Introduccion

Derivadas Parciales de Orden Superior

Si $f$ es una función de doas variables $x,y$ $\Rightarrow$ ${\displaystyle \frac{\partial f}{\partial x},{\displaystyle \frac{\partial f}{\partial y}}}$ son funciones de las mismas variables, cuando derivamos ${\displaystyle \frac{\partial f}{\partial x}}$ y ${\displaystyle \frac{\partial f}{\partial y}}$ obtenemos las derivadas parciales de segundo orden, las derivadas de ${\displaystyle \frac{\partial f}{\partial x}}$ están definidas por:

$${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}(x,y)={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\partial f}{\partial x}(x+h,y)-{\displaystyle \frac{\partial f}{\partial x}(x,y)}}}{h}}}}$$

$${\displaystyle \frac{{\partial }^{2}f}{\partial y\partial x}(x,y)={\displaystyle \underset{k\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\partial f}{\partial x}(x,y+k)-{\displaystyle \frac{\partial f}{\partial x}(x,y)}}}{k}}}}$$

Si $f$ es una función de dos variables entonces hay cuatro derivadas parciales de segundo orden.

Consideremos las diferentes notaciones para las derivadas parciales:

$$f_{1,1}={\displaystyle \frac{{\partial }^{2}f}{\partial x^2}=f_{xx}}$$

$$f_{1,2}={\displaystyle \frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\partial }{\partial y}(\frac{\partial f}{\partial x})=f_{xy}}$$

$$f_{2,1}={\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial y})=f_{yx}}$$

$$f_{2,2}={\displaystyle \frac{{\partial }^{2}f}{\partial y^2}=\frac{\partial }{\partial y}(\frac{\partial f}{\partial y})=f_{yy}}$$

Ejemplo. $z=x^3+3x^{2}y-2x^2{y}^2-y^4+3xy$ hallar ${\displaystyle \frac{\partial z}{\partial x},{\displaystyle \frac{\partial z}{\partial y},{\displaystyle \frac{{\partial }^{2}z}{\partial x^2},{\displaystyle \frac{{\partial }^{2}z}{\partial x\partial y},{\displaystyle \frac{{\partial }^{2}z}{\partial y\partial x},{\displaystyle \frac{{\partial }^{2}z}{\partial y^2}}}}}}}$

$${\displaystyle \frac{\partial z}{\partial x}=3x^2+6xy-4x{y}^2+3y}$$

$${\displaystyle \frac{\partial z}{\partial y}=3x^2-4x^{2}y-4y^3+3x}$$

$${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}=6x+6y-4y^2}$$

$${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=-4x^2-12y^2}$$

$${\displaystyle \frac{{\partial }^{2}z}{\partial y\partial x}=6x-8xy+3}$$

$${\displaystyle \frac{{\partial }^{2}z}{\partial x\partial y}=6x-8xy+3}$$

Teorema 1.Teorema de schwarz

Sea $f:A\subset {\mathbb{R}}^2\to \mathbb{R}$ una función definida en el abierto A de ${\mathbb{R}}^2$. Si las derivadas parciales

$$\frac{{\partial }^{2}f}{\partial y\partial x}y\frac{{\partial }^{2}f}{\partial x\partial y}$$

existen y son continuas en $A$, entonces

$$\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{{\partial }^{2}f}{\partial x\partial y}$$

Demostración. Sea

${\displaystyle M=f(x+h_1,y+h_2)-f(x+h_1,y)-f(x,y+h_2)+f(x,y)}$ y definimos $$\phi (x)=f(x,y+h_2)-f(x,y)$$de manera que
$$\phi (x+h_1)-\phi (x)=f(x+h_1,y+h_2)-f(x+h_1,y)-(f(x,y+h_2)-f(x,y))=M$$

