First, the form of the complementary solution must be determined in order to make sure that the particular solution does not have duplicate terms. $x''+2x'+4x=9\sin(t)$. See Figure \(\PageIndex{1}\). The code implementation is the intellectual property of the developers. 12. x +6x +13x = 10sin5t;x(0) = x(0) = 0 Previous question Next question For \(c>0\), the complementary solution \(x_c\) will decay as time goes by. \end{equation}, \begin{equation*} Passing negative parameters to a wolframscript. 0000004233 00000 n The temperature \(u\) satisfies the heat equation \(u_t=ku_{xx}\), where \(k\) is the diffusivity of the soil. We will not go into details here. First of all, what is a steady periodic solution? The steady periodic solution \(x_{sp}\) has the same period as \(F(t)\). 0000007965 00000 n \right) }\), \(\sin (\frac{\omega L}{a}) = 0\text{. \newcommand{\gt}{>} Here our assumption is fine as no terms are repeated in the complementary solution. I don't know how to begin. are almost the same (minimum step is 0.1), then start again. = \frac{2\pi}{31,557,341} \approx 1.99 \times {10}^{-7}\text{. which exponentially decays, so the homogeneous solution is a transient. Steady state solution for a differential equation, solving a PDE by first finding the solution to the steady-state, Natural-Forced and Transient-SteadyState pairs of solutions. 0000009344 00000 n That is because the RHS, f(t), is of the form $sin(\omega t)$. \frac{\cos (1) - 1}{\sin (1)} P - transition matrix, contains the probabilities to move from state i to state j in one step (p i,j) for every combination i, j. n - step number. Suppose that \( k=2\), and \( m=1\). We did not take that into account above. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. nor assume any liability for its use. \[\label{eq:1} \begin{array}{ll} y_{tt} = a^2 y_{xx} , & \\ y(0,t) = 0 , & y(L,t) = 0 , \\ y(x,0) = f(x) , & y_t(x,0) = g(x) . \cos (x) - To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Legal. The units are again the mks units (meters-kilograms-seconds). \left(\cos \left(\omega t - \sqrt{\frac{\omega}{2k}}\, x\right) + B \sin x $$D[x_{inhomogeneous}]= f(t)$$. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. 0000001950 00000 n & y(0,t) = 0 , \quad y(1,t) = 0 , \\ So $~ = -0.982793723 = 2.15879893059 ~$. Why did US v. Assange skip the court of appeal? Let \(u(x,t)\) be the temperature at a certain location at depth \(x\) underground at time \(t\text{. \end{equation}, \begin{equation*} Let us do the computation for specific values. which exponentially decays, so the homogeneous solution is a transient. }\) Find the particular solution. = \newcommand{\mybxbg}[1]{\boxed{#1}} Suppose \(h\) satisfies (5.12). The steady periodic solution is the particular solution of a differential equation with damping. \end{equation}, \begin{equation*} Then, \[ y_p(x,t)= \left( \cos(x)- \frac{ \cos(1)-1 }{ \sin(1)}\sin(x)-1 \right) \cos(t). \cos \left( \frac{\omega}{a} x \right) - You might also want to peruse the web for notes that deal with the above. in the form Similar resonance phenomena occur when you break a wine glass using human voice (yes this is possible, but not easy\(^{1}\)) if you happen to hit just the right frequency. Then our wave equation becomes (remember force is mass times acceleration). Hint: You may want to use result of Exercise5.3.5. The units are cgs (centimeters-grams-seconds). y = \newcommand{\unit}[2][\!