Efter videon följer ett problem som du kan lösa för att testa att du tillgodogjort dig innehållet.
Problem:
Låt \(z=f(x,y)\) vara en funktion av (klass \(C^2\) i planet och inför \(u=x^2-y^2\) och \(v=2xy\). Transformera uttrycket \[ \frac{1}{x^2+y^2}\left(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}\right) \] till de nya variablerna \(u,v\).Svara med koefficienten framför $$\left(\frac{\partial^2 f}{\partial u^2}+ \frac{\partial^2 f}{\partial v^2}\right).$$ (ganska svår)
Svar:
Vi räknar med kedjeregeln
\[
\frac{\partial f}{\partial x}=
\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+
\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=
2x\frac{\partial f}{\partial u}+2y\frac{\partial f}{\partial v},
\]
\[
\frac{\partial f}{\partial y}=
\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+
\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=
-2y\frac{\partial f}{\partial u}+2x\frac{\partial f}{\partial v}.
\]
Ytterligare en derivation ger
\[
\frac{\partial^2 f}{\partial x^2}=
\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)=
\frac{\partial}{\partial x}\left(
2x\frac{\partial f}{\partial u}+2y\frac{\partial f}{\partial v}\right)=
\]
\[
2\frac{\partial f}{\partial u}+
2x\left(\frac{\partial}{\partial x}\left(
\frac{\partial f}{\partial u}\right) \right)+
2y\left(\frac{\partial}{\partial x}\left(
\frac{\partial f}{\partial v}\right) \right)=
\]
\[
2\frac{\partial f}{\partial u}+
2x\left(2x\frac{\partial^2 f}{\partial u^2}+
2y\frac{\partial^2 f}{\partial v\partial u}\right)+
\]
\[
2y\left(2x\frac{\partial^2 f}{\partial v\partial u}+
2y\frac{\partial^2 f}{\partial v^2}\right)=
\]
\[
2\frac{\partial f}{\partial u}+
4x^2\frac{\partial^2 f}{\partial u^2}+
8xy\frac{\partial^2 f}{\partial u\partial v}+
4y^2\frac{\partial^2 f}{\partial v^2}.
\]
På samma sätt beräknas
\[
\frac{\partial^2 f}{\partial y^2}=
\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)=
\frac{\partial}{\partial y}\left(
-2y\frac{\partial f}{\partial u}+2x\frac{\partial f}{\partial v}\right)=
\]
\[
-2\frac{\partial f}{\partial u}-
2y\left(\frac{\partial}{\partial y}\left(
\frac{\partial f}{\partial u}\right) \right)+
2x\left(\frac{\partial}{\partial y}\left(
\frac{\partial f}{\partial v}\right) \right)=
\]
\[
-2\frac{\partial f}{\partial u}
-2y\left((-2y)\frac{\partial^2 f}{\partial u^2}+
2x\frac{\partial^2 f}{\partial u\partial v}\right)+
\]
\[
2x\left((-2y)\frac{\partial^2 f}{\partial v\partial u}+
2x\frac{\partial^2 f}{\partial v^2}\right)=
\]
\[
-2\frac{\partial f}{\partial u}+
4y^2\frac{\partial^2 f}{\partial u^2}-
8xy\frac{\partial^2 f}{\partial u\partial v}+
4x^2\frac{\partial^2 f}{\partial v^2}.
\]
Addition av de två uttrycken ger:
\[
\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=
4(x^2+y^2)\left(\frac{\partial^2 f}{\partial u^2}+
\frac{\partial^2 f}{\partial v^2}\right),
\]
dvs
\[
\frac{1}{(x^2+y^2)}
\left(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial
y^2}\right)=
4\left(\frac{\partial^2 f}{\partial u^2}+
\frac{\partial^2 f}{\partial v^2}\right).
\]
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