Aplicando el TVM a $\phi$ en el intervalo $[x,x+h_1]$ se tiene que existe $\theta \in (x,x+h_1)$ tal que

$$\phi (x+h_1)-\phi (x)={\phi }^{\prime }(\theta )h_1$$

por otro lado
$${\phi }^{\prime }(x)=\frac{\partial f}{\partial x}(x,y+h_2)-\frac{\partial f}{\partial x}(x,y)$$
por lo tanto
$${\phi }^{\prime }(\theta )=\frac{\partial f}{\partial x}(\theta ,y+h_2)-\frac{\partial f}{\partial x}(\theta ,y)$$
tenemos entonces que

$$M=\phi (x+h_1)-\phi (x)={\phi }^{\prime }(\theta )h_1=(\frac{\partial f}{\partial x}(\theta ,y+h_2)-\frac{\partial f}{\partial x}(\theta ,y))h_1$$
Consideremos ahora ${\displaystyle \psi (y)=\frac{\partial f}{\partial x}(x,y)}$. Aplicando el TVM a $\psi$ en el intervalo $[y,y+h_2]$ se tiene que existe $\eta \in (y,y+h_2)$ tal que
$$\psi (y+h_2)-\psi (y)={\psi }^{\prime }(\eta )h_2$$
por otro lado

$${\psi }^{\prime }(y)=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)(x,y)=\frac{{\partial }^{2}f}{\partial y\partial x}(x,y)$$
por lo tanto
$${\psi }^{\prime }(\eta )=\frac{{\partial }^{2}f}{\partial y\partial x}(x,\eta )$$
de esta manera

$$\psi (y+h_2)-\psi (y)={\psi }^{\prime }(\eta )h_2=(\frac{{\partial }^{2}f}{\partial y\partial x}(x,\eta ))h_2$$
y si $\theta \in (x,x+h_1)$ tenemos entonces que

$$\frac{\partial f}{\partial x}(\theta ,y+h_2)-\frac{\partial f}{\partial x}(\theta ,y)=(\frac{{\partial }^{2}f}{\partial y\partial x}(\theta ,\eta ))h_2$$
en consecuencia
$$M=(\frac{\partial f}{\partial x}(\theta ,y+h_2)-\frac{\partial f}{\partial x}(\theta ,y))h_1=(\frac{{\partial }^{2}f}{\partial y\partial x}(\theta ,\eta ))h_2{h}_1$$

Consideremos ahora $$\bar{\phi }(y)=f(x+h_1,y)-f(x,y)$$de manera que
$$\bar{\phi }(y+h_2)-\bar{\phi }(y)=f(x+h_1,y+h_2)-f(x+h_1,y)-(f(x,y+h_2)-f(x,y))=M$$

Aplicando el TVM a $\bar{\phi }$ en el intervalo $[y,y+h_2]$ se tiene que existe $\bar{\eta }\in (y,y+h_2)$ tal que
$$\bar{\phi }(y+h_2)-\bar{\phi }(y)={\bar{\phi }}^{\prime }(\bar{\eta })h_2$$
por otro lado
$${\bar{\phi }}^{\prime }(y)=\frac{\partial f}{\partial y}(x+h_1,y)-\frac{\partial f}{\partial y}(x,y)$$
por lo tanto

$${\bar{\phi }}^{\prime }(\bar{\eta })=\frac{\partial f}{\partial y}(x+h_1,\bar{\eta })-\frac{\partial f}{\partial y}(x,\bar{\eta })$$
tenemos entonces que
$$M=\bar{\phi }(y+h_2)-\bar{\phi }(y)={\bar{\phi }}^{\prime }(\bar{\eta })h_2=(\frac{\partial f}{\partial y}(x+h_1,\bar{\eta })-\frac{\partial f}{\partial y}(x,\bar{\eta }))h_2$$

Consideremos ahora ${\displaystyle \bar{\psi }(x)=\frac{\partial f}{\partial y}(x,y)}$. Aplicando el TVM a $\psi$ en el intervalo $[x,x+h_1]$ se tiene que existe $\bar{\theta }\in (x,x+h_1)$ tal que
$$\bar{\psi }(x+h_1)-\bar{\psi }(x)={\bar{\psi }}^{\prime }(\bar{\theta })h_1$$
por otro lado

$${\bar{\psi }}^{\prime }(x)=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)(x,y)=\frac{{\partial }^{2}f}{\partial x\partial y}(x,y)$$
por lo tanto
$${\bar{\psi }}^{\prime }(\bar{\theta })=\frac{{\partial }^{2}f}{\partial y\partial x}(\bar{\theta },y)$$
de esta manera

