Introduction to Partial Differential Equations (PDEs)
Partial Differential Equations (PDEs) are equations that involve multiple independent variables, an unknown function, and partial derivatives of that function. They are used to formulate problems involving functions of several variables and are fundamental in describing various physical phenomena such as heat conduction, wave propagation, and fluid dynamics.
Key Concepts
Formation of PDEs: A PDE can be formed by eliminating arbitrary constants or functions from an equation involving multiple variables. For example, from \(F(x, y, z, a, b) = 0\), eliminate \(a\) and \(b\) to obtain a PDE in \(x, y, z\).
Order of a PDE: The order is determined by the highest order of the partial derivative present in the equation. For instance, \(u_{xx} + u_{yy} = 0\) is a second-order PDE because the highest derivative is second order.
Linear vs Nonlinear PDEs: Linear PDEs involve linear terms of the unknown function and its derivatives, while nonlinear PDEs involve nonlinear terms. For example, \(u_t = \alpha u_{xx}\) is a linear PDE, and \(u_t + u u_x = \nu u_{xx}\) is a nonlinear PDE.
Homogeneous vs Inhomogeneous PDEs: Homogeneous PDEs have no terms independent of the unknown function and its derivatives, whereas inhomogeneous PDEs have additional terms. For instance, \(u_{xx} + u_{yy} = 0\) is homogeneous, while \(u_{xx} + u_{yy} = f(x, y)\) is inhomogeneous.
Formation of PDEs
There are three approaches to form PDE from a static equation.
- Elimination of arbitrary constants
- Elimination of arbitrary functions
- Elimination of arbitrary function \(\phi\) from the relation \(\phi(u,v)=0\), where \(u\) and \(v\) are functions of \(x\) ,\(y\) and \(z\).
Method-1
- Step 1: Differentiate partially given static equation.
- Step 2: Eliminate the arbitrary constants using the expression for partial differentials.
Example
- Form a PDE from \(z = ax + by\).
Solution:
Step 1: Find the partial derivatives: \[ \frac{\partial z}{\partial x} = a, \quad \frac{\partial z}{\partial y} = b \]
Step 2: Eliminating the arbitrary constants:
The required PDE is: \[\begin{align*} z&=\left(\dfrac{\partial z}{\partial x}\right) x+\left(\dfrac{\partial z}{\partial y}\right) y\\ \therefore z&=px+qy \end{align*}\]
- Form a PDE from \(z=ax+by+ab\).
Solution:
- Step 1: Find the partial derivatives: \[ \frac{\partial z}{\partial x} = a, \quad \frac{\partial z}{\partial y} = b \]
- Step 2: Eliminating the arbitrary constant: the required PDE is: \[\begin{align*} z&=\left(\dfrac{\partial z}{\partial x}\right) x+\left(\dfrac{\partial z}{\partial y}\right) y+\left(\dfrac{\partial z}{\partial x}\right)\cdot \left(\dfrac{\partial z}{\partial y}\right)\\ \therefore z&=px+qy+pq \end{align*}\]
- Form a PDE from \(x^2+y^2+(z-c)^2=a^2\).
Solution:
- Step 1: Differentiating both sides with respect to \(x\) and \(y\): \[\begin{align*} 2x+2(z-c)\dfrac{\partial z}{\partial x}&=0\\ 2y+2(z-c)\dfrac{\partial z}{\partial y}&=0 \end{align*}\] Or \[\begin{align} x&=(z-c)\dfrac{\partial z}{\partial x}\\ y&=(z-c)\dfrac{\partial z}{\partial y} \end{align}\]{#eq-one}
Solving \(\eqref{eq-one}\) , we get \[qx=yp\].
- Form a PDE from the static equation, \(z=(x-a)^2+(y-b)^2\).
Solution:
Given the static equation:
\[z = (x - a)^2 + (y - b)^2\]
we can find the corresponding PDE by following these steps:
- Step 1: Find Partial Derivatives
\[\frac{\partial z}{\partial x} = 2(x - a)\]
Let \(\frac{\partial z}{\partial x} = p\):
\[p = 2(x - a) \quad \text{or} \quad x - a = \frac{p}{2}\]
\[\frac{\partial z}{\partial y} = 2(y - b)\]
Let\(\frac{\partial z}{\partial y} = q\):
\[q = 2(y - b) \quad \text{or} \quad y - b = \frac{q}{2}\]
- Step 2: Replace the Arbitrary Constants \(a\) and \(b\)
From the partial derivatives, express \(a\) and \(b\) as follows:
\[a = x - \frac{p}{2}\]
\[b = y - \frac{q}{2}\]
- Step 3: Simplify the Representation for \(z\)
Substitute \(a\) and \(b\)back into the original equation:
\[z = \left(x - \left(x - \frac{p}{2}\right)\right)^2 + \left(y - \left(y - \frac{q}{2}\right)\right)^2\]
On Simplification, we get:
\[z = \left(\frac{p}{2}\right)^2 + \left(\frac{q}{2}\right)^2\]
\[z = \frac{p^2}{4} + \frac{q^2}{4}\]
\[4z = p^2 + q^2\]
Advanced problems
- Find the partial differential equation representing the family of spheres whose centers lies on \(z-\) axis.
