diff options
Diffstat (limited to 'source/know/concept')
30 files changed, 234 insertions, 527 deletions
diff --git a/source/know/concept/binomial-distribution/index.md b/source/know/concept/binomial-distribution/index.md index 1193a93..dc75221 100644 --- a/source/know/concept/binomial-distribution/index.md +++ b/source/know/concept/binomial-distribution/index.md @@ -44,11 +44,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-mean"/> -<label for="proof-mean">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-mean">Proof.</label> + +{% include proof/start.html id="proof-mean" -%} The trick is to treat $$p$$ and $$q$$ as independent until the last moment: $$\begin{aligned} @@ -62,8 +59,8 @@ $$\begin{aligned} \end{aligned}$$ Inserting $$q = 1 - p$$ then gives the desired result. -</div> -</div> +{% include proof/end.html id="proof-mean" %} + Meanwhile, we find the following variance $$\sigma^2$$, with $$\sigma$$ being the standard deviation: @@ -74,12 +71,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-var"/> -<label for="proof-var">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-var">Proof.</label> -We use the same trick to calculate $$\overline{n^2}$$ + +{% include proof/start.html id="proof-var" -%} (the mean squared number of successes): $$\begin{aligned} @@ -106,8 +99,8 @@ $$\begin{aligned} \end{aligned}$$ By inserting $$q = 1 - p$$, we arrive at the desired expression. -</div> -</div> +{% include proof/end.html id="proof-var" %} + As $$N \to \infty$$, the binomial distribution turns into the continuous normal distribution, @@ -119,11 +112,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-normal"/> -<label for="proof-normal">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-normal">Proof.</label> + +{% include proof/start.html id="proof-normal" -%} We take the Taylor expansion of $$\ln\!\big(P_N(n)\big)$$ around the mean $$\mu = Np$$: @@ -211,8 +201,7 @@ $$\begin{aligned} \end{aligned}$$ Taking $$\exp$$ of this expression then yields a normalized Gaussian distribution. -</div> -</div> +{% include proof/end.html id="proof-normal" %} ## References diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md index 9ed2fd2..d2631b2 100644 --- a/source/know/concept/boltzmann-equation/index.md +++ b/source/know/concept/boltzmann-equation/index.md @@ -145,11 +145,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-moment0"/> -<label for="proof-moment0">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-moment0">Proof.</label> + +{% include proof/start.html id="proof-moment0" -%} We insert $$Q = m$$ into our prototype, and since $$m$$ is constant, the rest is trivial: @@ -159,9 +156,8 @@ $$\begin{aligned} \\ &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0 \end{aligned}$$ +{% include proof/end.html id="proof-moment0" %} -</div> -</div> If we instead choose the momentum $$Q = m \vb{v}$$, we find that the **first moment** of the BTE describes conservation of momentum, @@ -174,11 +170,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-moment1"/> -<label for="proof-moment1">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-moment1">Proof.</label> + +{% include proof/start.html id="proof-moment1" -%} We insert $$Q = m \vb{v}$$ into our prototype and recognize $$\rho$$ wherever possible: $$\begin{aligned} @@ -220,9 +213,8 @@ $$\begin{aligned} 0 &= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F} \end{aligned}$$ +{% include proof/end.html id="proof-moment1" %} -</div> -</div> Finally, if we choose the kinetic energy $$Q = m |\vb{v}|^2 / 2$$, we find that the **second moment** gives conservation of energy, @@ -237,11 +229,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-moment2"/> -<label for="proof-moment2">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-moment2">Proof.</label> + +{% include proof/start.html id="proof-moment2" -%} We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible: $$\begin{aligned} @@ -349,9 +338,7 @@ $$\begin{aligned} \end{bmatrix} = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i \end{aligned}$$ - -</div> -</div> +{% include proof/end.html id="proof-moment2" %} diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md index 742c8ff..510417a 100644 --- a/source/know/concept/convolution-theorem/index.md +++ b/source/know/concept/convolution-theorem/index.md @@ -12,6 +12,8 @@ is equal to a product in the frequency domain. This is especially useful for computation, replacing an $$\mathcal{O}(n^2)$$ convolution with an $$\mathcal{O}(n \log(n))$$ transform and product. + + ## Fourier transform The convolution theorem is usually expressed as follows, where @@ -27,11 +29,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-fourier"/> -<label for="proof-fourier">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-fourier">Proof.