Consider the partial differential equation (PDE) \(x\left(\frac{\partial u}{\partial x}\right)^2+y\left(\frac{\partial u}{\partial y}\right)^2+(x+y) \frac{\partial u}{\partial x} \frac{\partial u}{\partial y}-u\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\right)+1=0\) Which of the following statements are true?
Partial Differential Equation Analysis
The question asks to verify statements about a given partial differential equation (PDE). The PDE is a first-order, non-linear equation.
The given PDE is:
\[x\left(\frac{\partial u}{\partial x}\right)^2+y\left(\frac{\partial u}{\partial y}\right)^2+(x+y) \frac{\partial u}{\partial x} \frac{\partial u}{\partial y}-u\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\right)+1=0\]We often use the notation \(p = \frac{\partial u}{\partial x}\) and \(q = \frac{\partial u}{\partial y}\). So the PDE can be written as:
\[x p^2 + y q^2 + (x+y) p q - u(p+q) + 1 = 0\]General Solution Form Verification
One statement proposes that the general solution can be expressed in the form \(u(x, y) = a x + b y + \frac{1}{a+b}\), where \(a\) and \(b\) are arbitrary constants. This form suggests looking for a complete integral. A complete integral of a first-order PDE is a solution involving as many arbitrary constants as there are independent variables (in this case, two: \(x\) and \(y\)).
If \(u(x, y) = a x + b y + \frac{1}{a+b}\) is a solution, then the partial derivatives would be:
- \(p = \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \left(a x + b y + \frac{1}{a+b}\right) = a\)
- \(q = \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \left(a x + b y + \frac{1}{a+b}\right) = b\)
Now substitute \(u = a x + b y + \frac{1}{a+b}\), \(p=a\), and \(q=b\) into the original PDE:
\[x (a)^2 + y (b)^2 + (x+y) (a)(b) - \left(a x + b y + \frac{1}{a+b}\right)(a+b) + 1 = 0\] \[x a^2 + y b^2 + (x+y) ab - (ax+by)(a+b) - \frac{1}{a+b}(a+b) + 1 = 0\] \[x a^2 + y b^2 + x ab + y ab - (ax(a+b) + by(a+b)) - 1 + 1 = 0\] \[x a^2 + y b^2 + x ab + y ab - (a^2 x + ab x + aby + b^2 y) = 0\] \[x a^2 + y b^2 + x ab + y ab - a^2 x - ab x - aby - b^2 y = 0\]Grouping terms with \(x\) and \(y\):
\[(xa^2 - a^2x) + (yb^2 - b^2y) + (xab - abx) + (yab - aby) = 0\] \[0 + 0 + 0 + 0 = 0\]Since the substitution satisfies the PDE for arbitrary constants \(a\) and \(b\), the expression \(u(x, y) = a x + b y + \frac{1}{a+b}\) is indeed a complete integral, and thus a valid form for the general solution in this context.
Charpit's Equations Derivation
The PDE is given by \(F(x, y, u, p, q) = x p^2 + y q^2 + (x+y) p q - u(p+q) + 1 = 0\).
Charpit's method is used to find a complete integral of a first-order non-linear PDE. The Charpit's auxiliary equations are given by:
\[\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{du}{p F_p + q F_q} = \frac{dp}{-(F_x + p F_u)} = \frac{dq}{-(F_y + q F_u)}\]We need to calculate the partial derivatives of \(F\) with respect to \(p, q, x, y, u\):
- \(F_p = \frac{\partial F}{\partial p} = \frac{\partial}{\partial p} (x p^2 + y q^2 + (x+y) p q - u(p+q) + 1) = 2xp + (x+y)q - u\)
- \(F_q = \frac{\partial F}{\partial q} = \frac{\partial}{\partial q} (x p^2 + y q^2 + (x+y) p q - u(p+q) + 1) = 2yq + (x+y)p - u\)
- \(F_x = \frac{\partial F}{\partial x} = \frac{\partial}{\partial x} (x p^2 + y q^2 + (x+y) p q - u(p+q) + 1) = p^2 + q p\)
- \(F_y = \frac{\partial F}{\partial y} = \frac{\partial}{\partial y} (x p^2 + y q^2 + (x+y) p q - u(p+q) + 1) = q^2 + p q\)
- \(F_u = \frac{\partial F}{\partial u} = \frac{\partial}{\partial u} (x p^2 + y q^2 + (x+y) p q - u(p+q) + 1) = -(p+q)\)
Now substitute these partial derivatives into the Charpit's equations:
\[\frac{dx}{2xp + (x+y)q - u} = \frac{dy}{2yq + (x+y)p - u}\] \[\frac{du}{p(2xp + (x+y)q - u) + q(2yq + (x+y)p - u)}\] \[\frac{dp}{-( (p^2 + pq) + p (-(p+q)) )} = \frac{dp}{-( p^2 + pq - p^2 - pq )} = \frac{dp}{0}\] \[\frac{dq}{-( (q^2 + pq) + q (-(p+q)) )} = \frac{dq}{-( q^2 + pq - pq - q^2 )} = \frac{dq}{0}\]Combining these, the Charpit's equations are:
\[\frac{d x}{2 p x+(x+y) q-u}=\frac{d y}{2 q y+(x+y) p-u} = \frac{d u}{p(2 p x+(x+y) q-u)+q(2 q y+(x+y) p-u)}=\frac{d p}{0}=\frac{d q}{0}\]This matches the form given in the second true statement.
Based on the verification, both the statement about the general solution form \(u(x, y) = a x + b y + \frac{1}{a+b}\) and the statement providing the specific Charpit's equations are true.
Important Questions from Lagrange and Charpit Methods
The general solution of the equation
\(x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0\) is
Which of the following is a solution to ux + x2uy = 0 with u(x, 0) = ex?
Let u ∶ ℝ2 → ℝ be the solution to the Cauchy problem:
\(\left\{\begin{aligned} \rm \partial_x u+2 \partial_y u & =0 & & \text { for } \rm (x, y) \in ℝ ^2, \\ \rm u(x, y) & =\rm \sin (x) & & \text { for } \rm y=3 x+1, x \in ℝ.\end{aligned}\right.\)
Let v ∶ ℝ2 → ℝ be the solution to the Cauchy problem:
\(\left\{\begin{aligned} \rm \partial_x v+2 \partial_y v & =0 & & \text { for } \rm (x, y) \in ℝ ^2, \\ \rm v(x, 0) & =\rm\sin (x) & & \text { for } \rm x \in ℝ.\end{aligned}\right.\)
Let δ = [0, 1] × [0, 1].
Which of the following statements are true?
Let u(x, y) be the solution of the Cauchy problem
uux + uy = 0, x ∈ ℝ, y > 0,
u(x, 0) = x, x ∈ ℝ.
Which of the following is the value of u(2, 3)?
Let u = u(x, y) be the solution to the following Cauchy problem
ux + uy = eu for (x, y) ∈ ℝ × \(\rm\left(0, \frac{1}{e}\right)\) and u(x, 0) = 1 for x ∈ ℝ.
Which of the following statements are true?
