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Thermal process in a thin homogeneous rod length $ l $ in the time interval (0;T] . The lateral surface of the rod is thermally insulated. The temperature of one of the ends of the rod is constant and equal $ u{}_0$ . At the other end, there is heat exchange with the environment described by Newton's law with the coefficient of heat transfer $ \alpha $.The initial temperature is distributed according to law $u(x, 0) = \eta(x) $.

My mathematical model: $$ u (x,t){}_t = \frac{k}{c} u(x,t){}_{xx}, 0 <x < l, 0 < t < T $$ initial condition: $$ u(x,0) = \eta(x) , 0 < x < l $$ Border conditions: $$ u(0,t){}_x = 0, 0 < t < T $$ $$ u(l,t){}_x = -\frac{\alpha}{k}(u - u{}_0), 0 < t < T $$ I'm doing a replacement $ w = u - u{}_0 $

I am trying to make a Fourier expansion in eigenfunctions of the Laplace operator. I got decomposition: $$ w(x,t) = \sum\nolimits_{n=1}^\infty cos(\sqrt{\lambda{}_n} x) A{}_n e^{\frac{-k \lambda{}_n t}{c}} $$ After substituting the initial conditions, I obtained:

$$ w(x,0) = \eta(x) - u{}_0 = \sum\nolimits_{n=1}^\infty cos(\sqrt{\lambda{}_n} x) A{}_n $$ is there any property or technique for finding the coefficients A? I apologize for the stupid question.

Lionell
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  • What's $\sqrt{\lambda_n}$? – zahbaz Sep 18 '18 at 17:20
  • @zahbaz eigenvalues. I find them numerically through Newton's method – Lionell Sep 18 '18 at 17:22
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    Is $\cos(\sqrt{\lambda_n}x)$ orthogonal to $\cos(\sqrt{\lambda_m}x)$ for $n\ne m$? (It is say if $\sqrt{\lambda_n}=n\omega$ for $n=\mathbb{Z}$) That is, can you use something like $$\int_0^{2\pi}w(x,0)cos(\sqrt{\lambda{}m} x)dx = \sum\nolimits{n=1}^\infty A{}_n\int_0^{2\pi}cos(\sqrt{\lambda{}_n} x)cos(\sqrt{\lambda{}_m} x) dx $$ to produce a Kronecker delta on the rhs? – zahbaz Sep 18 '18 at 17:31
  • you need to give the boundary conditions and initial conditions. –  Sep 18 '18 at 17:32
  • @zahbaz my $$ \lambda $$ this solution of equation $$ \frac{k\sqrt \lambda}{\alpha},\tan(\sqrt \lambda l) -1 = 0 $$ when they are orthogonal? – Lionell Sep 18 '18 at 17:50
  • @zahbaz added the full condition – Lionell Sep 18 '18 at 18:09
  • @zahbaz thank you, Could you check the correctness of the model? – Lionell Sep 18 '18 at 18:37

1 Answers1

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Your title is confusing. This the heat equation for a rod which is a classic example.

Setting up the problem

$$ \frac{\partial u}{\partial t} = \frac{k}{c}\frac{\partial^{2} u}{\partial x^{2}} \tag{1}$$

with boundaries given as

$$ BC1 : \frac{\partial u}{\partial x}(0,t) = 0 \\ BC2 : \frac{\partial u}{\partial x}(L,t) = \frac{-\alpha }{k}(u -u_{0}) \tag{2}$$

and initial condition $$ IC: u(x,0) = \eta(x) \tag{3} $$

If you solve it you should get

$$ \phi(x) = c_{1}\cos(\sqrt{\lambda}x) +c_{2}\sin(\sqrt{\lambda}x) \tag{4} $$ which gives

$$ \frac{d\phi}{dx} = \sqrt{\lambda} \big( -c_{1} \sin(\sqrt{\lambda} x) +c_{2}\cos(\sqrt{\lambda}x \tag{5} \big)$$

then

$$ \frac{d\phi(0)}{dx} = 0 \implies c_{2} = 0 \tag{6}$$

so your equation becomes

$$ \phi(x) = c_{1}\cos(\sqrt{\lambda}x) \tag{7} $$

to determine the eigenvalues we have

$$ \frac{d\phi(L)}{dx} = -c_{1}\sqrt{\lambda} \sin(\sqrt{\lambda} L) = -\frac{-\alpha}{k}(u-u_{0}) \tag{8}$$

If I interpret the second boundary condition you get something like this.

$$ \frac{d\phi(L)}{dx} = -c_{1}\sqrt{\lambda} \sin(\sqrt{\lambda} L) = -\frac{-\alpha}{k}\bigg(c_{1} \cos(\sqrt{\lambda} L) + \eta(L) \bigg) \tag{9}$$

For the next part

$$ u(x,t) = A_{0} + \sum_{n=1}^{\infty} \cos(\sqrt{\lambda}x) e^{- \lambda \frac{k}{c}t} \tag{10} $$

Getting the coefficients

There is something called Fourier's trick. It comes from the orthogonality of eigenfunctions with different eigenvalues. This is the original derivation, it is simply for homogenous boundaries.

