Comments
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I see that for this problem, the two are identical because $(x-y)$/$(y-x)$ is squared, so you get the same result. I was just curious about the general formula for convolution. I would imagine there are some functions where the convolution would yield different results depending on which definition you used.
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(1) As we have shown in numerous previous problems, separation of variables leads to the solution $u(x,t) = X(x)T(t),$ where $$T(t)=e^{-k\lambda t}$$ and $$X(x)=a\cos\left(\sqrt{\lambda}x\right) + b\sin\left(\sqrt{\lambda}x\right).$$ From our boundary conditions, we know $X(0) = a = 0,$ so…
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(1) The solution to the unbounded heat problem $u_t = ku_xx$, where $u(x,0) = f(x)$ is $$u(x,t) = \int_{-\infty}^{\infty}f(y)G(y-x,t)dy = \frac{1}{\sqrt{4\pi k}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{t}}f(y)e^{-\frac{(y-x)^2}{4kt}}dy .$$ For $$f(x) = \begin{cases} M & |x|<\varepsilon \\ 0 & \text{else}, \end{cases},$$…
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@mark Also, I just realized: The ObservableHQ page on unbounded heat flow (and our notes) says that the solution to an unbounded heat flow problem is $$u(x,t) = \int_{-\infty}^{\infty} f(y)G(y-x,t)dy,$$ while the book writes the same equation as $$u(x,t) = \int_{-\infty}^{\infty} f(y)G(x-y,t)dy.$$ Which is correct?
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@mark Sure. $$u_1'(t) = 125\left(u_2(t)-2u_1(t)\right),$$ $$u_2'(t) = 125\left(u_3(t)-2u_2(t) + u_1(t)\right),$$ $$u_3'(t) = 125\left(u_4(t)-2u_3(t) + u_2(t)\right),$$ and $$u_4'(t) = 125\left(2-2u_4(t) + u_3(t)\right).$$
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(1) The difference quotient for $u_t = 5u_{xx}$ is $$u_t = 5u_{xx} = 5u_{xx} \approx 5\frac{u(x_i + h, t) - 2u(x_i, t) + u(x_i-h)}{h^2}.$$ Hence, our system of equations is $$u_i'(t) = 5\frac{u_{i+1}(t) - 2u_i(t) + u_{i-1}(t)}{\left(\frac{1-0}{5}\right)^2}.$$ for $i = 1, 2, 3, 4$ with $$u_i(0) = 2(\frac{i}{5})^2.$$ (2)
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(1) We start by separating variables, and assume $u(r,t) = R(r)T(t)$. Our PDE then becomes $$(RT)_t = (RT)_{rr} + \frac{1}{r}(RT)_r$$ or $$RT' = R''T + \frac{1}{r}R'T.$$ Dividing both sides by $RT$, we get $$\frac{T'}{T} = \frac{R''}{R} +\frac{1}{r}\frac{R'}{R} = -\lambda.$$ Thus, $\frac{T'}{T} = -\lambda$, so $T' =…
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We start by separating variables, and assume $u(x,t) = X(x)T(t).$ Our partial differential equation then becomes $(XT)_{tt} = (9XT)_{xx}$ or $XT'' = 9X''T." Rearranging, $$\frac{T''}{9T} = \frac{X''}{X} = -\lambda.$$ We have solved the equation $X'' = -\lambda X$ , where $X(0) = X(L) = 0$ many times before, so we know the…
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My matrix was: matrix = [ [0,1,4,3,3], [2,0,2,4,3], [3,3,0,3,2], [2,1,1,0,2], [3,1,3,1,0] ] and my ranking was: Team 3: rating = 0.5054388063612655 Team 1: rating = 0.5016207712530641 Team 2: rating = 0.48157957363128245 Team 5: rating = 0.4108981576640922 Team 4: rating = 0.30356553355238997 Team 3 was the best.
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@mark Um, I dunno. The other problem had n=4, so I guess that works for this one too. Are you saying we can use any partition size we want?
