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UNIVERSITY OFCAL IF OR NI A, I RVINE

MAE200B - Homework No. 1

Laura Novoa

January 15, 2016

Problem1

The functionu(x,y) is governed by:

u

x=2xu

Subject to the BC:u(0,y)= si n(y). Calculate the Solution.

Solution

Assuming our "common clock" to be the new variables, from the boundary condition:

u(0,y)= si n(y)

u(s= 0)= si n()

Thus, we know that y(0)=

x(0)= 0

Using the method of characteristics:

A=1

2x

,B= 0, andF=u

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dx(s)

ds=A=

1

2x(s)

2x(s)dx(s)=

ds x2 = s+C1 s= x

2

0C1 1

BC: x(0)= 0 C1 = 0

dy(s)

ds=B= 0

dy= 0y(s)=C2 y(s)= 2

BC: y(0)= C2 =

du(s)

ds=F=u

du(s)

u(s)=

dsu(s)=C3e

s u(s)= si n()es 3

BC: u(0)= si n() C3 = si n()

Thus, from 1 and 2 into 3

u(x,y)= si n(y)ex2

Problem2Use the Method of characteristics to solve the following two equations foru(x,y) :

u

t+2t

u

x= 1 u(x, 0)=f(x)

u

t+x

u

x= si n(t) u(x, 0)= g(x)

In both cases:

(a) Use the rigorous approach to express the characteristics and the solutions in terms of an

independent variables.

(b) Derive and plot the shapes of the characteristics.

(c) Derive the solutionsu(x, t) for arbitraryf andg.

(d) Specialize the solutions for f(x)= ex andg(x)= co s(x). Make 3D plots of the respective so-

lutions.

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Solution

(I)u

t+2t

u

x= 1,u(x, 0)=f(x)

A= 1,B= 2t, andF= 1

Assumingsthe new variable, from the boundary condition:

u(x, 0)=f(x)

s= 0u(s= 0)=f()

Thus, we know that x(s) x(0)=

t(s) t(0)= 0

a)

dt(s)

ds=A= 1 t(s)= s+C1 s= t(s)

0C1 1

BC: t(0)= 0 C1 = 0

dx(s)

ds=B= 2t(s)= 2s x(s)= s2+C2 x(s)= s

2+ 2

BC: x(0)= C2 =

du(s)

ds=F=1u(s)= s+C3 u(s)= s+f() 3

BC: u(0)=f() C3 =f()

b) Eliminating s from the first two equation gives the characteristic equation:

Substituting 1 into 2 :

x(s)= t(s)2+

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Which is a parabola that crosses the x axis at a positive or negative

c) The solution for an arbitrary f(x) will be:

u(s)= s+f()

From 1 s= t(s)

From 2 = x(s) s2 = (x(s) t(s))2

Thus: u(x, t)= t+f(x t2)

d) Specialize the solution for f(x)= ex

u(x, t)= t+ext2

The surface plot of the function:

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(II)u

t+x

u

x= si n(t),u(x, 0)= g(x)

A= 1,B= x, andF= si n(t)

Assumingsthe new variable, from the boundary condition:

u(x, 0)= g(x)

s= 0u(s= 0)= g()

Thus, we know that x(s) x(0)=

t(s) t(0)= 0

a)

dt(s)

ds=A= 1 t(s)= s+C1 s= t(s)

0C1 1

BC: t(0)= 0 C1 = 0

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dx(s)

ds=B= x(s)

d x(s)

x(s)=

ds x(s)=C2e

s x(s)= es 2

BC: x(0)= C2 =

du(s)

ds=F= si n(t(s)) from 1 si n(s)

du(s)=

si n(s)

u(s)=cos(s)+C3 u(s)=co s(s)+g()+1 3

BC: u(0)= g()1+C3 = g() C3 = g()+1

b) Eliminating s from the first two equation gives the characteristic equation:

Substituting 1 into 2 :

x(s)= et(s)

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c) From 2 = x(s)es. Thus:

u(x, t)=co s(t)+g(xet)

d) Specialize the solution forg(x)= cos(x)

u(x, t)=co s(t)+cos(xet)

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Problem3

The potential, incompressible solution for 2D stagnation-point flow is:

u=Ax, v=Ay

Where u and v are the horizontal and vertical velocity components, respectively, and A is a con-

stant. At time t = 0 a blob of contaminants is introduced in this flow. The initial shape of the

blob is a circle centered at (x = 0, y = b) with radius R < b. You are asked to find the shape of the

blob at later times using the following steps.

