# Difference between $u_t + \Delta u = f$ and $u_t - \Delta u = f$?

What is the difference between these 2 equations? Instead of $\Delta$ change it to some general elliptic operator.

Do they have the same results? Which one is used for which?

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They correspond to each other via $t \mapsto -t$ and $f\mapsto -f$. – Fabian Jan 2 '13 at 17:56
@Fabian Sorry I made an error. I am wondering more about showing things like coercivity and existence. – hopo2 Jan 2 '13 at 17:58

The relation boils down to time-reversal, replacing $t$ by $-t$. This makes a lot of difference in the equations that model diffusion. The diffusion processes observed in nature are normally not reversible (2nd law of thermodynamics). In parallel to that, the backward heat equation $u_t=-\Delta u$ exhibits peculiar and undesirable features such as loss of regularity and non-existence of solution for generic data. Indeed, you probably know that the heat equation $u_t=\Delta u$ has a strong regularizing effect: for any integrable initial data $u(\cdot,0)$ the solution $u(x,t)$ is $C^\infty$ smooth, and also real-analytic with respect to $x$ for any fixed $t>0$. When the direction of time flow is reversed, this effect plays against you: there cannot be a solution unless you have real-analytic data to begin with. And even then the solution can suddenly blow up and cease to exist. Consider the fundamental solution of the heat equation, and trace it backward in time, from nice Gaussians to $\delta$-function singularity.

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Nice answer. So I must be very careful when I pick the sign of an elliptic operator in my parabolic PDE? "I can't show existence for this equation so I will change the sign of the operator"? I am confused because in some books the authors say we can solve the PDE $u' + Au = f$ where $A$ is some operator. But I don't know if I can choose $A = -B$ and/or $A = B$ for some elliptic operator $B$. – hopo2 Jan 2 '13 at 18:09
+1. " The diffusion processes observed in nature are normally not reversible (2nd law of thermodynamics). In parallel to that, the backward heat equation $u_t = - \nabla u$ exhibits peculiar and undesirable features such as loss of regularity and non-existence of solution for generic data." That is a very neat statement. Though I knew the two statements independently, I never thought about this connection. – user17762 Jan 2 '13 at 18:09
@hopo2 If you could give a precise reference to a confusing passage in a book, I might be able to help better. – user53153 Jan 2 '13 at 18:14
@PavelM It's ok, I got it. Such theorems (existence of solution) require some condition on the operator $A$ like $(Au,u) > 0$ which obviously do not hold when the sign is chosen differently. It just never struck me before that if I choose the sign wrongly I won't be able to show coercivity for example. – hopo2 Jan 2 '13 at 18:17

If the solution to the first equation i.e. $u_t + \nabla u = f$ is given by $u(x,t;f)$, then the solution to the second equation, $u_t - \nabla u = f$, is given by $u(x,-t;-f)$.

The equation $u_t - \nabla u = f$ is usually called the diffusion equation while the equation $u_t + \nabla u = f$ is usually called the backward diffusion equation.

The diffusion equation is typically used to model, not surprisingly, all diffusive processes like heat conduction for instance. The backward diffusion equation is popular in financial mathematics, and is used to determine the price of various instruments.

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Sorry I made an error. – hopo2 Jan 2 '13 at 17:57
Then showing coercivity for one will imply the other right? Or it depends on specific operator? – hopo2 Jan 2 '13 at 17:59