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

CompuChip
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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 us 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.
 
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CompuChip said:
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 us 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.
 
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