Solution:
Step 1. Equation of the Sphere
Consider a sphere with center \((0,0,c)\) and radius \(R\). The equation of the sphere is:
\[x^2 + y^2 + (z - c)^2 = R^2\]
Step 2. Differentiate with Respect to \(x\) and \(y\)
Differentiate with respect to \(x\):
\[ \frac{\partial}{\partial x}[x^2 + y^2 + (z - c)^2] = \frac{\partial}{\partial x}[R^2] \]
\[ 2x + 2(z - c) \frac{\partial z}{\partial x} = 0 \]
\[ x + (z - c)p = 0 \]
where \(p = \frac{\partial z}{\partial x}\).
Differentiate with respect to \(y\):
\[ \frac{\partial}{\partial y}[x^2 + y^2 + (z - c)^2] = \frac{\partial}{\partial y}[R^2] \]
\[ 2y + 2(z - c) \frac{\partial z}{\partial y} = 0 \]
\[ y + (z - c)q = 0 \]
where \(q = \frac{\partial z}{\partial y}\).
Step 3. Eliminate the Arbitrary Constant \(c\)
From the equations:
\[ x + (z - c)p = 0 \]
\[ y + (z - c)q = 0 \]
Solving for \(c\):
\[ z - c = -\frac{x}{p} \]
\[ z - c = -\frac{y}{q} \]
Equate the two expressions for \(z - c\):
\[ -\frac{x}{p} = -\frac{y}{q} \]
\[ \frac{x}{p} = \frac{y}{q} \]
Thus:
\[ xq = yp \]
Step 4. Final PDE
The partial differential equation representing the family of spheres whose centers lie on the \(z\)-axis is:
\[ xq - yp = 0 \]
where \(p = \frac{\partial z}{\partial x}\) and \(q = \frac{\partial z}{\partial y}\).
- Find the differential equation of all spheres of fixed radius having their centers in the xy-plane.
Solution:
Step 1: Equation of the Sphere
Consider a sphere with center \((a, b, 0)\) and radius \(R\). The equation of the sphere is:
\[ (x - a)^2 + (y - b)^2 + z^2 = R^2 \]
Step 2: Differentiate with Respect to \(x\) and \(y\)
Differentiate with respect to \(x\):
\[ \frac{\partial}{\partial x}[(x - a)^2 + (y - b)^2 + z^2] = \frac{\partial}{\partial x}[R^2] \]
\[ 2(x - a) + 2z \frac{\partial z}{\partial x} = 0 \]
\[ x - a + z p = 0 \]
where \(p = \frac{\partial z}{\partial x}\).
Differentiate with respect to \(y\):
\[ \frac{\partial}{\partial y}[(x - a)^2 + (y - b)^2 + z^2] = \frac{\partial}{\partial y}[R^2] \]
\[ 2(y - b) + 2z \frac{\partial z}{\partial y} = 0 \]
\[ y - b + z q = 0 \]
where \(q = \frac{\partial z}{\partial y}\).
Step 3: Eliminate the Arbitrary Constants \(a\) and \(b\)
From the equations:
\[ x - a + z p = 0 \]
\[ y - b + z q = 0 \]
Solve for \(a\) and \(b\):
\[ a = x + z p \]
\[ b = y + z q \]
Step 4: Substitute \(a\) and \(b\) Back into the Sphere Equation
Substitute \(a\) and \(b\) into the original sphere equation:
\[ (x - (x + z p))^2 + (y - (y + z q))^2 + z^2 = R^2 \]
Simplify:
\[ (-z p)^2 + (-z q)^2 + z^2 = R^2 \]
\[ z^2 p^2 + z^2 q^2 + z^2 = R^2 \]
Factor out \(z^2\):
\[ z^2 (p^2 + q^2 + 1) = R^2 \]
Final PDE:
The partial differential equation representing all spheres of fixed radius with centers in the \(xy\)-plane is:
\[ z^2 (p^2 + q^2 + 1) = R^2 \]
where \(p = \frac{\partial z}{\partial x}\) and \(q = \frac{\partial z}{\partial y}\).
- Find the partial differential equation of \((x-a)^2+(y-b)^2=z^2\cot^2\alpha\) by eliminating arbitrary constants.
