Differential Operators Solved Problems The Laplacian Operator And Curl Of A Vector

Differential Operators Solved Problems

Example. 1. If f = xy²i + 2x²yz j- 3yz²k find

1. div f,
2. curl f at the point (1, — 1,1)

Solution:

Given

f = xy²i + 2x²yz j- 3yz²k

1. \(\text{div} \mathbf{f}=\nabla \cdot \mathbf{f}=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}=\frac{\partial}{\partial x}\left(x y^2\right)+\frac{\partial}{\partial y}\left(2 x^2 y z\right)+\frac{\partial}{\partial z}\left(-3 y z^2\right)\)

= \(y^2+2 x^2 z-6 y z \quad \text { at }(1,-1,1) \text{div} \mathbf{f}=1+2+6=9\)

2. \(\text{curl} \mathbf{f}=\nabla \times \mathbf{f}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x y^2 & 2 x^2 y z & -3 y z^2
\end{array}\right|\)

= \(\mathbf{i}\left[\frac{\partial}{\partial y}\left(-3 y z^2\right)-\frac{\partial}{\partial z}\left(2 x^2 y z\right)\right]-\mathbf{j}\left[\frac{\partial}{\partial x}\left(-3 y z^2\right)-\frac{\partial}{\partial z}\left(x y^2\right)\right]\)

+ \(\mathbf{k}\left[\frac{\partial}{\partial x}\left(2 x^2 y z\right)-\frac{\partial}{\partial y}\left(x y^2\right)\right]-3 z^2-2 x^2 y) \mathbf{i}+(4 x y z-2 x y) \mathbf{k}\)

at (1,-1,1), \(\text{curl} \mathbf{f}=-\mathbf{i}-2 \mathbf{k}\)

Examples Of Laplacian And Curl In Vector Calculus

Example. 2. Find div f and curl f where f = grad(x³ + y³ + z³ – 3xyz)

Solution:

Given

f = grad(x³ + y³ + z³ – 3xyz)

f = \(\text{grad}\left(x^3+y^3+z^3-3 x y z\right)=\sum \mathbf{i} \frac{\partial}{\partial x}\left(x^3+y^3+z^3-3 x y z\right)\)

= \(\Sigma \mathbf{i}\left(3 x^2-3 y z\right)=\mathbf{i}\left(3 x^2-3 y z\right)+\mathbf{j}\left(3 y^2-3 z x\right)+\mathbf{k}\left(3 z^2-3 x y\right)\)

div \(\mathbf{f}=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}=6 x+6 y+6 z=6(x+y+z)\)

curl \(\mathbf{f}=\nabla \times \mathbf{f}=\left|\begin{array}{ccc}
\mathbf{1} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
3 x^2-3 y z & 3 y^2-3 z x & 3 z^2-3 x y
\end{array}\right|\)

= \(\mathbf{i}(-3 x+3 x)-\mathbf{j}(-3 y+3 y)+\mathbf{k}(-3 z+3 z)=\mathbf{0}\)

Example.3. Prove that grade(r.a)=a where a is a constant vector

Solution: Let a= a₁i + a₂j + a₃k . Then i.a = a₁; j.a=a₂ ;k .a = a₃

Grad \((\mathbf{r} \cdot \mathbf{a})=\nabla(\overline{\mathbf{r}} \cdot \overline{\mathbf{a}})=\Sigma \mathbf{i} \frac{\partial}{\partial x}(\mathbf{r} \cdot \mathbf{a})\)

= \(\Sigma \mathbf{i}\left(\frac{\partial \mathbf{r}}{\partial x} \cdot \mathbf{a}\right)=\Sigma \mathbf{i}(\mathbf{i} \cdot \mathbf{a})=\mathbf{i}(\mathbf{i} \cdot \mathbf{a})+\mathbf{j}(\mathbf{j} \cdot \mathbf{a})+\mathbf{k}(\mathbf{k} \cdot \mathbf{a})\)

Grad \((\mathbf{r} \cdot \mathbf{a})=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}=\mathbf{a}\)

Example.4. If a is a constant vector, prove that curl \(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\)=\(-\frac{\mathbf{a}}{r^3}+\frac{3 \mathbf{r}}{r^5}(\mathbf{a} \cdot \mathbf{r})\)

Solution:

