Differential Operators Solved Problems
Example.1. If f and g are two scaler point functions, prove that
- div ( f∇g) =f ∇2g+∇f .∇g
- div (f ∇g) =div (g∇f)=f ∇2-g ∇2f
Solution:
⇒ \(\nabla g=\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z}\)
f \(\nabla g=\mathbf{i} f \frac{\partial g}{\partial x}+\mathbf{j} f \frac{\partial g}{\partial y}+\mathbf{k} f \frac{\partial g}{\partial z}\)
∴ \(\nabla:(f \nabla g)=\frac{\partial}{\partial x}\left(f \frac{\partial g}{\partial x}\right)+\frac{\partial}{\partial y}\left(f \frac{\partial g}{\partial y}\right)+\frac{\partial}{\partial z}\left(f \frac{\partial g}{\partial z}\right)\)
= \(f\left(\frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}+\frac{\partial^2 g}{\partial z^2}\right)+\frac{\partial f}{\partial x} \frac{\partial g}{\partial x}+\frac{\partial f}{\partial y} \frac{\partial g}{\partial y}+\frac{\partial f}{\partial z} \frac{\partial g}{\partial z}\)
= \(f \nabla^2 g+\left(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\right) \cdot\left(\mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z}\right)\)
∴ \(\text{div}(f \nabla g)=f \nabla^2 g+\nabla f . \nabla g\)
Another Method: Let ∇g =A. By Theorem 1
⇒ \(\nabla \cdot(f \mathbf{A})=(\nabla f) \cdot \mathbf{A}+f(\nabla \cdot \mathbf{A})\)
= \((\nabla f) \cdot(\nabla g)+f[\nabla \cdot(\nabla g)\)
= \((\nabla f) \cdot(\nabla g)+f\left(\nabla^2 g\right)\)
We have proved that \(\text{div}(f \nabla g)=f \nabla^2 g+\nabla f . \nabla g\)…..(1)
Interchanging f and \(g \text{div}(g \nabla f)=g \nabla^2 f+\nabla g. \nabla f\)….(2)
(1) – (2) : \(\text{div}(f \nabla g) \text{div}(g \nabla f)=f \nabla^2 g-g \nabla^2 f\)
Example.2. Prove that div{(r×a) ×b)=-2(a.b) where a and b are constant vectors.
Solution: div{(r×a)× b}=div [(r.b)a-a(a.b)r=div(r.b)a-div(a.b)r By theorem 1
= \((\mathbf{r} \cdot \mathbf{b}) \text{div} \mathbf{a}+\mathbf{a} \cdot \text{grad}(\mathbf{r} \cdot \mathbf{b})-\{(\mathbf{a} \cdot \mathbf{b}) \text{div} \mathbf{r}+\mathbf{r} \cdot \text{grad}(\mathbf{a} \cdot \mathbf{b})\}\)
= \(0+\mathbf{a} \cdot \text{grad}(\mathbf{r} \cdot \mathbf{b})-(\mathbf{a} \cdot \mathbf{b}) 3+0=+\mathbf{a} \cdot \Sigma \mathbf{i} \frac{\partial}{\partial x}(\mathbf{r} \cdot \mathbf{b})-3(\mathbf{a} \cdot \mathbf{b})\)
= \(+\mathbf{a} \cdot \Sigma \mathbf{i}\left(\frac{\partial r}{\partial x} \cdot \mathbf{b}\right)-3(\mathbf{a} \cdot \mathbf{b})=+\mathbf{a} \Sigma \mathbf{i}(\mathbf{i} \cdot \mathbf{b})-3(\mathbf{a} \cdot \mathbf{b})\)
= \(\mathbf{a} \cdot \mathbf{b}-3(\mathbf{a} \cdot \mathbf{b})=-2(\mathbf{a} \cdot \mathbf{b})\)
Example.3. Prove that curl[(r×a)×b]=b×a where a and b are constant vectors.
Solution:
curl\([(\mathbf{r} \times \mathbf{a}) \times \mathbf{b}]=\text{curl}[(\mathbf{r} . \mathbf{b}) \mathbf{a}-(\mathbf{a} . \mathbf{b}) \mathbf{r}]\)
= \(\text{curl}(\mathbf{r} \cdot \mathbf{b}) \mathbf{a}-\text{curl}(\mathbf{a} \cdot \mathbf{b}) \mathbf{r}=(\mathbf{r} . \mathbf{b}) \text{curl} \mathbf{a}+\text{grad}(\mathbf{r} \cdot \mathbf{b}) \times \mathbf{a}-(\bar{a} \cdot \bar{b}) \text{curl} \bar{r}\)
= \(0+\nabla(r . \bar{b}) \times \bar{a}-0 \quad=\mathbf{b} \times \mathbf{a}\) (because \(\nabla(\bar{r} \cdot \bar{b})=\bar{b}\))
Example.4. Prove that curl (f grad Φ) =(grad f) × (grad Φ)
Solution: curl (f grad Φ)= ∇×(f ∇ Φ)=f curl (grad Φ) + (grad f) × (grad Φ) [by Theorem 2]
= grad f× grad Φ
∴ curl ∇ Φ=0
Example.5. Prove that div(∇Φ ×∇f) =0
Solution: div(A×B) = B . curl A-A. curl B
∴ div (∇Φ ×∇f)=∇f.curl (∇Φ)−∇Φ. curl (∇f) = 0
(since Curl grad Φ = 0 = Curl grad f)
Example.6. Prove that \(\nabla \cdot\left(\frac{\mathbf{r}}{r^3}\right)\)=0
Solution:
We know that \(\nabla \cdot(\phi \mathbf{a})=\nabla \phi \cdot \mathbf{a}+\phi(\nabla \cdot \mathbf{a})\)
Take \(\phi=\frac{1}{r^3}\). Then, \(\nabla \cdot\left(\frac{1}{r^3} \mathbf{r}\right)=\nabla\left(\frac{1}{r^3}\right) \cdot \mathbf{r}+\frac{1}{r^3}(\nabla \cdot \mathbf{r})=\left(-3 r^{-5}\right)(\mathbf{r} \cdot \mathbf{r})+\frac{1}{r^3}(3)\)
Using \(\nabla\left(r^n\right)=n r^{n-2} \mathbf{r} ; \nabla \cdot(\bar{r})=3\)
= \(\frac{-3}{r^5}\left(r^2\right)+\frac{3}{r^3}=\frac{-3}{r^3}+\frac{3}{r^3}=0\)
Example.7. If Φ(x,y,z) is a solution of the Laplace equation, Then ∇Φ is both solenoidal and irrotational.
