,APM3701 Assignment 1 (COMPLETE ANSWERS) 2025 (608471) -
DUE 29 May 2025; 100% TRUSTED Complete, trusted solutions
and explanations.
QUESTION 1 Solve the following (initial)-boundary value problem, a.
uxy (x, y) = xy3, x, y 0. u (x, 0) = f (x) , and uy (0, y) = g (y) .
Determine u (x, y) , if f (x) = cosx and g (y) = y+sin y. (Check your
answer by substituting, and explain all the steps clearly) (15 Marks) b.
xux + yuy = yu u (2x2, x) = x2 − 1. (1) (Check your answer by
substituting before applying the boundary condition, and explain all the
steps clearly) (15 Marks) [30 Marks]
Let's address each part one by one.
Part (a)
We are given the partial differential equation:
uxy(x,y)=xy3,x,y≥0u_{xy}(x, y) = xy^3, \quad x, y \geq 0uxy
(x,y)=xy3,x,y≥0
with boundary conditions:
u(x,0)=f(x)=cos(x)anduy(0,y)=g(y)=y+sin(y).u(x, 0) = f(x) =
\cos(x) \quad \text{and} \quad u_y(0, y) = g(y) = y +
\sin(y).u(x,0)=f(x)=cos(x)anduy(0,y)=g(y)=y+sin(y).
We need to determine u(x,y)u(x, y)u(x,y) by solving this problem step
by step.
Step 1: Integrating with respect to yyy
To solve for u(x,y)u(x, y)u(x,y), start by integrating the PDE with
respect to yyy.
, From the equation uxy=xy3u_{xy} = xy^3uxy=xy3, we first integrate
with respect to yyy:
ux(x,y)=∫xy3 dy=xy44+h(x),u_x(x, y) = \int xy^3 \, dy = \frac{x
y^4}{4} + h(x),ux(x,y)=∫xy3dy=4xy4+h(x),
where h(x)h(x)h(x) is an arbitrary function of xxx (because the
integration was with respect to yyy).
Step 2: Integrating with respect to xxx
Next, we integrate the expression for ux(x,y)u_x(x, y)ux(x,y) with
respect to xxx:
u(x,y)=∫(xy44+h(x))dx=x2y48+H(x)+C(y),u(x, y) = \int \left( \frac{x
y^4}{4} + h(x) \right) dx = \frac{x^2 y^4}{8} + H(x) +
C(y),u(x,y)=∫(4xy4+h(x))dx=8x2y4+H(x)+C(y),
where H(x)H(x)H(x) is another arbitrary function of xxx, and
C(y)C(y)C(y) is a function of yyy.
Step 3: Apply the boundary conditions
Now, use the boundary conditions to determine the unknown functions.
Boundary condition 1: u(x,0)=cos(x)u(x, 0) =
\cos(x)u(x,0)=cos(x).
Substitute y=0y = 0y=0 into u(x,y)u(x, y)u(x,y):
u(x,0)=x2⋅048+H(x)+C(0)=H(x)+C(0).u(x, 0) = \frac{x^2 \cdot 0^4}{8}
+ H(x) + C(0) = H(x) + C(0).u(x,0)=8x2⋅04+H(x)+C(0)=H(x)+C(0).
We are given that u(x,0)=cos(x)u(x, 0) = \cos(x)u(x,0)=cos(x), so:
H(x)+C(0)=cos(x).H(x) + C(0) = \cos(x).H(x)+C(0)=cos(x).
Thus, we can write:
DUE 29 May 2025; 100% TRUSTED Complete, trusted solutions
and explanations.
QUESTION 1 Solve the following (initial)-boundary value problem, a.
uxy (x, y) = xy3, x, y 0. u (x, 0) = f (x) , and uy (0, y) = g (y) .
Determine u (x, y) , if f (x) = cosx and g (y) = y+sin y. (Check your
answer by substituting, and explain all the steps clearly) (15 Marks) b.
xux + yuy = yu u (2x2, x) = x2 − 1. (1) (Check your answer by
substituting before applying the boundary condition, and explain all the
steps clearly) (15 Marks) [30 Marks]
Let's address each part one by one.
Part (a)
We are given the partial differential equation:
uxy(x,y)=xy3,x,y≥0u_{xy}(x, y) = xy^3, \quad x, y \geq 0uxy
(x,y)=xy3,x,y≥0
with boundary conditions:
u(x,0)=f(x)=cos(x)anduy(0,y)=g(y)=y+sin(y).u(x, 0) = f(x) =
\cos(x) \quad \text{and} \quad u_y(0, y) = g(y) = y +
\sin(y).u(x,0)=f(x)=cos(x)anduy(0,y)=g(y)=y+sin(y).
We need to determine u(x,y)u(x, y)u(x,y) by solving this problem step
by step.
Step 1: Integrating with respect to yyy
To solve for u(x,y)u(x, y)u(x,y), start by integrating the PDE with
respect to yyy.
, From the equation uxy=xy3u_{xy} = xy^3uxy=xy3, we first integrate
with respect to yyy:
ux(x,y)=∫xy3 dy=xy44+h(x),u_x(x, y) = \int xy^3 \, dy = \frac{x
y^4}{4} + h(x),ux(x,y)=∫xy3dy=4xy4+h(x),
where h(x)h(x)h(x) is an arbitrary function of xxx (because the
integration was with respect to yyy).
Step 2: Integrating with respect to xxx
Next, we integrate the expression for ux(x,y)u_x(x, y)ux(x,y) with
respect to xxx:
u(x,y)=∫(xy44+h(x))dx=x2y48+H(x)+C(y),u(x, y) = \int \left( \frac{x
y^4}{4} + h(x) \right) dx = \frac{x^2 y^4}{8} + H(x) +
C(y),u(x,y)=∫(4xy4+h(x))dx=8x2y4+H(x)+C(y),
where H(x)H(x)H(x) is another arbitrary function of xxx, and
C(y)C(y)C(y) is a function of yyy.
Step 3: Apply the boundary conditions
Now, use the boundary conditions to determine the unknown functions.
Boundary condition 1: u(x,0)=cos(x)u(x, 0) =
\cos(x)u(x,0)=cos(x).
Substitute y=0y = 0y=0 into u(x,y)u(x, y)u(x,y):
u(x,0)=x2⋅048+H(x)+C(0)=H(x)+C(0).u(x, 0) = \frac{x^2 \cdot 0^4}{8}
+ H(x) + C(0) = H(x) + C(0).u(x,0)=8x2⋅04+H(x)+C(0)=H(x)+C(0).
We are given that u(x,0)=cos(x)u(x, 0) = \cos(x)u(x,0)=cos(x), so:
H(x)+C(0)=cos(x).H(x) + C(0) = \cos(x).H(x)+C(0)=cos(x).
Thus, we can write:
