Lecture notes: Introduction to Partial Differential Equations

Transcription

1 Lecture notes: Introduction to Partial Differential Equations Sergei V. Shabanov Department of Mathematics, University of Florida, Gainesville, FL USA

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3 CHAPTER 1 Classification of Partial Differential Equations 1. What are Partial Differential Equations? 1.1. Basic notations. Let u be a function of several variables, denoted x, y, z, etc. Unless stated otherwise, r = (x, y, z,... or x = (x 1, x 2,..., x n is the position vector of a point in a Euclidean space R n relative to a particular point O and the components of the position vector are rectangular coordinates of the point with respect to some coordinate system with the origin at the point O. So, u is a rule that assigns a unique number, denoted u(r to every point r Ω R n : u : Ω R n R A function u : Ω R n C is called complex-valued. In accord with the definition of complex numbers, it can always be written as a linear combination of two realvalued functions u(r = v(r + iw(r, v(r = Reu(r, w(r = Im u(r A neighborhood of a point. The Euclidean distance between two points x = (x 1, x 2,..., x n and y = (y 1, y 2,..., y n is denoted as x y = (x 1 y (x 2 y (x n y n 2 A collection of all points whose distance is strictly less than a > 0 from a given point y, } B a (y = {x R n x y < a is called an open ball of radius a centered at y or a neighborhood of the point y. If y = 0, that is, the ball is centered at the origin, then it is denoted as B a. Sets in R n. A set Ω is said to be open if for every point x Ω there is a neighborhood B a (x that is contained in Ω. For example B a (y is open. A point y is a limit point of a set Ω if the intersection Ω B a (y for any neighborhood B a (y contains at least one point of Ω different from the point y. Note that y does not necessarily belongs to Ω. For 3

4 4 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS example, let Ω = B 1 (all x such that x < 1. Clearly, every point of the ball is a limit point because there is a ball of a sufficiently small radius centered at any point of Ω that is contained in Ω (the set B 1 is open as was noted above. Take any point y such that y = 1 (at a distance 1 from the center of the ball. Clearly, any ball B a (y has a non-empty intersection wit B 1. So all points on the sphere r = 1 are limit points of B 1 but none of them belong to B 1. The closure of a set Ω is the union of all limit points of Ω with Ω. It is denoted by Ω. A set Ω is called closed if it contains all its limit points so that Ω = Ω. For example, } B a (y = {x R n x y a is a closed ball of radius a centered at y. A point y Ω is said to be isolated if there is a neighborhood B a (y whose intersection with Ω contains only the point y. Clearly, a finite collection of isolated points is a closed set (as it has no limit points. An infinite set of isolated point points {x k } 0 (a sequence in R n may have a limit point that does not belong to the set. For example, if the sequence has the limit x, that is, lim x k x = 0 k then x is a limit point of the set (any ball B a (x contains x k for all k > N for some (large enough N The largest open set Ω o that is contained in Ω is called the interior of Ω. So, for any Ω Ω o Ω Ω One can prove that the closure Ω is the smallest closed set that contains Ω. Note that the interior Ω o can be empty. For example a set containing only finitely many points has an empty interior. The interior of an open set coincides with the set itself. The set Ω = Ω \ Ω o is called the boundary of Ω. For example, the boundary of a ball (closed or open } B a (y = B a (y \ B a (y = {x R n x y = a is the sphere of radius a centered at y. The whole space has no boundary. Indeed, the whole space Ω = R n is open and closed (because all its points are interior points and it contains all its limit points so that its interior coincides with its closure and the boundary is empty.