$$\bar{\psi }(x+h_1)-\bar{\psi }(x)={\bar{\psi }}^{\prime }(\bar{\theta })h_1=(\frac{{\partial }^{2}f}{\partial x\partial y}(\bar{\theta },y))h_1$$
es decir
$$\frac{\partial f}{\partial y}(x+h_1,y)-\frac{\partial f}{\partial y}(x,y)=(\frac{{\partial }^{2}f}{\partial x\partial y}(\bar{\theta },y))h_1$$
y si $\bar{\eta }\in (y,y+h_2)$ tenemos entonces que
$$\frac{\partial f}{\partial y}(x+h_1,\bar{\eta })-\frac{\partial f}{\partial y}(x,\bar{\eta })=(\frac{{\partial }^{2}f}{\partial x\partial y}(\bar{\theta },\bar{\eta }))h_1$$
en consecuencia

$$M=(\frac{\partial f}{\partial y}(x+h_1,\bar{\eta })-\frac{\partial f}{\partial y}(x,\bar{\eta }))h_1{h}_2=(\frac{{\partial }^{2}f}{\partial x\partial y}(\bar{\theta },\bar{\eta }))h_2{h}_1$$
igualando ambas expresiones de M se tiene
$$(\frac{{\partial }^{2}f}{\partial y\partial x}(\theta ,\eta ))h_2{h}_1=(\frac{{\partial }^{2}f}{\partial x\partial y}(\bar{\theta },\bar{\eta }))h_2{h}_1$$
donde
$$(\frac{{\partial }^{2}f}{\partial y\partial x}(\theta ,\eta ))=(\frac{{\partial }^{2}f}{\partial x\partial y}(\bar{\theta },\bar{\eta }))$$
Tomando limite cuando $h_1,h_2\to 0$ y usando la continuidad asumida de las parciales mixtas se tiene que $\theta ,\bar{\theta }\to x$ y $\eta ,\bar{\eta }\to y$ se concluye
$$\frac{{\partial }^{2}f}{\partial y\partial x}(x,y)=\frac{{\partial }^{2}f}{\partial x\partial y}(x,y)$$ $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/14.0.0/svg/25fb.svg">}$

Ejemplo. Sea $f:{\mathbb{R}}^2\to \mathbb{R}$ dada por $f(x,y)=x^3+3x^{2}y-2x^2{y}^2-y^4+3xy$\
En este caso
$$\frac{\partial f}{\partial x}=3x^2+6xy-4x{y}^2+3y$$
$$\frac{\partial f}{\partial y}=3x^2-4x^{2}y-4y^3+3x$$
$$\frac{{\partial }^{2}f}{\partial x^2}=6x+6y-4y^2$$
$$\frac{{\partial }^{2}f}{\partial y^2}=-4x^2-12y^2$$
$$\frac{{\partial }^{2}f}{\partial x\partial y}=6x-8xy+3$$
$$\frac{{\partial }^{2}f}{\partial y\partial x}=6x-8xy+3$$

Ejemplo. Dada la función

tenemos que para $(x,y)\ne (0,0)$
$$\frac{\partial f}{\partial x}=y\frac{x^4+4x^2{y}^2-y^4}{(x^2+y^2{)}^2}$$
$$\frac{\partial f}{\partial y}=x\frac{x^4-4x^2{y}^2-y^4}{(x^2+y^2{)}^2}$$
para el primer caso hacemos $x=0$ y tenemos
$$\frac{\partial f}{\partial x}=y\frac{x^4+4x^2{y}^2-y^4}{(x^2+y^2{)}^2}\underbrace{=}x=0-y$$ para el segundo caso hacemos $y=0$ y tenemos $$\frac{\partial f}{\partial y}=x\frac{x^4-4x^2{y}^2-y^4}{(x^2+y^2{)}^2}\underbrace{=}y=01$$
Calculamos ahora
$$\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{{\partial }^2(-y)}{\partial y\partial x}=-1$$
$$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{{\partial }^2(1)}{\partial x\partial y}=1$$
por lo tanto
$$\frac{{\partial }^{2}f}{\partial y\partial x}=-1\ne 1=\frac{{\partial }^{2}f}{\partial x\partial y}$$
En este caso las parciales segundas no son contiuas en $(0,0)$

Teorema. Caso General

Sea $f:A\subset {\mathbb{R}}^n\to \mathbb{R}$ definida en el abierto A de ${\mathbb{R}}^n$ tal que
$$\frac{{\partial }^{2}f}{\partial x_i\partial x_j}$$ sean continuas en A, entonces
$$\frac{{\partial }^{2}f}{\partial x_i\partial x_j}=\frac{{\partial }^{2}f}{\partial x_j\partial x_i}$$

Mas adelante

Tarea Moral

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