Mixed multidimensional integral equation
$begingroup$
1) Suppose
$$
sum_{alphasubset{1,...,N}}int_{[0,1)^{|alpha|}}u({bf x})d{bf x}_{alpha}=f({bf x}).
$$
Find $u({bf x})$, i.e. express it in terms of $f({bf x})$.
If $N=1$ then we have
$$
u(x_1)+int_0^1u(x_1)dx_1=f(x_1) Rightarrow u(x_1)=f(x_1)-frac12int_0^1f(x_1)dx_1.
$$
If $N=2$ then
$$
u(x_1,x_2)+int_0^1u(x_1,x_2)dx_1+int_0^1u(x_1,x_2)dx_2+int_0^1int_0^1u(x_1,x_2)dx_1dx_2=f(x_1,x_2)
$$
yields
$$
u(x_1,x_2)=f(x_1,x_2)-frac12int_0^1f(x_1,x_2)dx_1-frac12int_0^1f(x_1,x_2)dx_2+frac14int_0^1int_0^1f(x_1,x_2)dx_1dx_2.
$$
What happens in the case $N>2$?
2) What are the solution and the criteria of invertibility in the general case
$$
sum_{alphasubset{1,...,N}}A_{alpha}int_{[0,1)^{|alpha|}}u({bf x})d{bf x}_{alpha}=f({bf x}),
$$
where $A_{alpha}$ are constant numbers?
integration
asked Dec 5 '18 at 22:53
AAKAAK
365
$endgroup$
add a comment |
$begingroup$
1) Suppose
$$
sum_{alphasubset{1,...,N}}int_{[0,1)^{|alpha|}}u({bf x})d{bf x}_{alpha}=f({bf x}).
$$
Find $u({bf x})$, i.e. express it in terms of $f({bf x})$.
If $N=1$ then we have
$$
u(x_1)+int_0^1u(x_1)dx_1=f(x_1) Rightarrow u(x_1)=f(x_1)-frac12int_0^1f(x_1)dx_1.
$$
If $N=2$ then
$$
u(x_1,x_2)+int_0^1u(x_1,x_2)dx_1+int_0^1u(x_1,x_2)dx_2+int_0^1int_0^1u(x_1,x_2)dx_1dx_2=f(x_1,x_2)
$$
yields
$$
u(x_1,x_2)=f(x_1,x_2)-frac12int_0^1f(x_1,x_2)dx_1-frac12int_0^1f(x_1,x_2)dx_2+frac14int_0^1int_0^1f(x_1,x_2)dx_1dx_2.
$$
What happens in the case $N>2$?
2) What are the solution and the criteria of invertibility in the general case
$$
sum_{alphasubset{1,...,N}}A_{alpha}int_{[0,1)^{|alpha|}}u({bf x})d{bf x}_{alpha}=f({bf x}),
$$
where $A_{alpha}$ are constant numbers?
integration
asked Dec 5 '18 at 22:53
AAKAAK
365
$endgroup$
add a comment |
$begingroup$
1) Suppose
$$
sum_{alphasubset{1,...,N}}int_{[0,1)^{|alpha|}}u({bf x})d{bf x}_{alpha}=f({bf x}).
$$
Find $u({bf x})$, i.e. express it in terms of $f({bf x})$.
If $N=1$ then we have
$$
u(x_1)+int_0^1u(x_1)dx_1=f(x_1) Rightarrow u(x_1)=f(x_1)-frac12int_0^1f(x_1)dx_1.
$$
If $N=2$ then
$$
u(x_1,x_2)+int_0^1u(x_1,x_2)dx_1+int_0^1u(x_1,x_2)dx_2+int_0^1int_0^1u(x_1,x_2)dx_1dx_2=f(x_1,x_2)
$$
yields
$$
u(x_1,x_2)=f(x_1,x_2)-frac12int_0^1f(x_1,x_2)dx_1-frac12int_0^1f(x_1,x_2)dx_2+frac14int_0^1int_0^1f(x_1,x_2)dx_1dx_2.
$$
What happens in the case $N>2$?
2) What are the solution and the criteria of invertibility in the general case
$$
sum_{alphasubset{1,...,N}}A_{alpha}int_{[0,1)^{|alpha|}}u({bf x})d{bf x}_{alpha}=f({bf x}),
$$
where $A_{alpha}$ are constant numbers?
integration
asked Dec 5 '18 at 22:53
AAKAAK
365
$endgroup$
1) Suppose
$$
sum_{alphasubset{1,...,N}}int_{[0,1)^{|alpha|}}u({bf x})d{bf x}_{alpha}=f({bf x}).
$$
Find $u({bf x})$, i.e. express it in terms of $f({bf x})$.
If $N=1$ then we have
$$
u(x_1)+int_0^1u(x_1)dx_1=f(x_1) Rightarrow u(x_1)=f(x_1)-frac12int_0^1f(x_1)dx_1.
$$
If $N=2$ then
$$
u(x_1,x_2)+int_0^1u(x_1,x_2)dx_1+int_0^1u(x_1,x_2)dx_2+int_0^1int_0^1u(x_1,x_2)dx_1dx_2=f(x_1,x_2)
$$
yields
$$
u(x_1,x_2)=f(x_1,x_2)-frac12int_0^1f(x_1,x_2)dx_1-frac12int_0^1f(x_1,x_2)dx_2+frac14int_0^1int_0^1f(x_1,x_2)dx_1dx_2.
$$
What happens in the case $N>2$?
2) What are the solution and the criteria of invertibility in the general case
$$
sum_{alphasubset{1,...,N}}A_{alpha}int_{[0,1)^{|alpha|}}u({bf x})d{bf x}_{alpha}=f({bf x}),
$$
where $A_{alpha}$ are constant numbers?
integration
integration
asked Dec 5 '18 at 22:53
AAKAAK
365
asked Dec 5 '18 at 22:53
AAKAAK
365
edited Dec 6 '18 at 22:12
AAK
asked Dec 5 '18 at 22:53
AAKAAK
365
asked Dec 5 '18 at 22:53
AAKAAK
365
asked Dec 5 '18 at 22:53
AAKAAK
365
365
add a comment |
add a comment |
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