The linear equations can be classificed into the following cases, from easier to more difficult: 1. Linear: u y. u x

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1 Week 02 : Method of C haracteristics From now on we will stu one by one classical techniques of obtaining solution formulas for PDEs. The first one is the method of characteristics, which is particularly useful when solving first order equations. 1. C lassification of first-order equations. The general form of first-order PDE in R 2 : or in R 3 : F x, y, u,, = 0, x, y D R 2. 1 F x, y, z, u,,, u z = 0. 2 Often the following notation is used in 2D: thus the equation can be written as p =, q = 3 F x, y, u, p, q = 0. 4 The linear equations can be classificed into the following cases, from easier to more difficult: 1. Linear: 2. Semi-linear: 3. Quasi-linear: 4. General case. F x, y, u,, = a x, y + b x, y + c x, y u d x, y. 5 F x, y, u,, = a x, y + b x, y c x, y, u. 6 F x, y, u,, = a x, y, u + b x, y, u c x, y, u Method of characteristics. We try to find a method to solve the general first-order quasi-linear equation a x, y, u + b x, y, u = c x, y, u. 8 Let s start by thinking geometrically. Consider the 3 dimensional space with coordinates x, y, u. Assume that u = u x, y is a solution to the equation, it is clear that it represents a surface in the x, y, u space. Now we explore the geometrical meaning of the equation. Observe that the equation can be written in the form of an inner product a x, y, u b x, y, u c x, y, u = 0 a x, y, u b x, y, u c x, y, u Therefore all we need to do is to understand the relation between the vector First introduce a new function Ψ: R 3 R through. 9 and the solution. Φ x, y, u = u u x, y. 1 0 Note that in the RHS of the above, the first u is a variable, the second u is a function. For example, suppose u x, y = x 2 + y 2, then the corresponding Φ x, y, u = u x 2 + y 2. Now we easily see that = Φ x Φ y Φ u = Φ. 1 1

2 Recall that geometrically, Φ and also Φ is a normal vector of the surface Φ = 0. As a consequence means is perpendicular to the solution surface u = u x, y. On the other hand, from the equation we know that a b c must be tangent to the surface u = u x, y. a b c is perpendicular to the vector which Now we summarize. We have shown that the equation is equivalent to the geometrical requirement that the vector curve of a b c a x, y, u b x, y, u c x, y, u, that is any is tangent to the solution surface u = u x, y. As a consequence, any integral X s Y s U s satisfying dx = a X, Y, U 1 2 ds dy = b X, Y, U 1 3 ds du = c X, Y, U 1 4 ds must be contained in one of the solution surfaces. Conversely, any surface woven by such integral curves is a solution surface. The above understanding leads to the following method of characteristics due to Lagrange. Theorem 1. The general solution of a first- order, quasi- linear PDE satisfies a x, y, u + b x, y, u = c x, y, u 1 5 F φ, ψ = 0, 1 6 where F is an arbitrary function of φ x, y, u and ψ x, y, u, and any intersection of the level sets of φ and ψ is a solution of the characteristic equations a = b = du c. 1 7 Remark 2. As we will see soon, φ, ψ are obtained through solving the characteristic equations. And each F gives a solution to the original equation. Remark 3. The curves mentioned above are called the families of characteristic curves of the equation. Remark 4. As we will see, the main technique in getting φ and ψ is a b = c d a ± c b ± d = a b = c d. 1 8 In the following, we will show how to apply this method. We start with the simplest case. 3. Solving linear first-order equations Equations with constant coefficients. We start from the simplest case, where a, b, c and d are just constants. Example , 3 b Find the general solution of the equation a + b = 0; a, b are constant. 1 9 a = b = du 0. 20

3 What we need are two functions φ x, y, u and ψ x, y, u such that dφ = 0, dψ = 0 along the characteristics. Obviously we can take φ = u. For ψ, notice that thus we can take As a consequence, the solution satisfies for any function F. This means for an arbitrary function f. d a y b x = a b = 0, 21 ψ = a y b x. 22 F a y b x, u = 0 23 u = f a y b x. 24 Example 6. C auchy problem, 2. 8, 5 a Oftentimes, the value of the solution along some specific curve in the plane is prescribed. For example, solve = 0, u x, 0 = sin x. 25 Solution. From the above example we know that the general solution takes the form Now substituting this into the initial condition, we obtain u = f 2 x 3 y. 26 Therefore f 2 x = sin x f x = sin x u x, y = sin 2 x 3 y. 2 Example 7. c, d 0 What happens when the c, d are not 0? The method still works. We write down the characteristic equations a = b = du d c u. 28 We have a = du d c u As a consequence and gives a = d b x a y = 0, 29 b du = a d a c u u = C e a c x + c d d e a c x u c d = 0 30 φ = b x a y, ψ = e a c x u c d 31 F φ, ψ = 0 32 u = c d + e a c x f b x a y. 33 Remark 8. In the above, one may be tempted to conclude d u C e a c x = 0 34 and try to use u C e a c x as ψ. This is obviously wrong as d somehow disappeared. One should keep in mind that neither φ or ψ can involve arbitrary constants. It is their values that are arbitrary constants. Example 9. Some times people use the following method of characteristics :

