Differential Operators Exercise 5(b)

Differential Operators Exercise 5(b)

1. prove that

1. div r=3 
2. curl r=0 
3. div (r×a)=0
4. curl(r×a)=-2a
5. grad(r.a) =a

Solution:

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

div r = \(\frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y)+\frac{\partial}{\partial z}(z)=1+1+1=3\)

2. curl \(\bar{r}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & z\end{array}\right|=0\)

3. div \((\mathbf{r} \times \mathbf{a})=\sum \mathbf{i} \cdot \frac{\partial}{\partial x}(\mathbf{r} \times \mathbf{a})\)

⇒ \(\sum \mathbf{i} \cdot\left(\frac{\partial \mathbf{r}}{\partial x} \times \mathbf{a}\right)=\sum \mathbf{i} \cdot(\mathbf{i} \times \mathbf{a})=\sum[\mathbf{i} \mathbf{a}]=0\)

4. Let \(\mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k} ; \quad \mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

curl \((\mathbf{r} \times \mathbf{a})=\nabla \times(\mathbf{r} \times \mathbf{a})=\sum \mathbf{i} \times \frac{\partial}{\partial x}(\mathbf{r} \times \mathbf{a})\)

= \(\sum \mathbf{i} \times\left(\frac{\partial \mathbf{r}}{\partial x} \times \mathbf{a}\right)=\sum \mathbf{i} \times(\mathbf{i} \times \mathbf{a})=\sum[(\mathbf{i} \cdot \mathbf{a}) \mathbf{i}-(\mathbf{i} \cdot \mathbf{i}) \mathbf{a}]\)

= \(\sum\left[\left(a_1\right) \mathbf{i}-\mathbf{a}\right]=\left(a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}\right)-3 \mathbf{a}=\bar{a}-3 \bar{a}=-2 \mathbf{a}\)

Differential Operators Exercise Step-By-Step Solutions

2. If =x²yi-2xzj+2yzk,find

1. div f
2. curl f

Solution:

1. div f = \(\frac{\partial}{\partial x}\left(x^2 y\right)+\frac{\partial}{\partial y}(-2 x z)+\frac{\partial}{\partial z}(2 y z)=2 x y+2 y\)

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

curl(curl f) = \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 z+2 x & 0 & -2 z-x^2
\end{array}\right|=2 \mathrm{j}(1+x)\)

3.

1. Show that 3y⁴z²i+4x³z²j-3x²y²k is solenoidal.
2. if f= (x+3y)i+y-2z)j+(x+pz)k is a solenoidal find p?

Solution:

If f is a solenoidal then div f=0

∴ \(\frac{\partial}{\partial x}(x+3 y)+\frac{\partial}{\partial y}(y-2 z)+\frac{\partial}{\partial z}(x+p z)=0\)

1+1+p=0

⇒ p=-2

4. Prove that f= (sin y+z)i+ (x cosy-z)j+9x-y0k is irrotational.

Solution: If f is irrotational then curl f=0

5. If Φ =2x³y²z find div (grad Φ).

Solution:

Given

If Φ =2x³y²z

Grad \(\phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}=6 x^2 y^2 z^4 \mathbf{i}+4 x^3 y z^4 \mathbf{j}+8 x^3 y^2 z^3 \mathbf{k}\)

∴ div\((\text{grad} \phi)=\frac{\partial}{\partial x}\left(6 x^2 y^2 z^4\right)+\frac{\partial}{\partial y}\left(4 x^3 y z^4\right)+\frac{\partial}{\partial z}\left(8 x^3 y^2 z^3\right)\)

= \(12 x y^2 z^4+4 x^3 z^4+24 x^3 y^2 z^2\)

6. Find curl f if e ^x+y+z (i+j+k).

Solution:

f = \(e^{x+y+z}(\mathbf{i}+\mathbf{j}+\mathbf{k})\)

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

= \(\mathrm{i}\left(e^{x+y+z}-e^{x+y+z}\right)-e t c=0\)

Examples Of Differential Operator Problems

7. If f=(x+y+1) i+j-(x+y)k prove that f. curl f=0

Solution:

Given

f=(x+y+1) i+j-(x+y)k

curl \(\mathbf{f}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x+y+1 & -1 & -(x+y)
\end{array}\right|\)

= \(\mathbf{i}(0-1)-\mathbf{j}(-1-0)+\mathbf{k}(0-1)=-\mathbf{i}+\mathbf{j}-\mathbf{k}\)

