Chapter 1 Partial Differential Equations UniMAP Chapter 1 Partial Differential Equations Introduction Small Increments and Rates of Change Implicit Function Change of Variables Inverse Function: Determine Partial Derivatives Stationary Point Applied Partial Differential Equations EUT 203

Introduction Consider the following function f (x1, x2,…, xn) UniMAP Consider the following function f (x1, x2,…, xn) where x1, x2,…, xn are independent variables. If we differentiate f with respect to variable xi , then we assume a) xi as a single variable, b) as constants. EUT 203

Example 1.1 UniMAP Write down all partial derivatives of the following functions: EUT 203

1.1 Small Increments and Rates of Change UniMAP Notation for small increment is δ. Let z = f (x1,x2,…, xn). a) A small increment δz is given by where δx1, δx2,, δxn are small increments at variables stated. b) Rate of change z with respect to time t is given by where are rates of change for the variables with respect t. EUT 203

KUKUM Example 1.2 Given a cylinder with r = 5cm and h =10cm. Determine the small increment for its volume when r increases to 0.2cm and h decreases to 0.1cm. Solution Volume of a cylinder is given by V   r2h. Small increment is Given EUT 203

KUKUM EUT 203

Exercise 1.1 KUKUM The radius r of a cylinder is increasing at the rate of 0.2 cm s-1 while the height, h is decreasing at 0.5 cm s-1. Determine the rate of change for its volume when r = 8 cm and h = 12 cm. Answer EUT 203

1.2 Implicit Functions KUKUM Let f be a function of two independent variables x and y, given by To determine the derivative of this implicit function, let z = f (x, y) = c. Hence, EUT 203

Example 1.3 KUKUM Solution Let EUT 203

KUKUM Exercise 1.2 Answer EUT 203

1.3 Change of Variables KUKUM Let z be a function of two independent variables x and y. Here x and y are functions of two independent variables u and v. We write the derivatives of z with respect to u and v as follows EUT 203

KUKUM Example 1.4 Let z = exy, where x = 3u2 + v and y = 2u + v3. Find Answer EUT 203

KUKUM Exercise 1.3 Let where Find Answer EUT 203

Definition 1.1 (Jacobian) Let be n number of functions of n variables i.e. KUKUM Jacobian for this system of equations is given by the following determinant: EUT 203

Note that both Jacobians will give the same answer, since KUKUM Note that both Jacobians will give the same answer, since EUT 203

Example 1.5 Given determine the Jacobian for the system of equation? KUKUM Given determine the Jacobian for the system of equation? Answer EUT 203

KUKUM Exercise 1.4 Given u = xy and v = x + y, determine the Jacobian for the system of equations? Answer J = y – x EUT 203

1.4 Inverse Functions: Determine Partial Derivatives KUKUM Let u and v be two functions of two independent variables x and y, i.e. u = f (x, y) and v = g(x, y). Partial derivatives are given by where J is the Jacobian of the system of equations. EUT 203

Example 1.6 Given evaluate Answer KUKUM EUT 203

Example 1.7 Let z = 2x2 + 3xy + 4y2, u = x2 + y2 and v = x + 2y. Find KUKUM Example 1.7 Let z = 2x2 + 3xy + 4y2, u = x2 + y2 and v = x + 2y. Find Solution EUT 203

KUKUM Hence, The Jacobian is EUT 203

KUKUM Therefore, EUT 203

From (1) and (2), we have KUKUM EUT 203

KUKUM Exercise 1.5 Let z = x3 + 2xy + 3y2, u = 3x2 + 4y2 and v = 2x + 5y. Find Let z = x3 + 2xy + 3y2, x = 3u2 + 4v2 and y = 2u + 5v. Find EUT 203

Answer KUKUM EUT 203

Definition 1.2 (Hessian) Let f be a function of n number of variables Hessian of f is given by the following determinant : KUKUM EUT 203

Hessian of a Function of Two Variables KUKUM Hessian of a Function of Two Variables Let f be a function of two independent variables x and y. Then the Hessian of f is EUT 203

Since x and y are independent, we have KUKUM Since x and y are independent, we have The Hessian becomes EUT 203

KUKUM Example 1.8 EUT 203

1.5 Stationary Point Given a function f = f (x, y). KUKUM Given a function f = f (x, y). The stationary point of f = f (x, y) occurs when and EUT 203

Properties of Stationary Points : KUKUM If This stationary point is a SADDLE POINT. Figure 1.1 Saddle point EUT 203

This Stationary point is a MAXIMUM POINT. If and whether a) and This Stationary point is a MAXIMUM POINT. KUKUM Figure 1.2 Maximum point EUT 203

This stationary point is a MINIMUM POINT. and KUKUM b) This stationary point is a MINIMUM POINT. and Figure 1.3 Minimum point EUT 203

KUKUM If This test FAILS. EUT 203

KUKUM Example 1.9 Determine the stationary points of z = x3 – 3x + xy2 and types of the stationary points. Solution Step 1 : Determine the stationary points. EUT 203

Hence, z has four stationary points, i.e. KUKUM From (4), we have x = 0 or y = 0. When x = 0, from (3); When y = 0, from (3); Hence, z has four stationary points, i.e. EUT 203

Step 2 : Compute the Hessian of z. KUKUM Step 2 : Compute the Hessian of z. EUT 203

Step 3: Determine properties of the stationary points based on Hessian of z. KUKUM Point Hessian Conclusion Saddle Point Maximum Point Minimum Point EUT 203

Exercise 1.6 Determine the stationary points of KUKUM Exercise 1.6 Determine the stationary points of f (x, y) = x2 + x y + y2 + 5x – 5y + 3 and types of the stationary points. EUT 203

1.6 Partial Differential Equations What is a PDE? Given a function u = u(x1,x2,…,xn), a PDE in u is an equation which relates any of the partial derivatives of u to each other and/or to any of the variables x1,x2,…,xn and u. Notation EUT 203

Example of PDE EUT 203

Focus  first order with two variables PDEs We can already solve  By integration Example Solution EUT 203

(b). Solve PDE in (a) with initial condition u(0,y) = y Solution EUT 203

Separation of Variables Given a PDE in u = u (x,t). We say that u is a product solution if for functions X and T. How does the method work? Let’s look at the following example. EUT 203

KUKUM Example 1.10 EUT 203

Solution KUKUM EUT 203

KUKUM EUT 203

KUKUM EUT 203

KUKUM EUT 203

KUKUM EUT 203

KUKUM EUT 203

KUKUM Exercise 1.7 Answer EUT 203