This problem is related to this question. If you can answer this, then you might be able to also answer the other question, so please have a look at it.

I am trying to solve the following problem:

$$\dfrac{\partial{\phi}}{\partial{t}} = \kappa \dfrac{\partial^2{\phi}}{\partial{x}^2}, \ x > 0, t > 0$$

$$\phi(x, 0) = 0, \ x > 0$$

$$\phi(0, t) = T_0e^{-bt}, \ t > 0$$

Taking the Laplace transform (in $t$, of course) gives

$$\mathcal{L} \left\{ \dfrac{\partial{\phi}}{\partial{t}} \right\} = \kappa \mathcal{L} \left\{ \dfrac{\partial^2{\phi}}{\partial{x}^2} \right\}$$

$$\therefore s \mathcal{L}\{\phi\} - \phi(0) = \kappa \dfrac{d^2 \mathcal{L} \{ \phi \}}{dx^2}$$

And since $\phi(x, 0) = 0$, we have

$$s \mathcal{L} \{ \phi \} = \kappa \dfrac{d^2 \mathcal{L} \{ \phi \}}{dx^2}$$

This is an ODE with constant coefficients.

From now on, let $\mathcal{L}\{ \phi \} = \bar{\phi}$ (for the sake of conciseness).

So our constant-coefficient ODE is

$$\dfrac{d^2 \bar{\phi}}{dx^2} - \dfrac{s}{\kappa} \bar{\phi} = 0$$

This is a homogeneous ODE.

The characteristic polynomial is

$$m^2 - \dfrac{s}{\kappa} = 0$$

$$\therefore m = \pm \sqrt{\dfrac{s}{\kappa}}$$

Therefore, the solution to the constant-coefficient homogeneous ODE is

$$\bar{\phi}(x) = Ae^{x\sqrt{\dfrac{s}{\kappa}}} + Be^{-x\sqrt{\dfrac{s}{\kappa}}}$$

Here is where I am stumped.

One way that I have seen other problems proceed is by assuming another boundary condition:

$$\lim\limits_{x \to \infty} \phi(x, t) = 0$$

This makes sense from a physical standpoint, but it was not specified in the problem statement, so I am unsure if it is valid for me to use it?

Also, I don't have full solutions for these problems, so I have no way of checking the intermediate steps.

I would greatly appreciate it if people could please help me solve this problem.

If we proceed with assuming the extra boundary condition (that is, if we proceed with assuming that $\lim\limits_{x \to \infty} \phi(x, t) = 0$), then we continue from above as follows:

Now, since $\lim\limits_{x \to \infty} \phi(x, t) = 0$ and

$$\bar{\phi}(x, s) = \int_0^\infty \phi(x, t) e^{-st} \ dt,$$

we have that $\bar{\phi}(x, s) \to 0$ as $x \to \infty$, and hence $A = 0$, since $Be^{-x\sqrt{\dfrac{s}{\kappa}}} \to 0$ as $x \to \infty$.

Therefore, we now have

$$\bar{\phi}(x, s) = Be^{-x\sqrt{\dfrac{s}{\kappa}}}.$$

We now use the given boundary conditions.

Letting $x = 0$ in the Laplace transform of $\phi$ gives

$$\bar{\phi}(0, s) = \int_0^\infty \phi(0, t)e^{-st} \ dt = \int_0^\infty T_0 e^{-bt}e^{-st} \ dt = \int_0^\infty T_0 e^{-t(b + s)} \ dt$$

$$= \dfrac{-T_0}{b + s} \int_0^{-\infty} e^u \ du = \dfrac{T_0}{b + s},$$

since $\phi(0, t) = T_0 e^{-bt}$ for all $t > 0$.

Inserting this boundary condition for $\bar{\phi}$ gives

$$B = \dfrac{T_0}{b + s},$$

Is this correct?

Therefore, our solution for $\bar{\phi}$ is

$$\bar{\phi}(x, s) = \dfrac{T_0}{b + s} e^{-x \sqrt{\dfrac{s}{\kappa}}},$$

where $b > 0$

Again, is this correct?

  • $\begingroup$ The initial value theorem says that $$\lim_{t\to 0^{+}}f(t)=\lim_{s\to\infty}[s F(s)],$$ where $F(s)=\mathcal{L}[f(t)].$ You might try using that. If that fails, you might have to do the inverse Laplace transform before applying boundary conditions. $\endgroup$ – Adrian Keister Aug 28 '18 at 19:48
  • $\begingroup$ @AdrianKeister At what point do I apply this? $\endgroup$ – The Pointer Aug 28 '18 at 19:48
  • $\begingroup$ You can use that on your $\overline{\phi}(x,s)=Ae^{x\sqrt{s/\kappa}}+Be^{-x\sqrt{s/\kappa}}.$ $\endgroup$ – Adrian Keister Aug 28 '18 at 19:49
  • $\begingroup$ It may not help you much; you already got that $A=0$ another way. $\endgroup$ – Adrian Keister Aug 28 '18 at 19:55
  • $\begingroup$ @AdrianKeister I've posted two of these problems now. I can't figure out either of them. Maybe have a look at the other one and see if you can figure that one out? Maybe that will help in figuring this one out. Help is greatly appreciated. math.stackexchange.com/questions/2897049/… $\endgroup$ – The Pointer Aug 28 '18 at 19:56

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