مسئله مقدار مرزی

${\displaystyle y''(x)+y(x)=0\,}$

${\displaystyle y(0)=0,\ y(\pi /2)=2.}$

${\displaystyle y(x)=A\sin(x)+B\cos(x).\,}$

${\displaystyle 0=A\cdot 0+B\cdot 1}$

${\displaystyle 2=A\cdot 1}$

${\displaystyle y(x)=2\sin(x).\,}$

${\displaystyle Lu=f}$ در ناحیه ${\displaystyle \Omega \subset R^{n}}$

${\displaystyle u=g}$ روی ناحیه ${\displaystyle \partial \Omega }$

${\displaystyle y''(x)+A(x)y'(x)+B(x)y(x)=0}$

${\displaystyle y(a)=y_{0},y(b)=y_{1}}$

${\displaystyle y'(x)=C+\int _{a}^{x}F(x')dx'-A(x)y(x)+A(a)y_{0}+\int _{a}^{x}[A'(x')-B(x')]y(x')dx'}$

${\displaystyle y(x)-y_{0}=[C+A(a)y_{0}](x-a)+\int _{a}^{x}\int _{a}^{x_{2}}F(x')dx'dx_{2}-\int _{a}^{x}A(x')y(x')dx'+\int _{a}^{x}\int _{a}^{x_{2}}[A'(x')-B(x')]y(x')dx'dx_{2}}$

${\displaystyle y(x)-y_{0}=[C+A(a)y_{0}](x-a)+\int _{a}^{x}(x-t)F(t)dt-\int _{a}^{x}\{A(t)-(x-t)[A'(t)-B(t)]\}y(t)dt}$

${\displaystyle y(x)=y_{0}+\int _{a}^{x}(x-t)F(t)dt+[{\frac {x-a}{b-a}}][(y_{1}-y_{0})-\int (b-t)F(t)dt]-\int _{a}^{x}\{A(t)-(x-t)(A'(t)-B(t))\}y(t)dt+\int [{\frac {x-a}{b-a}}]\{A(t)-(b-t)[A'(t)-B(t)]\}y(t)dt}$

${\displaystyle f(x)=y_{0}+\int _{a}^{x}(x-t)F(t)dt+[{\frac {x-a}{b-a}}][(y_{1}-y_{0})-\int (b-t)F(t)dt]}$

${\displaystyle x<t:}$

${\displaystyle K(x,t)=({\frac {x-a}{b-a}})(A(t)-(b-t)[A'(t)-B(t)])}$

${\displaystyle x>t:}$

${\displaystyle K(x,t)=A(t)\{[{\frac {x-a}{b-a}}]-1\}-[A'(t)-B(t)][{\frac {(t-a)(b-x)}{b-a}}]}$