\! A home could be heated or cooled by taking advantage of the above fact. Let \(x\) be the position on the string, \(t\) the time, and \(y\) the displacement of the string. We have already seen this problem in chapter 2 with a simple \(F(t)\). We will also assume that our surface temperature swing is \(\pm 15^{\circ}\) Celsius, that is, \(A_0=15\). The general solution is, The endpoint conditions imply \(X(0) = X(L) = 0\text{. \begin{array}{ll} It is very important to be able to study how sensitive the particular model is to small perturbations or changes of initial conditions and of various paramters. \nonumber \], \[\label{eq:20} u_t=ku_{xx,}~~~~~~u(0,t)=A_0\cos(\omega t). \nonumber \]. for the problem ut = kuxx, u(0, t) = A0cos(t). We now plug into the left hand side of the differential equation. \end{equation*}, \begin{equation*} the authors of this website do not make any representation or warranty, %PDF-1.3 % Can you still use Commanders Strike if the only attack available to forego is an attack against an ally? The full solution involves elliptic integrals, whereas the small angle approximation creates a much simpler differential equation. - \cos x + Let us assume say air vibrations (noise), for example a second string. The calculation above explains why a string begins to vibrate if the identical string is plucked close by. \cos \left( \frac{\omega}{a} x \right) - 0000001664 00000 n If we add the two solutions, we find that \(y = y_c + y_p\) solves (5.7) with the initial conditions. u(x,t) = \operatorname{Re} h(x,t) = What differentiates living as mere roommates from living in a marriage-like relationship? Find the particular solution. That is why wines are kept in a cellar; you need consistent temperature. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. \cos(n \pi x ) - }\), \(g(x) = -\frac{\partial y_p}{\partial t}(x,0) = 0\text{. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. ~~} For \(k=0.005\text{,}\) \(\omega = 1.991 \times {10}^{-7}\text{,}\) \(A_0 = 20\text{. We also assume that our surface temperature swing is \(\pm {15}^\circ\) Celsius, that is, \(A_0 = 15\text{. Find more Education widgets in Wolfram|Alpha. See Figure5.3. 0000002614 00000 n 0000025477 00000 n and after differentiating in \(t\) we see that \(g(x) = -\frac{\partial y_p}{\partial t}(x,0) = 0\text{. That is, the hottest temperature is \(T_0 + A_0\) and the coldest is \(T_0 - A_0\text{. }\), \(e^{(1+i)\sqrt{\frac{\omega}{2k}} \, x}\), \(e^{-(1+i)\sqrt{\frac{\omega}{2k}} \, x}\), \(\omega = \frac{2\pi}{\text{seconds in a year}} See Figure \(\PageIndex{1}\) for the plot of this solution. I want to obtain $$x(t)=x_H(t)+x_p(t)$$ so to find homogeneous solution I let $x=e^{mt}$, and find. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Could a subterranean river or aquifer generate enough continuous momentum to power a waterwheel for the purpose of producing electricity? \(A_0\) gives the typical variation for the year. Does a password policy with a restriction of repeated characters increase security? As \(\sqrt{\frac{k}{m}}=\sqrt{\frac{18\pi ^{2}}{2}}=3\pi\), the solution to \(\eqref{eq:19}\) is, \[ x(t)= c_1 \cos(3 \pi t)+ c_2 \sin(3 \pi t)+x_p(t) \nonumber \], If we just try an \(x_{p}\) given as a Fourier series with \(\sin (n\pi t)\) as usual, the complementary equation, \(2x''+18\pi^{2}x=0\), eats our \(3^{\text{rd}}\) harmonic. \nonumber \]. So I've done the problem essentially here? We get approximately \(700\) centimeters, which is approximately \(23\) feet below ground. Again, these are periodic since