📝 The Laplacian operator ∇² (14 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 14 questions available

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What is The Laplacian operator ∇²?

The Laplacian operator $\nabla^2$ :
The Laplacian is $\nabla^2 f = \nabla \cdot \nabla f = f_{xx} + f_{yy} + f_{zz}$ , measuring the second-order dispersion of a scalar field.

Example:
For $f(x,y) = x^2 - y^2$ , $\nabla^2 f = 2 - 2 = 0$ , making it harmonic.

Reason:
The Laplacian appears in key PDEs like Laplace's equation ( $\nabla^2 f = 0$ ) and the heat equation, modeling diffusion and potentials.

Easy

Medium

Hard

📝 All The Laplacian operator ∇² MCQs

Q1. For a twice-differentiable scalar field $f(x,y,z)$ , which expression correctly represents the Laplacian at a point?

A.The vector

\nabla f

B.The scalar sum

f_x+f_y+f_z

C.The scalar sum

f_{xx}+f_{yy}+f_{zz}

✅

D.The magnitude

|\nabla f|

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The Laplacian of a scalar field is obtained by adding its second partial derivatives with respect to all spatial coordinates. Thus, $\nabla^2f=f_{xx}+f_{yy}+f_{zz}$ . It is a scalar quantity, whereas $\nabla f$ is a vector. The other choices confuse first derivatives, gradient magnitude, or vector quantities with the required second-derivative operation.

Q2. A temperature field satisfies $\nabla^2T=0$ at a point. Which interpretation is most appropriate?

A.The temperature must be zero at that point

B.The temperature has no spatial variation anywhere

C.The local second-order spatial effects balance so the point is not a local source or sink of temperature ✅

D.The temperature gradient must be zero at that point

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A zero Laplacian does not mean that the temperature itself or its gradient is zero. It means the sum of the second spatial derivatives vanishes. Positive curvature in some directions can balance negative curvature in others, producing no net local second-order tendency associated with the Laplacian.

Q3. Consider $f(x,y)=x^2-4y^2$ . What does the value of $\nabla^2f$ reveal about the field?

A.It is

8

, indicating positive net curvature

B.It is

-6

, indicating negative net curvature

C.It is

0

, because the curvatures cancel ✅

D.It is

2x-8y

, because the Laplacian uses first derivatives

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For $f=x^2-4y^2$ , the second derivatives are $f_{xx}=2$ and $f_{yy}=-8$ . Therefore, $\nabla^2f=2-8=-6$ , so option C is not correct. The correct answer is B because the curvatures do not cancel; their sum is negative, indicating net concavity in the Laplacian sense.

Q4. A student argues that if $\nabla^2f=0$ , then $f$ must be constant. Which counterexample most directly disproves the claim?

f(x,y)=x+y

f(x,y)=x^2+y^2

f(x,y)=e^{x+y}

f(x,y)=x^2-y^2

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The function $f=x^2-y^2$ has $f_{xx}=2$ and $f_{yy}=-2$ , so $\nabla^2f=0$ , yet the function clearly varies with both $x$ and $y$ . Therefore, a zero Laplacian does not imply constancy. It indicates a balance among second-order spatial variations.

Q5. For $f(x,y,z)=x^2+3y^2-2z^2$ , a modeler claims that the Laplacian is $2+3-2=3$ . What is the most important error?

A.The Laplacian requires first derivatives only

B.The coefficient of

y^2

contributes

6

, not

3

, after taking two derivatives ✅

C.The negative

z^2

term must be ignored

D.The Laplacian cannot be calculated for polynomial fields

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The second derivative of $x^2$ is $2$ , the second derivative of $3y^2$ is $6$ , and the second derivative of $-2z^2$ is $-4$ . Hence, $\nabla^2f=2+6-4=4$ . The student's error is treating the coefficients as if they were already second derivatives.

Q6. A pollutant concentration is modeled by $C(x,y)=100-x^2-2y^2$ . At the point $(1,2)$ , what is $\nabla^2C$ , and what does its sign indicate?

-6

, indicating negative net curvature ✅

6

, indicating positive net curvature

-5

, indicating zero curvature

100

, indicating the concentration itself controls the Laplacian

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The second derivatives are $C_{xx}=-2$ and $C_{yy}=-4$ , so $\nabla^2C=-6$ everywhere, including at $(1,2)$ . The negative value indicates that the field has negative net second-order spatial curvature. Notice that the point coordinates affect the concentration value but not the Laplacian in this particular quadratic model.

Q7. A numerical simulation gives $f_{xx}=5$ , $f_{yy}=-2$ , and $f_{zz}=-3$ at one grid point. Another analyst reports $\nabla^2f=10$ . Which conclusion is justified?

