Zweite Richtungsableitung/Gramsche Matrix/Direkt/sin x^2y-e^(x+y)/Aufgabe/Lösung

a) Es ist

${}{\frac {\partial f}{\partial x}}=2xy\cos x^{2}y-e^{x+y}\,$

und

${}{\frac {\partial f}{\partial y}}=x^{2}\cos x^{2}y-e^{x+y}\,.$

Somit ist

${}{\frac {\partial ^{2}f}{\partial ^{2}x}}=-4x^{2}y^{2}\sin x^{2}y+2y\cos x^{2}y-e^{x+y}\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial x}}=-2x^{3}y\sin x^{2}y+2x\cos x^{2}y-e^{x+y}\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}y}}=-x^{4}\sin x^{2}y-e^{x+y}\,.$

Die Hesse-Matrix in ${}P=(1,2)$ ist somit gleich

${\begin{pmatrix}-16\sin \left(2\right)+4\cos \left(2\right)-e^{3}&-4\sin \left(2\right)+2\cos \left(2\right)-e^{3}\\-4\sin \left(2\right)+2\cos \left(2\right)-e^{3}&-\sin \left(2\right)-e^{3}\end{pmatrix}}.$

b) Da die Hesse-Form bilinear ist, gilt

${}{\begin{aligned}D_{v}D_{w}f(P)&=\left(2,\,3\right){\begin{pmatrix}-16\sin \left(2\right)+4\cos \left(2\right)-e^{3}&-4\sin \left(2\right)+2\cos \left(2\right)-e^{3}\\-4\sin \left(2\right)+2\cos \left(2\right)-e^{3}&-\sin \left(2\right)-e^{3}\end{pmatrix}}{\begin{pmatrix}-1\\4\end{pmatrix}}\\&=\left(2,\,3\right){\begin{pmatrix}16\sin \left(2\right)-4\cos \left(2\right)+e^{3}-16\sin \left(2\right)+8\cos \left(2\right)-4e^{3}\\4\sin \left(2\right)-2\cos \left(2\right)+e^{3}-4\sin \left(2\right)-4e^{3}\end{pmatrix}}\\&=\left(2,\,3\right){\begin{pmatrix}4\cos \left(2\right)-3e^{3}\\-2\cos \left(2\right)-3e^{3}\end{pmatrix}}\\&=2\cos \left(2\right)-15e^{3}.\end{aligned}}$

c) Für einen beliebigen Punkt ${}(x,y)$ gilt die Beziehung

${}D_{w}f(x,y)=h'(0)\,$

mit

${}{\begin{aligned}h(t)&=f((x,y)+tw)\\&=\sin \left((x-t)^{2}(y+4t)\right)-e^{x+y+3t}\\&=\sin \left(x^{2}y+(4x^{2}-2xy)t+(-8x+y)t^{2}+4t^{3}\right)-e^{x+y+3t}.\end{aligned}}$

Also ist

${}h'(t)={\left(4x^{2}-2xy+2(-8x+y)t+12t^{2}\right)}\cos \left(x^{2}y+(4x^{2}-2xy)t+(-8x+y)t^{2}+4t^{3}\right)-3e^{x+y+3t}\,$

und somit für ${}t=0$

${}D_{w}f(x,y)=h'(0)={\left(4x^{2}-2xy\right)}\cos \left(x^{2}y\right)-3e^{x+y}=:g(x,y)\,.$

Das gleiche Verfahren, angewendet auf diese Funktion, den Vektor ${}v$ und die Hilfsfunktion

${}{\begin{aligned}p(t)&={\left(4(x+2t)^{2}-2(x+2t)(y+3t)\right)}\cos \left((x+2t)^{2}(y+3t)\right)-3e^{x+y+5t}\\&={\left(4x^{2}+16xt+16t^{2}-2xy-6xt-4yt-12t^{2}\right)}\cos \left(x^{2}y+(4xy+3x^{2})t+(12x+y)t^{2}+12t^{3}\right)-3e^{x+y+5t}\\&={\left(4x^{2}-2xy+(10x-4y)t+4t^{2}\right)}\cos \left(x^{2}y+(4xy+3x^{2})t+(12x+y)t^{2}+12t^{3}\right)-3e^{x+y+5t}\end{aligned}}$

ergibt

${}p'(t)=-{\left(4x^{2}-2xy+(10x-4y)t+4t^{2}\right)}{\left(4xy+3x^{2}+2(12x+y)t+36t^{2}\right)}\sin \left(x^{2}y+(4xy+3x^{2})t+(12x+y)t^{2}+12t^{3}\right)-{\left(10x-4y-8t\right)}\cos \left(x^{2}y+(4xy+3x^{2})t+(12x+y)t^{2}+12t^{3}\right)-15e^{x+y+5t}\,$

und somit

${}D_{v}g(x,y)=p'(0)=-{\left(4x^{2}-2xy\right)}{\left(4xy+3x^{2}\right)}\sin \left(xy\right)-{\left(10x-4y\right)}\cos \left(x^{2}y\right)-15e^{x+y}\,.$

Für ${}P=(x,y)=(1,2)$ ist daher

${}{\begin{aligned}D_{v}D_{w}f(1,2)&=D_{v}g(1,2)\\&=-{\left(4x^{2}-2xy\right)}{\left(4xy+3x^{2}\right)}\sin \left(2\right)+2\cos \left(2\right)-15e^{3}\\&=2\cos \left(2\right)-15e^{3}.\end{aligned}}$

Zur gelösten Aufgabe