∴ \(f . \text{curl} \mathbf{f}=-(x+y+1)+1+x+y=0\)

8. If A=2yzi-x²yj+xz²k and Φ=2x²yz³ find  (A×∇)Φ.

Solution:

Given

A=2yzi-x²yj+xz²k and Φ=2x²yz³

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

= \(\mathbf{A} \times\left(4 x y z^3 \mathbf{i}+2 x^2 z^3 \mathbf{j}+6 x^2 y z^2 \mathbf{k}\right)\)

= \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
.2 y z & -x^2 y & x z^2 \\
4 x y z^3 & 2 x^2 z^3 & 6 x^2 y z^2
\end{array}\right|\)

Simplify to get the answer.

9. If Φ=x²-y² show that ∇²Φ=0

Solution:

Given

Φ=x²-y²

⇒ \(\nabla^2 \phi=\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}+\frac{\partial^2 \phi}{\partial z^2}=\frac{\partial}{\partial x}(2 x)+\frac{\partial}{\partial y}(-2 y)+0=2-2=0\)

10. Show that ∇²(x/r³)=0

Solution:

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

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

= \(\frac{\partial}{\partial x}\left[\frac{1}{r^3}-\frac{3 x}{r^4}\left(\frac{x}{r}\right)\right] \cdot\)

(because \(r^2=x^2+y^2+z^2\) gives \(\frac{\partial r}{\partial x}=\frac{x}{r}\))

= \(\frac{\partial}{\partial x}\left[\frac{1}{r^3}-\frac{3 x^2}{r^5}\right]=-\frac{3}{r^4} \frac{\partial r}{\partial x}-\frac{6 x}{r^5}+\frac{15 x^2}{r^6} \cdot \frac{\partial r}{\partial x}\)

= \(-\frac{3}{r^4}\left(\frac{x}{r}\right)-\frac{6 x}{r^5}+\frac{15 x^2}{r^6}\left(\frac{x}{r}\right)=-\frac{9 x}{r^5}+\frac{15 x^3}{r^7}\)

Again \(\frac{\partial^2}{\partial y^2}\left(\frac{x}{r^3}\right)=\frac{\partial}{\partial y}\left[\frac{\partial}{\partial y}\left(\frac{x}{r^3}\right)\right]=\frac{\partial}{\partial y}\left[-\frac{3 x}{r^4} \cdot \frac{\partial r}{\partial y}\right]\)

= \(\frac{\partial}{\partial y}\left[-\frac{3 x}{r^4}\left(\frac{y}{r}\right)\right]\)

= \(\frac{\partial}{\partial y}\left[-\frac{3 x y}{r^5}\right]=-\frac{3 x}{r^5}+\frac{15 x y}{r^6} \cdot \frac{\partial r}{\partial y}=-\frac{3 x}{r^5}+\frac{15 x y^2}{r^7}\)

Similarly \(\frac{\partial^2}{\partial z^2}\left(\frac{x}{r^3}\right)=-\frac{3 x}{r^5}+\frac{15 x z^2}{r^7}\)

∴ \(\nabla^2\left(\frac{x}{r^3}\right)=\left(\frac{\partial^2}{\partial x}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)\left(\frac{x}{r^3}\right)\)

= \(-\frac{9 x}{r^5}+\frac{15 x^3}{r^7}-\frac{3 x}{r^5}+\frac{15 x y^2}{r^7}-\frac{3 x}{r^5}+\frac{15 x z^2}{r^7}\)

= \(-\frac{15 x}{r^5}+\frac{15 x}{r^7}\left(x^2+y^2+z^2\right)=-\frac{15 x}{r^5}+\frac{15 x}{r^7}\left(r^2\right)=0\)

Understanding Exercise 5(B) In Differential Operators

11. Show that ∇²(log r)=1/r².

Solution:

⇒ \(\nabla^2(\log r)=\sum \frac{\partial^2}{\partial x^2}(\log r)=\sum \frac{\partial}{\partial x}\left(\frac{1}{r} \frac{\partial r}{\partial x}\right)=\sum \frac{\partial}{\partial x}\left(\frac{1}{r} \cdot \frac{x}{r}\right)\)

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

= \(\sum\left(\frac{1}{r^2}-\frac{2 x^2}{r^4}\right)\)

= \(\frac{3}{r^2}-\frac{2}{r^4}\left(x^2+y^2+z^2\right)=\frac{3}{r^2}-\frac{2}{r^4}\left(r^2\right)=\frac{1}{r^2}\)