</label> + +{% include proof/start.html id="proof-fourier" -%} We expand the right-hand side of the theorem and rearrange the integrals: @@ -57,8 +56,8 @@ $$\begin{aligned} &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'} = B \cdot (\tilde{f} * \tilde{g})(k) \end{aligned}$$ -</div> -</div> +{% include proof/end.html id="proof-fourier" %} + ## Laplace transform @@ -79,11 +78,8 @@ $$\begin{aligned} \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)} \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-laplace"/> -<label for="proof-laplace">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-laplace">Proof.</label> + +{% include proof/start.html id="proof-laplace" -%} We expand the left-hand side. Note that the lower integration limit is 0 instead of $$-\infty$$, because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$: @@ -106,8 +102,7 @@ $$\begin{aligned} &= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'} = \tilde{f}(s) \: \tilde{g}(s) \end{aligned}$$ -</div> -</div> +{% include proof/end.html id="proof-laplace" %} diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md index cb22e43..48a5a72 100644 --- a/source/know/concept/curvilinear-coordinates/index.md +++ b/source/know/concept/curvilinear-coordinates/index.md @@ -48,6 +48,7 @@ we derive general formulae to convert expressions from Cartesian coordinates to the new orthogonal system $$(x_1, x_2, x_3)$$. + ## Basis vectors Consider the the vector form of the line element $$\dd{\ell}$$, @@ -86,6 +87,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Gradient In an orthogonal coordinate system, @@ -102,11 +104,8 @@ $$\begin{gathered} } \end{gathered}$$ -<div class="accordion"> -<input type="checkbox" id="proof-grad"/> -<label for="proof-grad">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-grad">Proof.</label> + +{% include proof/start.html id="proof-grad" -%} For a direction $$\dd{\ell}$$, we know that $$\idv{f}{\ell}$$ is the component of $$\nabla f$$ in that direction: @@ -127,9 +126,8 @@ $$\begin{gathered} + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2} + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3} \end{gathered}$$ +{% include proof/end.html id="proof-grad" %} -</div> -</div> ## Divergence @@ -145,11 +143,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-div"/> -<label for="proof-div">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-div">Proof.</label> + +{% include proof/start.html id="proof-div" -%} As preparation, we rewrite $$\vb{V}$$ as follows to introduce the scale factors: @@ -222,8 +217,8 @@ $$\begin{aligned} After repeating this procedure for the other components of $$\vb{V}$$, we get the desired general expression for the divergence. -</div> -</div> +{% include proof/end.html id="proof-div" %} + ## Laplacian @@ -246,6 +241,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Curl The curl of a vector $$\vb{V}$$ is as follows @@ -264,11 +260,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-curl"/> -<label for="proof-curl">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-curl">Proof.</label> + +{% include proof/start.html id="proof-curl" -%} The curl is found in a similar way as the divergence. We rewrite $$\vb{V}$$ like so: @@ -317,8 +310,8 @@ $$\begin{aligned} If we go through the same process for the other components of $$\vb{V}$$ and add up the results, we get the desired expression for the curl. -</div> -</div> +{% include proof/end.html id="proof-curl" %} + ## Differential elements diff --git a/source/know/concept/detailed-balance/index.md b/source/know/concept/detailed-balance/index.md index b89d5da..98f9bd3 100644 --- a/source/know/concept/detailed-balance/index.md +++ b/source/know/concept/detailed-balance/index.md @@ -103,11 +103,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-reversibility"/> -<label for="proof-reversibility">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-reversibility">Proof.