Lecture two January 17, 2019 We will learn how to solve rst-order linear equations in this lecture. Example 1. 1) Find all solutions satisfy the equation u x (x, y) = 0. 2) Find the solution if we know the value at x = x 0, u(x 0, y) = y 2. We can integrate once to get u = constant for any xed y. u(x 1, y) = u(x 2, y). Because u is independent with x. So the solutions are in the form of where f(y) is arbitrary. If u(x 0, y) = y 2, then u(x, y) = f(y), (1) u(x, y) = y 2. Example 2. Solve a constant coecient transport equation where a and b are constants not both zero. au x + bu y = 0, (2) Method one: Introducing new a parameter (t, s), such that Then we have x = a, = b. x(t, s) = at + g(s) y(t, s) = bt + h(s). 1
x We assume that b 0. x And If we choose g(s) = s and h(s) = 0 ( (x,y) 0), such that (x, y) and (t, s) have one to one corresponds Let's integrate in t to get x(t, s) = at + s y(t, s) = bt. au x + bu y = x u x + u u(x, y) = u(t, s) = f(s) = f( = u = 0. bx ay ). b (t,s) := Or make a change of coordinates which will give you the same form of solutions t = ax + by s = bx ay. Replace all x and y derivatives by t and s derivatives. By the Chain rule, and Hence u x = u x = u x + u x = au t + bu s, u y = u = u + u = bu t au s. au x + bu y = a(au t + bu s ) + b(bu t au s ) = (a 2 + b 2 )u t. Because a 2 + b 2 0, the equation takes the form u t (t, s) = 0. previous example, By the u(x, y) = u(t, s) = f(s) = f(bx ay), where f an arbitary function of one variable. Method two: Geometric Method. The quantity au x + bu y is the directional derivative of u in the direction of the vector V = (a, b). This means that u(x, y) must be constant in the direction of V. The lines parallel to V have the equations bx ay = constant. They are called the characteristic lines. On any xed line bx ay = c the solution u has a constant value. Thus the solution is u(x, y) = f(bx ay). 2
For example, if u(0, y) = y 3 and a 0 then Letting w = ay yields u(0, y) = y 3 = f( ay). So we have f(w) = w3 a 3. (bx ay)3 u(x, y) = a 3. Example 3. Solve the variable coecient equation (linear and homogeneous equation) u x + yu y = 0. The same as the geometric method before, the dirctional derivative in the direction of the vector (1.y) is zero. The (characteristic curves) curves in xy plane with (1, y) as its tangent vectors have slopes y. Their equations are This ODE has the solutions dy dx = y 1. y = se x. We can check that on each of the curves u(x, y) is a constant becuse d dx u(x, sex ) = u x + u sdex dx = u x + se x u y = u x + yu y = 0. The curves ll out the xy plane perfectly without intersecting, as s is changed. Thus suppose (x, y) lies at most and only in one curve (x, se x ) u(x, y) = u(x, se x ) = u(0, s) is independent of x. For any s if we know u(0, s) we know all the value of u(x, y). Putting s = e x y we have It follows that u(x, y) = u(0, e x y) u(x, y) = f(e x y) For example, if u(0, z) = z 2, we have u(x, y) = u(0, e x y) = e 2x y 2. 3
Example 4. Solve the inhomogeneous quation u x + 2u y + (2x y)u = 2x 2 + 3xy 2y 2. Make a change of coordinates Then we have by the chain rule Then t = x + 2y s = 2x y. u x = u t + 2u s u y = 2u t u s. u x + 2u y + (2x y)u = u t + 2u s + 2(2u t u s ) + su So we get a equation of the form So we have from the Example 6 u = t s where f is arbitrary. = u t + su = (2x y)(x + 2y) = st. u t + su = st. + f(s)e st = x + 2y 2x + f(2x 2 +3xy 2y 2 y)e, 2x y We review some methods of rst order ODE. If u satises the homogeneous linear ODE du du = P (t)q(u), then Q(u) = P (t) + C. Example. For example, du = 4 u. From the formula, we have log u = 4 t + c 4t c u = e. We can also solve the following form of inhomogeneous linear ODE. Example 6. du + p(t)u = q(t). 4
Find f(t) such that d(fu) f du + fp(t)u = fq(t) u df + fp(t)u = fq(t). If df = fp(t), we have df f = log f = p(t) + c p(t) + c. If we choose f = e p(t), So we have d(e p(t) u) u(t) = e p(t) [ = e p(t) q(t). e p(t) q(t) + c]. For example, if u t + su = st we have u(t, s) = e s [ e s st + f(s)] = e st [ e z z dz + f(s)] s = e st st [te s e st + f(s)] = t s + f(s)e st.