Let's consider the following Cauchy problem:
$$u_t=\alpha(x,t,u)u_{xx}+\beta(x,t,u)(u_x)^2+\gamma(x,t,u)u_x+\phi(x,t,u)$$
$$u(x,0)=u_0(x).$$
Assume that functions $\alpha,\beta,\gamma,\phi$ are regular (Hölder continous or even smooth but may have some sort of singularity in $t=0$ like $1/t$ term for example). Suppose that we can prove that this problem has unique solution $u=u(x,t)$ in some domain $D\subset\mathbb{R}\times\mathbb{R}_+$.
Im interested in short time symptotic expansions of solutions, i.e. im looking for methods of showing that
$$u(x,t)=\Sigma_{k=0}^mu^k(x)t^k+o(t^m)$$
as $t\rightarrow 0$.
(With description of functions $u^k$ in terms of simpler PDE's of course).
My question is: are there any papers or monographs dealing with such asymptotic expansions in case of complicated nonlinear PDE's (with mathematical rigour, not only clever Ansatz coming out of the blue and with no proof)?
Another problem I'm concerned with:
In various works i saw asymptotic expansion technique was used in following manner:
1) Write down a series $\Sigma_{i=0}^{\infty}\lambda^iu_i(x,t)$, where $\lambda$ is time variable or some synthetic parameter
2) Plug that formal series into equation, perform differential operations term by term, group everything by powers of $\lambda$
3)Observe that coefficients standing by powers of $\lambda$ are differential operators applied to unknown functions $u_0,u_1,u_2,…$
4)Solve system of equations obtained by equating those coefficients to $0$
5)Be happy that you solved initial equation
So here is a basic question:
how can we be sure without any further analysis between points 4) and 5) that obtained series is in any way related to actual solution?
In some papers i saw (physics and mathematical finance) authors just obtained that formal series and didn't even bother with proving that it has anythin to do with actual solution. Why?
Here is an example of a problem im trying to "assymptotically expand" in short time:
$$tu_t=Tr[\frac{u}{2}AD(\frac{x}{u})\otimes D(\frac{x}{u})]-t^2Tr[\frac{u^3}{8}ADu\otimes Du]+t<\alpha,Du>+t Tr[\frac{1}{2}AD^2u]-\frac{u}{2}$$
Where: $u=u(x,y,t)$ ($x,y>0$ $t\geq 0$) -solution (assume that it exists and that $u>0$),
$A=A(x,y,t)$ -is $2\times 2$ matrix for all $x,y,t$ and its is Hölder continous with exponent $\alpha$ in space and $\alpha/2$ in time, $\alpha=\alpha(x,y,t)\in \mathbb{R}^2$ also Hölder continous with exponent $\alpha$ in space variables and $\alpha/2$ in time variable. $Tr$ stand for trace of a matrix and $\otimes$ for tensor product of vectors: $(a_{i})\otimes (b_{i})=(a_i b_j)_{i,j}$ ($a,b$ -vectors, $a\otimes b$ -matrix).
Best Answer
The most direct way to get a small time expansion is using a gradient expansion. This can be worked out in the following way. Rescale your time variable as $t\rightarrow\lambda t$ being $\lambda$ a parameter taken to be arbitrary large and introduced to fix expansion order. Then, takes $\phi$ proportional to $\lambda$. You will get
$$\lambda u_\tau=\alpha(x,\frac{\tau}{\lambda},u)u_{xx}+\beta(x,\frac{\tau}{\lambda},u)(u_x)^2+\gamma(x,\frac{\tau}{\lambda},u)u_x+\lambda\phi(x,\frac{\tau}{\lambda},u).$$
Then, expand $u$ as
$$u(x,\tau)=\sum_{n=0}^\infty\frac{1}{\lambda^n}u_n(x,\tau).$$
At the end of the computation just put $\lambda=1$. Your solution will be given as polynomials in $t$.