Given

A is a constant vector

Now \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

⇒ \(\frac{\partial \mathbf{r}}{\partial x}=\mathbf{i}, \frac{\partial \mathbf{r}}{\partial y}=\mathbf{j}, \frac{\partial \mathbf{r}}{\partial z}=\mathbf{k} \left|\mathbf{r}^2\right|=r^2=x^2+y^2+z^2\)

⇒ \(\frac{\partial r}{\partial x}=\frac{x}{r}, \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r} \text{curl} \frac{\mathbf{a} \times \mathbf{r}}{r^3}=\Sigma \mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)\)

Now \(\frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=\mathbf{a} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{r}}{r^3}\right)=\mathbf{a} \times\left[\frac{1}{r^3} \frac{\partial \mathbf{r}}{\partial x}-\frac{3}{r^4} \frac{\partial r}{\partial x} \mathbf{r}\right]\)

= \(\mathbf{a} \times\left[\frac{1}{r^3} \mathbf{i}-\frac{3 x}{r^5} \mathbf{r}\right]\)

= \(\frac{\mathbf{a} \times \mathbf{i}}{r^3}-\frac{3 x(\mathbf{a} \times \mathbf{r})}{r^5}\)

∴ \(\mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=\mathbf{i} \times\left[\frac{\mathbf{a} \times \mathbf{i}}{r^3}-\frac{3 x}{r^5}(\mathbf{a} \times \mathbf{r})\right]=\frac{\mathbf{i} \times(\mathbf{a} \times \mathbf{i})}{r^3}-\frac{3 x}{r^5} \mathbf{i} \times(\mathbf{a} \times \mathbf{r})\)

= \(\frac{(\mathbf{i} . \mathbf{i}) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \mathbf{i}}{r^3}-\frac{3 x}{r^5}[(\mathbf{i} . \mathbf{r}) \mathbf{a}-(\mathbf{i} . \mathbf{a}) \mathbf{r}]\)

let \(\mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a \mathbf{k}\)

⇒ \(\mathbf{i} \cdot \mathbf{a}=a_1 ; \mathbf{j} \cdot \mathbf{a}=a_2 ; \mathbf{k} \cdot \mathbf{a}=a_3\)

∴ \(i \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=\frac{\left(\mathbf{a}-a_1 \mathbf{i}\right)}{r^3}-\frac{3 x}{r^5}\left(x \mathbf{a}-a_1 \mathbf{r}\right)\)

∴ \(\sum \mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=\sum\left(\frac{\mathbf{a}-a_1 \mathbf{i}}{r^3}\right)-\frac{3}{r^5} \sum\left(x^2 \mathbf{a}-a_1 x \mathbf{r}\right)\)

= \(\frac{3 \mathbf{a}-\mathbf{a}}{r^3}-\frac{3 \mathbf{a}}{r^5}\left(x^2+y^2+z^2\right)+\frac{3 \mathbf{r}}{r^5}\left(a_1 x+a_2 y+a_3 z\right)\)

= \(\frac{2 \mathbf{a}}{r^3}-\frac{3 \mathbf{a} r^2}{r^5}+\frac{3 \mathbf{r}}{r^3}(\mathbf{r} \cdot \mathbf{a})=-\frac{\mathbf{a}}{r^3}+\frac{3 \mathbf{r}}{r^5}(\mathbf{r} \cdot \mathbf{a})\)

Theorems Involving Laplacian Operator And Curl With Examples

Example.5. Prove thatt ∇ × f(r)r=0

Solution:

⇒ \(\nabla \times f(r) \mathbf{r}=\nabla \times\{f(r) x \mathbf{i}+f(r) y \mathbf{j}+f(r) z \mathbf{k}\}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
f(r) x & f(r) y & f(r) z
\end{array}\right|\)

= \(\sum \mathbf{i}\left\{\frac{\partial}{\partial y} f(r) z-\frac{\partial}{\partial z} f(r) y\right\}=\sum \mathbf{i}\left\{f^{\prime}(r) \frac{\partial r}{\partial y} z-f^{\prime}(r) \frac{\partial r}{\partial z} y\right\}\)

= \(\sum \mathbf{i} \frac{f^{\prime}(r)}{r}\{y z-z y\}=0\)

Example.6. Prove that (f × ∇).r =0

Solution:

