To have straightforward and homogeneous boundary conditions for solving $$u_t = \kappa u_{xx}$$ with boundary conditions of $$u_x(0,t)=A\\ u_x(L,t)=0$$ To solve the PDE with homogenous boundaries, $$u(x,t) = w(x) + \nu(x,t)$$ where $w(x)$ is time-independent steady state condition. Then solution for $\nu(x,t)$ with new boundary conditions of $$\nu(0)=\nu(L)=0$$ by the method of separation of the varaibles $\nu(x,t)=X(x)T(t)$, the general solution is $$X(x)=a \cos \lambda x + b \sin \lambda x$$ by applying the boundary conditions, the solution is

$$\nu(x,t)= \sum_{n=1}^\infty b_n e^{-t\kappa(n\pi/L)^2} \sin\left( \frac{n\pi x}{L}\right) \\ b_n=\frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \,\mathrm{d}x$$

For deriving $f(x)$, we use the boundary condition of $$u(x,0) = f(x) = \sum_{n=1}^\infty b_n \sin(n \pi x)$$

How do we find $u(x,0)$ from the original boundary condition of $u_x=A$?

  • $\begingroup$ You should derive the conditions for your coefficients yourself. See here, it might help clear up any confusion. Also, how did you get a sine series with a Neumann boundary condition? $\endgroup$ – Mattos Sep 3 '17 at 9:47
  • $\begingroup$ @Mattos I try to derive the coefficient similar to your approach in that answer, but I do not know $f(x)$. When the boundary conditions are simple as $u(0)=T_1,u(L)=T_2$, then, $f(x)=(T_2-T_1)x$, but what is $f(x)$ in this case? By separting the variable, $\nu(x,t)=a\sin\lambda x + b \sin \lambda x$, and with the boundary conditions of $\nu(0)=\nu(L)=0$, the sine series obtains. $\endgroup$ – Googlebot Sep 3 '17 at 10:00
  • $\begingroup$ $u(x,0) = w(0) + \nu(x,0) = f(x)$ i.e your initial condition. $f(x)$ is not the solution you have just stated. The form of $f(x)$ must be given to you explicitly in order to compute the Fourier coefficients explicitly. Otherwise, the coefficients are just given by the formula you stated in your post. $\endgroup$ – Mattos Sep 3 '17 at 11:31
  • $\begingroup$ I think you should start from the beginning of the problem again and apply your Neumann boundary conditions as you have done with your Dirichlet conditions. This might clear up any confusion. $\endgroup$ – Mattos Sep 3 '17 at 11:37
  • $\begingroup$ @Mattos I edited the question. Does it make sense now? It is the entire problem. How to do it from here? $\endgroup$ – Googlebot Sep 3 '17 at 12:04

Your Answer

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

Browse other questions tagged or ask your own question.