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Let $E(t)={\int }_V|\nabla w{|}^{2}dV$, using real-valued smooth $w$ and a fixed bounded region. Differentiation under the integral sign and Green's first identity yieldThe boundary term vanishes because $w_t=0$ on $S$, and the heat equation gives $w_t=\Delta w$ in $V$. ConsequentlyThe integrand is continuous and nonnegative. Thus at any fixed time, equality is equivalent to $\Delta w=0$ everywhere in $V$, also giving $w_t=0$ there at that time. This proves Dirichlet energy dissipation for the heat equation.

The boundary condition fixes the boundary temperature in time; it is not a zero normal derivative condition. No sign condition on ${\partial }_{n}w$ is used. The stated smoothness licenses the time derivative, spatial integration by parts and boundary trace; without that regularity an appropriate weak energy argument would be required.

Put $h=u-\varphi -C$. Then $h=0$ on $S$, and the constant does not change a gradient, so $\nabla u=\nabla \varphi +\nabla h$. Expand the unnormalized Dirichlet energy:The cross term is zero by Green's first identity:Thus the useful stronger identity isEquality holds exactly when $h$ has zero gradient. It is then constant on every connected component and zero on that component’s boundary, hence $h=0$. Therefore equality holds precisely when $u=\varphi +C$ throughout $V$. This is the Dirichlet principle; an additive boundary constant does not alter the minimizing energy. The energy here omits the optional factor $1/2$, which does not affect minimizers or the inequality.

Work with real-valued smooth functions on the bounded region enclosed by $S$, with the oriented surface element pointing outward. Set $\psi =\varphi -{\varphi }_1$. It is a harmonic function in $V$ and has zero Dirichlet boundary conditions on $S$. Applying Green's first identity to $\psi$ with itself givesBoth terms on the right vanish by the boundary condition and harmonicity. The integrand is nonnegative and continuous, so $\nabla \psi =0$ throughout $V$. Consequently $\psi$ is constant on each connected component; every bounded component meets the prescribed boundary, where that constant is zero. ThereforeThis is the energy proof of Uniqueness of the Dirichlet problem. Connectedness of $V$ is not essential if the boundary values are prescribed on every component. Boundedness, or suitable decay and integrability at infinity, is needed for the boundary/energy argument; the finite enclosed-volume interpretation is used here.

The directional derivative along the specified vector field givesUse the magnitude formula for the curvature of an integral curve of a vector field:It vanishes on the plane $z=0$ and on the $z$ axis away from the origin, consistent with straight field lines there. At the origin the vector field vanishes, so it fails the nowhere-zero hypothesis and this field-line curvature is not defined.

As an independent geometric check, an integral curve of a vector field satisfies $\dot{x}=x$, $\dot{y}=y$, $\dot{z}=-z$, giving $(x,y,z)=(a{e}^u,b{e}^u,c{e}^{-u})$. Its velocity is $(x,y,-z)$ and acceleration $(x,y,z)$; the ordinary curvature of a space curve formula $|r^{\prime }\times r^{\prime \prime }|/|r^{\prime }{|}^3$ gives the same expression.

Let $F=|\mathbf{F}|>0$ and choose an orientation of a regular curve following the vector field. Its unit tangent vector is $t=\sigma \mathbf{F}/F$, where $\sigma =\pm 1$ is constant along a connected regular segment. Arc length differentiation along that curve is $d/ds=(\sigma \mathbf{F}/F)\cdot \nabla$. ThereforeTaking the cross product with $t$ removes the term parallel to $\mathbf{F}$:Where the Frenet frame exists, $t\times t^{\prime }=\kappa (t\times n)=\kappa b$. ThusThe printed sign corresponds to orientation along or against the field. Taking magnitudes gives the orientation-independent curvature of an integral curve of a vector field, $\kappa =|\mathbf{F}\times (\mathbf{F}\cdot \nabla )\mathbf{F}|/F^3$. This scalar formula also applies at zero curvature, where $b$ itself may be undefined.

For an arc-length parametrization, the unit tangent vector is $t=r^{\prime }(s)$, with $|t|=1$. Its derivative is perpendicular to $t$. The curvature of a space curve is $\kappa =|t^{\prime }|$; where $\kappa >0$, its principal normal vector is $n=t^{\prime }/\kappa$, so $t^{\prime }=\kappa n$. The binormal vector $b=t\times n$ completes the positively oriented Frenet frame.