$$ \int_{0}^{L} \cos(\frac{n \pi x}{L} ) \cos(\frac{m \pi x}{L} )dx =\begin{align}\begin{cases} 0 & n \neq m \\ \frac{L}{2} & n = m \neq 0 \\ L & n = m =0 \end{cases} \end{align} \tag{11}$$

$$ \int_{0}^{L} \eta(x) \cos(\frac{m \pi x}{L}) dx = \sum_{n=0}^{\infty} A_{n} \int_{0}^{L} \cos(\frac{n \pi x}{L}) \cos(\frac{m \pi x}{L}) dx \tag{12}$$

using the orthogonality we get

$$ \int_{0}^{L} \eta(x) \cos(\frac{m \pi x}{L}) dx = A_{m} \int_{0}^{L} \cos^{2} \frac{m \pi x}{L} dx \tag{13} $$

when $m=n$, there are two cases $m=0$, $m\neq 0$ which yield the below results.

$$ A_{0} = \frac{1}{L} \int_{0}^{L} \eta(x) dx \tag{14}$$ $$ A_{m} = \frac{2}{L}\int_{0}^{L} \eta(x)\cos(\sqrt{\lambda}x) dx \tag{15}$$

If you note when we evaluate the right hand of the expression below we will get the equations, 14 and 15.

$$ \int_{0}^{L} \eta(x) \cos(\frac{0 \pi x}{L}) dx = A_{0} \int_{0}^{L} \cos^{2} \frac{0 \pi x}{L} dx \tag{16} $$

$$ \int_{0}^{L} \eta(x) dx = A_{0} \int_{0}^{L} 1 dx \tag{17} $$ $$ \int_{0}^{L} \eta(x) dx = A_{0} L \tag{18} $$ So we simply divide by $L$ giving $$ A_{0} = \frac{1}{L} \int_{0}^{L} \eta(x) dx \tag{19} $$

In order to get $A_{m}$ if you look at line $11$ there is case $2$ and we look down to line $13$. Then you would observe the right hand part of the expression will evaluate you to $\frac{L}{2}$

$$ \int_{0}^{L} \eta(x) \cos(\frac{m \pi x}{L}) dx = A_{m} \int_{0}^{L} \cos^{2} \frac{m \pi x}{L} dx \tag{20} $$ $$ \int_{0}^{L} \eta(x) \cos(\frac{m \pi x}{L}) dx = A_{m} \frac{L}{2}\tag{21} $$ If you move this around then we get $$ A_{m} = \frac{2}{L}\int_{0}^{L} \eta(x) \cos(\frac{m \pi x}{L}) dx \tag{22} $$

  • my $$ \lambda $$ this solution of equation $$ \frac{k\sqrt \lambda}{\alpha},\tan(\sqrt \lambda l) -1 = 0 $$ – Lionell Sep 18 '18 at 17:49
  • can you post your work and the original problem. –  Sep 18 '18 at 17:51
  • thank you. The task is given a cross-sectional area, but I do not know where to use it. – Lionell Sep 18 '18 at 18:53
  • I'm not sure what you mean by a cross sectional area. Your second boundary condition is odd. –  Sep 18 '18 at 18:58
  • but how did you get the formula for $ A_{m} $ ? – Lionell Sep 18 '18 at 19:20
  • one moment..I'll update it –  Sep 18 '18 at 19:29
  • for my task ratio $ A_{m} = \frac {2}{l} $ or are these other coefficients which are obtained by calculating the integral ? – Lionell Sep 25 '18 at 18:50
  • you should have $A_{m} = \frac{2}{L} \int_{0}^{L} \eta(x) \cos(\sqrt{\lambda}x) dx $ you gave the expression for the $\lambda $ I think you needed to use newtons method in your other question –  Sep 25 '18 at 18:53
  • the first coefficient $A_{0}$ is calculated from $\frac{1}{L} \int_{0}^{L} \eta(x) dx$ –  Sep 25 '18 at 18:58
  • why $ \frac {2}{l} $ if using $ \int_0^L cos^2((\sqrt{\lambda{}_m} x))dx $ gives a $ \frac {4 \sqrt{\lambda{}_m}}{2l \sqrt{\lambda{}_m} + sin(2 \sqrt{\lambda{}_m} l)} $ – Lionell Sep 25 '18 at 19:05
  • how'd you get that, I added some context –  Sep 25 '18 at 19:24
  • the point of the expression is to find where it is $1$. whatever that means for $\lambda_{m}$ –  Sep 25 '18 at 19:32
  • note that $\lambda_{n} $ in the homongeous case is $(\frac{n \pi }{L})^{2}$ you have $\lambda_{n}$ given above from your second boundary condition. –  Sep 25 '18 at 19:34
  • Is this a general case or specifically my task? I can not get $ (\frac{n \pi}{L})^2 $. My $ \lambda{}_m $ this is the solution of the transcendental equation. I do not understand whether the coefficient $ \frac{2}{L} $ or it will be different. The orthogonality property for my problem gives the coefficient that I wrote above. Or I am wrong and he $ \frac{2}{L} $ ? – Lionell Sep 25 '18 at 19:45
  • one second let me write something. –  Sep 25 '18 at 19:50
  • you are given an approximation that is nearly $\frac{\pi}{2}$ i've seen something like this before . Your $\lambda_{n}$ is the intersection of the tangent graph and a line which is the constant in your heat equation. https://math.stackexchange.com/questions/2917780/solving-for-lambda-in-frack-lambda-alpha-tan-lambda-r-1-0 –  Sep 25 '18 at 20:17
  • I think you're looking for some closed form expression which probably doesn't exist. –  Sep 25 '18 at 20:45