we have $e^{i\omega t}$, but they are not steady state solutions as they decay proportional to $e^{-t}$. }\) Find the depth at which the temperature variation is half (\(\pm 10\) degrees) of what it is on the surface. Below, we explore springs and pendulums. Hooke's Law states that the amount of force needed to compress or stretch a spring varies linearly with the displacement: The negative sign means that the force opposes the motion, such that a spring tends to return to its original or equilibrium state. Once you do this you can then use trig identities to re-write these in terms of c, $\omega$, and $\alpha$. Now we get to the point that we skipped. Free function periodicity calculator - find periodicity of periodic functions step-by-step We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. That is, we try, \[ x_p(t)= a_3 t \cos(3 \pi t) + b_3 t \sin(3 \pi t) + \sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } b_n \sin(n \pi t). \right) . \end{equation*}, \begin{equation*} B_n \sin \left( \frac{n\pi a}{L} t \right) \right) Now we can add notions of globally asymptoctically stable, regions of asymptotic stability and so forth. in the sense that future behavior is determinable, but it depends We will employ the complex exponential here to make calculations simpler. Just like when the forcing function was a simple cosine, resonance could still happen. The earth core makes the temperature higher the deeper you dig, although you need to dig somewhat deep to feel a difference. If we add the two solutions, we find that \(y=y_c+y_p\) solves \(\eqref{eq:3}\) with the initial conditions. }\), \(y(x,t) = \frac{F(x+t) + F(x-t)}{2} + \left( \cos (x) - Extracting arguments from a list of function calls. Extracting arguments from a list of function calls. The homogeneous form of the solution is actually \sum\limits_{\substack{n=1 \\ n \text{ odd}}}^\infty y_p(x,t) = \cos \left(\omega t - \sqrt{\frac{\omega}{2k}}\, x\right) . }\), \(\pm \sqrt{i} = \pm Could Muslims purchase slaves which were kidnapped by non-Muslims? Learn more about Stack Overflow the company, and our products. 0000004497 00000 n \end{aligned}\end{align} \nonumber \], \[ 2x_p'' +18 \pi^2 x= -12a_3 \pi \sin(3 \pi t)+ 12b_3 \pi \cos(3 \pi t) +\sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } (-2n^2 \pi^2 b_n+ 18 \pi^2 b_n) \sin(n \pi t.) \nonumber \]. 0000006517 00000 n We then find solution \(y_c\) of \(\eqref{eq:1}\). {{}_{#2}}} The amplitude of the temperature swings is \(A_0e^{- \sqrt{\frac{\omega}{2k}}x}\). periodic steady state solution i (r), with v (r) as input. ]{#1 \,\, {{}^{#2}}\!/\! \]. First we find a particular solution \(y_p\) of (5.7) that satisfies \(y(0,t) = y(L,t) = 0\text{. Since the force is constant, the higher values of k lead to less displacement. Let's see an example of how to do this. 0000008732 00000 n Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Find the steady periodic solution to the differential equation Let us assume for simplicity that, where \(T_0\) is the yearly mean temperature, and \(t=0\) is midsummer (you can put negative sign above to make it midwinter if you wish). Then the maximum temperature variation at \(700\) centimeters is only \(\pm 0.66^{\circ}\) Celsius. We define the functions \(f\) and \(g\) as, \[f(x)=-y_p(x,0),~~~~~g(x)=- \frac{\partial y_p}{\partial t}(x,0). We see that the homogeneous solution then has the form of decaying periodic functions: The natural frequencies of the system are the (angular) frequencies \(\frac{n \pi a}{L}\) for integers \(n \geq 1\text{. Find all the solution (s) if any exist. Similar resonance phenomena occur when you break a wine glass using human voice (yes this is possible, but not easy1) if you happen to hit just the right frequency. Connect and share knowledge within a single location that is structured and easy to search. \nonumber \]. Differential calculus is a branch of calculus that includes the study of rates of change and slopes of functions and involves the concept of a derivative. 