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I have the following equation: $u_t - u_{xx} = 0$ with initial data $u(x, 0) = e^{kx}$ for some constant $k$ and $x \in \mathbb{R}, t > 0$ I'm looking for the general solution $u(x, t)$.

So far, I have set up the solution as follows:

$$u(x, t) = \frac{1}{2\sqrt{\pi t}} \int_{-\infty}^{\infty} e^{\frac{-(x - y)^2}{4t}} e^{ky} dy$$

But I ran in to trouble evaluating it. Any help would be much appreciated.

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  • $\begingroup$ What is the domain of x ? $\endgroup$ – Rebellos Apr 20 at 19:28
  • $\begingroup$ What you have written is the general solution. Its a convolution of the heat kernel and the initial condition. Just so you know, an initial condition of the form $u(x,0) = e^{kx}$ doesn't really make sense if $k > 0$. $\endgroup$ – Mattos Apr 20 at 19:29
  • $\begingroup$ I have updated the question with the domain of X. Yes, I realize the initial condition is physically meaningless. $\endgroup$ – Raymond94 Apr 20 at 19:39
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You can directly solve the integral by completing the square: $$ -(y-x)^2 + Ky = -\big(y-x-\frac K2\big)^2+Kx+\frac{K^2}4.$$ E.g. use $a^2 - b^2 = (a-b)(a+b)$ with $a-b=\frac K2$ and $a+b=2y-2x-\frac K2$.

Setting $K = 4kt$, we recover DisintegratingByParts's solution,

$$ u(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{\mathbb R} \exp\left(\frac{-\big(y-x-\frac {4kt }2\big)^2}{4t}+kx+k^2t\right) dy = e^{kx+k^2t}.$$

By Widder's uniqueness result for non-negative solutions, this is the unique solution.

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$u(x,t)=e^{kx+k^2t}$ looks like it might work.

In order to evaluate using the integral solution, write the terms in $y$ as $e^p$ where $p$ is a quadratic in $y$; then complete the square in $y$ and shift the integral to reduce to $\alpha y^2+\beta$ in the exponent. Then you should be able to see the solution.

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  • $\begingroup$ How did you get that? $\endgroup$ – Raymond94 Apr 21 at 16:32
  • $\begingroup$ @Raymond94 I guessed that a separated solution $X(x)T(t)$ would work, and it did. $\endgroup$ – DisintegratingByParts Apr 21 at 17:13

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