5 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? 5 Partial derivatives. Let u : Ω R and Ω be an open set. Suppose that the function u has partial derivatives up to some order. The first order partial derivatives are denoted as x = u x, y = u y, z = u z,... The second order partial derivatives are denoted as 2 u x 2 = u xx, 2 u y 2 = u yy, 2 u z 2 = u zz,... There are also mixed partial derivatives of the second order: 2 u x y = x y = u xy, 2 u y x = y x = u yx and similarly for any other pair of variables, that is, xz, zx, etc. Partial derivatives of the third order and higher are denoted similarly. For example xyy = x 2 u y = 3 u 2 x y, 2 u xyz = x y z = 3 u x y z. Note that the order of variables in the subscript indicates the order in which partial derivatives are taken. In general, the value of partial derivatives of higher order depends on the order of differentiation. If, however, the partial derivatives are continuous functions, then the order of differentiation is irrelevant. This statement is known as Clairaut s theorem. In what follows, it will always be assumed that partial derivatives are continuous (up to the order of interest and, hence, xy = or xyz = xzy, etc. C p functions. In what follows, it will always be assumed that partial derivatives are continuous (up to the order of interest and, hence, xy = u yx, or u xyz = u xzy, etc. A function on an open set Ω whose partial derivatives up to order p are continuous on Ω is called a function from the class C p (Ω. All continuous functions on Ω form the class C 0 (Ω. The functions whose partial derivatives of any order are continuous form the class C (Ω. If Ω = R n, then C p (R n = C p for simplicity of notations. A collection of all partial derivatives of a function u C p (Ω of order q p is denoted by D q u. For example, for a function of two variables x and y Du = {u x, u y}, D 2 u = { xx, xy, yy} yx,

6 6 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS Similarly, for functions of n variables { } Du = j = 1, 2,..., n, x j { D 2 2 u } u = j k = 1, 2,..., n. x j x k The vector in the domain of a function u whose components are partial derivatives are called the gradient: R 2 : R 3 : u = (u x, u y, u = (u x, u y, u z, R n : u = (u x 1, u x 2,..., u x n. For any unit vector n, n = 1, in the domain of a C 1 function u, the dot product n u(x = n is called the directional derivative of u in the direction of n at a point x. It defines the rate of change of u in the direction of n at a point x. An extension of a function. Let Ω be open and u C 0 (Ω. Suppose that for every point y Ω the limit lim u(x x y exists. Then u is said to be continuously extendable to Ω so that u(y = lim x y u(x, y Ω The class of all functions continuous functions on Ω that are continuously extendable to Ω is denoted by C 0 ( Ω. The class C p ( Ω consists of all functions from C p (Ω whose all partial derivatives have continuous extensions to Ω so that D q u(y = lim x y D q u(x, y Ω, q = 0, 1,..., p For example, multivariable polynomials on any open set Ω are continuously extendable together with all partial derivatives to the boundary of Ω. So, polynomials are from the class C ( Ω for any open Ω.

7 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? Partial differential equations (PDEs. A partial differential equation is a relation between function u, its partial derivatives D q u up to some order q p, and the argument r of u. The highest order of partial derivatives involved into the relation is called the order of the partial differential equation: (1.1 F(r, u, Du, D 2 u,..., D p u = 0, r Ω R n. for an open set Ω. It indicates that a function u is not an arbitrary function but such that it and its partial derivatives satisfy a certain relation F that must hold for every point r in an open set of interest Ω R n. The integer p is called the order of the partial differential equation. A solution to a PDE. A particular function u(r is called a solution to a partial differential equation in an open set Ω if u and its partial derivatives satisfy the relation F = 0 for all r Ω. Example 1.1. Let t > 0 and x R. Find a constant k such that the function u(t, x = t 1/2 e kx2 /t is a solution of the equation u t = xx, t > 0, x R, or show that no such constant exists. Solution: In this case, Ω = (0, (, is an open half-plane in R 2. One has to calculate the partial derivatives u t and u xx, substitute them into the equation, and check whether there is a particular numerical value of k such that the relation between the partial derivatives is fulfilled for all positive t and all real x (the set Ω. The partial derivatives are u t = 1 2 t 3/2 e kx2 /t + kx 2 t 5/2 e kx2 /t u x = 2kxt 3/2 e kx2 /t xx = 2kt 3/2 e kx2 /t + 4k 2 x 2 t 5/2 e kx2 /t The substitution of the partial derivatives into the equation yields (after the cancellation of the positive factor e kx2 /t > t 3/2 + kx 2 t 5/2 = 2kt 3/2 + 4k 2 x 2 t 5/2 The terms t 3/2 and x 2 t 5/2 are independent and, hence, the equality is possible for all values of t > 0 and x R if and only if the coefficients

