5
$\begingroup$

I have been struggling with following Heat equation IBVP, \begin{equation} \frac{\partial v\left(x, t\right)}{\partial t} = \alpha \frac{\partial^2 v\left(x, t\right)}{\partial x^2}, \quad t \in \left(0, T\right], x \in \left(- \infty, 1\right]. \label{heatEqn} \end{equation} with the following initial and boundary conditions \begin{equation} v\left(x, 0\right) = \delta\left(x\right), \quad v\left(-\infty, t\right) = 0, \quad \int_{-\infty}^{1}v\left(x, T\right) dx = \beta T^{\alpha-1} \end{equation} where $\alpha \in \left(0, 1\right)$ and $\beta$ is some positive constant.

Some interesting observation about this problem, if we consider the semi-infinite plane $x \in \left(- \infty, 1\right]$ $\times$ $t \in \left[0, T\right]$, we know line integrals along three of the boundaries of this rectangle, as in $\int_{-\infty}^{1}v\left(x, 0\right) dx = 1$, $\int_{0}^{T}v\left(-\infty, t\right) dt = 0$ and the boundary condition with the integral equation above.

Also if we can deduce $v\left(1, t\right)$ and $\frac{\partial v\left(x, t\right)}{\partial x}|_{x = 1}$ then we can attack the problem using laplace transform.

I am wondering if there is version of Greens or Divergence Theorem which can be applied to tackle this problem.

Any help or guidance will be greatly appreciated. Thank you.

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ Is $T$ fixed or the integral condition should hold for any $T>0$? If the first, would it make any difference if to change the rhs to $C$? If the second how Laplace transform is supposed to be applied? And btw does $\Omega=(-\infty,1)$? $\endgroup$ – Andrew Jul 6 '17 at 10:23
  • $\begingroup$ @Andrew, $T$'s condition should hold for all $T>0$. I have replaced the $\Omega$, sorry about that. I could be mistaken about the Laplace transform, what i was thinking was that if we can work out $v(1, t)$ and $v$'s partial derivative wrt to $x$ at $x = 1$, based on the other boundary conditions, then these are the pieces needed to apply the Laplace transform to the PDE wrt to $x$. no ? $\endgroup$ – Comic Book Guy Jul 6 '17 at 14:44
  • $\begingroup$ The last formula holds at least if $\psi\in C^\gamma$, $\gamma>1/2$ and $\psi(0)=0$. $\endgroup$ – Andrew Jul 6 '17 at 20:46
  • 1
    $\begingroup$ For $T< 1$ $\log T$ is negative, what about it? The problem can be reduced to the zero initial condition considering function $u(x,t)=v(x,t)-Z(x,t)$ where $Z$ is the fundamental solution for the heat equation. The rhs in the integral condition will change somewhat, the change can be written down explicitly via $\mathtt{erf}$. The solution $u$ can be sought in the form of a double layer potential, the density $\psi$ of which is equal up to factor 2$\alpha$ the boundary values of the solution: $u(1,t)=2\alpha \psi(t)$, and it's derivative $u_x(1,t)=D^{1/2}_t\psi(t)$. $\endgroup$ – Andrew Jul 6 '17 at 20:49
  • 1
    $\begingroup$ Btw is $\alpha$ the same in the heat equation and the exponent in the rhs of the integral condition? In this form the integral condition very nicely go into (integrating $Z_x$ from the double layer potential) into $\int_{-\infty}^{1}v\left(x, 0\right) dx=C(\alpha)I^{1/2}_t\psi(t)$. Hence $\psi(t)=C(\alpha) D_t^(1/2)g(t)$. $\endgroup$ – Andrew Jul 6 '17 at 21:03
2
$\begingroup$

Here is the answer to your modified question via potential approach but simpler than proposed in the comments.

Denote $Z(x,t)=(4\alpha\pi t)^{-1/2} e^{-x^2/(4\alpha t)}$ the fundamental solution for the heat equation. I'll write $t$ instead of $T$ in the question. Subtracting $Z(x,t)$ from $v$ reduces the initial condition to zero and the integral to $$ \beta t^{\alpha-1}-\int_{-\infty}^1Z(x,t)\,dx= \beta t^{\alpha-1}-\frac{1}{2} \left(\text{erf}\left(\frac{1}{2 \sqrt{\alpha t}}\right)+1\right):=g(t). $$ Shifting the side boundary to $x=0$ (for convenience) the problem becomes $$ \left\{ \begin{array}{rccl} u_t&=&\alpha u_{xx},&t>0,\ x<0,\\ u|_{t=0}&=&0,\\ \int_{-\infty}^0u(x,t)\,dx&=&g(t),&t>0. \end{array} \right. $$ Its solution can be written down explicitly as a potential: $$ u(x,t)=2\alpha\int_{-\infty}^0Z_{xx}(x,t-\tau)g(\tau)\,d\tau. $$ To see that consider the double layer potential $$ Wg(x,t)=\int_0^tZ_x(x,t-\tau)g(\tau)\,d\tau. $$ It satisfies the heat equation for $x<0$ and for continuous densities $g$ has a property $$ \lim_{x\to0-}Wg(x,t)=\frac{g(t)}{2\alpha}. $$ Actually it's the jump relation for the double layer potential in this particular case.

For $\varepsilon>0$ we have $$ \int_{-\infty}^{-\varepsilon}u(x,t)\,dx= 2\alpha\int_0^t\int_{-\infty}^{-\varepsilon}Z_{xx}(x,t-\tau)g(\tau)\,dx d\tau= $$ $$ 2\alpha\int_0^tZ_{x}(-\varepsilon,t-\tau)g(\tau)\,d\tau= 2\alpha Wg(-\varepsilon,t). $$ Taking $\varepsilon\to0+$ gives the result.

The solution of the initial problem is $v(x,t)=u(x-1,t)+Z(x,t)$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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