Charpit Method

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### Charpit Method Overview The Charpit method is a general method for finding the complete integral of a non-linear first-order partial differential equation (PDE) of the form $f(x, y, z, p, q) = 0$, where $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$. #### Charpit's Auxiliary Equations The core of the method lies in solving the following system of ordinary differential equations (ODEs), known as Charpit's Auxiliary Equations: $$\frac{dp}{f_x + p f_z} = \frac{dq}{f_y + q f_z} = \frac{dz}{-p f_p - q f_q} = \frac{dx}{-f_p} = \frac{dy}{-f_q} = \frac{df}{0}$$ Here, $f_x = \frac{\partial f}{\partial x}$, $f_y = \frac{\partial f}{\partial y}$, $f_z = \frac{\partial f}{\partial z}$, $f_p = \frac{\partial f}{\partial p}$, $f_q = \frac{\partial f}{\partial q}$. #### Steps to Solve 1. **Formulate the PDE:** Write the given PDE in the form $f(x, y, z, p, q) = 0$. 2. **Calculate Partial Derivatives:** Find $f_x, f_y, f_z, f_p, f_q$. 3. **Write Charpit's Auxiliary Equations:** Substitute the partial derivatives into the auxiliary equations. 4. **Find a Relation between p and q:** Solve any two suitable fractions of the auxiliary equations to find a relation between $p, q$ and possibly $x, y, z$. This often involves finding a simpler integral. * The goal is to express $p$ and $q$ in terms of $x, y, z$ and an arbitrary constant (say, $a$). 5. **Substitute into dz:** Substitute the expressions for $p$ and $q$ into the total differential $dz = p dx + q dy$. 6. **Integrate:** Integrate the resulting equation to obtain the complete integral $z = F(x, y, a, b)$, where $b$ is another arbitrary constant. ### Example 1: $p^2 + q^2 = 1$ Solve the PDE $p^2 + q^2 = 1$ using Charpit's method. 1. **Formulate $f(x, y, z, p, q) = 0$:** $f = p^2 + q^2 - 1 = 0$ 2. **Calculate Partial Derivatives:** $f_x = 0$ $f_y = 0$ $f_z = 0$ $f_p = 2p$ $f_q = 2q$ 3. **Write Charpit's Auxiliary Equations:** $$\frac{dp}{0 + p(0)} = \frac{dq}{0 + q(0)} = \frac{dz}{-p(2p) - q(2q)} = \frac{dx}{-2p} = \frac{dy}{-2q}$$ Simplifying: $$\frac{dp}{0} = \frac{dq}{0} = \frac{dz}{-2(p^2 + q^2)} = \frac{dx}{-2p} = \frac{dy}{-2q}$$ 4. **Find a Relation between p and q:** From $\frac{dp}{0}$, we get $dp = 0$, which implies $p = a$, where $a$ is an arbitrary constant. Substitute $p=a$ into the original PDE: $a^2 + q^2 = 1 \implies q^2 = 1 - a^2 \implies q = \sqrt{1 - a^2}$ 5. **Substitute into $dz = p dx + q dy$:** $dz = a dx + \sqrt{1 - a^2} dy$ 6. **Integrate:** $\int dz = \int a dx + \int \sqrt{1 - a^2} dy$ $z = ax + \sqrt{1 - a^2} y + b$ This is the complete integral. #### Special Case: Clairaut's Equation If the PDE is of the form $z = px + qy + f(p, q)$, it is known as Clairaut's equation. Its complete integral is obtained by simply replacing $p$ with $a$ and $q$ with $b$: $z = ax + by + f(a, b)$. The previous example is not Clairaut's form. ### Example 2: $z = px + qy + p^2 + q^2$ Solve the PDE $z = px + qy + p^2 + q^2$ using Charpit's method. 1. **Formulate $f(x, y, z, p, q) = 0$:** $f = px + qy + p^2 + q^2 - z = 0$ 2. **Calculate Partial Derivatives:** $f_x = p$ $f_y = q$ $f_z = -1$ $f_p = x + 2p$ $f_q = y + 2q$ 3. **Write Charpit's Auxiliary Equations:** $$\frac{dp}{p + p(-1)} = \frac{dq}{q + q(-1)} = \frac{dz}{-p(x+2p) - q(y+2q)} = \frac{dx}{-(x+2p)} = \frac{dy}{-(y+2q)}$$ Simplifying: $$\frac{dp}{0} = \frac{dq}{0} = \frac{dz}{-(px+2p^2+qy+2q^2)} = \frac{dx}{-(x+2p)} = \frac{dy}{-(y+2q)}$$ 4. **Find a Relation between p and q:** From $\frac{dp}{0}$, we get $dp = 0$, so $p = a$. From $\frac{dq}{0}$, we get $dq = 0$, so $q = b$. 5. **Substitute into $dz = p dx + q dy$:** $dz = a dx + b dy$ 6. **Integrate:** $\int dz = \int a dx + \int b dy$ $z = ax + by + c$ (where $c$ is the integration constant) *Wait! This is not the complete integral yet.* We must use the original PDE to relate $c$ to $a$ and $b$. Substitute $p=a$ and $q=b$ into the original PDE $z = px + qy + p^2 + q^2$: $z = ax + by + a^2 + b^2$ This is the complete integral. Notice this is a Clairaut's equation, and the result aligns with the special case mentioned above.