8 8 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS at these terms in the left and right side match, which gives a system of two equations for the unknown k: { 1 2 = 2k k = 4k 2 k = 1 4 The first equation has the solution k = 1/4, while the second one has two solutions k = 0 and k = 1/4. The common solution is k = 1/4. Thus, the indicated function is a solution of the given partial differential equation only if k = 1/4. Example 1.2. If f and g are from the class C 2 (R, show that u(t, x = f(x ct + g(x ct is a solution to the wave equation in Ω = (0, (0, L Solution: By the chain rule tt c2 xx = 0, t > 0, 0 < x < L u t = cf (x ct + cg (x + ct, tt = ( c 2 f (x ct + c 2 g (x + ct u x = f (x ct + g (x + ct, xx = f (x ct + g (x + ct, tt c2 xx = c2 [f (x ct + g (x + ct] c 2 [f (x ct + g (x + ct] = 0 for all t > 0 and 0 < x < L. Note that a PDE may have infinitely many solutions. For example, a solution to the wave equation considered in Example 1.2 contains two arbitrary functions of a single variables. Linear and non-linear PDEs. A partial differential equation is called linear if the relation F is a linear function in u and its partial derivatives. For example, the equation xx + f(x, yu y + g(x, yu = h(x, y, where f, g, and h are some functions of two variables x and y, is a linear second order partial differential equation for a function u(x, y of two variable. PDEs discussed in Examples 1.1 and 1.2 are linear equations. The equation xx + f(x, y(u y 2 + g(x, yu 3 = h(x, y

9 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? 9 is also a second order partial differential equation, but it is not linear because it contains the square of the partial derivative u y and the cube of the function u. Superposition principle for linear homogeneous PDEs. Among all linear equations, consider homogeneous linear PDEs. The function F in (1.1 is linear in D q u and, in addition, has the property that the zero function u(r = 0 is a solution. For example, the following equation is linear tt(t, x + c 2 xx(t, x = f(t, x but it is not homogeneous because u = 0 is not a solution (f(x, t 0. It becomes homogeneous when the inhomogeneity f vanishes. A homogeneous linear PDE has the form of a vanishing linear combination of u and its partial derivatives: p F = a q D q u = a 0 u + a 1 Du + + a p D p u = 0, q=0 where a q D q u denotes a general linear combination of partial derivatives of order q (with coefficients being functions of r. Taking a partial derivative is a linear operation: D q (c 1 u 1 + c 2 u 2 = c 1 D q u 1 + c 2 D q u 2, where c 1 and c 2 are constants. For any two solutions, u 1 and u 2, to a linear homogeneous PDE, their linear combinations with constant coefficients u(r = c 1 u 1 (r + c 2 u 2 (r is also a solution. Indeed, p a q D q u = q=0 = p a q D q (c 1 u 1 + c 2 u 2 q=0 p a q (c 1 D q u 1 + c 2 D q u 2 q=0 p = c 1 a q D q u 1 + c 2 q=0 = = 0 p a q D q u 2 because u 1 and u 2 are solutions This is called the superposition principle for linear PDEs. In Example 1.2, put u 1 (t, x = f(x ct and u 2 (t, x = g(x + ct. They are solutions to the homogeneous wave equation and, by the q=0

10 10 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS superposition principle, their sum is also a solution (their general linear combinations with constant coefficients would also be a solution Boundary conditions. A partial differential equation may have infinitely many solutions. In order for a partial differential equation to have a unique solution, some additional conditions must be imposed on the solutions. Suppose u(x is a solution to (1.1 in an open set Ω. It is required that the solution has a continuous extension to the boundary. In other words, the solution is required to be from the class u C p (Ω C 0 ( Ω The more continuous partial derivatives exist, the smaller is the set of such functions. C p (Ω C 0 (Ω and every function from C 0 ( Ω is continuous in Ω, but not every continuous function in Ω has a continuous extension to the boundary: C 0 ( Ω C 0 (Ω So, not every function from the class C p (Ω has a continuous extension to the boundary Ω as depicted in Fig. 1.1 Figure 1.1. A smooth function and its partial derivatives in an open region Ω cannot in general be continuously extended to the boundary Ω. For example, the function u(x, t = t 1/2 e x2 /(4t, t > 0, x R is from the class C (t > 0 (it has continuous partial derivatives of any order in the (open half plane t > 0. It is also a solution to the heat equation discussed in Example 1.1. The boundary of the half-plane