Differentiate $b\cdot b=1$ to get $b^{\prime }\cdot b=0$. AlsoThus $b^{\prime }$ has only an $n$ component. Define the torsion of a space curve by $\tau =-b^{\prime }\cdot n$, giving $b^{\prime }=-\tau n$.

Next, $n^{\prime }\cdot n=0$, and differentiating the other inner products gives $n^{\prime }\cdot t=-\kappa$ and $n^{\prime }\cdot b=\tau$. Hence the Frenet-Serret formulas areThese statements require enough smoothness, for example a $C^3$ curve with positive curvature on the interval under discussion. If $\kappa =0$, the unit tangent vector is still defined, but the standard principal normal vector, binormal vector and torsion of a space curve need not be; their existence cannot be inferred just from smoothness of the curve.

Let ${\sigma }_N=\sqrt{Np(1-p)}$ and use the floor function to chooseInterpret an empty integer interval as an empty sum. These bounds give exactly the event $a<(K_N-Np)/{\sigma }_N\le b$, where $K_N$ has the binomial distribution of parameters $N,p$. Since $0<p<1$, the central limit theorem for a sum of independent Bernoulli distribution variables gives convergence in distribution of that standardized count to the standard normal distribution. Its cumulative distribution function is continuous at both endpoints, soThe TeX's additional malformed inequality involving $p/k_a(N)$ and $p/k_b(N)$ is absent from the PDF and is not a condition on the requested integers.

For fixed $k$ and sufficiently large $N$,The finite product and last factor tend to one, while $(1-\lambda /N{)}^N\to e^{-\lambda }$. Thus the Poisson limit theorem givesThese numbers form a probability mass function because they are nonnegative and their sum is $e^{-\lambda }{\sum }_{k\ge 0}{\lambda }^k/k!=1$, by the exponential series. They define the Poisson distribution of parameter $\lambda$. Only $N\ge \lambda$ is needed to make the binomial parameter valid, which is sufficient for the limit.

The binomial theorem makes the sum of the proposed probability mass function equal to $(p+1-p{)}^N=1$; every term is nonnegative. Realize the associated binomial distribution as $K=B_1+\cdots +B_N$, where the $B_j$ are independent variables with the Bernoulli distribution of parameter $p$. Each has expected value $p$ and variance $p(1-p)$. Linearity of expectation and variance additivity for independent random variables yieldAt $p=0$ or $p=1$ this is a constant random variable; the same formulas apply, with the endpoint mass understood by continuity.

Choose $X=x-2y$ and $T=x$, so $(\alpha ,\beta ,\gamma ,\delta )=(1,-2,1,0)$. The Jacobian determinant of the coordinate change is two, so it is an invertible linear map. By the chain rule, ${\partial }_x={\partial }_X+{\partial }_T$ and ${\partial }_y=-2{\partial }_X$. ConsequentlyThis is the wave equation with $c=1$. The D'Alembert formula is $F(X,T)=U(X+T)+V(X-T)$, with arbitrary twice differentiable functions $U,V$. Returning to the old coordinates and absorbing factors of two into the arbitrary functions givesDirect differentiation checks the cancellation. Equivalently, first integrating ${\partial }_x(f_x+f_y)=0$ produces the same two-function family.

For real initial data, separation of variables givesThis includes $z_0=0$, whose solution is identically zero. Every other real $z_0$ has a pole at $t=1/z_0$, so only $z_0=0$ gives a finite solution on the whole real line. If complex-valued initial data are allowed, a nonreal $z_0$ has no real-time pole; the whole-real-line conclusion here uses the usual real-valued differential-equation convention.

Along a characteristic curve $y=u+ax$, the chain rule gives $df(x,u+ax)/dx=f_x+a{f}_y$. For the homogeneous transport equation this is zero, so $f(x,y)=F(y-ax)$. For the nonlinear equation, the same method of characteristics reduces the problem to $df/dx=f^2$, with initial value $g(u)$. ThereforeFor differentiable real $g$, this is the classical solution wherever its denominator is nonzero, on the characteristic intervals containing the initial line. If $g(u)\ne 0$ for some $u$, that characteristic blows up at $x=1/g(u)$, $y=u+a/g(u)$. Thus the only real-valued solution bounded on all of ${\mathbb{R}}^2$ has $g\equiv 0$, giving $f\equiv 0$.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot{y}-y\dot{x})/r^2$. The polar coordinates satisfyHence the rotation is clockwise. The origin is an unstable equilibrium, while $r=1$ is a limit cycle of period $2\pi$. Writing $u=r^2$ gives the logistic differential equation $\dot{u}=2u(1-u)$. For $r(t_0)=r_0>0$,All nonzero solutions tend to the unit-circle attracting limit cycle as $t\to \infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to -\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes atThis finite-time blow-up of an ordinary differential equation occurs backward in time. Such exterior solutions have no $t\to -\infty$ behavior: their maximal time interval is $(t_{\ast },\infty )$. The zero solution exists for all time and has no defined polar angle.

