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I follow this course for solving the heat equation. However, I cannot understand a possible case of semi-infinite.

The general equation obtained from the separation of variables is

$$\phi(x) = c_1\cos (\sqrt{\lambda}x) + c_2 \sin(\sqrt{\lambda} x)$$

as we have to obtain $\lambda$ by applying the boundary conditions. All example are for $x=0$ and $x=L$. I am trying to figure out how to solve it for a simple set of boundary conditions:

$$u(x,0)=C_0, u(0,t)=C_1, u(\infty,t)=C_2$$

How do we apply $\phi(\infty) = 0$ to obtain $\lambda$?

In fact, I don't know how to apply the condition of $\infty$ to $\phi(x)$

I know that the common solution for the semi-infinite boundary conditions is Laplace transform, but I am curious what is the mathematical limitations for applying the semi-infinite boundary conditions to this solution by the approach of separation of variables.

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  • $\begingroup$ This is confusing. $u(x,t)=0$ is not a boundary condition. I assume you mean a specific $x$ and/or $t$ but you don't say which one. Also, $\Phi(x)=c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x)$ is not the general solution of the heat equation. Also, I've never seen extended real numbers used in the context of PDEs (although maybe there is a way to do it, I doubt it). $\endgroup$ – Ben W Jan 23 '19 at 15:00
  • $\begingroup$ This solution form only applies to a finite interval. For the half line you can solve the heat equation using Laplace transforms in the x variable then solving the resulting ODE in the t variable. $\endgroup$ – Paul Jan 23 '19 at 15:13
  • $\begingroup$ In addition to @Paul: In order to do the Laplace transform you need to specify $u_x(0,t)$. This shows that an asymptotic condition is generally speaking not a "full" replacement for a BC. $\endgroup$ – maxmilgram Jan 23 '19 at 15:51
  • $\begingroup$ @BenW sorry that was a typo. That was the initial condition as $u(x,t)$. $\endgroup$ – Kimia Jan 23 '19 at 15:59
  • $\begingroup$ @Paul this is exactly my aim. The solution of semi-infinite is usually given by Lapance transform, as it is easier. However, there should be possibility of using this approach for a boundary condition of$x = L \to \infty$ $\endgroup$ – Kimia Jan 23 '19 at 16:02
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The heat equation doesn't make good physical sense if the total amount of heat is infinite. The total heat at time $t$ is a constant multiple of $\int_{0}^{\infty}u(x,t)dx$. Assuming positive temperatures also makes sense, which makes $L^1[0,\infty)$ a natural space in $x$. These types of conditions replace explicit values of the solution at $\infty$. Then if you think of a limit of problems for $0 \le x \le L$ as $L\rightarrow\infty$, it might not surprise you do think in terms of integral combinations of solutions instead of sums. For example, a general type of solution of $u_t = c^2u_{xx}$ might look like $$ u(x,t)=\int_{0}^{\infty}e^{-c^2\lambda^2 t}\{C(\lambda)\sin(\lambda x)+D(\lambda)\cos(\lambda x)\}d\lambda. $$

The coefficients $C(\lambda),D(\lambda)$ in such a solution would then be determined from initial data through Fourier sine and cosine transforms: $$ u(x,0)=\int_{0}^{\infty}C(\lambda)\sin(\lambda x)+D(\lambda)\cos(\lambda x)dx. $$

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