4
$\begingroup$

Given the following PDE with an initial condition and two boundary conditions:

$$\begin{cases} \frac{\partial^2 u}{\partial Z^2} = \frac{\partial u}{\partial T_v} \\ u(Z,0) = u_t(Z) \\ u(0,T_v) = u_i \\ u(2,T_v)=u_f \end{cases}$$ I know that in the particular case in which:

$$u_i = u_f = 0, \; \; \; u_t(Z) = u_0$$ the Fourier series solution is as follows:

$$u(Z,T_V) = \sum_{n=0}^{\infty} \frac{2\,u_0}{m}\,\sin(m\,Z)\,e^{-m^2 T_v}\,, \; \; \; \text{with} \; \; m = \frac{(2\,n+1)\,\pi}{2}\,.$$

Unfortunately, I can not really understand how to get this last solution in Wolfram Mathematica 11 and therefore not even the general solution. Can anyone help me? Thank you!

Improve this question
$\endgroup$
1
  • $\begingroup$ Looks like a one-dimensional consolidation problem (or heat transfer). Do you really need to get the analytical solution in Mathematica? This can be solved manually assuming $u(z,t) = G(t) F(z)$. You can easily get a numerical solution using NDSolve and check your calculations. $\endgroup$
    – K.J.
    Commented Nov 24, 2017 at 19:55

1 Answer 1

Reset to default
3
$\begingroup$

Try

ClearAll[u,u0,t,z,L0,n];
pde  = D[u[z,t],t]== D[u[z,t],{z,2}];
bc   = {u[0,t]==0,u[L0,t]==0}
ic   = u[z,0]==u0

sol  = DSolve[{pde,bc,ic},u[z,t],{z,t}]

Mathematica graphics

For L=2

 %/.{L0->2,K[1]->n}

Mathematica graphics

Version 11.2 on windows.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Can match the fourier series answer by noting that (-1)^n - 1 is 0 for even n. Avoid calculating the 0 terms by changing n->2n+1 and changing the sum lower limit to n = 0. $\endgroup$
    – Bill Watts
    Commented Nov 25, 2017 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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