example 1

Let \(D^* \subset {\mathbb R}^2\) be the rectangle \(D^* = [0,1] \times [0,2 \pi]\). Then all points in \(D^*\) are of the form \((r, \theta)\), where \(0 \le r \le 1, 0 \le \theta \le 2 \pi\). Let \(T\) be the polar coordinate “change of variables” defined by \(T (r, \theta) = ( r \cos \theta, r \sin \theta)\). Find the image set \(D\).

solution Let \((x,y) = ( r \cos \theta, r \sin \theta)\). Because of the identity \(x^2 + y^2 = r^2 \cos^2 \theta + r^2 \sin^2 \theta = r^2 \le 1\), we see that the set of points \((x,y) \in {\mathbb R}^2\) such that \((x,y) \in D\) has the property that \(x^2 + y^2 \le 1\), and so \(D\) is contained in the unit disk. In addition, any point \((x,y)\) in the unit disk can be written as \((r \cos \theta , r \sin \theta)\) for some \(0 \le r \le 1\) and \(0 \le \theta \le 2 \pi\). Thus, \(D\) is the unit disk (see Figure 16.2).

Figure 16.2: \(T\) gives a change of variables between Euclidean and polar coordinates. The unit circle is the image of a rectangle.

example 2

Let \(T\) be defined by \(T (x,y) = (( x+ y) /2, (x-y) /2)\) and let \(D^* = [-1,1] \times [-1,1] \subset {\mathbb R}^2\) be a square with side of length 2 centered at the origin. Determine the image \(D\) obtained by applying \(T\) to \(D^*\).

solution Let us first determine the effect of \(T\) on the line \({\bf c}_1 (t) = (t,1)\), where \(-1 \le t \le 1\) (see Figure 16.3). We have \(T ( {\bf c}_1 (t))= ( ( t+1)/2, (t-1) /2)\). The map \(t \mapsto T ( {\bf c}_1 (t))\) is a parametrization of the line \(y=x-1, 0 \le x \le 1\), because \((t-1) / 2 = (t+1)/2 -1\). This is the straight line segment joining \((1,0)\) and \((0,-1)\).

Figure 16.3: Domain for the transformation \(T\) of Example 2.

Let \[ \begin{array}{l@{\qquad}l} {\bf c}_2 (t) = (1,t), & -1 \le t \le 1 \\[6pt] {\bf c}_3 (t) = (t, -1), & -1 \le t \le 1 \\[6pt] {\bf c}_4 (t) = (-1,t), & -1 \le t \le 1 \end{array} \] be parametrizations of the other edges of the square \(D^*\). Using the same argument as before, we see that \(T \circ {\bf c}_2\) is a parametrization of the line \(y= 1-x, 0 \le x \le 1\) [the straight line segment joining \((0,1)\) and \((1,0)\)]; \(T \circ {\bf c}_3\) is the line \(y=x+1, -1 \le x \le 0\) joining \((0,1)\) and \(({-}1,0)\); and \(T \circ {\bf c}_4\) is the line \(y = - x-1, -1 \le x \le 0\) joining \(({-}1,0)\) and \((0,{-}1)\). By this time it seems reasonable to guess that \(T\) “flips” the square \(D^*\) over and takes it to the square \(D\) whose vertices are \((1,0), (0,1), (-1,0), (0,-1)\) (Figure 16.4).

Figure 16.4: The effect of \(T\) on the region \(D^{\ast }\).

To prove that this is indeed the case, let \(-1 \le \alpha \le 1\) and let \(L_\alpha\) (Figure 16.3) be a fixed line parametrized by \({\bf c} (t) = ( \alpha, t), -1 \le t \le 1\); then \(T ( {\bf c} (t)) = (( \alpha \,{+}\, t) /2, (\alpha \,{-}\,t) /2)\) is a parametrization of the line \(y=-x \,{+}\, \alpha, (\alpha \,{-}\,1) /2 \le x \le (\alpha \,{+}\,1) /2\). This line begins, for \(t\,{=}\,-1\), at the point \(((\alpha \,{-}\,1)/2, (1\,{+}\, \alpha)/2)\) and ends at the point \(((1\,{+}\, \alpha) /2, ( \alpha \,{-}\,1) /2)\); as is easily checked, these points lie on the lines \(T \circ {\bf c}_3\) and \(T \circ {\bf c}_1\), respectively. Thus, as \(\alpha\) varies between \(-1\) and \(1\), \(L_\alpha\) sweeps out the square \(D^*\) while \(T ( L_\alpha)\) sweeps out the square \(D\) determined by the vertices \((-1,0), (0,1), (1,0),\) and \((0,-1)\).

Theorem 1

Let \(A\) be a \(2 \times 2\) matrix with det \(A \ne 0\) and let \(T\) be the linear mapping of \({\mathbb R}^2\) to \({\mathbb R}^2\) given by \(T ({\bf x})= A{\bf x}\) \((\)matrix multiplication\().\) Then \(T\) transforms parallelograms into parallelograms and vertices into vertices\(.\) Moreover, if \(T (D^*)\) is a parallelogram, \(D^*\) must be a parallelogram.