Use the complete system in the PDF: the TeX omits $\dot{y}=y(4x-1)/8$. Both coordinate axes are invariant, and the equilibrium points are $(0,0)$, $(1,0)$ and $(1/4,3/4)$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1,-1/8$, and at $(1,0)$ they are $-1,3/8$: both points are saddle equilibria. At the coexistence point,Thus the interior point is a stable spiral. The nontrivial nullclines are $x+y=1$ and $x=1/4$: $x$ increases below the first and decreases above it, while $y$ increases to the right of the second and decreases to the left. The rotation near coexistence is counterclockwise.

A local linearization by itself does not prove that every interior solution has the same limit. For this logistic predator-prey model, let $x_{\ast }=1/4,y_{\ast }=3/4$ and use the Lyapunov functionThe inequality $u-1-\log u\ge 0$ makes $V$ nonnegative, with equality only at coexistence. Its sublevel sets are compact inside the positive quadrant, because $V$ diverges both at the axes and at infinity. Direct differentiation cancels the predator-prey cross terms and givesOn $\{\dot{V}=0\}$ one has $x=x_{\ast }$, and remaining there requires $\dot{x}=-x_{\ast }(y-y_{\ast })=0$. The largest invariant subset is therefore the coexistence point. The LaSalle invariance principle provesOn $x=0$, $y$ decays as $e^{-t/8}$ toward the origin. On $y=0$, the logistic differential equation takes every $x(0)>0$ to one. The origin itself stays fixed. These boundary trajectories are the exceptions to the interior convergence statement.

The PDF gives $V=H(t)$, the Heaviside step function; the TeX incorrectly changes this to $V=I(t)$. A unit voltage step has derivative the Dirac delta function, so the current is the causal Green function for $L{d}^2/d{t}^2+Rd/dt+1/C$.

The current must be continuous at zero: a jump would create a delta derivative through $L{I}^{\prime \prime }$, absent on the right-hand side. Integrating across zero therefore gives $L[I^{\prime }]+R[I]=1$, so $I(0^{+})=0$ and $I^{\prime }(0^{+})=1/L$. PutThese distinct negative characteristic roots give the overdamped RLC responseTogether with $I(t)=0$ for $t<0$, this solves the step response. The eventual current is zero: the capacitor charges and blocks steady current in the series RLC circuit.

With no electrical resistance or applied electric potential difference, divide the equation by the inductance $L$ to get a simple harmonic oscillator with angular frequency ${\omega }_0=(LC{)}^{-1/2}$. Its general real solution isEach nonzero solution has period $2\pi \sqrt{LC}$. Here ${\omega }_0$ is an angular frequency; the ordinary frequency in cycles per unit time is ${\omega }_0/(2\pi )$.

The conditional density givesusing either integration by parts or the expected value of an exponential distribution. HenceThis is a finite value for almost every realization of $X$, but not an integrable random variable: $\mathbb{E}[1/X]={\int }_0^{\infty }{e}^{-x}/xdx=\infty$. Since $Y\ge 0$, its conditional expectation remains well-defined in the nonnegative, extended-expectation sense; the tower property of conditional expectation then gives $\mathbb{E}Y=\infty$.

Integrate the joint probability density over $y$ to obtain the marginal density of $X$:Thus $X$ has an exponential distribution of rate one. Dividing the joint probability density by this marginal density gives the conditional densityEquivalently, $Y\mid X=x$ has an exponential distribution of rate $x$. The value at $x=0$ can be assigned arbitrarily: it is a probability-zero conditioning value and the density ratio there is not meaningful.

Put $U_0=1$, so the same definition $z_n=x_n/U_{n-1}$ includes $n=1$. Since $U_n=a_n{U}_{n-1}$ and every $a_n$ is nonzero,The factor $U_n^{-1}$ is a discrete integrating factor for this variable-coefficient affine recurrence. Sum the increments as a telescoping series and multiply back by $U_n$:The identity requires no assumption that the coefficients have the same sign.