مسئله مقدار مرزی

${\displaystyle y^{″}(x)+y(x)=0}$

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

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

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

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

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

${\displaystyle Lu=f}$

${\displaystyle u=g}$

${\displaystyle y^{″}(x)+y(x)=0}$

${\displaystyle y(a)=\alpha }$

${\displaystyle y(b)=\beta }$

${\displaystyle {\mathrm{\nabla }}^{2}y+y=0}$

${\displaystyle y(x)=f(x)}$

${\displaystyle \frac{\mathrm{\partial }y(x)}{\mathrm{\partial }n}=f(x)}$

${\displaystyle \mathrm{\forall }x\in \mathrm{\partial }\mathrm{\Omega }}$

${\displaystyle \frac{\mathrm{\partial }y(x)}{\mathrm{\partial }n}=\mathrm{\nabla }y(x).\hat{n}(x)}$

${\displaystyle au+b\frac{\mathrm{\partial }u}{\mathrm{\partial }n}=g}$

${\displaystyle au(0)-b{u}^{\prime }(0)=g(0)}$

${\displaystyle au(1)+b{u}^{\prime }(1)=g(1)}$

${\displaystyle y^{″}(x)+A(x)y^{\prime }(x)+B(x)y(x)=0}$

${\displaystyle y(a)=y_0,y(b)=y_1}$

${\displaystyle y^{\prime }(x)=C+{\int }_a^{x}F(x^{\prime })d{x}^{\prime }-A(x)y(x)+A(a)y_0+{\int }_a^x[A^{\prime }(x^{\prime })-B(x^{\prime })]y(x^{\prime })d{x}^{\prime }}$

${\displaystyle y(x)-y_0=[C+A(a)y_0](x-a)+{\int }_a^x{\int }_a^{x_2}F(x^{\prime })d{x}^{\prime }d{x}_2-{\int }_a^{x}A(x^{\prime })y(x^{\prime })d{x}^{\prime }+{\int }_a^x{\int }_a^{x_2}[A^{\prime }(x^{\prime })-B(x^{\prime })]y(x^{\prime })d{x}^{\prime }d{x}_2}$

${\displaystyle y(x)-y_0=[C+A(a)y_0](x-a)+{\int }_a^x(x-t)F(t)dt-{\int }_a^x\{A(t)-(x-t)[A^{\prime }(t)-B(t)]\}y(t)dt}$

${\displaystyle y(x)=y_0+{\int }_a^x(x-t)F(t)dt+[\frac{x-a}{b-a}][(y_1-y_0)-\int (b-t)F(t)dt]-{\int }_a^x\{A(t)-(x-t)(A^{\prime }(t)-B(t))\}y(t)dt+\int [\frac{x-a}{b-a}]\{A(t)-(b-t)[A^{\prime }(t)-B(t)]\}y(t)dt}$

${\displaystyle f(x)=y_0+{\int }_a^x(x-t)F(t)dt+[\frac{x-a}{b-a}][(y_1-y_0)-\int (b-t)F(t)dt]}$

${\displaystyle K(x,t)=(\frac{x-a}{b-a})(A(t)-(b-t)[A^{\prime }(t)-B(t)])}$

${\displaystyle x>t:}$

${\displaystyle K(x,t)=A(t)\{[\frac{x-a}{b-a}]-1\}-[A^{\prime }(t)-B(t)][\frac{(t-a)(b-x)}{b-a}]}$