I am stuck on this problem I have to do for my PDE class.

Consider heat diffusion in a 1D object between x=0 and x=L. It can be shown that the heat flux at the surface is $-k\nabla{}T*\hat{n}$ i.e. its is proportional to the normal derivative at the surface. In our case, we specify adiabatic boundary conditions as follows : $$u_x(0,t)= 0 $$ $$ u_x(L,t)=0$$ If the initial temperature is $f(x)$ solve for the temperature as a function of x and t.

To be honest, I don't really know where to start. So far I have tried to solve the diff. eq : $$ \frac{\partial u}{\partial t} =-k\nabla{}u*\hat{n}$$ using separation of variables but I got nowhere. Could anyone help me understand how to properly start this problem ? Thank you !

  • $\begingroup$ What does the heat equation look like? How did you write down the equation you wrote down? In particular, why is there no time evolution in the equation you wrote down? It seems something may be wrong then! $\endgroup$
    – Matt
    Commented Feb 9, 2017 at 22:42
  • $\begingroup$ Thank you for your answer. I wrote the equation from what I've seen in class, I corrected the LHS of the equation, I don't know why I wrote d^2u/dx^2 before. $\endgroup$
    – Pal
    Commented Feb 9, 2017 at 22:51

1 Answer 1


I will omit the derivation of the heat equation, but I believe you should have $$ \frac{\partial u}{\partial t} = k\Delta u. $$ The idea is that for any small region $V$ we have $$\frac{d}{dt} \int_V u = \int_{\partial V} -k\nabla u \cdot {\bf n}dS.$$ This equation is simply saying that the energy change in the region $V$ (LHS) is given by the flux through the boundary (RHS). Using the divergence theorem on the right hand side you can derive the heat equation.

Now, in 1D, on the interval $[0,L]$ we have simply

$$u_t = k u_{xx}.$$

Assume that $u(x,0)= f(x)$ and $u_x(0,t)=u_x(L,t)=0$. You can use separation of variables from here: assume that $u(x,t) = X(x)T(t)$ and see that

$$XT' = kX''T.$$

Rearranging we have

$$\frac{T'}{kT} = \frac{X''}{X} = -\lambda.$$

This gives us two ODEs:

$$T'+k\lambda T=0$$


$$X''+\lambda X=0.$$

Note the boundary values: $u_x(0,t) = X'(0)T(t)=0$ and $u_x(L,t)=X'(L)T(t)=0$.

If $T(t) =0 $ then $u\equiv 0$ and we only recover the trivial solution. So, we may assume that $X'(0)=X'(L)=0$. I believe the rest of the problem should be calculations from ODEs.

  • $\begingroup$ Just saw that! Thank you very much, very helpful and clear ! $\endgroup$
    – Pal
    Commented Feb 20, 2017 at 1:43

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