What Conditions on f Ensure a Good Solution for u(s, t)?

[CompuChip]

Hi.

I'm not so well-versed in the topic of partial differential equations, but the following question has arisen.

Suppose that for some unknown function u(s, t) of two variables, we have a set of differential equations
$$\left\{ \begin{matrix} u_s(s, t) & {} = f(s, t, u(s, t)) \\ u_t(s, t) & {} = g(s, t, u(s, t)) \end{matrix} \right.$$
where u_s denotes the partial derivative of u(s, t) w.r.t. s.

My question is, what the conditions on f would have to be in order to have a good solution for u(s, t). For example, we can integrate the first one to get u(s, T) for fixed t = T, and similarly the second one will give u(S, t) for fixed s = S, but of course u(S, T) must be well-defined (i.e. single-valued).

I am particularly interested in the case where f and g do not depend on s and t explicitly (i.e. only through u(s, t)) and the case where they do not depend on u(s, t) explicitly.

Thanks for sharing your thoughts.

 

[Mathwebster]

Hi.

I'm not so well-versed in the topic of partial differential equations, but the following question has arisen.

Suppose that for some unknown function u(s, t) of two variables, we have a set of differential equations
$$\left\{ \begin{matrix} u_s(s, t) & {} = f(s, t, u(s, t)) \\ u_t(s, t) & {} = g(s, t, u(s, t)) \end{matrix} \right.$$
where u_s denotes the partial derivative of u(s, t) w.r.t. s.

My question is, what the conditions on f would have to be in order to have a good solution for u(s, t). For example, we can integrate the first one to get u(s, T) for fixed t = T, and similarly the second one will give u(S, t) for fixed s = S, but of course u(S, T) must be well-defined (i.e. single-valued).

I am particularly interested in the case where f and g do not depend on s and t explicitly (i.e. only through u(s, t)) and the case where they do not depend on u(s, t) explicitly.

Thanks for sharing your thoughts.

For a solution to exist you certainly want that your two PDEs are compatible, i.e.

$$u_{ts} = u_{st}$$ which gives $$f_t + g f_u = g_s + g_u f$$

The two cases your interested in

(i) f and g independent of u [text]f_t = g_s[/tex]
(ii) f and g independent of t and s so $$g f_u = f g_u$$ so $$f = c \;g$$ for some constant c.