</label> + +{% include proof/start.html id="proof-reversibility" -%} Consider the following weighted inner product, whose weight function is a stationary distribution $$\pi$$ satisfying detailed balance, @@ -222,8 +219,7 @@ $$\begin{aligned} Where the integral gave the expectation value at $$X_0$$, since $$\pi$$ does not change in time. -</div> -</div> +{% include proof/end.html id="proof-reversibility" %} diff --git a/source/know/concept/dirac-delta-function/index.md b/source/know/concept/dirac-delta-function/index.md index 518eba1..0185b78 100644 --- a/source/know/concept/dirac-delta-function/index.md +++ b/source/know/concept/dirac-delta-function/index.md @@ -65,11 +65,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-scale"/> -<label for="proof-scale">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-scale">Proof.</label> + +{% include proof/start.html id="proof-scale" -%} Because it is symmetric, $$\delta(s x) = \delta(|s| x)$$. Then by substituting $$\sigma = |s| x$$: @@ -77,9 +74,8 @@ $$\begin{aligned} \int \delta(|s| x) \dd{x} &= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|} \end{aligned}$$ +{% include proof/end.html id="proof-scale" %} -</div> -</div> An even more impressive property is the behaviour of the derivative of $$\delta(x)$$: @@ -89,11 +85,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-dv1"/> -<label for="proof-dv1">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-dv1">Proof.</label> + +{% include proof/start.html id="proof-dv1" -%} Note which variable is used for the differentiation, and that $$\delta'(x - \xi) = - \delta'(\xi - x)$$: @@ -102,9 +95,8 @@ $$\begin{aligned} &= \dv{}{x}\int f(\xi) \: \delta(x - \xi) \dd{x} = f'(x) \end{aligned}$$ +{% include proof/end.html id="proof-dv1" %} -</div> -</div> This property also generalizes nicely for the higher-order derivatives: diff --git a/source/know/concept/dynkins-formula/index.md b/source/know/concept/dynkins-formula/index.md index c0d20c5..307f098 100644 --- a/source/know/concept/dynkins-formula/index.md +++ b/source/know/concept/dynkins-formula/index.md @@ -39,11 +39,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-kolmogorov"/> -<label for="proof-kolmogorov">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-kolmogorov">Proof.</label> + +{% include proof/start.html id="proof-kolmogorov" -%} We define a new process $$Y_t \equiv h(X_t)$$, and then apply Itō's lemma, leading to: $$\begin{aligned} @@ -84,9 +81,8 @@ $$\begin{aligned} \hat{L}\{h(X_0)\} \approx \frac{1}{t} \mathbf{E}[Y_t - Y_0| X_0] \end{aligned}$$ +{% include proof/end.html id="proof-kolmogorov" %} -</div> -</div> The general definition of resembles that of a classical derivative, and indeed, the generator $$\hat{A}$$ can be thought of as a differential operator. @@ -104,11 +100,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-dynkin"/> -<label for="proof-dynkin">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-dynkin">Proof.</label> + +{% include proof/start.html id="proof-dynkin" -%} The proof is similar to the one above. Define $$Y_t = h(X_t)$$ and use Itō’s lemma: @@ -136,9 +129,9 @@ $$\begin{aligned} = \mathbf{E}\bigg[ Y_\tau - Y_0 - \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg] \end{aligned}$$ -Isolating this equation for $$\mathbf{E}[Y_\tau | X_0]$$ then gives Dynkin's formula. -</div> -</div> +Isolating this equation for $$\mathbf{E}[Y_\tau \!\mid\! X_0]$$ then gives Dynkin's formula. +{% include proof/end.html id="proof-dynkin" %} + A common application of Dynkin's formula is predicting when the stopping time $$\tau$$ occurs, and in what state $$X_\tau$$ this happens. diff --git a/source/know/concept/equation-of-motion-theory/index.md b/source/know/concept/equation-of-motion-theory/index.md index 02ed856..c1ed8da 100644 --- a/source/know/concept/equation-of-motion-theory/index.md +++ b/source/know/concept/equation-of-motion-theory/index.md @@ -63,11 +63,8 @@ $$\begin{aligned} = - \sum_{\nu''} u_{\nu \nu''} \hat{c}_{\nu''} \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-commH0"/> -<label for="proof-commH0">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-commH0">Proof.