Solution:
Step 1: Equation of the Surface
Consider the surface equation:
\[ (x - a)^2 + (y - b)^2 = z^2 \cot^2 \alpha \]
where \(\alpha\) is a constant angle.
Step 2: Differentiate with Respect to \(x\) and \(y\)
Differentiate with respect to \(x\):
\[ \frac{\partial}{\partial x}[(x - a)^2 + (y - b)^2] = \frac{\partial}{\partial x}[z^2 \cot^2 \alpha] \]
\[ 2(x - a) + 0 = 2z \cot^2 \alpha \frac{\partial z}{\partial x} \]
\[ x - a = z \cot^2 \alpha \frac{\partial z}{\partial x} \]
Let \(p = \frac{\partial z}{\partial x}\), then:
\[ x - a = z \cot^2 \alpha \cdot p \]
Differentiate with respect to \(y\):
\[ \frac{\partial}{\partial y}[(x - a)^2 + (y - b)^2] = \frac{\partial}{\partial y}[z^2 \cot^2 \alpha] \]
\[ 0 + 2(y - b) = 2z \cot^2 \alpha \frac{\partial z}{\partial y} \]
\[ y - b = z \cot^2 \alpha \frac{\partial z}{\partial y} \]
Let \(q = \frac{\partial z}{\partial y}\), then:
\[ y - b = z \cot^2 \alpha \cdot q \]
Step 3: Eliminate the Arbitrary Constants \(a\) and \(b\)
From the equations:
\[ x - a = z \cot^2 \alpha \cdot p \]
\[ y - b = z \cot^2 \alpha \cdot q \]
Solve for \(a\) and \(b\):
\[ a = x - z \cot^2 \alpha \cdot p \]
\[ b = y - z \cot^2 \alpha \cdot q \]
Step 4: Substitute \(a\) and \(b\) Back into the Surface Equation
Substitute \(a\) and \(b\) into the original surface equation:
\[ (x - (x - z \cot^2 \alpha \cdot p))^2 + (y - (y - z \cot^2 \alpha \cdot q))^2 = z^2 \cot^2 \alpha \]
Simplify:
\[ (z \cot^2 \alpha \cdot p)^2 + (z \cot^2 \alpha \cdot q)^2 = z^2 \cot^2 \alpha \]
\[ z^2 \cot^4 \alpha \cdot p^2 + z^2 \cot^4 \alpha \cdot q^2 = z^2 \cot^2 \alpha \]
Factor out \(z^2 \cot^4 \alpha\):
\[ z^2 \cot^4 \alpha (p^2 + q^2) = z^2 \cot^2 \alpha \]
Divide through by \(z^2 \cot^2 \alpha\):
\[ \cot^2 \alpha (p^2 + q^2) = 1 \]
Final PDE:
The partial differential equation representing the surface is:
\[ \cot^2 \alpha (p^2 + q^2) = 1 \]
where \(p = \frac{\partial z}{\partial x}\) and \(q = \frac{\partial z}{\partial y}\).
- Find the partial differential equation of \(z=ae^{bx}\sin(by)\) by eliminating arbitrary constants \(a\) and \(b\).
Solution:
Step 1: Function Definition
Consider the function:
\[ z = ae^{bx} \sin(by) \]
where \(a\) and \(b\) are arbitrary constants.
Step 2: Find the Partial Derivatives
First partial derivative with respect to \(x\):
\[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}[ae^{bx} \sin(by)] \]
Using the product rule:
\[ \frac{\partial z}{\partial x} = a b e^{bx} \sin(by) \]
Let \(p = \frac{\partial z}{\partial x}\):
\[ p = a b e^{bx} \sin(by) \]
First partial derivative with respect to \(y\):
\[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}[ae^{bx} \sin(by)] \]
Using the product rule:
\[ \frac{\partial z}{\partial y} = a e^{bx} \cdot b \cos(by) \]
Let \(q = \frac{\partial z}{\partial y}\):
\[ q = a b e^{bx} \cos(by) \]
Step 3: Eliminate the Arbitrary Constant \(b\)
From the expressions for \(p\) and \(q\):
\[ p = a b e^{bx} \sin(by) \]
or \[p=bz\] \[ q = a b e^{bx} \cos(by) \]
Divide \(p\) by \(q\):
\[ \frac{p}{q} = \frac{a b e^{bx} \sin(by)}{a b e^{bx} \cos(by)} \]
\[ \frac{p}{q} = \frac{\sin(by)}{\cos(by)} \]
\[ \frac{p}{q} = \tan(by) \] Using \(b=\frac{p}{z}\), we get
\[p=q\tan\left(\frac{py}{z}\right)\]