the product of the integrand's physical dimension and the integration variable's physical dimension: here $(\mathrm{m}/\mathrm{s})\mathrm{s}=\mathrm{m}$. An area on a rate graph need not be a spatial area.^[ref]

As an explicitly synthetic motion model, let $v(t)=(2\mathrm{m}{\mathrm{s}}^{-2})t-2\mathrm{m}{\mathrm{s}}^{-1}$ for $0\le t\le 3\mathrm{s}$, with $s(0)=0$. Using numerical values in seconds and metres, an antiderivative gives $s(t)=t^2-2t$. The velocity changes sign at $t=1$: the model first moves $1$ m in the negative direction, then $4$ m in the positive direction. Its displacement is $-1+4=3$ m, and its distance traveled from velocity is $1+4=5$ m. The initial position fixes the constant of integration, but does not change either accumulated quantity.

Orientation matters independently of whether the integrand is positive or negative. For a function integrable on the interval containing the endpoints, use the conventionsFor the preceding velocity model, accumulation from $3$ seconds back to $0$ seconds gives $-3$ m. Reversing the bounds of the speed integral gives $-5$ m, which is an oriented integral, not a negative distance traveled from velocity; physical elapsed-time totals use the earlier endpoint first. The splitting rule holds even when $b$ is outside the interval between $a$ and $c$, provided integrability holds on the interval containing all three endpoints.^[ref]

Existence needs hypotheses. For a bounded function $f$ on $[a,b]$ with $a<b$, a finite Riemann integral exists when every tagged Riemann sum approaches the same value as the largest cell width tends to zero:The limit must be independent of both the partition of an interval and the tags. Merely making the number of cells large can leave a wide cell unresolved. On a finite closed interval, every continuous function, and every bounded function that is continuous except at finitely many points, is Riemann integrable.^[ref] Boundedness alone is insufficient: the Dirichlet function on $[0,1]$ is $1$ on the rational numbers and $0$ on the irrational numbers. Tags can all be chosen rational or all irrational, giving Riemann sums of $1$ or $0$ at arbitrarily small cell widths.^[ref]

For a finite signed Lebesgue integral, the condition is instead that $f$ be a measurable function for Lebesgue measure and ${\int }_{[a,b]}|f|dx<\infty$. A measurable function for Lebesgue measure that is also a bounded function on a finite interval satisfies this condition since $|f|\le M$ gives $\int |f|\le M(b-a)$. The Dirichlet function therefore has Lebesgue integral zero: the rational numbers form a null set, so it equals zero Lebesgue almost everywhere. The Riemann integral and Lebesgue integral agree for the continuous functions used in the diagrams.^[ref] An unbounded integrand or infinite endpoint requires further analysis, for example the limits defining an improper integral; the notation alone does not establish convergence.

Integration by parts transfers a derivative from one factor of a product to the other, while keeping an endpoint contribution. Its usefulness comes from choosing which factor becomes simpler when differentiated, and which has an accessible antiderivative. The integral is then replaced by a boundary term minus another integral.

For real continuously differentiable functions $u,v$ on $[a,b]$, the precise definite integral formula isIn the shorter notation, $du=u^{\prime }(x)dx$ and $dv=v^{\prime }(x)dx$, giving $\int udv=uv-\int vdu$ for indefinite integrals, with a constant of integration understood. The product rule says $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$; the fundamental theorem of calculus turns its integral into the endpoint difference. Thus the formula packages a differentiation rule into a tool for integration.^[ref]

For example, choose $u=x$ and $dv=e^{x}dx$. Then $du=dx$ and $v=e^x$, since the exponential function is its own derivative:The boundary term is $e\approx 2.718282$; the remaining integral is $e-1\approx 1.718282$. Their difference is exactly one. The endpoint product alone would overcount, because its derivative includes $u^{\prime }v=e^x$ as well as the desired $u{v}^{\prime }=x{e}^x$.

In the figure, the area under $(x{e}^x{)}^{\prime }=(1+x)e^x$ is the endpoint contribution $e$. Its lower band has area $e-1$, and its upper band has thickness $x{e}^x$ and area one. This decomposition makes the subtraction visible. The endpoint contribution is a difference of product values; it need not equal the area under either factor separately.