</label> + +{% include proof/start.html id="proof-commutator" -%} Using the commutator identity for $$\comm{A B}{C}$$, we decompose it like so: @@ -105,9 +102,8 @@ $$\begin{aligned} - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big) = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''} \end{aligned}$$ +{% include proof/end.html id="proof-commutator" %} -</div> -</div> Substituting this into $$G_{\nu \nu'}^R$$'s equation of motion, we recognize another Green's function $$G_{\nu'' \nu'}^R$$: diff --git a/source/know/concept/euler-bernoulli-law/index.md b/source/know/concept/euler-bernoulli-law/index.md index dad67ca..5a6c38d 100644 --- a/source/know/concept/euler-bernoulli-law/index.md +++ b/source/know/concept/euler-bernoulli-law/index.md @@ -81,11 +81,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-field"/> -<label for="proof-field">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-field">Proof.</label> + +{% include proof/start.html id="proof-field" -%} By integrating the above strains $$u_{ii} = \ipdv{u_i}{i}$$, we get the components of $$\va{u}$$: @@ -171,8 +168,8 @@ $$\begin{aligned} Inserting this into the components $$u_x$$, $$u_y$$ and $$u_z$$ then yields the full displacement field. -</div> -</div> +{% include proof/end.html id="proof-field" %} + In any case, the beam experiences a bending torque with an $$x$$-component $$T_x$$ given by: diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md index 0bc849b..c86d997 100644 --- a/source/know/concept/fourier-transform/index.md +++ b/source/know/concept/fourier-transform/index.md @@ -67,6 +67,7 @@ on whether the analysis is for forward ($$s > 0$$) or backward-propagating ($$s < 0$$) waves. + ## Derivatives The FT of a derivative has a very useful property. @@ -113,6 +114,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Multiple dimensions The Fourier transform is straightforward to generalize to $$N$$ dimensions. @@ -150,11 +152,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-constants-ND"/> -<label for="proof-constants-ND">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-constants-ND">Proof.</label> + +{% include proof/start.html id="proof-constants-ndim" -%} The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so: $$\begin{aligned} @@ -180,9 +179,8 @@ $$\begin{aligned} &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'} = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x}) \end{aligned}$$ +{% include proof/end.html id="proof-constants-ndim" %} -</div> -</div> Differentiation is more complicated for $$N > 1$$, but the FT is still useful, @@ -197,11 +195,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-laplacian"/> -<label for="proof-laplacian">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-laplacian">Proof.</label> + +{% include proof/start.html id="proof-laplacian" -%} We insert $$\nabla^2 f$$ into the FT, decompose the exponential and the Laplacian, and then integrate by parts (limits $$\pm \infty$$ omitted): @@ -236,9 +231,7 @@ $$\begin{aligned} &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}} = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f} \end{aligned}$$ - -</div> -</div> +{% include proof/end.html id="proof-laplacian" %} diff --git a/source/know/concept/fundamental-solution/index.md b/source/know/concept/fundamental-solution/index.md index 312cc2e..947aada 100644 --- a/source/know/concept/fundamental-solution/index.md +++ b/source/know/concept/fundamental-solution/index.md @@ -42,11 +42,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-solution"/> -<label for="proof-solution">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-solution">Proof.</label> + +{% include proof/start.html id="proof-solution" -%} $$\hat{L}$$ only acts on $$x$$, so $$x' \in ]a, b[$$ is simply a parameter, meaning we are free to multiply the definition of $$G$$ by the constant $$f(x')$$ on both sides, @@ -72,8 +69,8 @@ $$\begin{aligned} By definition, $$\hat{L}$$'s response $$u(x)$$ to $$f(x)$$ satisfies $$\hat{L}\{ u(x) \} = f(x)$$, recognizable here. -</div> -</div> +{% include proof/end.html id="proof-solution" %} + While the impulse response is typically used for initial value problems, the fundamental solution $$G$$ is used for boundary value problems. @@ -117,11 +114,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-reciprocity"/> -<label for="proof-reciprocity">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-reciprocity">Proof.