Math 3335: Orthogonal Change of Variables., x(u) =

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1 Math 3335: Orthogonal Change of Variables In these exercises we will explore what are called orthogonal change of coordinates. In general a d dimensional change of coordinates is a mapping x (u,...,u d ) x : R d R d, x(u) =... x d (u,...,u d ) A particular three dimensional change of coordinates you are all familiar with is cylindrical coordinates (c c) u = (r,,ζ), x(u) = x(r,,ζ) y(r,,ζ) rcos rsin. z(r,,ζ) ζ This maps (r,,ζ) [, ) [,2π) (, ) onto R 3. You have also seen spherical coordinates written here as (s c) u = (r,,), x(u) = x(r,,) rcos sin y(r,,) rsin sin. z(r,,) rcos This maps (r,,) [, ) [,2π) [,π] onto R 3. (Here = denotes north and = π denotes south.) We ll see two other interesting coordinate transformations in the exercises below. We say the transformation x = x(u) is orthogonal if the vectors x ui u i x(u) and x uj u j x(u) are perpendicular to each other for all i j. In cylindrical coordinates, see (c c) above with u = r, u 2 =, u 3 = ζ, we have x r r x = cos sin, x x = rsin rcos, x ζ ζ x =. Since (x r x ) =, (x r x ζ ) = and (x x ζ ) = cylindrical coordinates therefore defines an orthogonal transformation. So do spherical coordinates. The lengths of the d vectors {x u,...,x ud } coming from an orthogonal change of coordinates can be normalized, i.e. e ui x ui / x ui, to obtain an orthonormal basis {e u,...,e ud }. For cylindrical coordinates you can compute that x r = e r = cos sin, x = r e = sin cos, x ζ = e ζ =. Note that in general these orthonormal basis vectors are not constant.

2 The d dimensional gradient operator was defined in terms of Cartesian coordinates only; = d i= e x i. In the first three exercises given below you will derive the following x i especially pleasing representation of the gradient operator in orthogonal coordinates. d = e ui. (Recall x ui = x and e ui = x ui / x ui.) x ui u i u i i= In particular this says that in cylindrical coordinates (r,, ζ) r +e r +e ζ ζ.. Suppose an orthogonal transformation x(u) is nonsingular at a point u R d in the sense that x ui (u) for each i =,...,d. Define e ui = x ui / x ui and consider the orthonormal basis {e ui } d i=. Derive the following relationship between the standard Cartesian basis {e xi } d i= and the basis {e u i } d i= : e xi = d j= x uj x i u j e uj. 2. Use the chain rule and the previous exercise to calculate that d u k d d d = = x i x i u k k= 3. Again use the chain rule to find that δ k,j = u k u j = i= j= k= d i= x i u j u k x i e uj x uj x i u j u k x i. u k. Use this fact combined with the result of the previous exercise to derive the following formula for the gradient in terms of the new orthogonal coordinates u,...,u d d = e uj. x uj u j j= 4. For spherical coordinates, see (s c) on the previous page, show that x r =, x = r and x = rsin and then compute e r = cossin sinsin, e = coscos sincos, e = sin cos. cos sin Finally show that in spherical coordinates the gradient is given by r +e r +e rsin. 2

3 Fig. x = cosh(µ)cos(), y = sinh(µ)sin(). Fig 2. x = u 2 v 2, y = 2uv. 5. The 2-d transformation x = cosh(µ)cos(), y = sinh(µ)sin(), µ, [,2π), is the usual form for what are called elliptical coordinates; see Figure above and for a brief discussion there is a wiki at []. (a) Determine the vectors x µ and x to conclude this is an orthogonal transformation. It is nonsingular when µ >. (b) Determine x µ and x and the orthonormal basis {e µ, e }. You may find the fact that cosh 2 (µ) sinh 2 (µ) = useful. x µ 2 = x 2 = sinh 2 (µ)+sin 2 (). (c) Determine the gradient operator in this coordinate system. 6. Consider the 2-d transformation x = u 2 v 2, y = 2uv, u,v R; see Figure 2. One might call these parabolic coordinates. (a) Determine the vectors x u and x v to conclude this is an orthogonal transformation. It is nonsingular when u 2 +v 2 >. (b) Determine x u and x v and the orthonormal basis {e u, e v }. (c) Determine the gradient operator in this coordinate system. Answers: e u = = u2 +v 2 (ue x +ve y ), e v = 2 u 2 +v 2 ( e u u +e v ). v u2 +v 2 ( ve x +ue y ) [] coordinate system 3

