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Let smooth functions $f(x) , g(t)$ are given
solve the heat equation on the semi infinite domain $(a,\infty) \times (0,T)$. for simplicity, we can let $a = 0$. \begin{eqnarray} &&u_t(x,t) = u_{xx}(x,t) \quad a<x<\infty , \quad 0<t<T \\ &&u(x,0) = f(x), \quad a<x\\ &&u(a,t) = g(t), \quad 0<t<T \\ && lim_{x\rightarrow \infty} u(x,t) = 0 . \end{eqnarray} i need the closed form solution to the problem subject to $f(x) , g(t)$

Thanks for any help in advance

moradipour
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  • Here is a solution for $f(x)=0$ http://math.stackexchange.com/questions/350742/find-the-greens-function-and-solution-of-a-heat-equation-on-the-half-line/350758#350758 For the general case, though, you may need a Green function approach. – Ron Gordon Dec 12 '13 at 17:52
  • Thanks, in my problem $f(x) \neq 0$ ,but $f$ has this property: $\lim_{x \rightarrow \infty} f(x) = 0.$ – moradipour Dec 13 '13 at 13:38

2 Answers2

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For the last condition to be true one has to make some assumptions on $f$. Such as $\lim_{x\to +\infty} f(x) = 0$. For example, if $f\equiv1$ then $\lim_{x\to +\infty} u(x,t) = 1$. The solution can be expressed as a sum of two potentials: $G*f+2Wg$, where $G$ is the Green function of the first boundary value problem and $Wg$ is double layer potential for the heat equation.

Andrew
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  • thanks for your help but I have difficulty to calculate $Wg$.
    From definition of double layer potential, we have \begin{eqnarray} W [g(x,t)] = \int_0^t \int_S \frac{\partial}{\partial n_y } \Phi(x-y , t- \tau) g(y,\tau) dS_y d \tau \quad\quad (1) \end{eqnarray} Here $$\Phi(x,t) = \frac{1}{\sqrt{4\pi t}}^3 e^{-x^2/4t} $$ In my problem $g(x,t)$ is only a function of time $t$ and the boundary $S$ contains two pints${a,\infty}.$ How to calculate the inner integral in (1)?     – moradipour Dec 13 '13 at 13:35
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    The boundary is one-dimensional. So $S$ is just a point and $Wg$ can be written as $$ W g(x,t) = -\int_0^t \frac{\partial}{\partial x} \Phi(x-a , t- \tau) g(\tau) d \tau, $$ $$ \Phi(x,t) = \frac{1}{\sqrt{4\pi t}} e^{-x^2/4t}. $$ – Andrew Dec 13 '13 at 14:14
  • But a bout term $Gf$, I consider the Green function $G(x,t) = (\frac{1}{4 \pi t}) ^{3/2} e^{-\frac{(x-a)^2}{4 t}} $ The convolution product can be written as \begin{eqnarray} G(x,t) f(x) &=& \int_0^x G(x-\lambda , t) f(\lambda) d\lambda \ &=& (\frac{1}{4 \pi t}) ^{3/2} \int_0^x e^{-\frac{(x-\lambda-a)^2}{4 t}} f(\lambda) d\lambda \end{eqnarray} But this convolution does not satisfy the PDE. Do you mean that the convolution of $G(x,t)$ must be calculated with respect to $t$? thanks for your guidance again. – moradipour Dec 13 '13 at 15:23
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    As is written in the books on PDF (for $a=0$) $$ G(x,t,y)=\Phi(x-y,t)-\Phi(x+y,t) $$ and the integral in question is $$ \int_0^\infty G(x,t,y)f(y),dy. $$ – Andrew Dec 13 '13 at 16:45
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    In fact we splited the problem into two parts.

    part $1$: \begin{eqnarray} &&u_t = u_{xx}\ &&u(x,0) = f(x)\ &&u(0,t) = 0\ &&\lim_{x\rightarrow \infty} u(x,0) = 0 \end{eqnarray} with the solution $$u_1(x,t) = \int_a^{\infty} (e^{ \frac{(x-y)^2}{4t}} - e^{\frac{(x+y)^2}{4t}}) f(y) dy$$

    and part $2$: \begin{eqnarray} &&u_t = u_{xx}\ &&u(x,0) = 0\ &&u(0,t) = g(t)\ &&\lim_{x\rightarrow \infty} u(x,0) = 0 \end{eqnarray} the solution of part $2$ is $$u_2(x,t) = \frac{x} {2\sqrt{\pi}} \int_0^t g(\lambda) (t-\lambda)^{-3/2} e^{-\frac{x^2}{4(t-\lambda)}} d\lambda.$$

    $u = u_1 + u_2$.

     – moradipour Dec 13 '13 at 20:29
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As a hint, I suspect you'll want to use Duhamel's principle where the solution to the problem with time-variant conditions is the convolution of the transient BC with the constant BC version of your problem...

Paul Safier
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