0000004192 00000 n u_t = k u_{xx}, \qquad u(0,t) = A_0 \cos ( \omega t) .\tag{5.11} \end{array} \right.\end{aligned}\end{align} \nonumber \], \[ F(t)= \dfrac{1}{2}+ \sum^{\infty}_{ \underset{n ~\rm{odd}}{n=1} }\dfrac{2}{\pi n} \sin(n \pi t). Practice your math skills and learn step by step with our math solver. \]. We get approximately 700 centimeters, which is approximately 23 feet below ground. 0000005765 00000 n \begin{aligned} \frac{1+i}{\sqrt{2}}\), \(\alpha = \pm (1+i)\sqrt{\frac{\omega}{2k}}\text{. The natural frequencies of the system are the (circular) frequencies \(\frac{n\pi a}{L}\) for integers \(n \geq 1\). $$X_H=c_1e^{-t}sin(5t)+c_2e^{-t}cos(5t)$$ \nonumber \], The particular solution \(y_p\) we are looking for is, \[ y_p(x,t)= \frac{F_0}{\omega^2} \left( \cos \left( \frac{\omega}{a}x \right)- \frac{ \cos \left( \frac{\omega L}{a} \right)-1 }{ \sin \left( \frac{\omega L}{a} \right)}\sin \left( \frac{\omega}{a}x \right)-1 \right) \cos(\omega t). We see that the homogeneous solution then has the form of decaying periodic functions: and after differentiating in \( t\) we see that \(g(x)=- \frac{\partial y_P}{\partial t}(x,0)=0\). The steady periodic solution is the particular solution of a differential equation with damping. \nonumber \], We will need to get the real part of \(h\), so we apply Eulers formula to get, \[ h(x,t)=A_0e^{- \sqrt{\frac{\omega}{2k}}x} \left( \cos \left( \omega t - \sqrt{\frac{\omega}{2k}x} \right) +i \sin \left( \omega t - \sqrt{\frac{\omega}{2k}x} \right) \right). \nonumber \]. \sum_{n=1}^\infty \left( A_n \cos \left( \frac{n\pi a}{L} t \right) + For example, it is very easy to have a computer do it, unlike a series solution. 11. it is more like a vibraphone, so there are far fewer resonance frequencies to hit. \end{equation}, \begin{equation} Suppose that \(L=1\text{,}\) \(a=1\text{. How to force Unity Editor/TestRunner to run at full speed when in background? 3.6 Transient and steady periodic solutions example Part 1 DarrenOngCL 2.67K subscribers Subscribe 43 8.1K views 8 years ago We work through an example of calculating transient and steady. h_t = k h_{xx}, \qquad h(0,t) = A_0 e^{i\omega t} .\tag{5.12} Symbolab is the best calculus calculator solving derivatives, integrals, limits, series, ODEs, and more. \newcommand{\amp}{&} You must define \(F\) to be the odd, 2-periodic extension of \(y(x,0)\text{. A_0 e^{-(1+i)\sqrt{\frac{\omega}{2k}} \, x + i \omega t} Note that there now may be infinitely many resonance frequencies to hit. \sin( n \pi x) 11. When \(c>0\), you will not have to worry about pure resonance. This page titled 5.3: Steady Periodic Solutions is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Ji Lebl via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. \end{equation*}, \begin{equation*} }\) Suppose that the forcing function is the square wave that is 1 on the interval \(0 < x < 1\) and \(-1\) on the interval \(-1 < x< 0\text{. + B \sin \left( \frac{\omega L}{a} \right) - Let us say \(F(t) = F_0 \cos (\omega t)\) as force per unit mass. The general solution is x = C1cos(0t) + C2sin(0t) + F0 m(2 0 2)cos(t) or written another way x = Ccos(0t y) + F0 m(2 0 2)cos(t) Hence it is a superposition of two cosine waves at different frequencies. Check that \(y=y_c+y_p\) solves \(\eqref{eq:3}\) and the side conditions \(\eqref{eq:4}\). 