(a) Neglecting diffusion, and assuming that the contaminant is a passive scalar, write the trans-

port equation for the contaminant concentration C(x, y, t).

Considering steady-state flow: Ct = 0 we have:

0C

t+u

C

x+v

C

y= 0

(b) and (c): Write differential evolution equations for the characteristics and the concentration

field, using as independent variable the time t measured by an observed moving with a fluidparticle.Integrate the above equations to obtain algebraic relations for the characteristics (x(t),

y(t)) and the concentration field c(t).

dx(t)

dt=Ax(t)

dx(t)

x(t)=A

dt x(t)=C1e

At 1 ln

x(t)

C1

=At

d y(s)dt

=Ay(t) y(t)=C2eAt 2

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dC(t)

dt= 0C(t)=C3 = constant

1 into 2 y(t)=C2exp lnx(t)

C1=C2expln

x(t)

C11

=C2C1

x(t) y(t)=

C3

x(t)

(d) Plot the shapes of the characteristics on thexyplane.

(e) Derive an analytic expression for the shape of the blob for t > 0. Characterize this shape,and

plot for A = 1, b = 1, R = 0.5, t = 1.

At t = 0, from the equation of the circumference with a b displacement in the y axis:

x20+ (y0b)2=R2

x0R

2+

y0b

R

2= 1 1

From part c), we had:

x(t)=C1eAt

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y(t)=C2eAt

Applying initial conditions:

x(0)= x0 C1 = xo x(t)= x0eAt 2

y(0)=y0 C2 =yo y(t)=y0eAt 3

Substituting 2 and 3 into 1 will give us an analytic expression for the shape of the blob for

t > 0.

x(t)

eAtR

2+

y(t)eAt

b

R

2= 1

x(t)

eAtR

2+

y(t)b

ReAt

2=1

Which describes an Ellipsoid.

Since we know that eAtR is increasing over time, we can expect that the ellipse radius in the xdirection will progressively increase over time.

Similarly, sinceeAtR is decreasing, the ellipse radius in the y direction will progressively de-

crease over time

Also,eAtbis decreasing over time, meaning the blob will flow downward in the y axis.

Eventually, whent=, the blob will become a flat "line" in the y = 0 position.

The plot below describes the evolution of the blob with time. The blue circle represents the

blob at t = 0 , the red circle is the blob at t = 1 and the yellow at t = 2.:

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MATLAB SOURCE CODE

PDE # 1

figure

[X,T] = meshgrid(-4:.2:4);

U = T + exp(X - T.^2);

surf(X,T,U)

xlabel('x')

ylabel('t')

zlabel('u')

%Plot characteristics

C = [-10:1:10];t = [-5:0.1:5];

figure

fori=1:length(C)

x = t.^2 + C(i);

plot(t,x)

ylim = ([0 100]);

hold on

end

xlabel('t(s)')

ylabel('x(s)')

PDE #2

[X,T] = meshgrid(-10:.5:25);

U = -1*cos(T) + cos(X.*exp(-T));

figure

surf(X,T,U)

xlabel('x')

ylabel('t')

zlabel('u')

%Plot characteristics

C = [-10:1:10];

t = [-5:0.1:5];

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figure

fori=1:length(C)

x = C(i).*exp(t);

plot(t,x)

ylim = ([0 100]);

hold on

end

xlabel('t(s)')

ylabel('x(s)')

Problem 3

clc

clear all

close all

%Plot characteristic lines

C = [-10:1:10];

x = [-5:0.1:5];

fori=1:length(C)

y = C(i)./x;

plot(x,y)

axis([ -inf inf 0 100]);

hold on

end

xlabel('x(t)')

ylabel('y(t)')

Plotting blob

%t = 0

r = 0.5; %radius

xc = 0; %0

yc = 1; % b

theta = 0 : .1 : 2*pi;

x = r * cos(theta) + xc;

y = r * sin(theta) + yc;

%figure

plot(x, y)

t = 1;

A = 1;

xt = x.*exp(A.*t);

yt = y.*exp(-1*A.*t);

hold on

plot(xt, yt)

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t = 2;

A = 1;

xt = x.*exp(A.*t);

yt = y.*exp(-1*A.*t);

hold on

plot(xt, yt)

xlabel('x(t)')

ylabel('y(t)')

axis([ -5 5 0 5]);

Published with MATLAB R2015a

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