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I've got a situation where I'm supposed to solve the 1-D heat equation $v_t = c^2 u_{xx}$ under boundary conditions and an initial condition, and so after some run-of-the-mill process, I attained the general solution $$v(x,t) = \sum_{n = 1}^{\infty} A_n\sin(n\pi x)e^{-\lambda_n^2 t}$$ where $\lambda_n = cn\pi$ and $n \in \mathbb{N}$.

However, I was given an initial condition of $v(x, 0) = f(x) = -(1 + x)$ with $0 < x < 1$, of which I understand is neither an odd or an even function. Thus, I'm not entirely sure that using Fourier series will be able to lead me to find values for $A_n$, and if it was able to be done then could someone explain how? I'm in a bit of a pickle now and so any help is much appreciated!

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If you apply the initial condition to your solution you will be able to solve for the coefficient $A_{n}$.

$$ v(x,0) = f(x) = \sum_{n=1}^{\infty} A_{n} \sin(n \pi x) = -(1+x) $$

You now use Fourier's Trick here.

$$ A_{n} = \int_{0}^{1} f(x) \sin(n \pi x) \textrm{d}x$$

$$ A_{n} = - \int_{0}^{1} (1+x) \sin(n \pi x) \textrm{d}x = - \frac{\pi n + \sin(\pi n) - 2\pi \cos( n \pi)}{\pi^{2} n^{2}}$$

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  • $\begingroup$ So it doesn't matter if the function is odd or not? $\endgroup$ – Maths Matador May 10 at 17:43
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    $\begingroup$ When the function is even or odd you get simpler representations I believe but this method for finding the coefficients works either way. The coefficients go to $0$ as $n \to \infty $ $\endgroup$ – user3417 May 10 at 17:44

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