# Heat equation initial value problem (General Solution)

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.

• What is the domain of x ? – Rebellos Apr 20 at 19:28
• 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 $k > 0$. – Mattos Apr 20 at 19:29
• I have updated the question with the domain of X. Yes, I realize the initial condition is physically meaningless. – Raymond94 Apr 20 at 19:39

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.

$$u(x,t)=e^{kx+k^2t}$$ 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.

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