# Modified Heat Equation

1. Aug 8, 2007

### 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.

2. Aug 8, 2007

### Kummer

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.

3. Aug 9, 2007

### AiRAVATA

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

4. Aug 9, 2007

### Kummer

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.

5. Aug 9, 2007

### bob321

The problem is over the entire real line so there are no boundary conditions, and the initial condition [tex]u(x,0)[\tex] can be an arbitrary [tex]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.

Thanks.

6. Aug 10, 2007

### AiRAVATA

It is linear. If $L=u_t+u_{xx},$ 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.

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