Question

Solve the following partial differential equations for \(u(x, y)\) with the boundary conditions given: (a) \(x \frac{\partial u}{\partial x}+x y=u, \quad u=2 y\) on the line \(x=1\) (b) \(1+x \frac{\partial u}{\partial y}=x u, \quad u(x, 0)=x\).

Short answer

(a) Solution: \(u(x,y)=2ye^{-x}\)(b) Solution: \(u(x,y)=x e^y\)

Step-by-step solution

Identify the Type of PDE

Both equations are first-order linear partial differential equations.

Rewrite Equation (a) in Standard Form

Rewrite equation (a): \[ x \frac{\partial u}{\partial x} + xy = u \] Divide both sides by x: \[ \frac{\partial u}{\partial x} + y = \frac{u}{x} \]

Solve the Homogeneous Part of (a)

Solve the homogeneous part \(\frac{du}{dx} + u = 0\): \[ \frac{du}{dx} = -u \] Integrate to get: \[ u = Ce^{-x} \]

Solve the Non-Homogeneous Part of (a)

For the non-homogeneous term, use variation of parameters: \[ u = Ce^{-x} + e^{-x} \int e^x y dx \] Evaluate the integral and solve to find the constant.

Apply Boundary Condition for (a)

Apply the boundary condition \(u = 2y\) when \(x = 1\): \[ u(1, y) = 2y = Ce^{-1} + e^{-1} \int_0^y e^{1} y dy \] Solve for the constant \(C\) and find the solution.

Rewrite Equation (b) in Standard Form

Rewrite equation (b): \[ 1 + x \frac{\partial u}{\partial y} = x u \] Divide both sides by x: \[ \frac{1}{x} + \frac{\partial u}{\partial y} = u \]

Solve the Homogeneous Part of (b)

Solve the homogeneous part \(\frac{\partial u}{\partial y} = u\): \[ \frac{du}{dy} = u \] Integrate to get: \[ u = C e^y \]

Solve the Non-Homogeneous Part of (b)

For the non-homogeneous term, use variation of parameters: \[ u = C e^y + e^y \int (\frac{1}{x}) e^{-y} dy \] Evaluate the integral and solve to find the constant.

Apply Boundary Condition for (b)

Apply the boundary condition \(u(x, 0) = x\): \[ u(x, 0) = x = C e^0 + e^0 \int (1/x) x dx \] Solve for the constant \(C\) and find the solution.

Key concepts

first-order linear PDE

A first-order linear partial differential equation (PDE) has the general form: A first-order linear partial differential equation (PDE) has the general form: \[ a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = c(x,y,u) \] The equation involves first-order partial derivatives of the unknown function, \(u\), with respect to the variables \(x\) and \(y\). In simpler terms, it tells how the function \(u\) changes as we move along the \(x\) and \(y\) directions.

boundary conditions

Boundary conditions specify values or behaviors of the solution at specific points or along specific lines. These are critical for finding a unique solution to a PDE. For example, in our exercise:

• For equation (a), the boundary condition is \(u = 2y\) on the line \(x = 1\).
• For equation (b), the boundary condition is \(u(x, 0) = x\).

These conditions help us determine integration constants and finalize our solutions.

variation of parameters

Variation of Parameters is a technique used to solve non-homogeneous PDEs. It extends the solution of a homogeneous equation to include the non-homogeneous part. For the exercise:

• In equation (a), we find the general solution of the homogeneous part and then adjust it by introducing an additional function to account for the non-homogeneous term.
• For equation (b), similarly, we define the general solution of the homogeneous part and vary the parameters to include the non-homogeneous term.

Mathematically, if the solution to the homogeneous equation is \(u_h\), then for the non-homogeneous equation, we look for a solution of the form \(u = u_h + u_p\).

homogeneous and non-homogeneous equations

A homogeneous PDE has no term that is only a function of the independent variables; it usually has the form: \[ L(u) = 0 \] where \(L\) represents some differential operator applied to \(u\). For example:

• \( \frac{\text{d}u}{\text{d}x} + u = 0 \ \frac{\text{d}u}{\text{d}y} = u \)

A non-homogeneous PDE contains additional terms that are functions of the independent variables, making the equation look like: \[ L(u) = f(x, y) \] , where \( \ L (u) = u_p \) is a particular solution of the non-homogeneous problem.

integration

Integration is critical in solving PDEs, especially for finding the particular solutions. For instance in our exercise:

• For equation (a), after isolating the homogeneous component, integrate with respect to \(x\) to find the general solution \( u = Ce^{-x} \).
• For the non-homogeneous part, we integrate the non-homogeneous term: \( e^{-x} \int e^x y dx \).
• For equation (b), after solving the homogeneous part and obtaining \( u = Ce^y \), integrate the non-homogeneous part by solving \( e^y \int (\frac{1}{x}) e^{-y} dy \).

In both cases, integrating involves finding the antiderivatives to determine the constants.