11 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? 11 t > 0 is the x axis whose equation is t = 0. The solution does not have a continuous extension to the boundary, that is, u / C 0 (t 0. Indeed, using the substitution z = 1/t: { lim u(x, t = lim t 0 + ze zx 2 0, x 0 /4 = z, x = 0 So, the limit has no numerical value at one point of the boundary and, hence, u cannot be continuously extended to the boundary of the half-plane. Among all solutions to a PDE in Ω, let us select only those that admit a continuous extension to the boundary Ω. Let u (y = lim u(x, y Ω, Ω x y denote the continuous extension of a solution to the boundary. If the solution u is required to have specific values u 0 on the boundary, then one says that the solution must satisfy a boundary condition: u (y = u 0 (y, y Ω Ω This boundary condition restricts the set of all solutions. Example 1.3. Find all constants A, ω, and k at which the function u(t, x = A cos(ωtsin(kx satisfies the wave equation in Example 1.2. Consider the following continuous function u 0 on the boundary of the open rectangle Ω = (0, (0, L: u 0 (t, 0 = u 0 (t, L = 0, t 0; ( 3πx u 0 (0, x = 2sin, 0 x L L Find all constants A, ω, and k for which the solution also satisfies the boundary conditions u (t, x = u 0 (t, x, (t, x Ω Ω Solution: The function u has continuous partial derivatives of any order everywhere in the tx plane for any choice of A, ω, and k. Therefore it is from the class C 2 (Ω C 0 ( Ω so that the problem in question makes sense. The second partial derivatives are tt(t, x = Aω 2 cos(ωtsin(kx = ω 2 u(t, x xx(t, x = Ak 2 cos(ωtsin(kx = k 2 u(t, x tt c2 xx = (ω2 c 2 k 2 u = 0 ω 2 = (ck 2

12 12 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS because u(t, x 0 for all (t, x Ω. Thus, the function u(t, x = A cos(kctsin(kx is a solution to the wave equation for any choice of k and A. The boundary consists of three line segments x = 0, t 0; x = L, t 0; and t = 0, 0 x L, on which the solution must coincide with the function u 0. One has lim u(t, x = u(t, 0 = u 0(t, 0, t 0 x = 0 lim u(t, x = u(t, L = u 0(t, L, t 0 x L A cos(kctsin(kl = 0, t 0 A = 0 or sin(kl = 0, A = 0 or k = πm, m = 0, ±1, ±2,... L In fact, the integer m can be taken positive because the sine function is odd so that m m yields only a change of sign of A, whereas the latter is arbitrary. Finally, lim u(t, x = u(0, x = u 0 (0, x, x 0 L, t 0 + ( 3πx A sin(kx = 2sin, x 0 L, L A = 2, k = 3π L The latter follows from linear independence of functions sin(πmx/l, that is, sin(3πx/l is not a (finite linear combination of any sin(πmx/l, m 3. Thus, the function ( 3cπt ( 3πx u(t, x = 2cos sin L L is a solution to the wave equation that satisfies the stated boundary conditions. Boundary conditions with partial derivatives. Boundary conditions may also include partial derivatives of the solution. Suppose all partial derivatives Du of a solution u to a PDE in an open Ω have continuous extensions to the boundary. In this case, it is said that u C p (Ω C 1 ( Ω If α, β, and u 1 are continuous functions on Ω, then a boundary condition involving u and Du can be written as a linear combination (with

13 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? 13 functional coefficients: α(yu (y + β(ydu (y = u 1 (y, Ω Ω y Ω As before βdu stands for a general linear combinations of first-order partial derivatives with functional coefficients: n β(ydu (y = β j (y Ω Ω y j Asymptotic boundary conditions. If Ω is not bounded, that is, it is not contained in a ball of sufficiently large radius, then one can also impose conditions on the asymptotic behavior of the solution as x. This condition is called an asymptotic boundary condition. For example, Or, more generally, j=1 u(x 0 as x, x Ω α(xu(x + β(xdu(x 0 as x, x Ω for some given continuous functions α and β in Ω. Smooth boundaries. A boundary of an open region Ω is called smooth if it is a level set of a function γ(x from the class C 1 (on any open set that contains Ω with the non-vanishing gradient: Ω = {x γ(x = C, γ 0} for some constant C. One of the main geometrical properties of the gradient is that it is normal to a level set of a function. Therefore, a smooth boundary has a continuous unit normal vector n = 1 γ γ For example, a vertical line x = x 0 in the xy plane can be viewed as a level set of the function γ(x, y = x from C 1. Its gradient γ = (γ x, γ y = (1, 0 does not vanish anywhere. So, the line is a smooth boundary of the half-plane x < x 0 (or x > x 0. The gradient vector is perpendicular to the line x = x 0. A circle x 2 + y 2 = a 2 of radius a centered at the origin can be viewed as a level set of the function γ(x, y = x 2 + y 2 from C 1. Its gradient γ = (γ x, γ y = (2x, 2y = 2(x, y

