then the substitution z = ax + by + c reduces this equation to the separable one.

Transcription

1 7 Substitutions II Some of the topics in this lecture are optional and will not be tested at the exams. However, for a curious student it should be useful to learn a few extra things about ordinary differential equations. 7.1 Examples If it is possible to solve an ODE, then it usually means that there is a smart substitution bringing this equation to the form, which we already studied. Here are a few examples: Example 1. Solve y = 4x + 2y 1. The substitution we can use here is zx) = 4x + 2y 1 such that z x) = 4 + 2y x), and This is a separable equation: 1 2 z 2 = z = z = 2 z + 4. dz dz = 2 z + 4 = dz 2 = dx = z + 4 dz 2 z + 4 = dx = u du u + 2 = x + C = u 2 log u + 2 = x + C = z 2 log z + 2 = x + C = ) 4x + 2y 1 2 log 4x 2y = x + C In general, if we are having an equation of the form y = fax + by + c), a, b, c R, then the substitution z = ax + by + c reduces this equation to the separable one. Example 2. Solve 2x 2 y = y 3 + xy. One way to tackle this problem is to consider a substitution y = z α, where α is a constant which has to be determined. After the substitution we get 2αx 2 z α 1 z = z 3α + xz α, or 2αx 2 z α 1 dz z 3α + xz α ) dx = 0, MATH266: Intro to ODE by Artem Novozhilov, artem.novozhilov@ndsu.edu. Fall

2 which will be homogeneous if 2 + α 1 = 3α = 1 + α. These are two equations with one unknown, but we are lucky here, and there is solution α = 1 2. Therefore, y = z: x 2 dz z + x)z dx = 0 = x du + u 2 dx = 0, if u = z/x. Finally we find 1 u + log x = C, or, for the original variables x + log x = C. y2 A similar trick is sometimes useful for looking for an integrating factor in the form µx, y) = x α y β to bring an ODE into the exact form. 7.2 Higher order equations Example 3. Consider the equation y e x + 1) + y = 0. First of all this equation has the order two, and we did not treat such equations systematically yet. However, this equation has an important feature that it does not involve the unknown function yx). Formally, we have that our equation is a representative of the general second order equation of the form F x, y, y ) = 0. In such cases we can always reduce the order of the equation by using the substitution zx) = y x), from which z x) = y x), and our equation becomes F x, z, z ) = 0, which is a first order ODE. For our example, we have z e x + 1) + z = 0, which is separable. We have the solution fill in the omitted steps): To find y we need to integrate once more y = z dx = z = C ex + 1 e x. C ex + 1 e x dx = Cx e x ) + C 1. Note that the general solution depends on two arbitrary constants, as should be expected since we solve the second order equation. Example 4 Equation for the catenary). Imagine a chain hanging freely such that the left end at the point a 1 and the right end at the point a 2 chosen such that the minimum of the chain has zero first coordinate. This curve bears the name catenary. Let us derive the equation of this line. Consider a small element of the chain, such that the left end of this element has the coordinate x and the right element has the coordinate x + x, and the length of the element is s. The force of gravity is applied to this elements is P = ρga s, 2

3 where ρ is the density of the material, g is the acceleration of a freely falling body, A is the area of the cross cut section of the chain. We also have the tension forces T x) and T x + x) in the directions corresponding to the tangent lines to the chain at the points x and x + x make a figure!). Since the chain is in equilibrium, we have that in both projections to coordinate axes we must have T x) cos αx) + T x + x) cos αx + x) = 0, T x) sin αx) + T x + x) sin αx + x) = P, where αx) is the angle of the tangent line to the x-axis at the coordinate x. From the first equation we see that the horizontal component of the tension is constant T x) cos αx) = T 0. From the second equation, considering x 0, we find or, using the horizontal component d T x) sin αx)) = dp x), d T 0 tan αx)) = dp x), which implies that T 0 d y ) = ρga ds. Using the expression for the length of the element of the curve ds = 1 + y ) 2 dx, we finally get y = a 1 + y ) 2, I lower the order of the equation by using z = y : a = ρga T 0. This equation is separable and its solution is z = a 1 + z 2. log z z 2) = ax + C. I chose the origin of the coordinate system such that the slope at x = 0 is zero, hence z0) = 0, and C = 0. We have z z 2 = e ax. If we multiply both sides of this equation by z 2, we get z z + z 2 = e ax. 3

