1
$\begingroup$

Peace be upon you,

I have the following system of partial differential equations \begin{align*} \begin{cases} \frac{\partial}{\partial a}S(a,b,c,d)=f_1(a)\\ \frac{\partial}{\partial b}S(a,b,c,d)=f_2(b)\\ \frac{\partial}{\partial c}S(a,b,c,d)=f_3(c)\\ \frac{\partial}{\partial d}S(a,b,c,d)=f_4(d)\\ \end{cases} \end{align*} where $f_i()$s are some nonlinear functions.

Does the above system have a unique answer(?) and if has can any one introduce a reference, explaining the techniques for analytic solutions?

Note: The usual PDE references (books, articles, webpages, etc.) speak about the systems for which the number of unknown functions and the number of system equations are equal.

$\endgroup$
3
$\begingroup$

Your problem can be reformulated as follows (upon the change of notation $a=x, b=y, c=z, d=t$).

You are assigning the differential form $$ \omega=f_1(x)dx +f_2(y) dy + f_3(z)dz+f_4(t)dt, $$ which is closed, hence exact on $\mathbb{R}^4$. You want to find a potential function, that is, a function $F=F(x, y, z, t)$ such that $dF=\omega$. One of them is given by the following integral, the others differ by an additive constant: $$ F(x,y,z,t)=\int_\gamma \omega,\qquad \gamma\colon[0, 1]\to \mathbb{R}^4,\ \gamma(0)=0,\ \gamma(1)=(x,y,z,t)$$ You can choose any curve $\gamma$, only its endpoints matter. Taking for instance the line segment $$\gamma(s)=(sx,sy,sz,st),$$ you obtain $$ F(x,y,z,t)=\int_0^1 f_1(sx)x\, ds + \int_0^1 f_2(sy)y\, ds+\int_0^1 f_3(sz)z\, ds+\int_0^1 f_4(st)t\, ds.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks so much. But, in fact, my main problem is \begin{align*} \begin{cases} \frac{\partial}{\partial x}F(x,y,z,t)=f_1(x,y,z,t)\\ \frac{\partial}{\partial y}F(x,y,z,t)=f_2(x,y,z,t)\\ \frac{\partial}{\partial z}F(x,y,z,t)=f_3(x,y,z,t)\\ \frac{\partial}{\partial t}F(x,y,z,t)=f_4(x,y,z,t) \end{cases} \end{align*} So, I am assigning the differential form \begin{align*} \omega=f_1(x,y,z,t)dx +f_2(x,y,z,t) dy + f_3(x,y,z,t)dz+f_4(x,y,z,t)dt, \end{align*} which is again closed and therefore exact on $\mathbb{R}^4$. So, it seems that again I am to find the $F=F(x, y, z, t)$ such that $dF=\omega$. $\endgroup$ – hossayni Oct 22 '14 at 13:15
  • $\begingroup$ But which $\gamma$ curve should I chose for my problem (regarding the point that the previous curve does not support the multivariate function argument)? $\endgroup$ – hossayni Oct 22 '14 at 13:16
  • $\begingroup$ In this more general formulation, the form $\omega$ needs not be closed. You need to impose the compatibility conditions $\partial_yf_1=\partial_x f_2, \partial_zf_1 =\partial_x f_3\ldots$ and so on. Note that if $\omega$ is not closed then you do not have existence of solutions to your PDE. $\endgroup$ – Giuseppe Negro Oct 22 '14 at 14:59
  • $\begingroup$ @hossayni: The last comment is addressed to you, perhaps you did not receive notification because I forgot the @... construct. $\endgroup$ – Giuseppe Negro Oct 22 '14 at 17:55
  • $\begingroup$ Thanks so much; could you also please introduce a reference for related topics? $\endgroup$ – hossayni Oct 22 '14 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.