14 14 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS does not vanish at any point on the circle. So, the circle is a smooth boundary of an open disk, x 2 +y 2 < a 2. The vector γ is parallel to a line segment from the origin to the point (x, y on the circle and, hence, is perpendicular to the tangent line to the circle at (x, y which means that γ is normal to the circle at (x, y. The unit normal vector n = 1 γ γ = 1 x2 + y (x, y = 1 (x, y 2 a is continuous on the circle. Similarly, a sphere x 2 + y 2 + z 2 = a 2 is a level of γ(x, y, z = x 2 + y 2 + z 2 C 1. Its gradient γ = (γ x, γ y, γ z = (2x, 2y, 2z = 2(x, y, z does not vanish at any point of the sphere. So, an open ball x 2 + y 2 + z 2 < a 2 has a smooth boundary. The gradient is also normal to the sphere at any point (x, y, z so that the sphere has a continuous normal unit vector. Piecewise smooth boundaries. The boundary Ω is said to be piecewise smooth if it consists of finitely many smooth pieces. For example, the boundary of a rectangle in a plane Ω = (a, b (c, d is piecewise smooth because it consists of four smooth pieces (straight line segments. Oriented boundaries. If n is a unit normal to Ω, then n is also a unit normal. So, a unit normal may be chosen as the inward or outward normal. The outward normal point in the direction from the region Ω. A particular choice of n is called an orientation of Ω and, in this case, one says that Ω is oriented. In what follows, Ω always means the boundary of Ω oriented outward. Boundary conditions with the normal derivative. Suppose that u is a solution to a PDE in Ω and Ω is smooth oriented outward by the unit normal n. If u C p (Ω C 1 ( Ω, then the following boundary condition can be imposed on the solution u: (1.2 = n u = u 1 (x, x Ω n Ω Ω where u 1 is a continuous function on the boundary Ω. The quantity in the left side is called the normal derivative of u at the boundary Ω. It defines the rate of change of u in the outward normal direction at the boundary of Ω. So, the boundary condition states that the solution

15 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? 15 u must have a prescribed rate of change along the outward normal direction at the boundary. Figure 1.2. The boundary of a rectangle is not smooth at the corners A, B, C, and D. The unit outward normal has jump discontinuities at these points. The normal derivative of a C 1 function in a plane can have jump discontinuities at these points. Suppose a function u(x, y has continuous partial derivatives everywhere. In particular, u x and u y are continuous along any curve, e.g., a boundary of a region Ω. If Ω has a piecewise smooth boundary, then the directional derivatives may not be continuous at a point where the boundary is not smooth, that is, a unit normal vector is not continuous and suffers a jump discontinuity. For example, let Ω = (0, a (0, b. The boundary Ω contains 4 smooth pieces, the edges of the rectangle as depicted in Fig The outward unit normal vector n is continuous along the edges and suffers jump discontinuities at the corners: n = ( 1, 0, n = (1, 0, n = (0, 1, n = (0, 1. x=0 x=a y=0 x=0 Let us investigate the normal derivative at the corner (0, 0. The normal derivatives on the line segments x = 0 and y = 0 are: = ( 1, 0 (u n x=0 x, u y = u x=0 x(0, y, y > 0, = (0, 1 (u n y=0 x, u y = u y=0 y(x, 0, x > 0.