The same choice repeatedly lowers the degree of a polynomial when integrating a polynomial times an exponential function, while choosing $u=\ln x$ gives $\int \ln xdx=x\ln x-x+C$ for $x>0$. More broadly, moving a derivative onto a convenient test function is the mechanism behind a weak derivative. Always evaluate the boundary term: it disappears only when the relevant endpoint values justify doing so, often through a boundary condition.

For an invertible fundamental matrix, differentiating $Y_{\lambda }=AY$ in $z$ and $Y_z=BY$ in $\lambda$ gives $Y_{\lambda z}=(A_z+AB)Y$ and $Y_{z\lambda }=(B_{\lambda }+BA)Y$. Thus the necessary and sufficient compatibility identity isHere both derivatives must use the actual deformation variable $z$; the printed $t$ has not been defined. The isomonodromic deformation identity can be checked as a polynomial in $\lambda$, without assuming the conclusion.

First retain the coefficients actually printed in the PDF. Write $p=u^{\prime }$, $d=w+z/2$, and introduce the two residualsDirect matrix multiplication and differentiation giveFor example, the quadratic lower entry is the difference between $[D,B_0{]}_{21}=2w/u$ and the contribution $(A_1{)}_{21}=-zw/u$; it does not cancel. The constant lower entry follows by differentiating $2u^{\prime }w/u^2$ and collecting it as $2wH/u^2+2u^{\prime }K/u^2$. These entries also provide independent checks of the two most important cancellations.

Work on an open connected deformation domain where $u\ne 0$, as the matrices require. The quadratic entry forces $(2-z)w=0$ throughout the domain, hence $w\equiv 0$ by continuity, including a possible point $z=2$. All remaining entries then vanish exactly when $u^{\prime \prime }+(z/2)u=0$. Thus the literal answer isThis is a rescaled Airy function equation: with $\xi =-2^{-1/3}z$, the equation is $u_{\xi \xi }-\xi u=0$, so $u=c_1\mathrm{Ai}(\xi )+c_2\mathrm{Bi}(\xi )$ on a domain avoiding its zeros. In particular, choose the local solution with $u(1)=u^{\prime }(1)=1$ and $w=0$. It gives compatible matrices but violates the printed $w=u^{\prime }$. This proves that the requested assertion is false for the original coefficients, independently of the undefined $t/z$ in the printed differential equation.

For completeness, use the explicitly corrected lower coefficient $(A_1{)}_{21}=-2w/u$ from the preceding solution. The same multiplication gives the much simpler residualConsequently compatibility is equivalent to $K=H=0$. The first equation integrates without dividing by $w$:The constant of integration $C$ therefore satisfies $w=C{u}^2$, and the second equation becomesConversely these two equations give $K=H=0$, and substitution makes every residual entry vanish, proving sufficiency as well as necessity. This is the quadratic spectral Lax pair for zero-parameter Painlevé II. For $C\ne 0$, choose constants $b,a$ with $b^3=-2$ and $C{a}^2{b}^2=-2$, set $z=bx$, $u=aq(x)$, and obtain $q^{\prime \prime }=2q^3+xq$, the zero-parameter Painlevé II equation. For $C=0$ it reduces to the Airy function equation. Even with this correction, the conclusions are $w=C{u}^2$ and the coefficient $z/2$, rather than the incompatible expressions printed in the question.

The Painlevé property says that the general solution has no movable critical singularities: analytic continuation around a singularity whose location depends on the initial data must not create a different branch. A fixed singularity of a complex differential equation has its location prescribed by the coefficients; a movable singularity of a complex differential equation has a location which varies with the initial data. A commonly used stronger form requires movable singularities to be poles, so that the solution is locally a meromorphic function there. Movable branch points are excluded; a movable pole is permitted. We will prove the stronger form for this equation by finding its complete general solution family as rational functions.

Use the linearization of a Riccati equation. If $w^{\prime }=a{w}^2+bw+c$, put $w=-v^{\prime }/(av)$ wherever $a,v\ne 0$. Differentiating this expression and cancelling the quadratic terms yieldsConversely a nonzero solution $v$ of this linear ordinary differential equation gives a solution $w$, wherever the quotient is defined. Every local finite solution $w$ at an ordinary point arises this way: solve $v^{\prime }=-awv$ with a nonzero initial value and differentiate to get the second-order equation. Multiplying $v$ by a nonzero constant leaves $w$ unchanged, giving the one effective constant of integration expected for a first-order Riccati equation.

