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At zero surface shear stress, the solution connected to part (a) isIndeed, it has $f_0^{\prime }=1$, $f_0^{\prime \prime }=f_0^{\prime \prime \prime }=0$ and $f_0(0)=0$. Insert the regular perturbation $f=\zeta +\gamma f_1+O({\gamma }^2)$ into the nonlinear ordinary differential equation. The terms areThe coefficient of $\gamma$ therefore givesExpanding the surface boundary conditions and the far-field matching condition givesThe printed $+\infty$ limit is a sign error. It is present in the native PDF on paper page 3, not introduced by transcription. The fluid occupies $z<0$, the inner variable is $\zeta =z/\delta <0$, and part (b)(iii) correctly matches at $-\infty$. We retain that physical domain throughout. A change to the positive coordinate $\eta =-\zeta$ would also change the differential equation and surface derivative signs; it cannot be made by changing only the far-field sign.

The inequality ${\delta }^2\ll \gamma$ ensures that effects discarded at order ${\delta }^2$ in the boundary-layer scaling are smaller than the retained order-$\gamma$ perturbation. The inequality $\gamma \ll 1$ justifies this regular perturbation. The two conditions control different approximations.

For $y_1=(1-x{)}^{-1}$, the derivatives are $y_1^{\prime }=(1-x{)}^{-2}$ and $y_1^{\prime \prime }=2(1-x{)}^{-3}$. Consequently the left-hand side of the ordinary differential equation iswhere $x\ne 1$.

Use reduction of order, writing $y=u/(1-x)$. A direct application of the product rule givesFor $v=u^{\prime }$ the resulting first-order ordinary differential equation is $v^{\prime }/v=2/(1-x)$ whenever $v\ne 0$. Integration yields $v=C(1-x{)}^{-2}$ and $u=A+B/(1-x)$; the case $v=0$ is already included. Henceon either interval $x<1$ or $x>1$. The two displayed solutions of a differential equation are linearly independent, so they span the solution space of a homogeneous linear differential equation on either interval. The point $x=1$ is a singular point of a differential equation and cannot be crossed by a nonzero member of this family that remains finite there.

The repeated root of the characteristic equation suggests factoring out $e^x$. Set $y=e^{x}u$. The product rule givesThus the ordinary differential equation reduces to $u^{\prime \prime }=1$, with general solution $u=x^2/2+Ax+B$. The first initial condition gives $B=1$. Since $y^{\prime }(0)=A+B=2$, the other initial condition gives $A=1$. ThereforeSubstitution into the reduced ordinary differential equation and the two initial conditions verifies the result.

Multiplying the ordinary differential equation by two suggests using the dependent variable $Y=y^2$, whose derivative is $Y^{\prime }=2y{y}^{\prime }$. On the interval through $x=1$, this gives the linear ordinary differential equationAfter integration, $Y=x^2+Cx$. The initial condition $Y(1)=4$ gives $C=3$. As $y(1)>0$, continuity selects the positive branch:The interval $x>0$ is the maximal real interval through the given initial condition: at $x=0$ the formula approaches $0$, but its derivative is unbounded, so it does not extend through zero as a differentiable solution of a differential equation.

For $y_1=e^x/x$, on an interval avoiding zero,Substitution into the ordinary differential equation givesTo find an independent solution, put $v=xy$. The product rule gives $v^{\prime \prime }=x{y}^{\prime \prime }+2y^{\prime }$, so the ordinary differential equation becomes $v^{\prime \prime }-v=0$. Its independent solutions are $e^x$ and $e^{-x}$. HenceThe two displayed solutions are linearly independent, since their ratio $e^{2x}$ is nonconstant on any open interval. Constants can be chosen separately on the two sides of the singular point $x=0$.

The logarithm requires $y>0$, and $x>0$ is given. Set $z=\log (y/x)$, so $y=x{e}^z$ and the chain rule yields $x{y}^{\prime }/y=1+x{z}^{\prime }$. Dividing the ordinary differential equation by $y$ therefore givesSince $(z/x{)}^{\prime }=(x{z}^{\prime }-z)/x^2=0$, we obtain $z=Cx$, including the case $z=0$. Thus the general positive solution is $y=x{e}^{Cx}$. The initial condition $y(1)=e$ gives $C=1$:For the separate decay requirement, $x{e}^{Cx}\to 0$ precisely when $C<0$: if $C=0$, the result is $x$, and if $C>0$, it grows. Consequently the full family with the stated limit isThis latter family does not also satisfy $y(1)=e$; the solution selected by that initial condition grows. The identically zero function is excluded by the original logarithm.

