characterstic of partial diffrential equation Show that the line $\lbrace t = 0\rbrace$ is the characteristic for the heat equation $u_t = u_{xx}$. Show that
there does not exist an analytic solution $u$ of the heat equation on $\mathbb{R}\times\mathbb{R}$, with $u =
1/(1+x^2)$
on $\lbrace t = 0\rbrace$.
I know that $u(x,0)=\phi(x)$ then then $dt(u(x,0))= \phi''(x)$ then how would we say it has characterstic zero
 A: Depending on your text's definition of a characteristic surface, this may look a bit different.  First, in order to define a characteristic surface, we will rewrite the heat equation as a first order system: $$ u_x=v \quad u_t=v_x$$  Then we note that the equation can be rewritten in matrix form as $$A \binom{u_x}{v_x} + B \binom{u_t}{v_t} + \binom{-v}{0} = \binom{0}{0}$$ where $$ A = \begin{pmatrix} 1 &0\\ 0 & 1 \end{pmatrix} \mbox{ and } B=\begin{pmatrix} 0 &0 \\-1&0 \end{pmatrix}.$$  The reason I rewrite it this way is to conform with Wikipedia's definition of a characteristic surface.  We then want to write the surface we want to check as charateristic in the form $\phi(x,t)=0$.  This is really easy, we just let $\phi(x,t)=t$.  Plugging this into the definition from Wikipedia we get $$ \det \left( A\phi_x + B\phi_t\right) = \det{B} = 0$$ which shows that $t=0$ is characteristic.
Now, to show that there is no analytic solution, let's suppose that one exists and derive a contradiction.  We say that $$ u(x,t) = \sum_{i,j \geq 0} a_{i,j}x^it^j$$
Compute the derivatives (term by term) and equate coefficients to get the following relationship $$i(i-1)a_{i,j-1}=ja_{i-2,j}.$$  Now doing some index shifting, we can get that $$a_{i,j} = \frac{(i+2)(i+1)}{j} a_{i+2,j-1}.$$
Now, let's use that initial condition to find some coefficients too: $$\sum_{i \geq 0} a_{i,0} x^i= u(x,0) = \frac{1}{1+x^2} = \sum_{i\geq 0} (-x^2)^i = \sum_{i \geq 0} (-1)^ix^{2i}.$$  This means that in our formal series for $u$ we must have that $$ a_{i,0} = \begin{cases} 0 & i \mbox{ odd.}\\ (-1)^{\frac{i}{2}} & i \mbox{ even.} \end{cases}$$
Now, at this point I want to show that our formal series will not converge for any positive value of $t$ and $x=0$.  Notice that $$u(0,t) = \sum_{j \geq 0} a_{0,j} t^j.$$  We are going to show that this series diverges. From our recurrence relation we have that $$a_{0,j} = \frac{2}{j} a_{2,j-1}.$$  We can repeat this $j$ times to get that $$a_{0,j} = \frac{l_j}{j!} a_{2j,0}$$ where $l_j$ is the $j$th element of the sequence generated by $l_0=0$ and $l_{n+1} = (l_n+2)(l_n+1)$.  Now we have reduced it to coefficients that we know that value of from the IC!  So then $$a_{0,j} = \frac{(-1)^j l_j}{j!}.$$ Now, let's apply the ratio test $$ \left| \frac{a_{0,j+1}}{a_{0,j}} \right| = \frac{l_{j+1}}{(j+1){l_j}} = \frac{(l_j+2)(l_j+1)}{(j+1)l_j} = \left(1+\frac{2}{l_j}\right)\frac{l_j+1}{(j+1)}$$.  Now, by induction you can show that $l_j > j^2$.  Hence $$ \lim_{j \to \infty}  \left(1+\frac{2}{l_j}\right)\frac{l_j+1}{(j+1)} = \infty>1.$$  The series then fails the ratio test, and thus must diverge.
So, we have shown that if an analytic solution existed with the prescribed IC, then it will not converge at $x=0$ for any positive value of $t$.  This is a contradiction, and thus no analytic solution can exist.
