1

I have a PDE problem (namely the heat equation for a semi-infinite solid) and I'm trying to solve it using the Laplace Transform.

$$\frac{1}{\alpha} \frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2}$$

The conditions are:

  • $T(x,0)=0$
  • $T(0,t)=A\cos(\omega t$)

I can find the solution in the $s$ domain, $F(x,s)$, but I cannot antitransform it to the time domain.

However, I'm not interested in the solution $f(x,t)$ but rather on the envelope of the function, that is, a function $f_e(x)$ that gives the maximum temperature reached at each point of the solid. I kown from the numerical solution that the envelope is a decaying exponential.

Is there any way to compute this envelope mathematically from the Laplace Transform $F(x,s)$?

Francisco
  • 484

1 Answers1

1

If you view this question by power series method approach, you will find that it is easier to solve.

Similar to Model for soil temperature underground using the Heat Equation:

Let $T(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^n}{n!}\dfrac{\partial^nT(0,t)}{\partial x^n}$ ,

Then $T(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^{2n}T(0,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}T(0,t)}{\partial x^{2n+1}}$

$=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{\alpha^n(2n)!}\dfrac{\partial^nT(0,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{\alpha^n(2n+1)!}\dfrac{\partial^{n+1}T_x(0,t)}{\partial t^n}$

$=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{\alpha^n(2n)!}\dfrac{\partial^n(A\cos\omega t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{\alpha^n(2n+1)!}\dfrac{\partial^{n+1}T_x(0,t)}{\partial t^n}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nA\omega^{2n}x^{4n}\cos\omega t}{\alpha^n(4n)!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nA\omega^{2n+1}x^{4n+2}\sin\omega t}{\alpha^n(4n+2)!}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{\alpha^n(2n+1)!}\dfrac{\partial^{n+1}T_x(0,t)}{\partial t^n}$

doraemonpaul
  • 16,488