16 16 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS Since the partial derivatives are continuous, the normal derivatives can be extended to the corner point (0, 0: lim = lim y 0 + n u x=0 y 0 + x (0, y = u x (0, 0 lim = lim u x 0 + n y=0 x 0 + y (x, 0 = u y (0, 0. For a general C 1 function, u x(0, 0 u y(0, 0 and, hence, its normal derivative can have jump discontinuities at points where the boundary Ω of a region Ω is not smooth. This analysis suggests that the function u 1 in (1.2 may be piecewise continuous if the boundary Ω is piecewise smooth. That u 1 is not continuous everywhere on the boundary does not imply that partial derivatives of the solution u have no continuous extension to the boundary. The following interpretation of (1.2 is adopted for piecewise smooth boundaries. Let S 1 and S 2 be two smooth pieces of Ω and x 0 be a point from the set at which S 1 and S 2 are joining (the outward unit normal n has a jump discontinuity at x 0 So, x 0 is an isolated point if Ω R 2 so that Ω is a curve, or x 0 is a point of a curve along which two smooth surfaces S 1 and S 2 are joining if Ω R 3. The condition (1.2 is imposed on S 1 and S 2. However the normal derivative is generally allowed to have a jump discontinuity at x 0 : lim lim x x 0 n S1 x x 0 n S2 In the first limit x is approaching x 0 from the interior of S 1, while in the second limit from the interior of S 2. The condition (1.2 requires that the jump discontinuity of the normal derivative at any point where Ω is not smooth is the same as the jump discontinuity of u 1. The following example illustrates this. Example 1.4. Find all constants A, B, α, and β for which the function u(x, y = sin(αx (Ae βy + Be βy is a solution to the Laplace equation in a rectangle Ω = ( L, L (0, xx (x, y + u yy (x, y = 0, (x, y Ω Consider the following piecewise continuous function u 1 on the boundary Ω: u 1 (±L, y = 0, y > 0, ( 3πx u 1 (x, 0 = 4sin, L < x < L. 2L

17 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? 17 Note that u 1 has jump discontinuities at the corners (±L, 0. Find all values of the constants at which the solution also satisfies the following boundary conditions lim y = u 1 (x, y, (x, y Ω, n Ω u(x, y = 0, L < x < L. where n is the unit outward normal to Ω. Solution: The function and all its partial derivatives are continuous in the whole R 2. Therefore u C 2 (Ω C 1 ( Ω and the stated problem makes sense. The needed partial derivatives are: xx = α 2 sin(αx (Ae βy + Be βy = α 2 u, yy = β 2 sin(αx (Ae βy + Be βy = β 2 u, xx + yy = ( α 2 + β 2 u = 0 α 2 = β 2 because u(x, y 0 for all (x, y Ω. Since A and B are arbitrary, and the sine function is odd, one can set α = β 0. Thus u(x, y = sin(αx (Ae αy + Be αy, α 0. is a solution to the Laplace equation. The asymptotic (last boundary condition requires that A = 0 (note that e αy as y if α > 0 or α = 0 (in the latter case, u(x, y = 0. The remaining constants α and B have to be chosen so that the solution has the specified value of the normal derivative. The boundary has three smooth pieces Ω = S 1 S 2 S 3 oriented outward by the following unit normals S 1 = {(x, y x = L, y 0}, n 1 = (1, 0, S 2 = {(x, y x = L, y 0}, n 2 = ( 1, 0, S 3 = {(x, y L x L, y = 0}, n 3 = (0, 1,

18 18 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS Therefore the normal derivative is = n 1 u = u x (L, y, n S1 x=l = n 2 u = u x ( L, y, n S2 x= L = n 3 u = u y (x, 0. n S3 y=0 Thus, the boundary conditions are u x (±L, y = 0, y 0, ( 3πx u y (x, 0 = 4sin, L x L. 2L Note that the corner points are included into these conditions. This means that the normal derivative is required to have the same jump discontinuities as u 1 at the points where the boundary is not smooth. The first two conditions yield: u x(±l, y = αb cos(αle αy = 0, y 0, α = 0 or B = 0 or cos(αl = 0 α = α m = π (2m + 1, m = 0, 1, 2,... 2L because α 0 (if α = 0 or B = 0, then u(x, y = 0. The third condition yields ( 3πx u y (x, 0 = Bα m sin(α m x = 4sin, L x L, 2L m = 1 and B = 4 α 1 because the functions sin(α m x are linearly independent in the interval [ L, L]. Thus, the function e 3πy 2L u(x, y = 8L ( 3πx 3π sin 2L is a solution to the Laplace equation in the rectangle Ω satisfying the stated boundary conditions Exercises. 1. Is the function u(x, y = ln(e x + e y a solution to the equation xx yy ( xy 2 = 0 in some open region in a plane?