∴ \((\mathbf{f} \times \nabla) \cdot \mathbf{r}=\left\{\mathbf{f} \times \Sigma \mathbf{i} \frac{\partial}{\partial x}\right\} \cdot \mathbf{r}=\mathbf{f} \times\left\{\Sigma \mathbf{i} \cdot \frac{\partial \mathbf{r}}{\partial x}\right\}=\left\{\Sigma(\mathbf{f} \times \mathbf{i}) \cdot \frac{\partial \mathbf{r}}{\partial x}\right\}=\Sigma(\mathbf{f} \times \mathbf{i}) \cdot \mathbf{i}=0\)

Example.7. Prove that (f ×∇)×r=-2f

Solution:

⇒ \((\mathbf{f} \times \nabla)=(\mathbf{f} \times \mathbf{i}) \frac{\partial}{\partial x}+(\mathbf{f} \times \mathbf{j}) \frac{\partial}{\partial y}+(\mathbf{f} \times \mathbf{k}) \frac{\partial}{\partial z}\)

⇒ \((\mathbf{f} \times \nabla) \times \mathbf{r}=(\mathbf{f} \times \mathbf{i}) \times \frac{\partial \mathbf{r}}{\partial x}+(\mathbf{f} \times \mathbf{j}) \times \frac{\partial \mathbf{r}}{\partial y}+(\mathbf{f} \times \mathbf{k}) \frac{\partial \mathbf{r}}{\partial z}\)

= \(\Sigma(\mathbf{f} \times \mathbf{i}) \times \mathbf{i}=\Sigma\{(\mathbf{f} . \mathbf{i}) \mathbf{i}-\mathbf{f}\}=(\mathbf{f} . \mathbf{i}) \mathbf{i}+(\mathbf{f} . \mathbf{j}) \mathbf{j}+(\mathbf{f} . \mathbf{k}) \mathbf{k}-3 \mathbf{f}\)

= \(\mathbf{f}-3 \mathbf{f}=-2 \mathbf{f}\)

Example.8. Show that \(\nabla^2\left(\frac{1}{r}\right)\)=0

Solution:

r = \(|\mathbf{r}|=\sqrt{x^2+\dot{y}^2+z^2} . \quad 2 r \frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}, \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r} \text {. }\)

⇒ \(\nabla^2\left(\frac{1}{r}\right)=\sum \frac{\partial^2}{\partial x^2}\left(\frac{1}{r}\right)=\sum \frac{\partial}{\partial x}\left[\left(\frac{-1}{r^2}\right)\left(\frac{\partial r}{\partial x}\right)\right]=\sum \frac{\partial}{\partial x}\left[\left(\frac{-1}{r^2}\right)\left(\frac{x}{r}\right)\right]=\sum \frac{\partial}{\partial x}\left[\frac{-x}{r^3}\right]\)

= \(\sum \frac{r^3(-1)+x\left(3 r^2\right) \frac{\partial r}{\partial x}}{r^6}=\sum \frac{-r^3+x\left(3 r^2\right) \frac{x}{r}}{r^6}=\sum \frac{-r^3+3 x^2 r}{r^6}\)

= \(\sum\left(\frac{-1}{r^3}+\frac{3 x^2}{r^5}\right)=\frac{-3}{r^3}+\frac{3 r^2}{r^5}=\frac{-3}{r^3}+\frac{3}{r^3}=0\)

Note: \(\frac{1}{r}\) satisfies the Laplace equation \(\nabla^2 f=0\) Thus \(\frac{1}{r}\) is a harmonic function.

Laplacian And Curl Operator Detailed Examples With Solutions

Example.9. Find div F and Curl F where F= x²zi- 2y³z³j + xy²zk at (1,—1,1)

Solution:

Given \(\mathbf{F}=x^2 z \mathbf{i}-2 y^3 z^3 \mathbf{j}+x y^2 z \mathbf{k}\)

div \(\mathbf{F}=\frac{\partial}{\partial x}\left(x^2 z\right)+\frac{\partial}{\partial y}\left(-2 y^3 z^3\right)+\frac{\partial}{\partial z}\left(x y^2 z\right)=2 x z-6 y^2 z^3+x y^2\)

div \(\mathbf{F}\) at (1,-1,1) = \(2(1)(1)-6(-1)^2(1)^3+1(-1)^2=2-6+1=-3\)

Curl \(\mathbf{F}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 z & -2 y^3 z^3 & x y^2 z\end{array}\right|=\mathbf{i}\left(2 x y z+6 y^3 z^2\right)-\mathbf{j}\left(y^2 z-x^2\right)+\mathbf{k}(0-0)\)

