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!


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.

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

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

  • $\begingroup$ So it doesn't matter if the function is odd or not? $\endgroup$ – Maths Matador May 10 at 17:43
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.