1. 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
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2. 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.
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3. Example 1.1
UniMAP
Write down all partial derivatives of the following functions:
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4. 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.
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5. 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
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6. KUKUM
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7. Exercise 1.1
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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
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8. 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,
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9. Example 1.3
KUKUM
Solution
Let
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10. KUKUM
Exercise 1.2
Answer
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11. 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
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12. KUKUM
Example 1.4
Let z = exy, where x = 3u2 + v and y = 2u + v3. Find
Answer
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13. KUKUM
Exercise 1.3
Let where
Find
Answer
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14. Definition 1.1 (Jacobian)
Let be n number of functions of n variables i.e.
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Jacobian for this system of equations is given by the following determinant:
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15. Note that both Jacobians will give the same answer, since
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Note that both Jacobians will give the same answer, since
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16. Example 1.5 Given determine the Jacobian for the system of equation?
KUKUM
Given determine the Jacobian for the system of equation?
Answer
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17. KUKUM
Exercise 1.4
Given u = xy and v = x + y, determine the Jacobian for the system of equations?
Answer
J = y – x
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18. 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.
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19. Example 1.6
Given
evaluate
Answer
KUKUM
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20. Example 1.7 Let z = 2x2 + 3xy + 4y2, u = x2 + y2 and v = x + 2y. Find
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Example 1.7
Let z = 2x2 + 3xy + 4y2, u = x2 + y2 and
v = x + 2y. Find
Solution
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21. KUKUM
Hence,
The Jacobian is
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22. KUKUM
Therefore,
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23. From (1) and (2), we have
KUKUM
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24. 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
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25. Answer
KUKUM
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26. Definition 1.2 (Hessian) Let f be a function of n number of variables
Hessian of f is given by the following determinant :
KUKUM
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27. Hessian of a Function of Two Variables
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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
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28. Since x and y are independent, we have
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Since x and y are independent, we have
The Hessian becomes
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29. KUKUM
Example 1.8
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30. 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
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31. Properties of Stationary Points :
KUKUM If
This stationary point is a SADDLE POINT.
Figure 1.1 Saddle point
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32. 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
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33. This stationary point is a MINIMUM POINT. and
KUKUM b)
This stationary point is a MINIMUM POINT.
and
Figure 1.3 Minimum point
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34. KUKUM If
This test FAILS.
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35. 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.
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36. 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.
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37. Step 2 : Compute the Hessian of z.
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Step 2 : Compute the Hessian of z.
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38. Step 3: Determine properties of the stationary
points based on Hessian of z.
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Point
Hessian
Conclusion
Saddle Point
Maximum Point
Minimum Point
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39. 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.
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40. 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
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41. Example of PDE
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42. Focus  first order with two variables
PDEs We can already solve
 By integration
Example
Solution
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43. (b). Solve PDE in (a) with initial condition u(0,y) = y
Solution
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44. 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.
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45. KUKUM
Example 1.10
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46. Solution
KUKUM
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47. KUKUM
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48. KUKUM
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49. KUKUM
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50. KUKUM
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51. KUKUM
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52. KUKUM
Exercise 1.7
Answer
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