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I am thinking about the definition of viscosity solutions. One is interested in pde's of the form

$F(x,u(x),Du(x),D^2u(x))=0$ in $\Omega$

for some $\Omega\subset\mathbb{R}^n$. Later in most books there arises at some point the discussion of viscosity solutions for

$F(x,u(x),Du(x),D^2u(x))=f(x)$ in $\Omega$

for some $f$. But in the definition of viscosity solutions there is a "$0$" required on the right hand side. Is this because of

$F(x,u(x),Du(x),D^2u(x))=f(x)$ $\Longleftrightarrow$ $F(x,u(x),Du(x),D^2u(x))-f(x)=0$ $\Longleftrightarrow$ $\hat{F}(x,u(x),Du(x),D^2u(x)) =0$

and the $f(x)$ term is just seen as an expression of "x" and thus one gets a new function $\hat{F}$?

Xander Henderson
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HelloEveryone
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1 Answers1

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Which books are you referring to? It would be peculiar to see $$F(x,u(x),Du(x),D^2u(x))=f(x) \ in \ \Omega,$$ Since the $f(x)$ on the right hand side is redundant, as $F$ is allowed to depend on $x$.

If you put $f(x)$ on the right hand side, usually you are specifying how the $x$-dependence arises, and that it is additive and does not involve $u$ or its derivatives. I would expect you to see $$H(u(x),Du(x),D^2u(x))=f(x) \ \ in \ \Omega.$$ Note $H$ does not depend on $x$. This is equivalent to $F=0$ where $F$ is given by $$F(x,u(x),Du(x),D^2u(x)) = H(u(x),Du(x),D^2u(x))-f(x).$$

Jeff
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