∴ Curl \(\mathbf{F}\) at (1,-1,1) = \(\mathbf{i}[2(-1)(1))]-\mathbf{j}(1-1)=-8 \mathbf{i}\)

Example.10. Show that \(\nabla^2[f(r)]\)=\(f^{\prime \prime}(r)+\frac{2}{r} f^{\prime}(r)=\frac{d^2 f}{d r^2}+\frac{2}{r} \frac{d f}{d r}\)

Solution:

⇒ \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} ; r=|\mathbf{r}|=\sqrt{x^2+y^2+z^2}\)

⇒ \(r^2=x^2+ y^2+z^2 ; 2 r \frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}\)

Similarly, \(\frac{\partial r}{\partial y}=\frac{y}{r} ; \frac{\partial r}{\partial z}=\frac{z}{r}\)

⇒ \(\nabla^2[f(r)]=\sum \frac{\partial^2}{\partial x^2}[f(r)]=\sum \frac{\partial}{\partial x}\left[f^{\prime}(r) \frac{\partial r}{\partial x}\right]=\sum \frac{\partial}{\partial x}\left[f^{\prime}(r) \frac{x}{r}\right]\)

= \(\Sigma \frac{r \frac{\partial}{\partial x}\left(f^{\prime}(r) x\right)-x f^{\prime}(r) \frac{\partial r}{\partial x}}{r^2}=\Sigma \frac{r\left(f^{\prime \prime}(r) \frac{\partial r}{\partial x} x+f^{\prime}(r)\right)-x f^{\prime}(r) \frac{x}{r}}{r^2}\)

= \(\Sigma \frac{r\left(f^{\prime \prime}(r) \frac{x}{r} x+f^{\prime}(r)\right)-\frac{x^2 f^{\prime}(r)}{r}}{r^2}=\Sigma \frac{f^{\prime \prime}(r) x^2+r f^{\prime}(r)-\frac{x^2 f^{\prime}(r)}{r}}{r^2}\)

= \(\frac{f^{\prime \prime}(r)}{r^2} \Sigma x^2+\frac{f^{\prime}(r)}{r} \Sigma 1-\frac{f^{\prime}(r)}{r^3} \Sigma x^2=\frac{f^{\prime \prime}(r)}{r^2} r^2+\frac{3 f^{\prime}(r)}{r}-\frac{f^{\prime}(r)}{r^3} r^2\)

= \(f^{\prime \prime \prime}(r)+2 \frac{f^{\prime}(r)}{r}\)

Example.11.(1) Find div F and curl F where F= rⁿ r̄.
(2) Show that rⁿ r is irrotational. Find for which value of n it is solenoidal.

Solution:

We have \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) and \(r^2=x^2+y^2+z^2\) where \(r=|\mathbf{r}|\)

D.V. r to x partially, \(2 r \frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}\).

Similarly \(\frac{\partial r}{\partial y}=\frac{y}{r}\) and \(\frac{\partial r}{\partial z}=\frac{z}{r}\)

div \(\mathbf{F}=\text{div}\left(r^n \mathbf{r}\right)=\sum i \cdot \frac{\partial}{\partial x}\left(r^n \mathbf{r}\right)\)

= \(\sum i \cdot\left[r^n \frac{\partial \mathbf{r}}{\partial x}+n r^{n-1} \frac{\partial r}{\partial x} \mathbf{r}\right]=\sum i \cdot\left[r^n(i)+n \cdot r^{n-1} \frac{x}{r} \mathbf{r}\right]\)

= \(\sum\left[r^n+n r^{n-2} x^2\right]\)

(because \(i \cdot \mathbf{i}=1\) and \( i \cdot \mathbf{r}=x)\)

= \(3 r^n+n r^n=(n+3) r^n\)

If \(\mathbf{F}\) is solenoidal, we have div f{F}=0

(n+3) \(r^n=0 \Rightarrow n=-3\)

Put n=-3, we get \(\mathbf{F}=r^n \mathbf{r}=r^{-3} \mathbf{r}=\frac{\mathbf{r}}{r^3}\) is solenoidal.

