A step in deriving the solution to heat equation

Question

I'm trying to understand how to solve the heat equation. I'm stuck on this particular step:

$$Q(x,t) = \dfrac{1}{2} + \dfrac{1}{\sqrt\pi}\int_{0}^{\dfrac{x}{\sqrt{4kt}}}e^{-s^2}ds, t > 0.$$

I want to compute $\frac{\partial Q}{\partial x}.$ This should turn out to be $$ \dfrac{1}{\sqrt {4k\pi}t}e^{-\dfrac{x^2}{4kt}}.$$ I think we have to use Leibniz integral rule but not sure how to do it. Any help appreciated.

Answer 1

Never forget the fundamental theorem of calculus! $$\mathrm{\partial}_x\left[\int_{a(x)}^{b(x)}f(s)\mathrm{d}s\right]=f(b(x))b'(x)-f(a(x))a'(x)$$ So in your example, $$\partial_x Q(x,t)=\partial_x\left[\frac{1}{\sqrt{\pi}}\int\limits_0^{x/\sqrt{4 kt}}\exp(-s^2)\mathrm{d}s\right]=\frac{1}{\sqrt{4\pi kt}}\exp\left(-\frac{x^2}{4kt}\right)$$

Answer 2

Let $$H(x,t) = Q(x\sqrt{4kt},t) = \frac{1}{2}+\frac{1}{\sqrt{\pi}}\int\limits_0^x e^{-s^2}ds $$

So $$ \frac{\partial H}{\partial x} = \frac{1}{\sqrt{\pi}}e^{-x^2} $$ by Leibnitz or just plain differentiation under the integral sign.

Can you finish from this point on?