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I have ran into the following Heat equation IBVP, but I am not quite sure how to solve it as it has these time dependent boundary conditions

$$ v_t = kv_{xx} \ \ \ \ \ \ ( 0 \le x \le \infty, \ \ 0 < t < \infty) ,$$ $$ v(x,0) = \delta (x - x_0) \ \ \ \ \ \ \mathrm{for} \ \ t = 0, \ \ x_0 > 0 $$ $$ v(x = ((2kt) ^{1 /2k}), ((2kt) ^{1 /2k})) = 0 $$ $$ \lim_{x\to\infty} v(x, ((2kt) ^{1 /2k})) = 0 . $$

How to solve this problem ? Thanks.

Edit: How I got to this problem.

Consider the IBVP $$v_p = v_q + kp^{2k - 1}v_{qq}$$ $$ v(q,0) = \delta (q - q_0) \ \ \ \ \ \ \mathrm{for} \ \ p = 0, \ \ q_0 > 0 $$ $$ v(0, p) = 0 $$ $$ \lim_{q\to\infty} v(q, p) = 0 . $$

Let us make transformation to eliminate the first term in the right-hand side $$ x = q + p.% $$ After that Eq. reads: $$ v_p = k p^{2k - 1}v_{xx}.% $$ Introduce now new variable $t= p^{2k}/2k$, then our Eqn reduces to the conventional heat equation $$ v_t = kv_{xx}$$

The initial condition $v(q,0) = \delta(q - q_0)$ in new variables reads $v(x,0) = \delta(x - q_0)$. Then, the boundary condition $v(0,p) = 0$ in the new variables becomes: $$ v\left[x = (2kt)^{1/2k},(2kt)^{1/2k}\right] = 0. $$ I hope this helps in resolving any ambiguities, in my OP.

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  • $\begingroup$ Does the third line mean $v(y, ((2kt) ^{1 /2k})) = 0$, $y>0$? $\endgroup$ – Andrew Jun 4 '16 at 5:23
  • $\begingroup$ @Andrew, I have revised the third equation, I hope this representation is clearer. $\endgroup$ – Comic Book Guy Jun 4 '16 at 6:04
  • $\begingroup$ I meant this line: $v(x + ((2kt) ^{1 /2k}), ((2kt) ^{1 /2k})) = 0$. $\endgroup$ – Andrew Jun 4 '16 at 6:08
  • $\begingroup$ @Andrew, Yes indeed, sorry about that, the first parameter is the x counter part. $\endgroup$ – Comic Book Guy Jun 4 '16 at 6:10
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    $\begingroup$ The problem with solution existence here is as follows. Subtracting from $v$ the fundamental solution for the heat equation $Z(x−x0,t)$ leads to the first BVP, which has a unique solution. Solutions of the heat equation are analytic wrt to $x$ variable. So if you demand that $v$ is a solution in the domain $x>0$ it means it can be analytically continued in a larger domain. Locally it could be done via Cauchy-Covalevsky theorem. But it just can happen that there is no global continuation. It's the same situation as for analytical functions which are not necessarily continued to an given domain. $\endgroup$ – Andrew Jun 4 '16 at 7:05
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Let $Z(x,t)=(4\pi k t)^{-1/2}e^{-x^2/(4kt)}$ be the fundamental solution of the heat equation and denote $$G(x,x_0,t)=Z(x-x_0,t)-Z(x+x_0,t)$$ the Green's function of the first BVP in the domain $x>0$. Then $$ v(q,p)= e^{\large q_0-q-\frac{p^{2 k}}{4 k}} G\left(q,q_0,\frac{p^{2 k}}{2 k}\right). $$ is the required solution.

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  • $\begingroup$ Except the BC is not $v(t,t)=0$ anymore (now that the OP has been edited). $\endgroup$ – Ian Jun 4 '16 at 13:26
  • $\begingroup$ @Andrew, I am not sure how the boundary condition can be written like so, I am sorry if I misunderstood your comment and mislead you. The reason there is an x and t term in the initial condition is due to a Galilean transformation. I will update my post to show my workings, how I got to the heat equation, hopefully that will remove any ambiguities. $\endgroup$ – Comic Book Guy Jun 4 '16 at 21:15
  • $\begingroup$ @Andrew, I have updated my post, please have a look. $\endgroup$ – Comic Book Guy Jun 4 '16 at 21:59
  • $\begingroup$ Hi Andrew, Can you please also explain why the boundary condition can be written like that? $\endgroup$ – Comic Book Guy Jun 5 '16 at 10:49
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    $\begingroup$ After your change $t= p^{2k}/2k$ turns into a parabolic one with constant coefficients. To satisfy the boundary condition Green's function of the first BVP was taken. To satisfy the equation solution was sought in the form $v(q,q_0,t)=G(q,q_0,t)u(q,q_0,t)$. To satisfy the initial condition one should impose on $u$ an additional condition $u(q_0,q_0,0)=1$. It turns out that $u$ has a closed form. If your problem is different just in a couple of constants, perhaps the solution can be sought in the form $Ge^{\alpha (q-q_0)+\beta t}$. $\endgroup$ – Andrew Jun 5 '16 at 18:22

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