Comments
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As can be observed in the screenshot above, the temperature near the midpoint of the insulated edge of the triangle (see my randomly generated problem for details of the model) is 1.04558.
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The temperature just above the door, u(0.15223,-0.54970) = .08239
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Let $w$ be a scalar field and let $\vec{\varphi}$ be a vector field. Then $$\nabla\cdot\left(w\vec{\varphi}\right) = \left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle\cdot w\left\langle\varphi_x,\varphi_y,\varphi_z\right\rangle.$$ Since $w$ is a scalar, we…
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The solution to the PDE is $$u = u_0(x - 2t),$$ so at each time interval, the graph above will shift to the right by two units. Here are sketches of the graph at each of the three times: If we changed the PDE to $u_t + 2u_x = -u$, the solution becomes $$u = u_0(x - 2t)e^{-t},$$ so $u$ will decay over time, approaching zero…
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This is an example of as advection equation of the form $$u_t + cu_x +au = f(x,t)$$ $$u(x,0) = u_0,$$ where $c = 1$, $a = -3$, $f(x,t) = 0$, and $u_0 = x^2$. To solve this equation, we rewrite $u(x,t)$ as $U(\xi,\tau)$, where $\xi = x-ct = x-t$ and $\tau = t.$ By applying the chain rule, the PDE becomes $$(U_{\xi}(-1) +…
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A metal bar of length 1 lies along the unit interval. Its temperature distribution is given by $$g(x) = 4x^2 - 2x.$$ At time $t =0 $, its left end is set to temperature 1 and its right end to 0. Here is a sketch of the the temperature distribution $u(x,t)$ at times $t = 0$, $t = 0.01$, $t = 0.1$, and $t = 10$.
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Recall our basic conservation law expressed as a PDE: $$u_t + \varphi_x = f.$$ Here we assume the source $f = 0$ and the flux $\varphi = \frac{1}{2}u^2$. Taking the derivative of the flux with respect to $x$ by using the chain rule, we find $\varphi_x = uu_x$. Plugging these variables back into the original equation, we…
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Consider the diffusion-like equation $$u_t = k(t)u_{xx}.$$ We are going to substitute $u(x,\tau)$ for $u(x,t)$ on both sides of the equation. We are allowed to do this because the function $u$ remains unchanged; it is simply rewritten in terms of a different variable. Note that $\tau$ is not a function of x, so…
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Heat flow with a constant internal heat source is governed by $$u_t = 2u_{xx} + 6, u(0,t) = -1, u(1,t) = 4.$$ For a steady state temperature distribution, $u(x,t)$ does not change with respect to time, so $u_t = 0$. Hence, $2u_{xx} + 6 = 0$. Solving this equation for $u_{xx}$ yields $u_{xx} = -3$, and then integrating…
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Thanks for the answer @flappy_bird. I don’t see how $g(x)$ can tell you $u(x,0)$, because if you plug in $x = 0$ and $x = 1$ into $g(x)$, you won’t get the values the problem gives you for the endpoints. What I’m thinking is that it’s supposed to be the distribution before $t = 0$, that is, before the endpoints are set at…
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The displacement of the midpoint from equilibrium at time t=1.1s is 0.010.