4 Using these expressions, we find that zx) = eax e ax 2 = sinh ax. By integrating once more, we find that the equation for the catenary takes the form yx) = eax + e ax 2a + C = 1 cosh ax + C, a where the constant C should be determined by the coordinates of the chain s ends. Assuming that C = 0 I present the final answer is that the freely hanging rope or chain of the constant density takes the form of the hyperbolic cosine: yx) = 1 cosh ax. a I advise you to make a few graphs of this function for different values of the parameter a. Example 5. Consider the equation y 3 y = 1. Again, this is a second order equation, but this time it does not involve the independent variable, its general form is F y, y, y ) = 0. The substitution here is to assume that y x) = uy), i.e., y becomes an independent variable. To find y x) in term of uy) we will need the chain rule: y x) = d2 y dx 2 x) = d dx y x) = d dx uy) = d dy uy) dy dx = u y)y = u y)uy), hence, in short, y = u u, and instead of the second order equation F y, y, y ) = 0 we end up with a first order ODE F y, u, u u) = 0. A very good exercise to make sure you understood how the substitution works is to find out what is y x) in terms of uy). The answer is y = uu ) 2 + u 2 u ). Returning to our example, we have, by making y = u and y = uu y 3 uu = 1, 4

5 which is a separable equation. u du = dy y 3 = u 2 2 = 1 2y 2 + C = u 2 = 1 y 2 + C = u = ± C 1 ) 1/2 y 2 = y = ± C 1 ) 1/2 y 2 = dy ±C y 2 = dx = ) 1/2 y dy ±Cy 2 1) 1/2 = x + C 1 = 1 dcy 2 ) 2C ±Cy 2 1) = x + C 1 = Cy 2 1) 1 2 = Cx + C Riccati equation The general Riccati equation has the form y = ax)y 2 + bx)y + cx). If we know some particular any) solution ỹx) to the Riccati equation, this means we can find all the solutions, because the substitution y = ỹ + z puts the equation into the form of the Bernoulli s equation prove this) z 2aỹ + b)z az 2 = 0. Which is more important for us here is that there is a theorem which states that in general the Riccati equation cannot be integrated, i.e., its solution cannot be expressed in the form of a finite combination of elementary functions and integrals of them. This theorem is true already for y = y 2 x. Again, it is proved that there is no way we can express the solution to the Riccati equation as an analytical formula involving only elementary functions and integrals of them. It is not that we are not smart enough to come up with this formula, this formula simply does not exist. There is a very good analogy with polynomial equations: You probably know the formula to find the roots of a quadratic equation. There is so-called Cardano s formula for cubic equation. It is still possible to present a formula for the equation of degree four. And very famous theorem is algebra Ruffini Abel Galois theorem) claims that the roots of some polynomials of degree five cannot be expressed in terms of radicals of the coefficients of the polynomial. The situation is analogous for ODE. It does not mean that there is no solution! Think about the geometric interpretation of ODE: If the right hand side is smooth enough, we always have a solution. However, our repertoire of elementary functions is too small to write down a formula for this solution. That is why it is so important to study other methods 5

6 to analyze ODE. In particular, we will also talk about qualitative methods of studying ODE and numerical methods of solving ODE. 7.4 Non-dimensional problems Very often substitutions are used to reduce the number of parameters involved in the problem under consideration. In physics this process is frequently referred to as non-dimensionalization. As an example consider the logistic equation in the form Ṅ = rn 1 N ), K where r > 0 and K > 0 are parameters of the problem. The former is the rate of growth and the latter is the carrying capacity. Nt) is the size of population at time moment t. Let us assume that here we are talking about elephants, hence Nt) is the number of elephants in the population at the time moment t. What are the dimensions of the parameters in this problem? First note that on the left hand side we have [ the derivative ] Ṅ, i.e., the rate of change, which means that the units on the left have to be [Ṅ] = elephants time the square parenthesis mean dimensions ), which implies that the right hand side has to have the same units, and we have no choice as to consider that [K] = [elephants] and [r] = [ 1 time]. Now assume that we would like to make the substitutions Nt) = Anτ), t = T τ, where τ is a new independent variable, nτ) is a new dependent variable, and A and T are constants to be determined. Using the chain rule hence A T dnt) dt = A dnτ) dt dn dτ = ran 1 An ) K Since we are free to choose A and T, we can set = A dnτ) dτ dτ dt = A T dnτ), dτ = dn dτ = rt N 1 An ). K T = 1, [T ] = [time], A = K, [A] = [elephants], r and the change of variables nτ) = Nt) K, τ = rt, reduces our equation to the form without any parameters: ṅ = n1 n), where now ṅ means the derivative with respect to τ. Note that new variables are dimensionless: nτ) is the proportion of the population at time τ, and time is measured in some abstract non-dimensional units. 6