4 The next 2 exercises and numbered C. C.6 (C for cylindrical) and S. S.6 (S for spherical). Problems C,S. and C,S.2 are about changing variables for vector fields. Make sure to do these. In problems C,S.3 C,S.6 you will compute the formulae for the divergence, Laplacian and curl in cylindrical and spherical coordinates. These are straight forward exercises in cylindrical coordinates but a bit more involved in spherical coordinates. Recall the cylindrical coordinates change of variables x = rcos y = rsin z = ζ with associated orthonormal basis vectors e r = cos sin, e = sin cos, e ζ =. Make sure you can compute that r + e r + e ζ ζ. C.. Convert the following vector fields to cylindrical coordinates. (a) f(x,y,z) = x 2 e x +xye y Answers: (a) r 2 cose r. (b) r 2 sine. (b) f(x,y,z) = y 2 e x +xye y C.2. Convert the following given in cylindrical coordinates to Cartesian coordinates. (a) f(r,,ζ) = rtane r (b) f(r,,ζ) = r 2 cose Answers: (a) ye x +(y 2 /x)e y. (b) xye x +x 2 e y. C.3. Compute the /r, / and /ζ partial derivatives of e r, e and e ζ and write these quantities in the basis {e r, e, e ζ }. Answers: r e r = e r = e ζ e r = r e = e = e r ζ e = r e ζ = e ζ = ζ e ζ = C.4. Use your work above to derive the formula for f in cylindrical coordinates where the vector field f is given as f = f r e r +f e +f ζ e ζ. f = f r r + r f r + f r + f ζ ζ. 4

5 C.5. Use your work above to derive the formula for 2 f in cylindrical coordinates. 2 f = 2 f r 2 + f r r + 2 f r f ζ 2 C.6. Finally derive the formula for f in cylindrical coordinates where as before the vector field f is written as f = f r e r +f e +f ζ e ζ. f = ( f ζ r f ) ( fr e r + ζ ζ f ) ( ζ f e + r r + r f r Recall that for spherical coordinates we have x = rcossin y = rsinsin, < 2π, π. z = rcos with associated orthonormal basis vectors e r = cossin sinsin, e = coscos sincos, e = sin cos. cos sin Make sure you can compute that r + e r + e rsin. S.. Convert the following vector fields to spherical coordinates. ) f r e ζ. (a) f(x,y,z) = x e x +y e y +z e z Answers: (a) r e r. (b) r sin e. (b) f(x,y,z) = y e x x e y S.2. Convert the following written in spherical coordinates into Cartesian coordinates. (a) f(r,,) = rsincos e +rcos e (b) f(r,,) = r 2 e r Answers: (a) z e x x e z. (b) x 2 +y 2 +z 2 (x e x +y e y +z e z ). S.3. Compute the /r, / and / partial derivatives of e r, e and e and write these quantities in the basis {e r, e, e }. Hint: To get these derivatives in the indicated basis use projection. Answers: r e r = e r = e e r = sine r e = e = e r e = cose 5 r e = e = e = sine r cose

6 S.4. Use your work above to derive the formula for f in spherical coordinates where the vector field f is given by f = f r e r +f e +f e. f = ( fr r + 2 ) r f r + f rsin + ( sin f ) rsin +cosf S.5. Use your work above to derive the formula for 2 f in spherical coordinates. S.6. Derive the formula for f in spherical coordinates where again the vector field f is given by f = f r e r +f e +f e. f = ( sin f rsin +cosf f ) + ( fr rsin rsin f ) r sinf + ( r f r r +f f ) r e. e r e 6