Modified Heat Equation Solutions with Asymptotic Decay

[bob321]

Hi folks,

Given the following heat equation

u_t = u_{xx} + t - x^2,

I'd like to find all solutions u(x,t)\in C^2(\mathbb{R}^2) such that the quotient

|u(x,t)| / (|x|^5 + |t|^5)

goes to zero as the sum |x| + |t| goes to infinity.

I know how to do the same problem with the usual heat equation u_t = u_{xx}, but I'm not entirely sure how to deal with this extra t - x^2 term. I suspect I can still start by taking the Fourier transform (in x) of each side to get something like:

\partial_t \hat{u}(\xi,t) = \widehat{\partial_t u}(\xi, t) = -|\xi|^2 \hat{u}(\xi,t) + \widehat{t-x^2}(\xi) = -|\xi|^2\hat{u}(\xi, t) + t\delta(\xi) - \delta^{\prime\prime}(\xi),

which gives me an ODE in t that is easy enough to solve. The issue is that I think this method only gives me smooth (by which I mean infinitely differentiable) solutions. Are there other C^2 solutions that I am missing with this approach?

Thanks in advance for any help.

 

[Kummer]

Hi folks,

Given the following heat equation

u_t = u_{xx} + t - x^2,

This solves "nicely" by a modified techinique using Seperation of Variables. If the boundary value problem is on a string of finite length I can post all the steps where are required to solve this analytically.

 

[AiRAVATA]

Your equation is linear, so, as long as you have nice boundary conditions, the solution is unique. Check your notes.

 

[Kummer]

Your equation is linear, so, as long as you have nice boundary conditions, the solution is unique.

Why is it linear? Perhaps you mean to say it is quasi-linear. It cannot be linear for if u_1 and u_2 are solutions does not mean that u_1 + u_2 are solutions.


@bob321. I will post complete steps, but I am unable to since you do not provide a boundary and initial value problems.

 

[bob321]

The problem is over the entire real line so there are no boundary conditions, and the initial condition u(x,0)[\tex] can be an arbitrary C^2[\tex] function. I've actually since worked out the general solution using the Fourier transform, as I started to do in my original post.<br /> <br /> Thanks.

 

[AiRAVATA]

Why is it linear? Perhaps you mean to say it is quasi-linear. It cannot be linear for if u_1 and u_2 are solutions does not mean that u_1 + u_2 are solutions.

It is linear. If L<u>=u_t+u_{xx},</u> then L[u_1+u_2]=L[u_1]+L[u_2]. What is not is homogeneous. A PDE is said to be quasilinear when is linear in the higher derivative term, but not necesarily in the terms of lower order.

The problem is over the entire real line so there are no boundary conditions

That is a boundary condition. You want your solutions to converge at \pm \infty, so \lim_{x\rightarrow \pm \infty}u(x,t)=0.