Question

If \(u(x, y)\) satisfies $$ \frac{\partial^{2} u}{\partial x^{2}}-3 \frac{\partial^{2} u}{\partial x \partial y}+2 \frac{\partial^{2} u}{\partial y^{2}}=0 $$ and \(u=-x^{2}\) and \(\partial u / \partial y=0\) for \(y=0\) and all \(x\), find the value of \(u(0,1)\).

Short answer

Plugging this u(x.y) we have u(0,1)=solve system equations.

Step-by-step solution

Understand the Partial Differential Equation (PDE)

The given PDE is: \[ \frac{\frac{\text{{partial}}^2 u}{\text{{partial}} x^2}}-\frac{3}{\text{{partial}}^2 u}{\text{{partial}} x \text{{partial}}}y + \frac{\text{{2}}}{\text{{partial}}^2 u}{\text{{partial}} y^2} \text{{=}}0 \]. This equation must be satisfied by the potential solution u(x,y).

Apply Given Boundary Conditions at y=0

For \(y = 0\), \(u = -x^2\) and \({\frac{\text{{partial}}u}{\text{{partial}} y}}= 0\). This information will help in reducing the PDE to solve for the more specific conditions.

Solve for the General Form of u(x,y)

Since \({\frac{\text{{partial}} u}{\text{{partial x^2}}}} = 2\), \({\frac{\text{{partial}} u}{\text{{partial}} x \text{{partial}}}y} - 3 {\text{{partial}}^{2}} n = 2 {\text{{partial}}^{2}} B \text{{partial}}^{2} y^{1.5}} = 0\).

Consider u(x,y) = ax^2+bxy+cy^2+d

Plugging this form back into PDE and using boundary conditions solve System of Equations to find a, b, c, and d.

Evaluate at given point u(0,1)

To find u(0,1) from the solution derived.

Key concepts

boundary conditions

Boundary conditions are essential in solving partial differential equations (PDEs). They provide additional constraints that help determine a specific solution from the general solutions of the PDE.

In our example, we have two boundary conditions at y = 0:

• u = -x^2
• ∂u/∂y = 0

These conditions specify the function value and its derivative at particular points. Without them, the PDE could have infinite possible solutions. Boundary conditions guide us to the one solution that fits the given physical or geometrical context.

When applying boundary conditions:

• First, incorporate them into the PDE to reduce complexity.
• Then, use these simplified forms to solve for specific constants or functions.

general solution of PDE

The general solution of a PDE includes all possible solutions before any boundary conditions are applied. In our case, the PDE is:

$$\frac{\frac{\text{{partial}}^2 u}{\text{{partial}} x^2}}-3 \frac{\frac{\text{{partial}}^2 u}{\text{{partial}} x \text{{partial}} y}}+2 \frac{\frac{\text{{partial}}^2 u}{\text{{partial}} y^2}}=0$$.

To find the general solution, we first need to propose a general form for u(x, y). One common approach is to assume a polynomial form, say u(x, y) = ax^2 + bxy + cy^2 + dx + ey + f. This form includes all possible terms up to the second order, which should encapsulate all potential solutions to our second-order PDE.

Plug this general form back into the PDE, and you will get a system of equations for the coefficients a, b, c, d, e, and f. Solving these equations yields expressions for these coefficients, which gives the general solution.

Next, enforce boundary conditions upon this general solution, which leads us to a specific solution that satisfies both the PDE and the boundary conditions.

method of characteristics

The method of characteristics is a technique to solve certain types of PDEs, particularly useful in first-order linear and quasilinear PDEs. It transforms a PDE into a set of ordinary differential equations (ODEs) by tracing the paths (characteristics) along which the PDE becomes an ODE.

Here's a simplified overview of the technique:

• Identify characteristic curves in the domain where the PDE can be simplified.
• Transform the PDE into a set of ODEs along these curves.
• Solve these ODEs to find the solution of the original PDE.

The method is valuable because it leverages the characteristic directions in the solution space where the problem simplifies, making it easier to solve complex PDEs.

In our given PDE, you might not directly use the method of characteristics since it typically looks for simpler first-order PDEs. However, the underlying principle of transforming a complex problem into simpler forms still holds true and guides the approach in solving higher-order PDEs by reducing them step by step.

boundary value problems

Boundary value problems (BVPs) involve finding a solution to a differential equation that satisfies specific conditions (boundary conditions) at the boundaries of the domain.

In our case, the boundary conditions at y = 0 provided:

• u = -x^2
• ∂u/∂y = 0

These constraints must hold true in the solution at the boundaries (in this case, at y = 0 for all x). Solving BVPs typically involves these steps:

• Formulating the general solution to the PDE without boundary conditions.
• Applying the boundary conditions to the general solution.
• Solving the resulting equations to determine all unknown constants or functions.

The given exercise is a typical boundary value problem where we need to find a specific solution that satisfies the overall PDE and the given boundary conditions.

The steps include:

• Writing the general form of u(x, y).
• Plugging in the boundary conditions to solve for specific constants a, b, c, etc.
• Thus, identifying the unique solution that fits the problem context.