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Practical-6 Solution of Cauchy Problem For First Order PDE: Question-1

The document presents solutions to various Cauchy problems for first-order partial differential equations (PDEs) using symbolic computation. It includes multiple equations, their corresponding solutions, and 3D plots to visualize the results. Each question outlines a different PDE scenario with initial conditions and provides the derived solutions step-by-step.

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Rohit Sharma
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204 views4 pages

Practical-6 Solution of Cauchy Problem For First Order PDE: Question-1

The document presents solutions to various Cauchy problems for first-order partial differential equations (PDEs) using symbolic computation. It includes multiple equations, their corresponding solutions, and 3D plots to visualize the results. Each question outlines a different PDE scenario with initial conditions and provides the derived solutions step-by-step.

Uploaded by

Rohit Sharma
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Practical-6

Solution of Cauchy problem for first order PDE


Question-1
In[1]:= eqn1 = D[u[x, y], x] + x * D[u[x, y], y] ⩵ 0
Out[1]= x u0,1  [x, y] + u1,0  [x, y] ⩵ 0

In[2]:= sol1 = u[x, y] / . DSolve [{eqn1, u[0, y] ⩵ Sin [y]}, u[x, y], {x, y}]
x2
Out[2]= - Sin  - y
2

In[3]:= Plot3D [sol1, {x, - 5, 5}, {y, - 5, 5}]

Out[3]=

Question-2
In[4]:= eqn2 = 3 * D[u[x, y], x] + 2 * D[u[x, y], y] ⩵ 0
Out[4]= 2 u0,1  [x, y] + 3 u1,0  [x, y] ⩵ 0

In[5]:= sol2 = u[x, y] / . DSolve [{eqn2, u[x, 0] ⩵ Sin [x]}, u[x, y], {x, y}]
3 2x
Out[5]= - Sin  - + y 
2 3
2

In[6]:= Plot3D [sol2, {x, - 5, 5}, {y, - 5, 5}]

Out[6]=

Question-3
In[7]:= eqn3 = y * D[u[x, y], x] + x * D[u[x, y], y] ⩵ 0
Out[7]= x u0,1  [x, y] + y u1,0  [x, y] ⩵ 0

In[11]:= sol3 = u[x, y] / . DSolve [{eqn3 , u[0, y] ⩵ Exp [- y ^ 2]}, u[x, y], {x, y}]

ⅇ 
x2 - y2
Out[11]=

In[9]:= Plot3D [sol3, {x, - 5, 5}, {y, - 5, 5}]

Out[9]=

Question 4 : ∂x u[x,y] + 2* ∂y u[x,y] = 1 + u[x, y]


u[x, y] = Sin[x] on y = 3*x + 1;
3

In[12]:= A = D[u[x, y], x] + 2 * D[u[x, y], y] ⩵ 1 + u[x, y]


Out[12]= 2 u0,1  [x, y] + u1,0  [x, y] ⩵ 1 + u[x, y]

In[13]:= sol = DSolve [{A, u[x, 3 * x + 1] ⩵ Sin [x]}, u[x, y], {x, y}]
u[x, y] → - ⅇ - ⅇ +ⅇ +ⅇ
-y 1+ 3 x y 1+ 3 x
Out[13]= Sin [1 + 2 x - y]

In[17]:= Plot3D [u[x, y] / . sol, {x, - 2, 2}, {y, - 7, 8}, AxesLabel → {Automatic }]

Out[17]=

Question 5. Solve the PDE Subscript[uu, x]+ Subscript[u, y]=1/2. With the initial
condition u(s,s)=s/4,
0<=s<=1.
Solution: x=s+st/4=t^2/4, y=s+t, u=s/4+t/2.
In[18]:= sol = DSolve [{x '[t] ⩵ u[t], y '[t] ⩵ 1, u '[t] ⩵ 1 / 2, x[0] ⩵ s,

y[0] ⩵ s, u[0] ⩵ s / 4}, {x[t], y[t], u[t]}, t]


1 1
 u[t] → (s + 2 t), x[t] → × 4 s + s t + t , y[t] → s + t
2
Out[18]=
4 4

In[19]:= Print ["u[t]= ", sol 〚 1, 1, 2〛]


1
u [ t ]= (s + 2 t)
4

In[20]:= Print ["y[t]= ", sol 〚 1, 2, 2〛]


1
× 4 s + s t + t 
2
y [ t ]=
4

In[21]:= Print ["x[t]= ", sol 〚 1, 3, 2〛]


x [ t ]= s + t
4

In[22]:= map = ParametricPlot3D [{sol 〚 1, 1, 2〛 , sol 〚 1, 2, 2〛 ,

sol 〚 1, 3, 2〛}, {t, - 1, 1}, {s, 0, 1}, PlotPoints → 10 ]

Out[22]=

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