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(1) First, we find the exact solution to compare our numerical approximation to. Integrating $u''(x)=6$ twice yields $$u(x)=3x^2+Cx+D.$$ Now we plug in our initial conditions: $$u(0)=D=0$$, and $$u(2)=3(2)^2+2C=0.$$ Therefore, $$u(x)=3x^2-6x.$$ Now we find an approximate numerical solution using the difference quotient…
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To show that two functions $f(x)$ and $g(x)$ are orthogonal over an interval $[-L,L],$ we must show that $$\left\langle f(x),g(x) \right\rangle = \int_{-L}^Lf(x)g(x)dxdx = 0.$$ $$\left\langle 1,x \right\rangle = \int_{-1}^1(1)xdx = \left[\frac{x^2}{2}\right]_{-1}^1 = \frac{1}{2}-\frac{1}{2} = 0.$$ So $1$ and $x$ are…
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The full Fourier series for $f(x)=x^2$ on $[-\pi,\pi]$ is $$\frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos(nx)+b_n\sin(nx)\right),$$ where $$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx$$ and $$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\sin(nx)dx.$$ First, we find $\frac{a_0}{2}$, taking advantage of the fact that for…
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Using separation of variables, our partial differential equation becomes $$(XT)_t = 2(XT)_{xx},$$ or $$XT' = 2X''T.$$ Rearranging, $$\frac{T'}{2T} = \frac{X''}{X} = -\lambda.$$ Thus, $T' = -2\lambda T,$ so $$T= e^{-2\lambda t},$$ while $X'' = -\lambda X,$ so $$X = a\cos\left(\sqrt{\lambda}x\right) +…
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$$b_n = 2\int_{0}^{L} 4(x - x^2)\sin(n\pi x)dx$$ should be $$b_n = 2\int_{0}^{1} 4(x - x^2)\sin(n\pi x)dx.$$
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According to our Fourier cosine series calculator, $$\sin(x) \approx \frac{2}{\pi} + \sum_{n=1}^{\infty}\left(\begin{cases} \frac{2\left((-1)^n +1\right)}{\pi \left(n^2 - 1\right)} & \text{for $n\neq 1$} \\ 0 & \text{otherwise} \end{cases}\right)\cos(nx) = \frac{2}{\pi} - \frac{2}{\pi}\sum_{n=1}^{\infty}\left(\begin{cases}…
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$u_t = u_{xx}$ is the classic diffusion equation, representing flow from areas of higher density to areas of lower density. The $-ku$ term represents decay, with the rate of decay of $u$ directly proportional to $u$ by a factor of $k.$ See the attached image below. (Unfortunately, the website does not seem to allow placing…
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And $$u(x,t) = \sum_{n=1}^{\infty}b_ne^{-\frac{n^2\pi^2}{8}t}\sin\left(\frac{x\pi x}{4}\right)$$ should be $$u(x,t) = \sum_{n=1}^{\infty}b_ne^{-\frac{n^2\pi^2}{8}t}\sin\left(\frac{n\pi x}{4}\right).$$
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Solving the differential equation $y'' = -\lambda y$, $$y = a\cos\left(\sqrt{\lambda}x\right) + b\sin\left(\sqrt{\lambda}x\right)$$ and $$y' = -a\sqrt{\lambda}\sin\left(\sqrt{\lambda}x\right) + b\sqrt{\lambda}\cos\left(\sqrt{\lambda}x\right).$$ From our initial conditions, $$y(0) + y'(0) = a + b\sqrt{\lambda} = 0,$$ so $$a…
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The Fourier series for $f(x)$ over the interval $[0,\frac{1}{2}]$ is $$\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n\cos(2\pi nx),$$ where $$a_n = 4\int_{0}^{\frac{1}{2}}x^2(1-x)^2\cos(2\pi nx)dx.$$ By using Mathematica to evaluate this integral for us, we find that $a_0 = \frac{1}{15}$ and $a_n = -\frac{3}{n^4\pi^4}$. Hence,…
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Whoops, the middle term in the penultimate line should be multiplied by $\frac{1}{2}$.
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We start by separating variables, and assume $u(x,t) = X(x)T(t).$ Our partial differential equation then becomes $(XT)_t = (2XT)_{xx}$, or $XT' = 2X''T.$ Rearranging, $$\frac{T'}{2T} = \frac{X''}{X}.$$ For this equation to be true, both sides must equal a constant, which we denote $-\lambda.$ Thus, $\frac{T'}{2T} =…
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The Fourier sine series of $f(x)$ over $[0,L]$ is $$f(x) = \sum_{n=1}^{\infty} b_n\sin\left(\frac{n\pi x}{L}\right),$$ where $$b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right).$$ For $f(x) = 1$ and $L = 1$, $$b_n = 2\int_{0}^{1} \sin(n\pi x) = -\frac{2}{n\pi}\cos(n\pi x)\Bigr|_{0}^{1} =$$ $$=…
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@dualrey I think so. If we know that definite integrals of nonnegative integrands are nonnegative, than it follows that for the original equation to be true, $\lambda$ must be negative for both sides to have the same sign.
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Thanks for the great explanation, @dualrey !
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@dualrey It depends on the domain, I would think. The integral of $x^2$, which is always positive, is $\frac{x^3}{3}$, which is negative for negative values of $x$.
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With $\kappa = 0$ and $f = 0$, the steady state temperature at the lower right corner of the bar is 0.84905.
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Let $u:\mathbb R^3\xrightarrow{}\mathbb R$. Then $$\nabla\cdot\nabla u = \left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle \cdot \left\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle u$$ $$ = \left\langle…
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An answer close to the point of maximum temperature can be obtained using the Heat Explorer tool, but it may be difficult to locate the precise point using the graph. Here was the point of greatest temperature that I could locate on the graph, $(0.55508, 0.83179)$: Here, we will calculate the exact point through calculus,…
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Yes, that seemed to be the problem. Thanks, @mark