For the present coefficients, $a^{\prime }/a=1/z+1/(z+1)$, so the linear ordinary differential equation simplifies toTwo direct substitutions give solutions $v_1=z-1$ and $v_2=1/(z+1)$. Their Wronskian iswhich is nonzero at every ordinary point $z\notin \{0,-1\}$. They therefore form a fundamental pair there, and all local solutions have the formTaking its logarithmic derivative givesThus, for $C_1\ne 0$, writing $k=(C_2-C_1)/C_1$, the general family isThe remaining member, corresponding to $C_1=0$, isThe completeness of the linearizing pair proves that no other local solution is omitted. These are rational functions, single-valued meromorphic functions; hence their continuation has no branch points. Their only possible moving singularities are zeros of $z^2+k$. A root $z_0$ outside $\{0,-1\}$ is simple, and its residue isEquivalently, poles of a Riccati solution from zeros of its linearizing solution are simple at ordinary points, because a nontrivial solution of a linear ordinary differential equation cannot have both $v(z_0)=0$ and $v^{\prime }(z_0)=0$. The exceptional collisions $k=0$ or $k=-1$ occur at the fixed coefficient singularities, and still give rational functions. Consequently every movable singularity is a pole, establishing the Painlevé property in its stronger meromorphic sense.

Take $x$ downslope and $z$ normal to the plane. To leading order in the lubrication approximation, Newtonian gravity gives the hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. The downslope Stokes equation is therefore $\mu u_{zz}=\rho g(h_x-\theta )$. Imposing the no-slip boundary condition at $z=0$ and the stress-free boundary condition at $z=h$ givesHere $\nu =\mu /\rho$ is the kinematic viscosity; $u$ is the downslope velocity. Applying mass conservation to this lubrication gravity-current flux,For typical thickness $H=\hat{h}$ and length $L=\hat{x}$, the transverse velocity is $O(UH/L)$ by incompressibility. Both convective terms are then $O(U^2/L)$, while transverse viscous acceleration is $O(\nu U/H^2)$. Thus the inertia criterion for an inclined viscous film is $U{H}^2/(\nu L)\ll 1$, in addition to $H/L\ll 1$. With evolution time $T\sim L/U$ the same condition controls unsteady inertia. If a faster time is independently imposed, also require $H^2/(\nu T)\ll 1$. Estimating $U$ from the velocity profile gives the two requested limits:

At long times, $h\sim A/x_N$ and the ratio of hydrostatic-gradient forcing to downslope Newtonian gravity in the bulk is $h_x/\theta =O(A/(\theta x_N^2))\ll 1$. Put $K=g\theta /(3\nu )$; the bulk conservation law is $h_t+(K{h}^3{)}_x=0$. The finite-volume gravity-driven thin-film current scales as $x\sim t^{1/3}$, $h\sim t^{-1/3}$. More explicitly, writing $h=t^{-1/3}f(\xi )$, $\xi =x/t^{1/3}$, givesThe constant of integration is zero for the zero-flux dry tail. ConsequentlyThe reduced bulk solution is zero outside this interval. Its thin nose region is resolved below. Integrating to impose the conserved area,The bulk front height is $h_N=3A/(2x_N)$, with $h_N^2=x_N/(3Kt)$. Differentiation of the front law givesEquivalently, this is the Rankine-Hugoniot condition for the bulk shock wave, whose speed is flux divided by height.

In moving coordinate $y=x-x_N(t)$, the full thin-film equation becomes $h_t{|}_y-{\dot{x}}_N{h}_y+q_y=0$. A nose thickness $h_N$ and slope $\theta$ give axial scale ${\ell }_N=h_N/\theta$. Since ${\ell }_N/x_N=O(A/(\theta x_N^2))\ll 1$, the explicit time derivative is smaller than translation by that ratio. The travelling front of a gravity-driven thin film is therefore quasisteady. Integrate once with dry-tip flux zero to obtain $q={\dot{x}}_{N}h$, orWith $f=h/h_N$, separation of variables and the convention $y=0$ at $f=0$ giveThe nose approaches $h_N$ exponentially far behind its tip, then falls to zero with a vertical tangent in this small-slope model. The requested sketch shows this monotone profile:

Expanding the inverse hyperbolic tangent, $f-\mathrm{atanh}f=-f^3/3+O(f^5)$, so $h\sim (-3\theta h_N^{2}y{)}^{1/3}$. Its slope has magnitude $\theta /f^2$ near the tip. The small-slope breakdown at an inclined-current nose occurs at $f=O(\sqrt{\theta })$, which givesFinally, surface tension produces a pressure-gradient scale $\gamma h_N/{\ell }_N^3$, whereas downslope Newtonian gravity gives $\rho g\theta$. Their ratio is $\gamma {\theta }^2/(\rho g{h}_N^2)$. Thus surface tension changes the principal nose length whenThis is a bulk-nose crossover; capillarity and contact-line physics can already affect a smaller region very close to the dry tip even when the bulk nose is Newtonian gravity dominated.