0000001972 00000 n A plot is given in Figure5.4. X(x) = A \cos \left( \frac{\omega}{a} x \right) So the big issue here is to find the particular solution \(y_p\). The above calculation explains why a string will begin to vibrate if the identical string is plucked close by. }\), \(A_0 e^{-\sqrt{\frac{\omega}{2k}} x}\text{. What is this brick with a round back and a stud on the side used for? Episode about a group who book passage on a space ship controlled by an AI, who turns out to be a human who can't leave his ship? x_p''(t) &= -A\sin(t) - B\cos(t)\cr}$$, $$(-A - 2B + 4A)\sin(t) + (-B + 2A + 4B)\cos(t) = 9\sin(t)$$, $$\eqalign{3A - 2B &= 1\cr What is Wario dropping at the end of Super Mario Land 2 and why? }\) See Figure5.5. 4.1.8 Suppose x + x = 0 and x(0) = 0, x () = 1. \end{equation*}, \begin{equation} To a differential equation you have two types of solutions to consider: homogeneous and inhomogeneous solutions. \], We will employ the complex exponential here to make calculations simpler. The equation, \[ x(t)= A \cos(\omega_0 t)+ B \sin(\omega_0 t), \nonumber \]. We then find solution \(y_c\) of (5.6). \[ i \omega Xe^{i \omega t}=kX''e^{i \omega t}. & y_{tt} = y_{xx} , \\ [Graphing Calculator] In each of Problems 11 through 14, find and plot both the steady periodic solution xsp(t)= C cos(t) of the given differential equation and the actual solution x(t)= xsp(t)+xtr(t) that satisfies the given initial conditions. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. }\) For example if \(t\) is in years, then \(\omega = 2\pi\text{. y(0,t) = 0 , & y(L,t) = 0 , \\ \], \[ X(x)= \frac{F_0}{\omega^2} \left( \cos \left( \frac{\omega}{a}x \right)- \frac{ \cos \left( \frac{\omega L}{a} \right)-1 }{ \sin \left( \frac{\omega L}{a} \right)}\sin \left( \frac{\omega}{a}x \right)-1 \right). Therefore, we are mostly interested in a particular solution \(x_p\) that does not decay and is periodic with the same period as \(F(t)\). \end{equation*}, \begin{equation*} \right) \end{equation*}, \begin{equation} Taking the tried and true approach of method of characteristics then assuming that $x~e^{rt}$ we have: See Figure \(\PageIndex{3}\). And how would I begin solving this problem? \nonumber \], The steady periodic solution has the Fourier series, \[ x_{sp}(t)= \dfrac{1}{4}+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \dfrac{2}{\pi n(2-n^2 \pi^2)} \sin(n \pi t). \nonumber \], \[ x_p''(t)= -6a_3 \pi \sin(3 \pi t) -9 \pi^2 a_3 t \cos(3 \pi t) + 6b_3 \pi \cos(3 \pi t) -9 \pi^2 b_3 t \sin(3 \pi t) +\sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } (-n^2 \pi^2 b_n) \sin(n \pi t). }\) We studied this setup in Section4.7. original spring code from html5canvastutorials. it is more like a vibraphone, so there are far fewer resonance frequencies to hit. Comparing we have $$A=-\frac{18}{13},~~~~B=\frac{27}{13}$$ +1 , Contact | Let \(u(x,t)\) be the temperature at a certain location at depth \(x\) underground at time \(t\). This function decays very quickly as \(x\) (the depth) grows. We want a theory to study the qualitative properties of solutions of differential equations, without solving the equations explicitly. \frac{F(x+t) + F(x-t)}{2} + The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Is there a generic term for these trajectories? 