pnas201120955 1..6
... pnas201120955 1.. 6 Bacterial rheotaxis Marcosa,b, Henry C. Fuc,d, Thomas R. Powersd, and Roman ... ) (Materials and Methods) in an H = 90-μm deep microfluidic channel. This strain has a 1 × 3-μm sausage-shaped ... pnas201120955 1.. 6 ... pnas201120955 1.. 6 Bacterial rheotaxis Marcosa,b, Henry C. Fuc,d, Thomas R. Powersd, and Roman ... pnas201120955 1.. 6 ...
pcbi.1000623 1..6
... ). Equations [ 1]–[ 6] describe the dynamics of the model: [ 3]/[4] dynein binding/unbinding and decrease/increase ... pcbi.1000623 1.. 6 A Stochastic Model for Microtubule Motors Describes the In Vivo Cytoplasmic ... ). Equations [ 1]–[ 6] describe the dynamics of the model: [ 3]/[4] dynein binding/unbinding and decrease/increase ... pcbi.1000623 1.. 6 ... pcbi.1000623 1.. 6 ...
Serie 0
... 3 4 5 Answer F T T T F In this exercise, ∆ = ∂xx + ∂yy is the Laplace operator in R2. 1. If ∆u = 0 ... , then u( 0, 0) = 1pi . 1/ 6 ETH Zürich HS 2021 Mathematik III Solutions of problem set 12 d-chem Prof. Dr ... 3 4 5 Answer F T T T F In this exercise, ∆ = ∂xx + ∂yy is the Laplace operator in R2. 1. If ∆u = 0 ... , then u( 0, 0) = 1pi . 1/ 6 ETH Zürich HS 2021 Mathematik III Solutions of problem set 12 d-chem Prof. Dr ... Serie 0 ...
Serie 0
... := cos ( npi 3 )− cos (2npi3 ) is periodic with period 6 and its values are a1 = 1, a2 = 0, a3 = −2, a4 ... (pi, t) = 0 t > 0 , u(x, 0) = { 1 if pi3 ≤ x ≤ 2pi3 , 0 if x < pi3 or 2pi 3 < x . Solution: Since we ... := cos ( npi 3 )− cos (2npi3 ) is periodic with period 6 and its values are a1 = 1, a2 = 0, a3 = −2, a4 ... (pi, t) = 0 t > 0 , u(x, 0) = { 1 if pi3 ≤ x ≤ 2pi3 , 0 if x < pi3 or 2pi 3 < x . Solution: Since we ... Serie 0 ...
Serie 0
... solution is y(x) = Ae− 3 2x sin (√ 7 2 x ) +Be− 3 2x cos (√ 7 2 x ) + 16 sin(2x) . 1/ 6 ETH Zürich HS 2021 ... for the non-constant solutions of the ODE: y − log | 1 + y| = x 3 3 + C, C ∈ R. 3/ 6 ETH Zürich HS 2021 ... solution is y(x) = Ae− 3 2x sin (√ 7 2 x ) +Be− 3 2x cos (√ 7 2 x ) + 16 sin(2x) . 1/ 6 ETH Zürich HS 2021 ... for the non-constant solutions of the ODE: y − log | 1 + y| = x 3 3 + C, C ∈ R. 3/ 6 ETH Zürich HS 2021 ... Serie 0 ...
Serie 0
... ambiguity. You do not have to provide any justification for your answers. Question 1 2 3 Answer 25vηη ... PDE satisfied by v(ξ, η) = u(ξ(x, t), η(x, t)). 1. ξ = 4x, η = 5t. 2. ξ = x+ 3t, η = t. 3. ξ = 2x, η ... ambiguity. You do not have to provide any justification for your answers. Question 1 2 3 Answer 25vηη ... PDE satisfied by v(ξ, η) = u(ξ(x, t), η(x, t)). 1. ξ = 4x, η = 5t. 2. ξ = x+ 3t, η = t. 3. ξ = 2x, η ... Serie 0 ...
Sheet 0
... of solutions of (♠) satisfying y( 0) = 1 is a 3-dimensional vector space. � The space of solutions of ... modified: 6. Oktober 2022 1/ 3 ETH Zürich HS 2022 Analysis II Exercise Sheet 3 d-infk Prof. Dr. Özlem ... (♠) satisfying y( 0) = 1 is a 3-dimensional vector space. The space of solutions of (♠) satisfying limt→∞ y(t ... = 0, which satisfies the initial conditions y( 0) = y′( 0) = y′′( 0) = 0, y( 3)( 0) = 1 . *3.4. ODE with ... Sheet 0 ...
Serie 0
... ) = w(x− 2t, 0) + w(x+ 2t, 0)2 + 1 4 ∫ x+2t x−2t ( 6 cos(u) + 14 ) du = 12 ( 2(x− 2t)2 + (x− 2t) 3 24 ... + 2u(x+ ct)2 2 = 2x 2 + 8t2, 1 4 ∫ 2+2t x−2t 6 cos(u) du = 3 cos(x) sin(2t), and 1 4 ∫ t 0 ∫ x+2(t−τ) x ... ) = w(x− 2t, 0) + w(x+ 2t, 0)2 + 1 4 ∫ x+2t x−2t ( 6 cos(u) + 14 ) du = 12 ( 2(x− 2t)2 + (x− 2t) 3 24 ... + 2u(x+ ct)2 2 = 2x 2 + 8t2, 1 4 ∫ 2+2t x−2t 6 cos(u) du = 3 cos(x) sin(2t), and 1 4 ∫ t 0 ∫ x+2(t−τ) x ... Serie 0 ...
Serie 0
... series of fe takes the form a0 + ∞∑ n= 1 an cos(npix). Then a0 = ∫ 1 0 x( 1− x) dx = (x22 − x 3 3 )∣∣x= 1 x ... = 0 = 1 6 and for n ≥ 1, we have an = 2 ∫ 1 0 x( 1− x) cos(npix) dx = 2 ∫ 1 0 x cos(npix) dx− 2 ∫ 1 0 ... series of fe takes the form a0 + ∞∑ n= 1 an cos(npix). Then a0 = ∫ 1 0 x( 1− x) dx = (x22 − x 3 3 )∣∣x= 1 x ... = 0 = 1 6 and for n ≥ 1, we have an = 2 ∫ 1 0 x( 1− x) cos(npix) dx = 2 ∫ 1 0 x cos(npix) dx− 2 ∫ 1 0 ... Serie 0 ...
Serie 0
... ,∫ R x2f(x) dx = − d 2 dξ2 fˆ( 0) = 6 125 . 1/ 3 ETH Zürich HS 2021 Mathematik III Solutions of problem ... →+∞ e−aR−iξR −a− iξ︸ ︷︷ ︸ = 0 + 1 a+ iξ = − 1 a− iξ + 1 a+ iξ = − 2iξ a2 + ξ2 . 3/ 3 ... ,∫ R x2f(x) dx = − d 2 dξ2 fˆ( 0) = 6 125 . 1/ 3 ETH Zürich HS 2021 Mathematik III Solutions of problem ... →+∞ e−aR−iξR −a− iξ︸ ︷︷ ︸ = 0 + 1 a+ iξ = − 1 a− iξ + 1 a+ iξ = − 2iξ a2 + ξ2 . 3/ 3 ... Serie 0 ...