19 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? If u = u(x, y, z, find the most general solution to each of the following equations in R 3 : xx = 0 xy = 0 u (n zz z = 0 xyz = 0 u x = yu 2 3. Let f be a twice continuously differentiable function of a real variable. Under what conditions on the n dimensional vector k = (k 1, k 2,..., k n is the function u(t,x = f(ct + k x, x R n, where k x = k 1 x 1 + k 2 x k n x n (the dot product in R n, a solution to the n dimensional wave equation in R n : c 2 tt x 1 x 1 x 2 x 2 x nx n = 0? 4. Let x = (x 1, x 2,..., x n R n and a x = a 1 x 1 + a 2 x a n x n (the dot product in R n. Under what conditions on the vector a is the function u(x = exp(a x a solution to the equation x 1 x 1 + x 2 x x nx n = u, x R n 5. Let r = r = x 2 + y 2 + z 2. Find a value of the constant k, if any, such that the function u(x, y, z = sin(kr r is a solution of the (Helmholtz equation xx + yy + zz + a 2 u = 0, r 0 where a > 0 is a given constant. Can the function u be extended to the origin r = 0 so that it becomes a solution in the whole space R 3? Hint: Consider the power series representation of the sine function. 6. Show that the function u(x, y = ln(x 2 + y 2 is a solution to the two-dimensional Laplace equation xx + yy = 0, (x, y (0, 0 7. Let r = x = (x x x 2 n 1/2 (the distance from the origin in R n. Show that the function u(x = r 2 n, n > 2, is a solution to the n dimensional Laplace equation x 1 x 1 + x 2 x x nx n = 0, x 0

20 20 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS 8. Find all functions T(t and constants k for which u(x, t = T(tsin(kx is a solution to the wave equation tt xx = 0, 0 < x < L, t > 0 that also satisfies the boundary conditions u = 4sin(2πx/L, u t = 0, 0 x L, t=0 t=0 u = u, t 0 x=0 x=l 9. Find all functions T(t and constants k for which u(x, t = T(tsin(kx is a solution to the heat equation u t xx = 0, 0 < x < L, t > 0 that also satisfies the boundary conditions u = 4sin(2πx/L, 0 x L; u = u, t 0 t=0 x=0 x=l 10. Find all functions T(t and constants k, A, and B for which u(x, t = T(te kx is a solution to the heat equation u t xx = 0, x > 0, t > 0 that also satisfies the boundary conditions u = Ae 2x, x 0; t=0 u = Bu x, lim u(t, x = 0, t 0 x=0 x=0 x 11. Find all function T(t and constants k, A, and B for which the function ( u(x, t = T(t A cos(kx + B sin(kx is a solution to the telegraph equation tt + 2γu t c2 xx = 0, t > 0, where γ > 0, that satisfy the boundary condition u = cos(kx sin(kx, = 0 t=0 t=0 12. Find all constants k, p, A, B, and µ for which the function u(x, y, z = cos(πx/2sin(πy (Ae µz + Be µz is a solution to the Laplace equation xx + yy + zz = 0, (x, y, z ( 1, 1 ( 2, 2 (0, u t

21 1. WHAT ARE PARTIAL DIFFERENTIAL EQUATIONS? 21 that satisfies the boundary conditions u = 0, u = 0, x=±1 y=±2 u = 4cos(πx/2sin(πy, z=0 lim u(x, y, z = 0 z

22 22 1. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS Answers. 1. Yes. 2. In the order of appearance, u = a(y, zx + b(y, x; u = f(x + g(y; u = a 0 (x, y + a 1 (x, yz + + a n 1 (x, yz n 1 ; u = f(x, y + g(y, z + h(x, z; u = (C xy k = a = k = a and the extension is u(0 = k, the power series representation u = k ( 1 n (kr2n (2n 1! = k ( 1 nk2n (x 2 + y 2 + z 2 n (2n 1! n=0 n=0 has infinite radius of convergence and, hence, u C (R u = 4 cos(2πt/l sin(2πx/l. 9. u = 4e k2t sin(kx, k = 2π/L. 10. u = Ae 4t 2x (, B = A/ u = e γt cos(ωt + (cos(kx γ sin(ωt sin(kx ω where ω = c 2 k 2 γ ( 2 if γ 2 < (ck 2 u = (ae γ+t +be γ t cos(kx sin(kx where γ ± = γ ± γ 2 c 2 k 2 and a = γ /(γ( + γ, b = γ + /(γ + γ if γ 2 > (ck 2 u = e γt (1 + γt cos(kx sin(kx if γ = ck. 12. A = 0, B = 4, µ = π 5/2.