0000001526 00000 n Let us say \(F(t)=F_0 \cos(\omega t)\) as force per unit mass. The frequency \(\omega\) is picked depending on the units of \(t\), such that when \(t=1\), then \(\omega t=2\pi\). }\) Then the maximum temperature variation at 700 centimeters is only \(\pm {0.66}^\circ\) Celsius. If you want steady state calculator click here Steady state vector calculator. \cos ( \omega t) . ]{#1 \,\, #2} Accessibility StatementFor more information contact us atinfo@libretexts.org. \nonumber \], \[ F(t)= \sum^{\infty}_{ \underset{n ~\rm{odd}}{n=1} } \dfrac{4}{\pi n} \sin(n \pi t). \nonumber \], We plug into the differential equation and obtain, \[\begin{align}\begin{aligned} x''+2x &= \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \left[ -b_n n^2 \pi^2 \sin(n \pi t) \right] +a_0+2 \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \left[ b_n \sin(n \pi t) \right] \\ &= a_0+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} b_n(2-n^2 \pi^2) \sin(n \pi t) \\ &= F(t)= \dfrac{1}{2}+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \dfrac{2}{\pi n} \sin(n \pi t).\end{aligned}\end{align} \nonumber \], So \(a_0= \dfrac{1}{2}\), \(b_n= 0\) for even \(n\), and for odd \(n\) we get, \[ b_n= \dfrac{2}{\pi n(2-n^2 \pi^2)}. \sin (x) & y(x,0) = - \cos x + B \sin x +1 , \\ h(x,t) = Hb```f``k``c``bd@ (.k? o0AO T @1?3l +x#0030\``w``J``:"A{uE '/%xfa``KL|& b)@k Z wD#h endstream endobj 511 0 obj 179 endobj 474 0 obj << /Type /Page /Parent 470 0 R /Resources << /ColorSpace << /CS2 481 0 R /CS3 483 0 R >> /ExtGState << /GS2 505 0 R /GS3 506 0 R >> /Font << /TT3 484 0 R /TT4 477 0 R /TT5 479 0 R /C2_1 476 0 R >> /ProcSet [ /PDF /Text ] >> /Contents [ 486 0 R 488 0 R 490 0 R 492 0 R 494 0 R 496 0 R 498 0 R 500 0 R ] /MediaBox [ 0 0 612 792 ] /CropBox [ 0 0 612 792 ] /Rotate 0 /StructParents 0 >> endobj 475 0 obj << /Type /FontDescriptor /Ascent 891 /CapHeight 656 /Descent -216 /Flags 34 /FontBBox [ -568 -307 2028 1007 ] /FontName /DEDPPC+TimesNewRoman /ItalicAngle 0 /StemV 94 /XHeight 0 /FontFile2 503 0 R >> endobj 476 0 obj << /Type /Font /Subtype /Type0 /BaseFont /DEEBJA+SymbolMT /Encoding /Identity-H /DescendantFonts [ 509 0 R ] /ToUnicode 480 0 R >> endobj 477 0 obj << /Type /Font /Subtype /TrueType /FirstChar 32 /LastChar 126 /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 0 250 278 500 500 500 500 500 500 500 500 500 500 278 278 0 0 564 0 0 722 667 667 0 611 556 0 722 333 0 0 0 0 722 0 0 0 0 556 611 0 0 0 0 0 0 0 0 0 0 0 0 444 500 444 500 444 333 500 500 278 0 500 278 778 500 500 500 500 333 389 278 500 500 722 500 500 444 0 0 0 541 ] /Encoding /WinAnsiEncoding /BaseFont /DEDPPC+TimesNewRoman /FontDescriptor 475 0 R >> endobj 478 0 obj << /Type /FontDescriptor /Ascent 891 /CapHeight 0 /Descent -216 /Flags 98 /FontBBox [ -498 -307 1120 1023 ] /FontName /DEEBIF+TimesNewRoman,Italic /ItalicAngle -15 /StemV 0 /XHeight 0 /FontFile2 501 0 R >> endobj 479 0 obj << /Type /Font /Subtype /TrueType /FirstChar 65 /LastChar 120 /Widths [ 611 611 667 0 0 611 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 500 0 444 500 444 0 0 500 278 0 444 0 722 500 500 500 0 389 389 278 0 444 667 444 ] /Encoding /WinAnsiEncoding /BaseFont /DEEBIF+TimesNewRoman,Italic /FontDescriptor 478 0 R >> endobj 480 0 obj << /Filter /FlateDecode /Length 270 >> stream
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