Definition

A mapping \(T\) is one-to-one on \(D^*\) if for \((u,v)\) and \((u',v') \in D^*\), \(T (u,v)\) \(=\) \(T (u', v')\) implies that \(u=u'\) and \(v=v'\).

example 3

Consider the polar coordinate mapping function \(T \colon\, {\mathbb R}^2 \to {\mathbb R}^2\) described in Example 1, defined by \(T (r, \theta)= ( r \cos \theta, r \sin \theta)\). Show that \(T\) is not one-to-one if its domain is all of \({\mathbb R}^2\).

solution If \(\theta_1\ne \theta_2\), then \(T (0,\theta_1) = T (0, \theta_2)\), and so \(T\) cannot be one-to-one. This observation implies that if \(L\) is the side of the rectangle \(D^* = [0,1] \times [0, 2 \pi]\) where \(0 \le \theta \le 2 \pi\) and \(r=0\) (Figure 16.5), then \(T\) maps all of \(L\) into a single point, the center of the unit disk \(D\). However, if we consider the set \(S^* = (0,1] \times [0, 2 \pi)\), then \(T \colon\, S^* \to S\) is one-to-one (see Exercise 5). Evidently, in determining whether a function is one-to-one, the domain chosen must be carefully considered.

Figure 16.5: The polar coordinate transformation \(T\) takes the line \(L\) to the point (0, 0).

example 4

Show that the function \(T \colon\, {\mathbb R}^2 \to {\mathbb R}^2\) of Example 2 is one-to-one.

solution Suppose \(T (x,y) = T (x', y')\); then \[ \left( \frac{x+y}{2}, \frac{x-y}{2} \right) = \left( \frac{x^\prime+ y'}{2}, \frac{x' - y'}{2} \right) \] and we have \begin{eqnarray*} x+ y &=& x' + y' ,\\ x - y &=& x' - y' .\\[-17pt] \end{eqnarray*}

Adding, we have \[ 2x =2x'. \]

Thus, \(x = x'\) and, similarly, subtracting gives \(y=y'\), which shows that \(T\) is one-to-one (with domain all of \({\mathbb R}^2\)). Actually, because \(T\) is linear and \(T ( {\bf x}) = A {\bf x}\), where \(A\) is a \(2 \times 2\) matrix, it would also suffice to show that det \(A \ne 0\) (see Exercise 12).

Definition

The mapping \(T\) is onto \(D\) if for every point \((x,y) \in D\) there exists at least one point \((u,v)\) in the domain of \(T\) such that \(T (u,v) = (x,y)\).

example 5

Let \(T \colon\, {\mathbb R}^2 \to {\mathbb R}^2\) be given by \(T (u,v) = (u,0)\). Let \(D\) be the square, \(D= [0,1] \times [0,1].\) Because \(T\) takes all of \({\mathbb R}^2\) to one axis, it is impossible to find a \(D^*\) such that \(T ( D^*) =D\).

example 6

Let \(T\) be defined as in Example 2 and let \(D\) be the square whose vertices are \((1,0),(0,1), (-1,0),(0,-1)\). Find a \(D^*\) with \(T ( D^*) =D\).

solution Because \(T\) is linear and \(T ( {\bf x}) = A {\bf x}\), where \(A\) is a \(2 \times 2\) matrix satisfying det \(A \ne 0\), we know that \(T \colon\, {\mathbb R}^2 \to {\mathbb R}^2\) is onto (see Exercises 12 and 13), and thus \(D^*\) can be found. By Theorem 1, \(D^*\) must be a parallelogram. In order to find \(D^*\), it suffices to find the four points that are mapped onto vertices of \(D\); then, by connecting these points, we will have found \(D^*\). For the vertex \((1, 0)\) of \(D\), we must solve \(T (x,y) = (1,0)= ((x+y) /2, (x-y)/2)\), so that \((x+y)/2 =1, (x-y) /2 =0\). Thus, \((x,y) = (1,1)\) is a vertex of \(D^*\). Solving for the other vertices, we find that \(D^* = [-1,1] \times [-1,1]\). This is in agreement with what we found more laboriously in Example 2.

example 7

Let \(D\) be the region in the first quadrant lying between the arcs of the circles \(x^2 + y^2 = a^2 , x^2 + y^2 = b^2 , 0 < a < b\) \((\)see Figure 16.6). These circles have equations \(r=a\) and \(r=b\) in polar coordinates. Let \(T\) be the polar-coordinate transformation given by \(T ( r, \theta) = (r \cos \theta, r \sin \theta) = (x,y)\). Find \(D^*\) such that \(T ( D^*) = D\).

solution In the region \(D, a^2 \le x^2 + y^2 \le b^2\); and because \(r^2 = x^2 + y^2\), we see that \(a \le r \le b\). Clearly, for this region \(\theta\) varies between \(0 \le \theta \le \pi /2\). Thus, if \(D^* = [a,b] \times [0, \pi /2]\), we have \(T ( D^*) = D\) and \(T\) is one-to-one.

remark

The inverse function theorem discussed in Section 14.5 is relevant to the material here. It states that if the determinant of \({\bf D} T ( u_0, v_0)\) [which is the matrix of partial derivatives of \(T\) evaluated at \((u_0, v_0)\)] is not zero, then for \((u, v)\) near \((u_0, v_0)\) and \((x,y)\) near \((x_0, y_0) = T (u_0, v_0)\), the equation \(T (u, v) = (x,y)\) can be uniquely solved for \((u, v)\) as functions of \((x,y)\). In particular, by uniqueness, \(T\) is one-to-one near \((u_0, v_0)\); also, \(T\) is onto a neighborhood of \((x_0, y_0)\), because \(T (u,v) = (x,y)\) is solvable for \((u,v)\) if \((x,y)\) is near \((x_0, y_0)\).

However, even if \(T\) is one-to-one near every point, and also onto, \(T\) need not be globally one-to-one. Thus, we must exercise caution (see Exercise 17).

One-to-One and Onto Mappings

A mapping \(T{:}\, D^{*}\to D\) is one-to-one when it maps distinct points to distinct points. It is onto when the image of \(D^*\) under \(T\) is all of \(D\).

A linear transformation of \({\mathbb R}^n\) to \({\mathbb R}^n\) given by multiplication by a matrix \(A\) is one-to-one and onto when and only when det \(A\neq 0\).