For the suggested substitution, put $y=xv(x)$. Then the product rule gives $y^{\prime }=v+x{v}^{\prime }$, while the right side of the ordinary differential equation becomes $v+v^{-3}$. Cancelling $v$ leaves the separable differential equationIntegration yields $v^4/4=\log x+C$. The initial condition gives $v(1)=1$, so $v^4=1+4\log x$. The positive root is forced by the initial value and continuity:This is the maximal connected interval containing $1$ on which the real solution exists. At the left endpoint $y=0$, where the original ordinary differential equation is undefined; extending the fourth-root formula further left would also cease to be real.

For the homogeneous linear ordinary differential equation, the characteristic equation isIts real solution is $y_h=e^{-x}(C_1\cos x+C_2\sin x)$. For a particular solution to the full ordinary differential equation, try $y_p=ax+b$. Substitution givesso $a=4$ and $b=-2$. Adding the homogeneous and particular solutions yieldsThe two arbitrary real constants account for the second order of the ordinary differential equation.

For the spatial mode with eigenvalue ${\kappa }_n=(n\pi /L{)}^2$, the temporal ordinary differential equation isIts general solution isThe printed reference to “each solution $X(x)$ ... in part (b)” refers to (b)(i). The zero spatial mode gives $u=0$ irrespective of $T$ and contributes no additional nonzero solution.^[ref]

For $y=1+x^2$, the derivatives are $y^{\prime }=2x$ and $y^{\prime \prime }=2$. Its endpoint values are $y(0)=1$ and $y(2)=5$, so it satisfies both the ordinary differential equation and the boundary conditions. Moreover any solution of $y^{\prime \prime }=2$ has the form $x^2+Ax+B$; the boundary conditions force $B=1$ and $A=0$. Thus the unique solution of the boundary value problem is

Apply separation of variables to each mode of the damped wave equation $u_{tt}+2k^2{u}_t=c^2{u}_{xx}$, assuming $L,c>0$. The spatial ordinary differential equation is $X_n^{\prime \prime }+({\lambda }_n/c{)}^2{X}_n=0$. Its general solution is $C\sin ({\lambda }_{n}x/c)+D\cos ({\lambda }_{n}x/c)$. The boundary condition at $0$ gives $D=0$; a nonzero mode at $L$ requiresOnly odd $n=2m+1$ occur in the initial Fourier sine series. Put ${\omega }_m=(2m+1)\pi c/L$ and $d_m=\sqrt{k^4-{\omega }_m^2}$. The temporal equation $T_m^{\prime \prime }+2k^2{T}_m^{\prime }+{\omega }_m^2{T}_m=0$ has roots $-k^2\pm d_m$. DefineFor distinct roots, let the full amplitude be $q_m(t)=A_m{e}^{-{\alpha }_{m}t}+B_m{e}^{-{\beta }_{m}t}$. The initial displacement and zero initial velocity impose $A_m+B_m=F_m$ and ${\alpha }_m{A}_m+{\beta }_m{B}_m=0$. Solving givesThus the physically consistent superposition isFor ${\omega }_m<k^2$ there is overdamping, with two real decaying exponentials. For ${\omega }_m>k^2$ there is underdamping; set ${\nu }_m=\sqrt{{\omega }_m^2-k^4}$; the equivalent real expression isAt critical damping, the repeated-root case ${\omega }_m=k^2$, take the limit, giving $q_m(t)=F_m(1+k^{2}t)e^{-k^{2}t}$. These formulas all satisfy $q_m(0)=F_m$ and $q_m^{\prime }(0)=0$.

Three source qualifications matter. First, the sum must start at $m=0$, for the same reason as in (b). Second, the printed $X(x)=u_0(x)$ cannot hold for a single separated mode: the tent has $X^{\prime \prime }=0$ on each open linear segment but $X\ne 0$, contradicting $X^{\prime \prime }=-({\lambda }_n/c{)}^{2}X$ for positive ${\lambda }_n$, and its peak is not twice differentiable. Separation applies mode by mode, with the initial tent recovered by superposition. Third, no finite “large” $k$ makes every mode overdamped, because ${\omega }_m\to \infty$. The two-exponential expression permits complex conjugate roots, while a repeated root requires its displayed limiting form. The tent initial data are naturally interpreted through the convergent modal solution, rather than as a single smooth separated product.

