I am currently trying to find another solution to the Diffusion Equation through another method. Now, the problem begins with showing that given the solution: $$u(x,t) = \int_{-\infty}^\infty \Phi(x-y,t)g(y) dy\>\>\>\>(*)$$ solves the Heat Equation $u_t=u_{xx}, \>\>u(x,0)=g(x)$, where $\Phi$ is the Heat Kernel: $$\Phi(x,t) = \frac{1}{\sqrt{4\pi t}}e^{-x^2/4ct}$$ We have the following questions:
1) Show that if $u(x,t)$ is a smooth solution to the Diffusion Equation $u_t=u_{xx}$, then so is $u(\lambda x, \lambda^2 t)$
2)This scaling invariance suggests that the ratio $x^2/t$ might play an important role. We look for a solution of the form $v(x^2/t) = u(x,t)$.
3) Show that if $u$ with the form $(*)$ solves the PDE $u_t=u_{xx}$, then $v$ solves the ODE $4sv''(s)+(2+s)v'(s)=0$.
4)Find $v(s)$, the general solution to the ODE.
5)Show that if $u(x,t)$ is a $C^\infty$ solution to the Diffusion Equation, then so is $u_x(x,t)$. Use this fact to differentiate $u(x,t)=v(x^2/t)$ with respect to $x$ and discover a new solution to the Diffusion Equation.
Now, I have shown that $u_{\lambda}(x,t) = u(\lambda x,\lambda ^2 t)$ solves the Heat Equation as well, along with being smooth. Then I considered $u(x,t)=v(s)$, where $s=\frac{x^2}{t}$. This gave me the ODE: $$4sv''(s)+(2+s)v'(s)=0$$ Solving this ODE, I found: $$v(s) = c_1\int\frac{e^{-s/4}}{\sqrt s} ds+c_2\>\>\>\>(1)$$ Where $c_1$ and $c_2$ are just constants. I also know that the complete solution is supposed to be along the lines of:$$u(x,t) =\frac{1}{2}+\frac{1}{\sqrt{\pi}}\int_0^{\frac{x}{\sqrt{4ct}}}e^{-y^2}dy\>\>\>\>\>\>(2)$$ But how in the world do I get $(2)$ from $(1)$? I tried substituting in $s=\frac{x^2}{t}$ and it didn't work. If someone could enlighten me it would be appreciated. Thank you for your time.