Now, let us consider your equation
$$tu_t=Tr[\frac{u}{2}AD(\frac{x}{u})\otimes D(\frac{x}{u})]-t^2Tr[\frac{u^3}{8}ADu\otimes Du]+t<\alpha,Du>$$
$$+t Tr[\frac{1}{2}AD^2u]-\frac{u}{2}.$$
We firstly note that $D(\frac{x}{u})=\frac{e_x}{u}-x\frac{Du}{u^2}$ and so we can write down this equation in the form
$$tu_t=\frac{1}{2u}[A_{xx}+x^2Tr[A Du\otimes Du]]-t^2Tr[\frac{u^3}{8}ADu\otimes Du]+t<\alpha,Du>$$
$$+t Tr[\frac{1}{2}AD^2u]-\frac{u}{2}.$$
I hope I have interpreted correctly your notation. Now, multiply by $u$ both sides and you will get
$$tuu_t=\frac{1}{2}[A_{xx}+x^2Tr[A Du\otimes Du]]-t^2uTr[\frac{u^3}{8}ADu\otimes Du]+t<\alpha,Du>$$
$$+t uTr[\frac{1}{2}AD^2u]-\frac{u^2}{2}.$$
Now, change the variable as $\tau=\lambda t$. You will get the new equation
$$\tau uu_\tau=\frac{1}{2}[A_{xx}+x^2Tr[A Du\otimes Du]]-\frac{1}{\lambda^2}\tau^2uTr[\frac{u^3}{8}ADu\otimes Du]+\frac{1}{\lambda}\tau <\alpha,Du>$$
$$+\frac{1}{\lambda}\tau uTr[\frac{1}{2}AD^2u]-\frac{u^2}{2}.$$
From this you can read off immediately the leading order being
$$\tau u_0u_{0\tau}=\frac{1}{2}[A_{xx}+x^2Tr[A Du_0\otimes Du_0]]-\frac{u_0^2}{2}.$$
This is a Hamilton-Jacobi equation that can be solved by the characteristic method. Higher orders can be obtained straightforwardly as an expansion in $\frac{1}{\lambda}$.
We can spend a few words about the next-to-leading order by substituting into the scaled equation
$$u=u_0+\frac{1}{\lambda}u_1+O\left(\frac{1}{\lambda^2}\right)$$
Then, the equation becomes
$$\tau (u_0+\frac{1}{\lambda}u_1)(u_{0\tau}+\frac{1}{\lambda}u_{1\tau})=\frac{1}{2}[A_{xx}+x^2Tr[A D(u_0+\frac{1}{\lambda}u_1)\otimes D(u_0+\frac{1}{\lambda}u_1)]]-$$
$$\frac{1}{\lambda^2}\tau^2(u_0+\frac{1}{\lambda}u_1)Tr[\frac{(u_0+\frac{1}{\lambda}u_1)^3}{8}AD(u_0+\frac{1}{\lambda}u_1)\otimes D(u_0+\frac{1}{\lambda}u_1)]$$
$$+\frac{1}{\lambda}\tau <\alpha,D(u_0+\frac{1}{\lambda}u_1)>$$
$$+\frac{1}{\lambda}\tau (u_0+\frac{1}{\lambda}u_1)Tr[\frac{1}{2}AD^2(u_0+\frac{1}{\lambda}u_1)]-\frac{1}{2}\left(u_0+\frac{1}{\lambda}u_1\right)^2+O\left(\frac{1}{\lambda^2}\right).$$
So, the equation to compute $u_1$ is
$$\tau (u_0u_{1\tau}+u_1u_{0\tau})=x^2Tr[A Du_0\otimes Du_1]$$
$$-\tau <\alpha,Du_0>+\tau u_0Tr[\frac{1}{2}AD^2u_0]-u_0u_1.$$
One can repeat this procedure at any order and, finally, gets the solution into the form given at the beginning of this post.