Put $h=\Delta x$ and suppose $L=(n+1)h$, so there are $n$ interior grid points $x_j=jh$, $j=1,\dots ,n$. Write $u_j(t)\simeq u(x_j,t)$. The fixed ends impose homogeneous Dirichlet boundary conditions,For a sufficiently smooth displacement field, the two Taylor series at fixed time givewhere all displayed derivatives on the right are evaluated at $(x_j,t)$. Adding cancels the odd terms, leaving the second-order central differenceSubstitute this finite difference into the wave equation and retain continuous time. The resulting method of lines gives coupled second-order ordinary differential equations,In particular, the first and last equations are ${\ddot{u}}_1=(c^2/h^2)(-2u_1+u_2)$ and ${\ddot{u}}_n=(c^2/h^2)(u_{n-1}-2u_n)$ for $n\ge 2$. With the interior displacement vector only,Equivalently, $A_{jj}=-2c^2/h^2$, $A_{j,j+1}=A_{j+1,j}=c^2/h^2$, and all other entries vanish. This entrywise definition also covers small $n$, including $A=(-2c^2/h^2)$ when $n=1$. The matrix is a symmetric matrix and a tridiagonal matrix. The spatial approximation has error $O(h^2)$ in Big O notation; time has not yet been discretized.

The null space consists of concentration patterns on which transport produces no change. Here it is the familyEvery box has the same concentration. Ocean circulation may continue, but inflow and outflow of the chemical tracer balance in each box. With $\mathbf{b}=0$ these are exactly the steady states; physically admissible members have $\alpha \ge 0$. With nonzero $\mathbf{b}$, adding $\alpha 1$ to any particular steady state leaves its transport balance unchanged, but a null space vector alone need not balance the sources and sinks.

To identify the tracer inventory, apply conserved inventory from a left null vector. Directly summing each column givesConsequently, for the supplied ordinary differential equations,For $\mathbf{b}=0$, $M$ is a conserved quantity. The connected box model equalizes concentrations, approaching the unique member of the null space with the initial inventory:The arithmetic mean is appropriate here because the supplied matrix has zero column sums. In a general box model the conserved quantity is $w^T\mathbf{c}$, where $w$ is a positive left null space vector; for physical concentrations expressed per unit volume, these weights represent the box volumes up to a common factor. One must establish the weights from the equations, rather than infer equal volumes from a schematic drawing.

Nonnegative initial concentrations remain nonnegative under transport alone. Indeed $A+2I$ is a nonnegative matrix, sohas nonnegative entries for $t\ge 0$: it is a positive matrix semigroup. Its column sums are one by $1^{T}A=0$. To see why equalization occurs, set $d=\mathbf{c}-(M/4)1$ for $\mathbf{b}=0$; then $1^{T}d=0$ andThe right side vanishes only for constant $d$, which in the zero-sum vector subspace means $d=0$. Its strict negative bound on the unit sphere in that vector subspace gives decay to zero. Thus the matrix exponential redistributes inventory without creating negative concentrations; arbitrary prescribed sinks can still exhaust the tracer if they are allowed to act indefinitely.

Assume sufficient decay and differentiability to justify integration by parts and exchange of integrals; Schwartz functions are one convenient class. With the stated Fourier transform convention, integrate first in $x$:The boundary contribution vanishes under the decay assumption. The Fourier transform of a derivative consequently turns constant-coefficient spatial differential operators into multipliers. Repeating it gives $\mathcal{F}[u_{xx}]=-k_x^2\hat{u}$, so the heat equation becomes an independent temporal ordinary differential equation at each wavevector.

The wave equation is second order in time. Even one normal mode obeys a second-order ordinary differential equation, so its initial displacement and initial velocity are independent data. Physically they specify stored elastic deformation and the initial motion.

The heat equation is first order in time; its initial temperature already fixes the instantaneous time derivative through the spatial Laplacian. It needs one initial field. In either case spatial boundary conditions remain necessary and are not substitutes for initial data. Classical fixed-end string data must satisfy the appropriate endpoint compatibility, including $f(0)=f(L)=0$ and, when endpoint velocities exist, $g(0)=g(L)=0$.

Patch-clamp series resistance, also called patch-clamp access resistance, is the effective electrical resistance between the pipette's sensed voltage and the cell interior. It includes tip access and any remaining obstruction after break-in. For injected electric current $I$ into the biological cell, an ideal resistive access path givesThus $1\mathrm{nA}$ through $20\mathrm{M\Omega }$ produces a $20\mathrm{mV}$ difference: a substantial possible voltage clamp error. The electric current in this equation is the access-path current, excluding any bypass seal leak current. In current clamp, the same access drop contaminates the pipette voltage during injection. Bridge compensation subtracts an estimated $I{R}_s$; an estimate that changes during recording leaves an error.

