3

The identity you're missing from Gauss' divergence theorem is: $$ \int_\Omega \nabla \varphi \cdot\mathbf{v} = -\int_\Omega \varphi\nabla\cdot\mathbf{v} +\int_{\partial\Omega}\varphi\mathbf{v\cdot n} $$ where I've written $\varphi$ as an arbitrary scalar field. So, using the divergence of $\mathbf{u}$ as the scalar field you'd get $$ -\int_\Omega(\lambda+\mu)...


3

The general form of the equation is $$ \frac{\partial \sigma_{ij}}{\partial x_j} + F_i = \rho \frac{\partial^2 U_i}{\partial t^2} $$ where the stress is given by $$ \sigma_{ij} = \sigma_{ij}(U) = 2 \mu \varepsilon_{ij} + \lambda \varepsilon_{kk} \delta_{ij}, \qquad \varepsilon_{ij} = \varepsilon_{ij}(U) = \frac{1}{2}\left( \frac{\partial U_i}{\partial x_j} + ...


2

The given $$ \int_\Omega v \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\right) dx dy = 0 $$ becomes $$ \int_\Omega \frac{\partial}{\partial x}\left[v\left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\right) \right] dx dy - \int_\Omega \frac{\partial v}{\partial x}\left( \frac{\partial u}{\...


1

You are approximating a definite integral with cumtrapz it won't give you the same result as the integrated equation unless you add a constant and plot with the given x coordinates: import numpy as np import matplotlib.pyplot as plt import scipy.integrate as it x = np.arange(-10,10, 0.01) # start,stop,step f = x**2 f_int=it.cumtrapz(f,x, initial=0) plt....


1

First one should verify the numerical solution for $a(t)$ against the analytic solution. $ d_t a = 1/a + 1/a^2 = \frac{a+1}{a^2} $ Use $b=a+1$, then the ODE becomes $ \frac{(b-1)^2}{b} db = dt, $ from which we find the solution $ b^2/2 - 2 b - \ln(b) = t $ This relation (defined up to a constant to satisfy the initial conditions) is not invertible but ...


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