2. Curl \(\left(r^n \mathbf{r}\right)=\sum \mathbf{i} \times \frac{\partial}{\partial x}\left(r^2 \mathbf{r}\right)=\sum \mathbf{i} \times\left(\frac{r^n \partial \mathbf{r}}{\partial x}+\mathbf{r} n r^{n-1} \frac{\partial r}{\partial x}\right)\)

= \(r^n \sum \mathbf{i} \times \mathbf{i}+n r^{n-1} \sum \mathbf{i} \times \mathbf{r}\left(\frac{x}{r}\right)=r^n(\mathbf{0})+n r^{n-2} \sum \mathbf{i} \times(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) x\)

= \(n r^{n-2} \sum x y \mathbf{k}-x z \mathbf{i}=\mathbf{0}+\mathbf{0}=\mathbf{0}\)

∴\( r^n \mathbf{r}\) is irrotational.

Examples Of Laplacian And Curl In Vector Calculus

Example.12. Find l,m,n so that the vector F=(2x+3y+lz)i+(mx+2y+3z)j=(2x+xy+3z)k is irrotational

Solution:

We have Curl \(\mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x+3 y+l z & m x+2 y+3 z & 2 x+n y+3 z
\end{array}\right|\)

= \(\mathbf{i}[n-3]-\mathbf{j}[2-l]+\mathbf{k}[m-3]\)

Since f{F} is irrotational curl \(\mathbf{F}=\mathbf{O} \Rightarrow n=3, l=2, m=3\)

Example.13. If F= xy²i +2x²yzj-34z²k find div F at (1,—1,1)

Solution:

Given \(\mathbf{F}=x y^2 \mathbf{i}+2 x^2 y z \mathbf{j}-3 y z \mathbf{k}=\mathrm{F}_1 \mathbf{i}+\mathrm{F}_2 \mathbf{j}+\mathrm{F}_3 \mathbf{k}\)

div \(\mathbf{F}=\frac{\partial \mathrm{F}_1}{\partial x}+\frac{\partial \mathrm{F}_2}{\partial y}+\frac{\partial \mathrm{F}_3}{\partial z}=\left(y^2+2 x^2 z-6 y z\right)\)

(div F) at (1,-1,1) = \((-1)^2+2(1)^2(1)-6(-1)(1)=1+2+6=9\)

Example.14. Prove that F = y³z²i-3x²z⁵j-15x⁵y⁴k is solenoidal 

Solution:

Let \(\mathbf{F}=\mathrm{F}_1 \mathbf{i}+\mathrm{F}_2 \mathbf{j}+\mathrm{F}_3 \mathbf{k}\)

where \(\mathrm{F}_1=y^3 z^2, \mathrm{~F}_2=-3 x^2 z^2, \mathrm{~F}_3=-15 x^5 y^4\)

div \(\mathbf{F}=\frac{\partial \mathrm{F}_1}{\partial x}+\frac{\partial \mathrm{F}_2}{\partial y}+\frac{\partial \mathrm{F}_3}{\partial z}=0\)

∴ f{F} is a solenoidal vector.

Example.15. Show that ∇²rⁿ=n(n+1)r^n-2

Solution:

We have r=xi+yj+zk= and r\(=|\mathbf{r}|\)=\(\sqrt{x^2+y^2+z^2}\)

⇒ \(r^2=x^2+y^2+z^2\)

D. W. r. to x partially

2r \(\frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r} \text {. Similarly } \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r}\)

We have \(\nabla^2 r^n=+\sum \frac{\partial^2}{\partial x^2}\left(r^n\right)\)

= \(\sum \frac{\partial}{\partial x}\left[n r^{n-1} \frac{\partial r}{\partial x}\right]=\sum \frac{\partial}{\partial x} n r^{n-1}\left(\frac{x}{r}\right)=n \sum \frac{\partial}{\partial x}\left(x r^{n-2}\right)\)

= \(n\left[\sum\left\{x(n-2) r^{n-3} \frac{\partial r}{\partial x}+r^{n-2}(1)\right\}\right]\)

= \(n\left[\sum\left\{(n-2) x r^{n-3}\left(\frac{x}{r}\right)+r^{n-2}\right\}\right]\)

= \(n\left[\sum\left\{(n-2) r^{n-4} x^2+r^{n-2}\right\}\right]=n\left[(n-2) r^{n-4} r^2+3 r^{n-2}\right]\)

= \(n\left[(n-2+3) r^{n-2}\right]=n(n+1) r^{n-2}\)

Hence the result.

Note : When n=-1, \(\nabla^2\left(r^n\right)=(-1)(-1+1) r^{-1-2}=0\) .

∴ \(\nabla^2\left(\frac{1}{r}\right)=0\)