Separation of variables: Cartesian coordinates

Transcription

1 Separation of variables: Cartesian coordinates October 3, 15 1 Separation of variables in Cartesian coordinates The separation of variables technique is more powerful than the methods we have studied so far. The approach begins with a simplifying assumption, that the potential may be written as a product or in some cases, the sum) of some simpler functions. The procedure is the same for other coordinate systems. For the case of Cartesian coordinates, the solution is particularly simple. The aplace equation in Cartesian coordiates is V x + V y V x, y, z) + V z 1.1 Step 1: We assume a solution of the form Substitution gives V x, y, z) X x) Y y) Z z) Y Z d X + XZ d Y dy where the partial derivatives now become ordinary. 1. Step : Next, divide by V XY Z. This gives 1 d X X + 1 Y d Y dy + XY d Z dz + 1 Z d Z dz and each terms is now a function of only one variable. Taking the partial derivative of the whole expression with respect to x gives 1 d ) X x X + 1 d ) Y x Y dy + 1 d ) Z x Z dz showing that 1 X d X x 1 X d X ) + + must be independent of x, hence constant. The same holds if we take y- or z-derivatives: 1

2 1 y Y z 1 Z d ) Y dy ) d Z dz This means that each of the three terms must be constant, and the constants must add to zero: 1 d X X a 1 d X X b 1 d Z Z dz c a + b + c 1.3 Step 3: Solve the equations. It is easiest to choose the constants to be in the form ±α ce the solutions are either exponential or usoidal. If we choose a α then and we immediately have the familiar solution, If we choose a +α instead then the equation is and the solution is 1 d X X α d X + α X X A αx + B cos αx d X α X X Ae αx + Be αx Because a + b + c, both signs must occur. If two of the constants are negative giving oscillating solutions) then the third must be positive and will have exponential solutions. If two of the signs are positive then there are two directions with exponential solutions. Then the third constant is negative and has oscillating solutions. Choose the signs so that the solutions match your boundary conditions. Suppose we choose a α b β c α + β Then X A αx + B cos αx

3 and Y C βy + D cos βy while Z satisfies d Z dz α + β ) Z For convenience, define γ α + β, so that d Z dz γ Z It is easy to see that e ±γz both give solutions, so the general solution is the arbitrary linear combination Z E e γz + F e γz It is often more useful to use the symmetric and antisymmetric combinations, Then and our solution is cosh γz eγz + e γz h γz eγz e γz e γz cosh γz + h γz e γz cosh γz h γz Z E cosh γz + h γz) + F cosh γz h γz) E + F ) cosh γz + E F ) h γz Renaming the constants, E E + F and F E F, this is just 1.4 Step 4: Z E cosh γz + F h γz Write the full solution as a linear combination. For any particular values of α and β the solution is V α,β x, y, z) A αx + B cos αx) C βy + D cos βy) E cosh γz + F h γz) Since the aplace equation is linear, we may build the general solution for V as a sum of such solutions for different α and β. The leading constants may be different for each choice of α and β, so we have V x, y, z) α,β A α,β αx + B α,β cos αx) C α,β βy + D α,β cos βy) E α,β cosh γz + F α,β h γz) However, this latter form is too general for most problems. It is simpler to choose some of the constants to satisfy the boundary conditions automatically. Fitting boundary conditions in Cartesian coordinates Suppose we wish to find the potential everywhere inside a conducting cube of side. et the boundary be held at zero potential except for the top of the box, which is held at V. This means that in the x-direction, the potential starts and ends at zero: V, y, z) V, y, z) 3

4 .1 Satisfying the first five conditions The easiest way to satisfy this condition is to choose the usoidal solutions in the x-direction, Then the boundary conditions require for all y and z. Therefore, With B we look at the second condition, X x) A αx + B cos αx V, y, z) X ) Y y) Z z) V, y, z) X ) Y y) Z z) X ) A + B cos B X ) A α + B cos α A α Since A would make the entire potential vanish, we must restrict the possible values of α instead. The boundary condition is satisfied if and only if α nπ α nπ for any integer n. We conclude X n x) A n nπx Notice that the constant A n may depend on n. The y-direction is completely analogous. We require vanishing potential on both sides, for all x and z, and this is achieved only if V x,, z) V x,, z) Y ) Y ) Again, we choose an oscillating solution and we find that Y m y) C m The integer m is independent of the integer n. Finally, we fit the boundary condition for Z z). Remembering that the solution in this direction must be exponential, we write Z E cosh γz + F h γz 4