Solution: Given∇2 (Φ) =0 ⇒∇ .(∇Φ) = 0.
∴ ∇Φ is solenoidal.
We know that ∇×(∇Φ) = 0 is always true. Thus ∇Φ is always irrational.
Example.8. Show that the vector (x2−yz)i + (y2− zx)j +(z2 − xy)k is irrotational and find its scalar potential.
Solution:
Let \(\mathbf{f}=\left(x^2-y z\right) \mathbf{i}+\left(y^2-z x\right) \mathbf{j}+\left(z^2-x y\right) \mathbf{k}\)
Then curl \(\mathbf{f}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2-y z & y^2-z x & z^2-x y\end{array}\right|=\sum i(-x+x)=0\)
∴ \(\mathbf{f}\) is irrotational. Then there exists \(\phi\) such that \(\mathbf{f}=\nabla \phi\).
⇒ \(\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}=\left(x^2-y z\right) \mathbf{i}+\left(y^2-z x\right) \mathbf{j}+\left(z^2-x y\right) \mathbf{k}\)
Comparing components, we get
⇒ \(\frac{\partial \phi}{\partial x}=x^2-y z \Rightarrow \phi=\int\left(x^2-y z\right) d x=\frac{x^3}{3}-x y z+f_1(y, z)\)……(1)
⇒ \(\frac{\partial \phi}{\partial y}=y^2-z x \Rightarrow \phi=\frac{y^3}{3}-x y z+f_2(z, x)\)….(2)
⇒ \(\frac{\partial \phi}{\partial z}=z^2-x y \Rightarrow \phi=\frac{z^3}{3}-x y z+f_3(x, y)\)……(3)
From (1), (2), (3), \(\phi=\frac{1}{3}\left(x^3+y^3+z^3\right)-x y z+\) constant, which is the required scalar potential.
Example.9. Find constants a,b,c, so that the vector A= (x+2y+az)i+(bx-3y-z)j+(4x+cy+2z)k= is irrotational. Also, find Φ such that A=∇Φ.
Solution:
Given vector is \(\mathbf{A}=(x+2 y+a z) \mathbf{i}+(b x-3 y-z) \mathbf{j}+(4 x+c y+2 z) \mathbf{k}\)
vector \(\mathbf{A}\) is irrotational
⇒ \(\text{curl} \mathbf{A}=\mathbf{0}\)
⇒ \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\partial / \partial x & \partial / \partial y & \partial / \partial z \\
x+2 y+a z & b x-3 y-z & 4 x+c y+2 z
\end{array}\right|\)= 0
⇒ \(\mathbf{i}(c+1)+\mathbf{j}(a-4)+\mathbf{k}(b-2)=0\)
⇒ \(\mathbf{i}(c+1)+\mathbf{j}(a-4)+\mathbf{k}(b-2)=0 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k}\)
Comparing both sides, c+1=0, a-4=0, b-2=0
∴ c=-1, a=4, b=2
Now \(\mathbf{A}=(x+2 y+4 z) \mathbf{i}+(2 x-3 y-z) \mathbf{j}+(4 x-y-2 z) \mathbf{k}\), on substituting the values of a, b, c
We have \(\mathbf{A}=\nabla \phi\)
(x+2 y+4 z) \(\mathbf{i}+(2 x-3 y-z) \mathbf{j}+(4 x-y+2 z) \mathbf{k}=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\)
Comparing both sides, we have
⇒ \(\frac{\partial \phi}{\partial x}=x+2 y+4 z \Rightarrow \phi=\frac{x^2}{2}+2 x y+4 z x+f_1(y, z)\)
⇒ \(\frac{\partial \phi}{\partial y}=2 x-3 y-z \Rightarrow \phi=2 x y-\frac{3 y^2}{2}-y z+f_2(z, x)\)
⇒ \(\frac{\partial \phi}{\partial z}=4 x-y+2 z \Rightarrow \phi=4 x z-y z+z^2+f_3(x, y)\)
Hence \(\phi=\frac{x^2}{2}-\frac{3 y^2}{2}+z^2+2 x y+4 z x-y z+c\) or \(\phi=\frac{x^2}{2}-\frac{3 y^2}{2}+z^2+2 x y-y z+4 z x+c\)