An independent passive voltage clamp calculation makes both speed and error visible. Write $v=V-E_L$ and $v_p=V_p-E_L$, with fixed leak reversal potential $E_L$, constant membrane electrical resistance $R_m$ and membrane capacitance $C_m$. Assume an ideal command at the pipette node and omit seal leak current and pipette capacitance. Electric charge conservation givesFor a step $v_p=\Delta V$ from $v(0)=0$, the ordinary differential equation givesWith the figure's parameters and $\Delta V=10\mathrm{mV}$, the time constant is $0.909\mathrm{ms}$, the final membrane rise is $9.091\mathrm{mV}$, and access-path electric current falls from $500$ to $45.45\mathrm{pA}$. Only when $R_m\gg R_s$ does the time constant approach $R_s{C}_m$ and the final voltage approach the command. Actual voltage clamp also depends on amplifier feedback control and the space clamp problem.

For an unstable Taylor–Goldstein equation mode in a finite channel, take $a=1$ in the power-transformed Taylor–Goldstein energy identity. Its imaginary part givesThus the real phase velocity lies strictly between the extremes of a nonconstant smooth shear profile. Equality would force the regular eigenfunction to vanish on an interval, hence everywhere by uniqueness for its ordinary differential equation.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set and $f:U\to {\mathbb{R}}^d$ a continuous function. Suppose $f$ is locally Lipschitz in the state $y$, with a Lipschitz constant uniform on a sufficiently small time-state neighborhood of each point. Then each $(t_0,y_0)\in U$ determines a unique local solution of a differential equation to the displayed initial value problem, on an interval around $t_0$. This is the local theorem in Ordinary Differential Equations and Dynamical Systems by Gerald Teschl.

For example, $y^{\prime }=y$, $y(0)=1$, has the solution of a differential equation $y(t)=e^t$: its derivative is itself and its starting value is $1$. The Lipschitz constant is $L=1$, since $|u-v|=|f(t,u)-f(t,v)|$. Thus uniqueness follows even before computing $y(1)=e\approx 2.71828183$. Picard iteration below constructs this exponential function one polynomial at a time.

The Lipschitz condition has real content. For the continuous function involving a square root,each waiting time $c\ge 0$ gives a different forward solution of a differential equation:The constant solution of a differential equation $y(t)=0$ is another choice. At $t=2$, delays $c=0,1,2$ give $y=4,1,0$ respectively. Before the join, both sides of the ordinary differential equation are zero; afterwards, $y_c^{\prime }=2(t-c)=2\sqrt{y_c}$. Both one-sided derivatives at $c$ are zero, so the join is valid. All these trajectories can also be extended by zero to $t<0$. The failed Lipschitz condition is visible in the quotientFor $\epsilon =10^{-2},10^{-4},10^{-6}$ this quotient is $20,200,2000$. Continuity still allows local existence by the Peano existence theorem; it cannot choose a waiting time. MIT's 18.034 Lecture 2, Example 2.3 develops the same waiting-time mechanism for a different ordinary differential equation, $y^{\prime }=3y^{2/3}$.

The useful promise is that an initial condition and a sufficiently regular local rule select one trajectory. It justifies comparing models and following trajectories without an arbitrary choice at each instant. A locally Lipschitz continuous rule need not keep that trajectory bounded: $y^{\prime }=1+y^2$ meets the theorem's hypotheses everywhere but has finite-time blowup. Longer existence is governed by the finite-dimensional continuation criterion.

Suppose $f$ is continuous and Lipschitz in its state variable on a closed cylinder $|t-a|\le h$, $\Vert y-b\Vert \le k$, with bounds $\Vert f\Vert \le M$ and Lipschitz constant $L_f$. On a smaller interval of radius $\eta$, the displayed Picard operator maps continuous curves in that closed ball into itself when $\eta M\le k$, and it is a contraction mapping in the supremum norm when $\eta L_f<1$. This function space is complete because it is a closed subset of a Banach space of continuous vector-valued functions. The contraction mapping theorem gives a unique fixed point, and the fundamental theorem of calculus makes it a solution of the ordinary differential equation. Choosing $\eta M<k$ also prevents any solution with the same initial data from first exiting the cylinder during that interval.