5 where γ + nπ ) + mπ ) + π n + m. With the potential on the top of the box equal to V, the boundary conditions for z are V x, y, ) V x, y, ) V Since these relations hold for all x and y, they must be satisfied by our choice of Z ) and Z ). The z condition is completely satisfied by This leaves our complete solution in the form Z ) E V x, y, z) A n nπx C m E m,n h γz A n C m E m,n ) nπx A nm nπx h γz h γz where it is sufficient to set A nm A n C m E m,n ce we only have a gle overall constant for each pair m, n.. The final solution Now comes the tricky part. We have one final boundary condition, V x, y, ) V, and we no longer have the freedom to satisfy it by choog γ. Instead, we must choose the constants A mn so that V A nm h γ) nπx for all x and all y. This is a double Fourier series. To solve, we use the orthogonality property of the e function see below for the explicit integrations), ˆ n 1πx n πx δ n 1n Consider the x-direction first. Multiply both sides of the boundary condition by x from to, ˆ V x A nm h γ) ˆ nπx x The left side is easy to integrate in this example!), while on the right we use orthogonality: V cos x A nm h γ) δ kn and integrate x 5

6 Performing the sum over all n, only one term survives, V cos + V cos V 1 1) k) m1 m1 Now repeat the procedure for the y direction. Multiply by jπy V V V 1 1) k) ˆ 1 1) k) [ jπy cos jπ jπy dy ] m1 A km h ) k + m A km h ) k + m and integrate, A km h k + m ) m1 ) A kj h k + j 1 1) k) jπ ) cos jπ jπ 4V j 1 1) k) 1 1) j) A kj h k + j ˆ ) A km h k + m ) δ jm jπy dy Since we can carry out these steps for any values of j and k, we have found all of the coefficients A kj, A kj Substituting these constant!) values into the potential, 4V 1 j h 1) k) 1 1) j) k + j V x, y, z) 4V mnπ h n + m ) 1 1)n ) 1 1) m ) nπx h γz This simplifies to sums over the odd terms ce and we may write This is our final potential. V x, y, z) 16V π 1 1) n ) nodd modd 1 mn { n even n odd h n + m z h n + m nπx.3 Orthogonality of e functions We evaluate the integrals ˆ n 1πx n πx 6

7 First, when n 1 and n are different, we use the sum and difference formulas to write ˆ n 1πx n πx 1 ˆ [ n1 πx cos 1 [ πx π n 1 n ) 1 n ) πx n1 πx cos ) n 1 n ) π n 1 n ) π n 1 n )) + n )] πx πx ) ] π n 1 + n ) n 1 + n ) π n 1 + n ) π n 1 + n )) ) π n 1 n ) π n 1 + so the integral vanishes. However, when n 1 n, ˆ n 1πx n πx ˆ n 1πx ˆ ˆ 1 cos n 1πx 1 cos n ) 1πx [ x n 1 π n ] 1πx + n ) 1πx We summarize these integrals by writing ˆ n 1πx n πx δ n 1n 3 Example Consider the same cube as in the example above, and let V on the sides and bottom as before, but suppose the potential on the top is V x, y, ) V πx 4 ) πx 3 3 The steps are the same as above up to the final integration. At z, we must satisfy V πx 4 ) πx 3 3 A nm h γ) nπx and the Fourier series in the y-direction is found as above to reduce this to V 1 1) k) πx 4 ) πx 3 3 n1 A nk h γ) nπx 7

8 At this point we could multiply by lx and integrate. However, in this case we may write the left hand side ug trigonometric identities as Since the functions nπx πx 4 πx πx 3 are mutually orthogonal, we immediately see that V 1 1) k) 3πx 3 n1 requires all of the A nk to vanish except for n 3. For A 3k, we have and the unique solution for the potential is A nk h γ) nπx V 1 1) k) 3πx A 3k h γ 3πx 3 V A 3k 1 1) k) 3 h γ V x, y, z) m odd 4V h γz 3πx 3mπ h γ This simplification works any time the potential on the left can be easily written in terms of Fourier series. The same approach works in other coordinate systems when the potential on a given boundary can be written in terms of the orthogonal functions of the solution. 4 Exercise Repeat the problem of a cube, but this time center the cube in the z-direction and let both the top and the bottom have potential V. Now the boundary conditions are: V, y, z) V, y, z) V x,, z) V x,, z) V x, y, ) V V x, y, + ) V Notice that cosh γz is symmetrical in z, so that cosh γz) cosh γz). Suggestion: work through all the details rather than just copying the calculations above. 8