</label> + +{% include proof/start.html id="proof-reciprocity" -%} Consider two parameters $$x_1'$$ and $$x_2'$$. The self-adjointness of $$\hat{L}$$ means that: @@ -135,9 +129,7 @@ $$\begin{aligned} G^*(x_2', x_1') &= G(x_1', x_2') \end{aligned}$$ - -</div> -</div> +{% include proof/end.html id="proof-reciprocity" %} diff --git a/source/know/concept/greens-functions/index.md b/source/know/concept/greens-functions/index.md index ddba2cd..eda5671 100644 --- a/source/know/concept/greens-functions/index.md +++ b/source/know/concept/greens-functions/index.md @@ -21,6 +21,7 @@ but in general they are not the same, except in a special case, see below. + ## Single-particle functions If the two operators are single-particle creation/annihilation operators, @@ -146,11 +147,8 @@ $$\begin{gathered} G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t') \end{gathered}$$ -<div class="accordion"> -<input type="checkbox" id="proof-time-diff"/> -<label for="proof-time-diff">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-time-diff">Proof.</label> + +{% include proof/start.html id="proof-time-delta" -%} We will prove that the thermal expectation value $$\expval{\hat{A}(t) \hat{B}(t')}$$ only depends on $$t - t'$$ for arbitrary $$\hat{A}$$ and $$\hat{B}$$, @@ -189,8 +187,7 @@ because $$\hat{H}$$ is time-independent by assumption. Note that thermodynamic equilibrium is crucial: intuitively, if the system is not in equilibrium, then it evolves in some transient time-dependent way. -</div> -</div> +{% include proof/end.html id="proof-time-delta" %} If the Hamiltonian is both time-independent and non-interacting, then the time-dependence of $$\hat{c}_\nu$$ @@ -214,6 +211,7 @@ $$\begin{aligned} \end{aligned}$$ + ## As fundamental solutions In the absence of interactions, @@ -237,11 +235,8 @@ $$\begin{aligned} = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r}) \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-commH0"/> -<label for="proof-commH0">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-commH0">Proof.</label> + +{% include proof/start.html id="proof-commutator" -%} In the second quantization, the Hamiltonian $$\hat{H}_0$$ is written like so: @@ -307,9 +302,8 @@ $$\begin{aligned} &= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r}) = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r}) \end{aligned}$$ +{% include proof/end.html id="proof-commutator" %} -</div> -</div> After substituting this into the equation of motion, we recognize $$G^R(\vb{r}, t; \vb{r}', t')$$ itself: diff --git a/source/know/concept/gronwall-bellman-inequality/index.md b/source/know/concept/gronwall-bellman-inequality/index.md index 8096aaf..da1bcad 100644 --- a/source/know/concept/gronwall-bellman-inequality/index.md +++ b/source/know/concept/gronwall-bellman-inequality/index.md @@ -26,11 +26,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-original"/> -<label for="proof-original">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-original">Proof.</label> + +{% include proof/start.html id="proof-original" -%} We define $$w(t)$$ to equal the upper bounds above on both $$w'(t)$$ and $$w(t)$$ itself: @@ -63,8 +60,8 @@ $$\begin{aligned} Since $$u' \le \beta u$$ as a condition, the above derivative is always negative. -</div> -</div> +{% include proof/end.html id="proof-original" %} + Grönwall's inequality can be generalized to non-differentiable functions. Suppose we know: @@ -84,11 +81,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-integral"/> -<label for="proof-integral">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-integral">Proof.</label> + +{% include proof/start.html id="proof-integral" -%} We start by defining $$w(t)$$ as follows, which will act as shorthand: @@ -138,8 +132,8 @@ $$\begin{aligned} \end{aligned}$$ Insert this into the condition under which the Grönwall-Bellman inequality holds. -</div> -</div> +{% include proof/end.html id="proof-integral" %} + In the special case where $$\alpha(t)$$ is non-decreasing with $$t$$, the inequality reduces to: @@ -151,11 +145,8 @@ $$\begin{aligned} } \end{aligned}$$ -<div class="accordion"> -<input type="checkbox" id="proof-special"/> -<label for="proof-special">Proof</label> -<div class="hidden" markdown="1"> -<label for="proof-special">Proof.</label> + +{% include proof/start.html id="proof-special" -%} Starting from the "ordinary" Grönwall-Bellman inequality, the fact that $$\alpha(t)$$ is non-decreasing tells us that $$\alpha(s) \le \alpha(t)$$ for all $$s \le t$$, so: @@ -194,9 +185,7 @@ |