1617-257/Classnotes for Friday October 14: Difference between revisions

Latest revision as of 22:50, 19 October 2016

DBN: Classes: 2016-17: 1617-257 / Navigation

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$\int _{M}d\omega =\int _{\partial M}\omega$

Dror's notes above / Students' notes below

Symmetry of Second Partial Derivatives

(Note: This is based off of the proof in the textbook, and may be slightly different from how it was presented in lecture.)

Theorem

Let $A$ be an open subset of $\mathbb {R} ^{2}$ , and let $f:A\rightarrow \mathbb {R}$ be of class $C^{2}$ . Then, for all $c\in A$ , we have that $D_{1}D_{2}f(c)=D_{2}D_{1}f(c)$ .

Proof

Step 1

We begin by defining a function that will help us in our proof. Let $(a,b)\in A\subseteq \mathbb {R} ^{2}$ be an arbitrary point. We then define the function $\lambda$ as follows:

$\lambda (h,k)=f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)$ .

Why do we define such a function? In fact, we can define both of the partial derivatives in question in terms of $\lambda$ :

${\begin{aligned}D_{1}D_{2}f(a,b)&={\frac {\partial }{\partial x}}{\frac {\partial f}{\partial y}}(a,b)\\&=\lim _{h\to 0}({\frac {1}{h}})[{\frac {\partial f}{\partial y}}(a+h,b)]-[{\frac {\partial f}{\partial y}}(a,b)]\\&=\lim _{h\to 0}({\frac {1}{h}})[\lim _{k\to 0}({\frac {1}{k}})f(a+h,b+k)-f(a+h,b)]-[\lim _{k\to 0}({\frac {1}{k}})f(a,b+k)-f(a,b)]\\&=\lim _{h\to 0}({\frac {1}{h}})[\lim _{k\to 0}({\frac {1}{k}})f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)]\\&=\lim _{h\to 0}\lim _{k\to 0}({\frac {1}{hk}})[f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)]\\&=\lim _{h\to 0}\lim _{k\to 0}({\frac {1}{hk}})\lambda (h,k)\end{aligned}}$

And similarly for $D_{2}D_{1}f(a,b)$ , but with the two limits taken in the opposite order.

Step 2

Now, let's make use of this function. Let us consider the square $Q=[a,a+h]\times [b,b+k]$ , where $h$ and $k$ are so small that $Q\subseteq A$ . We will show that there exist points $p,q\in Q$ such that:

${\begin{aligned}&\lambda (h,k)=D_{2}D_{1}f(p)\cdot hk\\&\lambda (h,k)=D_{1}D_{2}f(q)\cdot hk\end{aligned}}$

The proofs of the two equalities are symmetric, and thus we only explicitly prove the first one. We will prove this through a double application of the mean value theorem.

Consider the function $\phi$ , defined such that $\phi (s)=f(s,b+k)-f(s,b)$ . Then $\phi (a+h)-\phi (a)=\lambda (h,k)$ . By hypothesis, $D_{1}f$ exists at all points of $A$ , so we can differentiate $\phi$ with respect to the first variable of $f$ on the interval $[a,a+h]$ . By the mean value theorem, this means that there exists a point $s_{0}\in [a,a+h]$ such that $\lambda (h,k)=\phi (a+h)-\phi (a)=\phi ^{\prime }(s_{0})\cdot h=[D_{1}f(s_{0},b+k)-D_{1}f(s_{0},b)]\cdot h$ .

Now we want to apply the mean value theorem one more time. Consider the function $\gamma$ , defined such that $\gamma (t)=D_{1}f(s_{0},t)$ . By hypothesis, $D_{2}D_{1}f$ exists at all points of $A$ , so we can differentiate $\gamma$ with respect to the second variable of $f$ on the interval $[b,b+k]$ . So, by the mean value theorem, there exists a point $t_{0}\in [b,b+k]$ such that $\gamma (b+k)-\gamma (b)=\gamma ^{\prime }(t_{0})\cdot k$ . Thus:

${\begin{aligned}\lambda (h,k)&=[D_{1}f(s_{0},b+k)-D_{1}f(s_{0},b)]\cdot h\\&=[\gamma (b+k)-\gamma (b)]\cdot h\\&=\gamma ^{\prime }(t_{0})\cdot hk\\&=D_{2}D_{1}f(s_{0},t_{0})\cdot hk\end{aligned}}$

Step 3

Now we can prove the theorem. Let $c=(a,b)\in A$ , and let $t>0$ be so small that $Q_{t}=[a,a+t]\times [b,b+t]\subseteq A$ . By what we have just shown, $\lambda (t,t)=D_{2}D_{1}f(p_{t})\cdot t^{2}$ , for some $p_{t}\in Q_{t}$ .

$t$ is the length of the sides of the rectangle $Q_{t}$ , so as $t\rightarrow 0$ , $p_{t}\rightarrow c$ . By hypothesis, $D_{2}D_{1}f$ is continuous, and so as $p_{t}\rightarrow c$ , $D_{2}D_{1}f(p_{t})\rightarrow D_{2}D_{1}f(c)$ . Thus:

$\lim _{t\to 0}{\frac {\lambda (t,t)}{t^{2}}}=D_{2}D_{1}f(c)$ .

We can use the same argument, and the second equality from step 2, to show that:

$\lim _{t\to 0}{\frac {\lambda (t,t)}{t^{2}}}=D_{1}D_{2}f(c)$ .

