Prove by direct computation

$(1)$ $K(x,t)=t^{-\frac n2}$ $e^{-\frac {|x|^2}{4t}}$ the heat equation for $t\gt0.$

$(2)$ For any $\alpha \gt0$,$G(x,t)=(1-4\alpha t)^{-\frac n2} e^{\frac {\alpha|x|^2}{1-4\alpha t}}$ satisfies the heat equation for $t\lt \frac 1{4\alpha}$.

I tried lot but i don't know where i messed up so please could u show me the computation. I would be grateful. thanks in advance...

  • $\begingroup$ Context please. $\endgroup$ – Ron Gordon Apr 28 '14 at 1:59
  • $\begingroup$ This is all given in the problem, i am sorry what kind of context, for an example? $\endgroup$ – Toeplitz Apr 28 '14 at 2:14
  • $\begingroup$ For example, what is $K(x,t)$? $\endgroup$ – Ron Gordon Apr 28 '14 at 2:16
  • $\begingroup$ That is just function depends on x and t. Now we have to show it satisfies heat equation. $\endgroup$ – Toeplitz Apr 28 '14 at 2:19
  • $\begingroup$ Heat equation is $K_t -\Delta K=0$ $\endgroup$ – Toeplitz Apr 28 '14 at 2:22

Since $|x|^2=x_1^2+\dots+x_n^2\,$ and $\,e^{\beta|x|^2}=e^{\beta x_1^2}\cdots e^{\beta x_n^2}$, the check reduces to the one-dimensional case $n=1$.

  1. So, let $K(x,t)=\frac{e^{-\frac{x^2}{4t}}}{\sqrt{t}}$. Then for $t>0$ we find $$ K_t=-\frac{e^{-\frac{x^2}{4t}}}{2t^{3/2}}+\frac{x^2e^{-\frac{x^2}{4t}}}{4t^{5/2}},\quad K_x=-\frac{xe^{-\frac{x^2}{4t}}}{2t^{3/2}},\quad K_{xx}=-\frac{e^{-\frac{x^2}{4t}}}{2t^{3/2}}+ \frac{x^2e^{-\frac{x^2}{4t}}}{4t^{5/2}}, $$ whence follows $K_t=K_{xx}$.

  2. Let $G(x,t)=\frac{e^{\frac{\alpha x^2}{1-4\alpha t}}}{\sqrt{1-4\alpha t}}$ with some $\alpha>0$. Then for $\,t<\frac{1}{4\alpha}\,$ we find $$ G_t=\frac{2\alpha e^{\frac{\alpha x^2}{1-4\alpha t}}}{(1-4\alpha t)^{3/2}}+ \frac{4\alpha x^2e^{\frac{\alpha x^2}{1-4\alpha t}}}{(1-4\alpha t)^{5/2}},\quad G_x=\frac{2\alpha xe^{\frac{\alpha x^2}{1-4\alpha t}}}{(1-4\alpha t)^{3/2}}, $$ $$ G_{xx}=\frac{2\alpha e^{\frac{x^2}{1-4\alpha t}}}{(1-4\alpha t)^{3/2}}+ \frac{4\alpha x^2e^{\frac{\alpha x^2}{1-4\alpha t}}}{(1-4\alpha t)^{5/2}}, $$ whence follows $G_t=G_{xx}$. The latter solution represents a blowup example in Cauchy problem for the linear heat equation $$ \begin{cases} u_t=u_{xx}, \;\;x\in\mathbb{R},\;\;0<t<1/4,\\ u|_{t=0}=e^{x^2},\;\;x\in\mathbb{R}, \end{cases} $$ that admits a classical blowup solution $$ u(x,t)=\frac{e^{\frac{x^2}{1-4t}}}{\sqrt{1-4t}},\quad x\in\mathbb{R},\;\;0<t<1/4. $$ For more details see Example of linear parabolic PDE that blows up


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