4 First solve then solve ds = a, ds = b 35 u s + c u = d 36 to obtain the solution u in the form u = u s, t. Finally represent s, t by x, y and obtain the solution. This is equivalent to our method but considerably more complicated to use. Anyone who does not believe this should try using this method to the following examples with non-constant coefficients. Remark 1 0. One may wonder, how to find out φ and ψ efficiently? Unfortunately there may not be any short-cut. One way to systematically find φ and ψ is the following. The characteristic equations consists of three equations. Pick any two of them. If you can find general solutions, then you have φ and ψ. But this fails when any coefficient involves all other variables Equations with non-constant coefficients. Example , 3 h Find the general solution of Using we obtain Thus On the other hand, from we obtain As a consequence we can take Putting these together we obtain which gives y x = x = x = y 39 y + x = 0 d x y = φ = x y; du = d log y d u log y = ψ = u log y. 44 F x y, u log y = 0 45 u = log y + f x y. 46 Example , 5 c Find the solution of the following Cauchy problem: x + y = 2 x y, with u = 2 on y = x Solution. We need to first find the general solution, then using the value on y = x 2 to determine the arbitrary function involved. Find the general solution The characteristic equations are Using x = 2 x y. 48 x = y 49

5 we obtain On the other hand, we have therefore d = 0 can take φ = y x x. 50 du = 2 x = 2 y = x + y = d x y d u x y = 0 51 The general solution satisfies F x, u x y Determine the solution. We have u = 2 along y = x 2, that is ψ = u x y. 52 = 0 u = x y + f. 53 x u x, x 2 = Using the formula for the general solution, we have As a consequence x 3 + f x = 2 f x = 2 x u x, y = x y x 4. Solving semi-linear first-order equations. Example , 3 g Find the general solution of the following equation: From we have On the other hand, we have Thus we take Now gives x x u 2 y Example , 5 g Solve y 2 x y = x u 2 y. 57 y 2 = Solution. The characteritic equations are x y = y 2 = x y du x u 2 y d x 2 + y 2 = 0 φ = x 2 + y du = + 2 d u y = u y y d u y y = ψ = y u y. 62 F x 2 + y 2, y u y = 0 63 u = y + y f x 2 + y x + y = u + 1 with u x, y = x 2 on y = x x = u

6 which easily lead to Thus Now the Cauchy data implies thus As a consequence 5. Solving quasi-linear first-order equations. φ = y x, ψ = u + 1 x. 67 u = x f. x 68 x f x = u x, x 2 = x 2 69 f x = x + x. 70 x u x, y = x + x 2 = y + x x y. 71 Example , 3 f Find the general solution of From this we have Thus we can take On the other hand, we have d + u = y + d x y u + u + y = x y. 72 y + u = x y. 73 = du x y d x 2 + u 2 = φ = x u d u 2 x y 2 = As a consequence, the solution is given by F x 2 + u 2, u 2 x y 2 = Remark 1 6. Note that, in the above we can also use d x + + u = d = const. 78 y x + u y Thus the formula may not be unique. Example , 5 h solve Solution. The characteristic equation are u u = u 2 + x + y 2 with u = 1 on y = We have Then we have u = u u = u = du u 2 + x + y d x + y = 0 φ = x + y. 81 u = du u 2 + φ 2 ψ = e 2 y u 2 + φ 2. 82

7 Now from the Cauchy data we have Therefore effectively F has to be and the solution satisfies which leads to F φ, 1 + φ F φ, ψ = ψ φ e 2 y u 2 + x + y 2 + x + y 2 = 0 85 [ { u = ± 1 + x + y 2} ] e 2 y x + y 2 1 / Equations with more than two variables. The method of characteristics can be applied to higher dimensional problems with no difficulty in principle it indeed becomes more difficult in practice! Example , 8 d Solve the following equation y z x z + x y x 2 + y 2 u z = y z = x z = This time we need three invariants, let s denote them by φ, ψ and η. Clearly one can take For the second invariant, we observe y z = x z y = x The last invariant can be obtained through x z = dz x y φ z = dz y φ Therefore the solutions are obtained by setting dz x y x 2 + y 2 = du φ = u. 89 d x 2 + y 2 = 0 ψ = x 2 + y d z 2 + y 2 φ = 0 η = z 2 + y 2 φ. 91 which gives for arbitrary f. 1 F u, x 2 + y 2, z 2 + y 2 x 2 + y 2 = 0 92 u = f x 2 + y 2, z 2 + y 2 x 2 + y Keep in mind that this f will be determined once some Cauchy data is given.