Therefore, by the uniqueness of limits, the two quantities must be equal. $\square$

Handwritten Lecture Notes in PDF

MAT257 - Lecture14 (Oct 14)

@@ Line 2: / Line 2: @@
 {{1617-257:Dror/Students Divider}}
+== Symmetry of Second Partial Derivatives ==
+'''(Note: This is based off of the proof in the textbook, and may be slightly different from how it was presented in lecture.)'''
+=== Theorem ===
+Let <math>A</math> be an open subset of <math>\mathbb{R}^2</math>, and let <math>f: A \rightarrow \mathbb{R}</math> be of class <math>C^2</math>. Then, for all <math>c \in A</math>, we have that <math>D_1 D_2 f(c) = D_2 D_1 f(c)</math>.
+=== Proof ===
+==== Step 1 ====
+We begin by defining a function that will help us in our proof. Let <math>(a, b) \in A \subseteq \mathbb{R}^2</math> be an arbitrary point. We then define the function <math>\lambda</math> as follows:
+<math>\lambda(h, k) = f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)</math>.
+Why do we define such a function? In fact, we can define both of the partial derivatives in question in terms of <math>\lambda</math>:
+<math>
+\begin{align}
+D_1 D_2 f(a, b) & = \frac{\partial}{\partial x} \frac{\partial f}{\partial y} (a, b) \\
+& = \lim_{h \to 0} (\frac{1}{h}) [\frac{\partial f}{\partial y} (a + h, b)] - [\frac{\partial f}{\partial y} (a, b)] \\
+& = \lim_{h \to 0} (\frac{1}{h}) [\lim_{k \to 0} (\frac{1}{k}) f(a + h, b + k) - f(a + h, b)] - [\lim_{k \to 0} (\frac{1}{k}) f(a, b + k) - f(a, b)] \\
+& = \lim_{h \to 0} (\frac{1}{h}) [\lim_{k \to 0} (\frac{1}{k}) f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)] \\
+& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) [f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)] \\
+& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) \lambda(h, k)
+\end{align}
+</math>
+And similarly for <math>D_2 D_1 f(a, b)</math>, but with the two limits taken in the opposite order.
+==== Step 2 ====
+Now, let's make use of this function. Let us consider the square <math>Q = [a, a + h] \times [b, b + k]</math>, where <math>h</math> and <math>k</math> are so small that <math>Q \subseteq A</math>. We will show that there exist points <math>p, q \in Q</math> such that:
+<math>
+\begin{align}
+& \lambda(h, k) = D_2 D_1 f(p) \cdot hk \\
+& \lambda(h, k) = D_1 D_2 f(q) \cdot hk
+\end{align}
+</math>
+The proofs of the two equalities are symmetric, and thus we only explicitly prove the first one. We will prove this through a double application of the mean value theorem.
+Consider the function <math>\phi</math>, defined such that <math>\phi(s) = f(s, b + k) - f(s, b)</math>. Then <math>\phi(a + h) - \phi(a) = \lambda(h, k)</math>. By hypothesis, <math>D_1 f</math> exists at all points of <math>A</math>, so we can differentiate <math>\phi</math> with respect to the first variable of <math>f</math> on the interval <math>[a, a + h]</math>. By the mean value theorem, this means that there exists a point <math>s_0 \in [a, a + h]</math> such that <math>\lambda(h, k) = \phi(a + h) - \phi(a) = \phi^\prime(s_0) \cdot h = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h</math>.
+Now we want to apply the mean value theorem one more time. Consider the function <math>\gamma</math>, defined such that <math>\gamma(t) = D_1 f(s_0, t)</math>. By hypothesis, <math>D_2 D_1 f</math> exists at all points of <math>A</math>, so we can differentiate <math>\gamma</math> with respect to the second variable of <math>f</math> on the interval <math>[b, b + k]</math>. So, by the mean value theorem, there exists a point <math>t_0 \in [b, b + k]</math> such that <math>\gamma(b + k) - \gamma(b) = \gamma^\prime(t_0) \cdot k</math>. Thus:
+<math>
+\begin{align}
+\lambda(h, k) & = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h \\
+& = [\gamma(b + k) - \gamma(b)] \cdot h \\
+& = \gamma^\prime(t_0) \cdot hk \\
+& = D_2 D_1 f(s_0, t_0) \cdot hk
+\end{align}
+</math>
+==== Step 3 ====
+Now we can prove the theorem. Let <math>c = (a, b) \in A</math>, and let <math>t > 0</math> be so small that <math>Q_t = [a, a + t] \times [b, b + t] \subseteq A</math>. By what we have just shown, <math>\lambda(t, t) = D_2 D_1 f(p_t) \cdot t^2</math>, for some <math>p_t \in Q_t</math>.
+<math>t</math> is the length of the sides of the rectangle <math>Q_t</math>, so as <math>t \rightarrow 0</math>, <math>p_t \rightarrow c</math>. By hypothesis, <math>D_2 D_1 f</math> is continuous, and so as <math>p_t \rightarrow c</math>, <math>D_2 D_1 f(p_t) \rightarrow D_2 D_1 f(c)</math>. Thus:
+<math>\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_2 D_1 f(c)</math>.
+We can use the same argument, and the second equality from step 2, to show that:
+<math>\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_1 D_2 f(c)</math>.
+Therefore, by the uniqueness of limits, the two quantities must be equal. <math>\square</math>

1617-257/Classnotes for Friday October 14: Difference between revisions

Latest revision as of 22:50, 19 October 2016

Contents

Symmetry of Second Partial Derivatives

Theorem

Proof

Step 1

Step 2

Step 3

Handwritten Lecture Notes in PDF

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