A.The analyst is correct because

5-(-2)-(-3)=10

B.The analyst is incorrect because the Laplacian is

0

after summing signed second derivatives ✅

C.The Laplacian is

6

because only positive curvature counts

D.The Laplacian is

-10

because negative derivatives dominate

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The Laplacian preserves the signs of the second derivatives and simply adds them: $\nabla^2f=5+(-2)+(-3)=0$ . The common error is changing subtraction signs incorrectly. Negative second derivatives are not subtracted again; they are included algebraically with their existing negative signs.

Q8. Two surfaces have the same value $f=10$ at a point. Surface A has $\nabla^2f=8$ , while Surface B has $\nabla^2f=-8$ . Which comparison is most defensible?

A.Both surfaces must have identical local shapes because their function values match

B.Surface A has greater positive net second-order curvature than Surface B ✅

C.Surface B must have a larger gradient magnitude

D.The Laplacian values cannot be compared because both function values are equal

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The value of a function at a point does not determine its local curvature. A positive Laplacian indicates positive net second-order curvature, while a negative Laplacian indicates negative net second-order curvature. Therefore, Surface A and Surface B can have the same height but substantially different local curvature behavior.

Q9. A contour map shows a scalar field whose contours are nearly straight and equally spaced in one region. A student concludes that the Laplacian must be large because the field changes rapidly. What is the best assessment?

A.Correct, because any large gradient creates a large Laplacian

B.Correct, because contour spacing directly measures second derivatives

C.Incorrect, because rapid first-order change does not necessarily imply large second-order curvature ✅

D.Incorrect, because the Laplacian is always zero on contour maps

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Contour spacing primarily gives information about the magnitude of the gradient, not directly about second derivatives. If contours are straight and evenly spaced, the field can vary rapidly while maintaining nearly constant slope. In such a case, second-order derivatives may be small or zero even though the gradient is substantial.

Q10. A graph of $f(x,y)$ shows a smooth bowl-shaped surface near a point, with the surface curving upward in both coordinate directions. Which sign of the Laplacian would you expect locally?

A.Negative

B.Positive ✅

C.Exactly zero

D.Impossible to determine from curvature

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: If the surface curves upward in both independent coordinate directions, the corresponding second partial derivatives are positive. Their sum, the Laplacian, is therefore positive. The conclusion does not require knowing the exact numerical values. This distinguishes the Laplacian from the gradient, which describes first-order directional change rather than curvature.

Q11. Suppose $f(x,y)=g(x)+h(y)$ , where $g''(x)=4$ and $h''(y)=-7$ throughout a region. Without finding $g$ or $h$ , what can be concluded?

\nabla^2f=-3

throughout the region ✅

\nabla^2f=11

throughout the region

\nabla^2f=3

only where

x=y

D.The Laplacian cannot be determined without knowing

g

and

h

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Because $f(x,y)=g(x)+h(y)$ , its second derivatives separate naturally: $f_{xx}=g''(x)=4$ and $f_{yy}=h''(y)=-7$ . Therefore, $\nabla^2f=4-7=-3$ . No explicit formulas for $g$ and $h$ are required because their second derivatives are already provided.

Q12. A model uses $f(x,y)=x^2+ay^2$ . Engineers want the field to satisfy $\nabla^2f=0$ everywhere. What value of $a$ should they choose?

a=-1

✅

a=0

a=1

a=2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The second derivatives are $f_{xx}=2$ and $f_{yy}=2a$ . Thus, $\nabla^2f=2+2a$ . Requiring this to vanish gives $2+2a=0$ , so $a=-1$ . The result illustrates how positive curvature in one coordinate direction can exactly balance negative curvature in another.

Q13. A researcher compares $f=x^2+y^2$ with $g=2x^2+2y^2$ . At corresponding points, which statement correctly compares their Laplacians?

A.They are equal because both are quadratic

B.The Laplacian of

g

is twice the Laplacian of

f

✅

C.The Laplacian of

g

is four times the Laplacian of

f

D.Both Laplacians are zero because the fields are symmetric

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $f=x^2+y^2$ , the second derivatives are $2$ and $2$ , giving $\nabla^2f=4$ . For $g=2x^2+2y^2$ , the second derivatives are $4$ and $4$ , giving $\nabla^2g=8$ . Thus, scaling the entire field by $2$ scales its Laplacian by $2$ as well.

Q14. Let $f(x,y)=x^4-6x^2y^2+y^4$ . A student claims that $\nabla^2f=12x^2-12y^2$ because they differentiate each term twice independently and ignore mixed-variable effects. What is the correct conclusion?

A.The student's result is correct

B.The correct Laplacian is

12x^2+12y^2

C.The correct Laplacian is

12x^2-12y^2

, so the student's reasoning is valid

D.The Laplacian is

0

because the positive and negative terms cancel ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Compute $f_{xx}=12x^2-12y^2$ and $f_{yy}=-12x^2+12y^2$ . Adding them gives $\nabla^2f=0$ . The student's displayed expression is actually just $f_{xx}$ , not the full Laplacian. This example emphasizes that the Laplacian requires summing the second derivatives with respect to every coordinate.

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