The Gaussian integral over the full vertical velocity line is $\sqrt{2\pi }{\sigma }_z$. Thus the mass density of the self-gravitating isothermal slab isNeglecting radial derivatives makes the Poisson equation for Newtonian gravity ${\Phi }_{zz}=4\pi G\rho (z)$. Substituting $\Phi ={\sigma }_z^2\varphi$ and $z=z_0\zeta$, with $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, givesMultiply by ${\varphi }^{\prime }$ and integrate. The central value and symmetry under $z\mapsto -z$ fix the constant of integration:On the positive-$\zeta$ side, Newtonian gravity makes ${\varphi }^{\prime }\ge 0$. Put $y=e^{\varphi /2}$; then $4(y^{\prime }{)}^2=y^2-1$ with $y(0)=1$. Integration gives $2\mathrm{arcosh}y=\zeta$, and even reflection supplies the other side. ThereforeDirect differentiation verifies both the equation and central conditions. Uniqueness of the smooth initial value problem rules out other branches meeting those conditions. The surface density is $4{\rho }_0{z}_0$, and the conventional slab scale height is $2z_0$; the factor of two is only a definition of the length scale.

Outside the emitting layer, take $m\simeq M_c$, $L_r\simeq L$ and $P=\mathcal{R}\rho T/\mu$. The Kramers' opacity law then reads $\kappa =({\kappa }_0\mu /\mathcal{R})P{T}^{-9/2}$. Division of the radiative diffusion in a star equation by the stellar hydrostatic equation givesWith the outer pressure constant of integration neglected, integration yieldsSubstitute this into hydrostatic equilibrium:Define $B=4\mu G{M}_c/(17\mathcal{R})$. The general antiderivative is $T=B/r+T_0$. The Kramers radiative-zero envelope around a stellar core sets $T_0=0$, givingThis is exact for the idealized boundary $T\to 0$ as $r\to \infty$. With a finite outer radius $R_s$ and temperature $T_s$, instead $T=T_s+B(1/r-1/R_s)$; the displayed profile is the deep-envelope approximation when $R_s\gg R_c$ and $|T_s-B/R_s|\ll B/R_c$. Small outer pressure alone does not eliminate this temperature constant of integration.

The specific hydrogen burning rate becomesConsequentlyIts local radial e-folding length is $R_c/20$, so burning is concentrated within a small fraction of the stellar core radius. The volume heating is even steeper: $\rho ϵ\propto C^2{T}^{93/4}$. Using the exterior envelope profile to estimate the thin shell's luminosity givesA finite but very extended upper limit changes the factor by $1-(R_c/R_s{)}^{81/4}$. The large exponent makes this correction small and further verifies the hydrogen burning thin-shell approximation. Since $C^2\propto M_c/L$ and $B\propto M_c$, the integral implies $L^2\propto M_c^{97/4}{R}_c^{-81/4}$. HenceThis is the fixed-radius-core shell-burning luminosity relation. The proportionality coefficient is a thin-shell estimate: luminosity rises from its value beneath the shell to $L$ through the emitting layer, so treating it as constant there uses the exterior profile rather than resolving the burning region. The scaling holds while its dimensionless shell structure and stellar composition remain fixed.

For consistency, $M_c/R_c$ must provide a hot hydrogen burning shell while the stellar core remains inert to core helium burning; the radiative envelope mass must be much less than $M_c$, and $R_c$ must be well inside an extended radiative envelope. At the shell base the gas must remain a nondegenerate gas and dominated by gas pressure, in particular $a{T}_b^4/3\ll C{T}_b^{17/4}$ with $T_b=B/R_c$. The assumed stellar core sequence must genuinely have nearly fixed $R_c$ over the mass interval considered. A cold helium stellar core supported by a nonrelativistic degenerate electron gas instead has the mass-radius law from part (a), so substituting that law would describe a different model and change the luminosity exponent.